Fun fact, based on this definition you can actually rearrange the equation to arrive at the definition of e based on compound interest: lim(h-->0)((e^h - 1)/h) = 1 Let h_0 be a small number such that (e^h_0 - 1)/h_0 = 1 e^h_0 - 1 = h_0 e^h_0 = 1 + h_0 e = (1 + h_0)^(1/h_0) Convert back to limit e = lim(h-->0)(1+h)^(1/h) i.e. compound interest recursion formula with infinitesimally small interest rate for an infinitely large number of compounding periods.
When you write "let h_0 be a small number such that: (e^(h_0)-1)/h_0=1" How can you tell that such h_0 exist? And in particular if you rearrange the equation one may get e^(h_0)=h_0+1, and if I'm not mistaken there exist an inequality which tells that e^x>=x+1, whit equality at x=0
@@Evan-ne5bu I suppose it is equivalent of using the property that limit of f(g(x)) is equal to f(limit of g(x)) where f=(e^h - 1)/h and g=h. So h_0 is just a simpler way of writing lim(h-->0)(h). Also, didn't know about that inequality, nice! It is true for x>0 if you graph it. In this case, the equation e^(h_0)=1+h_0 is true if we are talking about limits since h_0 is a small number approaching 0
@@Evan-ne5bu Also fun fact #2, using base 'a' in the exponential instead of base 'e' you can show that lim(x-->0)((a^x - 1)/x) = ln(a). If anyone knows a way to directly approach this limit let me know, I've been curious haha
It's really important to understand that it's not that the derivate of e^x is just e^x, but that's WHY the number e exists. Videos like yours really inspire me to share my own videos as well!
The reason e exists is because when multiples I think it was 1.0001 the values of whole numbers were interesting to Euler and he wanted to calculate what the base was
There are many definitions of e it is not the only reason e exists 😂 For example lim h->infinite (1+1/h)^h. You can come up with this definition without calculus for example compound interest where you have an infinite amount of compound interest and every interest is 1+lim h->infinity(1/h) times higher than the previous interest.
As a veteran, one can conclude this is circular reasoning which meant euler's number has NOTHING to do with reality. Which meant the actual value of pi is completely different. It is also ignorant reasoning to presuppose the perimeter of circumscribed polygon is always longer than the circumference of the unit circle at different number of sides. There must be ways to prove that and yet nobody even attempted it. Clown world this is.
@@lukiepoole9254 this dude thinks that he's more knowledgeable about math than the collective efforts of the smartest mathematicians over thousands of years
@@Maple_MK Why are there not a single absolutely undeniably irrefutable rigorous proof for circumference of circle? The pi based on circumference of circle isn't 3.14159265; It is literally 4sqrt(1/golden ratio).
I love how this shows the process of discovering the derivatives and defining things along the way; it makes math much more interesting when you see how these things might’ve been originally figured out.
You can actually evaluate the limit of (2^h-1)/h by using the variable substitution u=2^h-1, turning it into the limit as u->0 u/log2(u+1) = 1/((1/u)log2(u+1)) =1/log2((1+u)^(1/u)). Change of variables to t=1/u gives lim t->inf 1/log2((1+1/t)^t). This includes the limit definition of e which can be shown to exist by the monotone sequence theorem. Then you have 1/log2(e) =log2(2)/log2(e) = ln(2) by the change of base formula. I found out about this from Khan Academy.
2:15 Here's how the ancient people probably did it: By using u substitution and letting 2^h=u, we can change the limit as h approaches 0 to the limit as u approaches 1 We can say that h is log base 2 of u from this statement. From there we can take the derivative with respect to u. By using the definition of the derivative, we can find the generalized derivative for any function log base n of x, using logarithm properties and the definition of e as a limit. From there we can find that d/dx log n of x= 1/(x*ln(n)). using that derivative we can plug in ln(2) as the solution of that limit and get 2^x*ln(2). Sorry if the text formatting on this is confusing, i could post a video about it
But before defining e in such a way, you need to prove that there exists a unique number with the described property and I can't say that that's quite easier then proving that the derivative of e^x is e^x using some other definition of e.
That was also my reaction when he did it. The existence and uniqueness of such a number are not guaranteed. You could perhaps build a fonction that for all input a outputs the limit of (a^h-1)/h as h tends to 0, then prove it's continuous and strictly increasing, and finally apply the intermediate value theorem. That could work but that sounds more complicated than using an other definition.
Yeah it's a flawed but intuitive way of defining it. I guess you could look at the function f(t) = d/dx (t^x) and see that f is continuous, and apply the Intermediate Value Theorem to t=2 and t=3
Excellent question! Good discussion, would like to see what more people say. Let’s take a hypothetical number (N^h - 1) ________. h-->0 h Where this equals 1. How do we know if it doesn’t settle in a value smaller than 1? Maybe it tends to a value like .85327 or something? I’m assuming people can just say If we plug in 2 for N, it’s lower than 1 If we plug in 3 for N, it’s more than 1 -> Therefore it’s somewhere between the two. But plugging in number isn’t that satisfying, it’s not really a proof :/ What we can do is compare the outputs as the inputs get bigger (a^h-1)/h Vs ((2a)^h-1)/h. The input is twice as big, but what happens to the output? Both have a common denominator so let’s take that out. Now what is bigger? (a^h-1) or ((2a^h)-1) Clearly ((2a^h)-1) is always gonna be larger than the previous value i.e (a^h - 1) as “a” gets larger. Example: (5^a)-1 < [(2*5)^a]-1 But the question is, how much? There’s probably a way that proves it isn’t asymptotic and extends to infinity, I’ll try and think of a proof, but this is an excellent discussion and a reminder to always question our intuition. Good job guys!
@@skylardeslypere9909 The problem with that is that this relies on t^x being well-defined, which itself relies on the exponential function and the logarithmic function also being well-defined a priori.
We can use the definition given here to prove that e = 1/0! + 1/1! + 1/2! +1/3! + ... then we can prove that the sum converges hence proving the existence of e and it's uniqueness
We can even evaluate that limit by using the expansion 2^x = e^(xln2) So e^(xln2) = 1+ xln2 +.... We plug that in And get lim h->0 1+hln2...-1/h Lim h->0 hln2+.../h We get ln2 as limit as all other terms would give zero
for the exponential function b^x, the derivative by first principles becomes (b^x) lim h→0 ((b^h) - 1)/h = (b^x) lim h→0 ((b^h) - (b^0))/h, which is the instantaneous gradient at x = 0, and b = e is the only case where the gradient at x = 0 is 1 idk, I think it's a fun visualization of what that weird limit is actually representing
I think when I learned it at school (in the math power lessons at grammar school with a teacher who had a math PhD), we started by pointing out that f(x)=1/x must have an antiderivative F (I'm avoiding to use the terms ln, e etc.). This cannot be calculated by using the (x^n)' = n*x^(n-1) formula because it would involve a division by 0, respectively it doesn't work for n=0. We chose to calculate the one with the constant leading to F(1)=0 and then made deductions about its inverse function G. We showed that G'=G, that G is of the form G(x) = Z^x by proving that for every x1 and x2, G(x1+x2)= G(x1)*G(x2). Finally we calculated G(1) = Z^1 = Z (which ie 2.71.....) and voila, we were done.
I think one intuitive way to define e is by defining e^x generally: What we want is that e^0=1, and that the derivative is the same as the original function. Lets set f_0 (x)=1 to achieve f(0)=1. Now to get closer to f(x)=f ' (x), we set f_1 (x)=1+x, and f_2 (x)=1+x+x^2 /2 and so on. In the limit, we get the formula that is the way to define e^x where c is a matrix or a complex number. Now with enough analytical skills, you can prove that this function f behaves like the following: f(x)=(f(1))^x. So if we set f(1)=e, then f(x)=e^x. This even works in the limit version of the derivative: ( f(x+h)-f(x) )/h=(e^x e^h -e^x)/h=(e^x (1+h+ O(h^2))-e^x)/h=(e^x +x e^x -e^x)/h=e^x h/h=e^x Now to prove that this definition of e is the same as the others (for example ((n+1)/n)^n for n = infinity) is not simple, but not noteworthy hard.
I'm preparing for math exams in may and this is the kind of content I need right now, I'll pray for more blackpenredpen content suggested for me by the youtube algorithm 🙏
My preferred definition of e^x is it's the solution to the differential equation y'(x)=y(x) with the initial condition y(0)=1. I like this definition because it's simplest sounding definition, you get the differentiability of e^x for free and I think it's an easy starting point to show the other 3 common definitions on wikipedia are equivalent without making circular arguments. It also makes defining the inverse, ln(x), in terms of the integral of 1/x relatively straight forward. All you have to do is start with y^-1(y(x))=x, differentiate both sides, use chain rule and solve for dy^-1/dx. You can also easily use the initial condition to deduce y^-1(1)=0 or ln(1)=0. Using seperation of variables, we can define y=exp(x) to satisfy the integral equation from 1 to y of dt/t=x. Dividing the interval [1, x] into n equal parts and making successive linear approximations you can deduce y(x) is approximately (1+x/n)^n. You can then make this exact by taking the limit as n approaches infinity, which is the compound interest definition of exp(x). Assuming you can interchange the limit and derivative it's easy to check this limit satisfies the differential equation. After that you can apply the binomial theorem and evaluate the limit as n approaches infinity to get the power series representation of exp(x). To prove that exp(x) is in fact an exponential function, you can use the power series representation to show that exp(x+y)=exp(x)exp(y), a property that is only applies to exponential functions, assuming the function is continuous and maybe meets a few other criteria. You could then define the limit of (1+1/n)^n as n approaches infinity to be e and use that as your base. As you can probably tell it takes a significant amount of work to rigorously justify the e^x notation using my preferred definition and BPRP's definition is arguably more intuitive. However some other people in the comments pointed out some problems BPRP's definition, so I think I prefer this one.
Theorem: There exists one unique function verifying f'=f and f(0)=1. We define this to be the exponential function exp(x). Therefore [exp(x)]'=exp(x) by definition
@@xinpingdonohoe3978 sure but it's probably the easiest way to define the exponential function. We work with this one in real analysis. Even for the bprp case you need to prove unicity which he doesn't.
@@SimsHacks it's the easiest but it's hardly doing much. You're just listing criteria, naming it e^x and claiming it's correct by definition. I do admit that sometimes definitions can be useful, like proving the integral from 0 to ∞ of e^(-x^2) - (√x)e^(-x) dx = 0, but there has to be some work going into it otherwise it's almost arbitrary.
@@SingaporeSkaterSam Wow, you are seriously arrogant. Do you think people with Ph.D's in mathematics do not know the history behind e? I am baffled by this comments section. No respect for mathematicians in here, I guess.
Finally! Someone explained this in a way that didn't seem like circular logic. I asked my professor about this when I first took calculus my freshman year of college. The answer I got never explained how we knew that d/dx(e^x) = e^x. It was frustrating because I felt like I was supposed to take it on faith. Your first demonstration opened the door to understanding why all the e and ln related rules exist. Thank you.
Answer to Q1 & Q2: . . . . . Q1: What is log base b of x? Let's name that value a, then a is log base b of x b^a=x by definition of the logarithm. Let's write that with base e instead (i.e. b= e^(ln b) ) : x=b^a=(e^ln(b))^a=e^(a*ln(b)) Take the natural logarithm of this: ln(x)=ln(b^a)=a*ln(b) a=ln(x)/ln(b) Q2: What is the derivative of log base b of x? Just differentiate the answer of Q1: d/dx(a)=d/dx(ln(x)/ln(b))=(1/x)/ln(b)=1/(x*ln(b))
wow! what a superb video. crazy to think there are nearly half a million or so people in the same situation: their teachers just gave them some generic compound interest problem to find how exponential growth works (if lucky), told e is 2.718……… and then told that e^x differentiates to itself, very few show the why.
We cant really use that to prove the derivative of e^x, because we need the derivative the get the taylor series expansion, which would result in circular reasoning.
@@Ninja20704 Ye the taylor Series expansion is also e^x’s definition in summation form right? Cause then when you take the derivative it ends up the same, since infinity-1 is simply infinity
@@Ninja20704 nope. Taylor series is a way to find the expansion but you don't need it to prove that sum from 0 to infinity of (x^n)/n! converges. We can just define e^x as that series
Let y=e^x Taking log both the sides, Ln(y)=xln(e) Ln(y)=x Differentiating both the sides with respect to x, 1/y(dy/dx)=1 Dy/dx=y But y=e^x Hence dy/dx= e^x
Funny enough if you arrange the limit (e^h-1)/h =1 as h->0, you get e=(1+h)^(1/h) as h->0. e=(1+h)^(1/h) as h->0 is a generally used interest formula, that is continuous growth.
I worked out that the area under the curve of 1/x (or any value b/x) must be logarithmic because you can set up proportional intervals where the trapezoids or rectangles that you use to approximate the area have constant areas. (Example: If you take trapezoids over [1, 2] that go 1, 1.1, 1.342, 1.7, 1.901, 2, then you can take trapezoids over [2, 4] that go 2, 2.2, 2.684, etc., and they'll have the same areas.) That means that the area from 1 to x^n is n times the area for 1 to x. We can approximate the areas from 1 to 2 and 1 to 3 and find them significantly less than 1 and somewhat greater than 1, respectively, so there must be some number I'll call ê that's a little less than 3 where the area is exactly 1. I haven't worked the rest of it out, but it involves the fact that dy/dx and dx/dy are reciprocals and the usual derivation of e.
After defining e it's easy to differentiate a^x just take log of both sides. ln(y) = xln(a) and now differentiate. (dy/dx)1/y= ln(a) so dy/dx=yln(a) = a^x ln(a)
Blackpenredpen the way to evaluate the limit of 2^h -1/h surely would be, by doing a substitution where u = 2^h - 1 Therefore evaluating lim u-> 0 of u/log2(1+u), then reicprocating and and doing 1 over allows you to take 1/u to be the coefficient of the logarithim to use as an exponential. 😊
If we look at the first principle i.e. f(x+h)-f(x)/h, when we take e^x as our function the taylor series expansion is basically e^x = (1 + x + x^2/2! ....... x^n/n!) - (1) when we apply this in the first principle and take tend h to 0 we find lim h->0 e^x(e^h-1)/h using (1) and replacing x -> h lim h->0 e^x(h + h^2/2! .... h^n/n!) dividing by h we get a 1+h/2! + h^2/3! .... h^n-1/n! applying our tentative value, d(e^x)/dx = e^x
From this definition of e we can directly get another one: We need x^h -1 to be h soo x^h is 1+h and one way is to simply put the exponent in the definition of e to be 1/h and the base 1+h soo we get limit as h goes to 0 of (1+h)^(1/h)
As illustrative as that was, the only way one really knows to start at 2^x is either using calculator to determine e is approximately 2.718 OR using the lim n approaches infinity (1+1/n)^n definition of e which really more of a result that follows from some not so obvious calculus
lim e^x-1 _____ x->0 x approaches 1, this becomes quite clear if you think about it graphically, the instantaneous slope of e^x at x=0, is it's derivative at x=0 or just 1, so when you increase x by some tiny dx, the change that takes place in e^x or y(direction) is also that same value or dx due to the slope being 1at that instant, the difference is that e^x changes from an initial value of 1 where as x changes from an initial value of 0, so when we subtract that '1' we get the same value in the numerator and in the denominator both being dx in this case, and the ratio is 1
@@angelmendez-rivera351 This is in no way a proof of any sort, it literally comes from the fact that e^x is it's own derivative, it's much more of an interesting consequence of the property of e, people generally( at least what I've seen) are focused on finding a proof for this in limits, to know why this limit approaches 1 when x approaches 0, but I think it's much easier and makes a lot of sense graphically why this limit approaches 1.
@@lightspeed2014 The fact that you are not grasping why people focus on finding proof for this limit equation tells me that you are missing the point of the entire problem. You cannot use the fact that the derivative of the exponential function is itself to prove the limit is 1, because you need to prove the limit is 1 to prove the function is its own derivative. This is why you can either start by defining e such that lim (e^h - 1)/h (h -> 0) = 1, in which case the equality is true by definition, or you can start with another definition of e, and use this definition to prove the equality. You cannot prove it without using the definition of e. Appealing to graphs does nothing, because the graphs are based on the fact that those proofs exist in the first place. Of course, if your course is not interested in explaining the proofs, but instead wants you take these statements as facts on as a matter of faith and memorize those facts, and merely wants you to be convinced that accepting these facts is not counterintuitive, then there is no issue with using invalid logic, and in that context, there is nothing wrong with using graphs for visual aid. However, this video is not such a context. This video is literally about a formal proof of why the exponential function is its own derivative, and about why the constant e relates to this.
@@angelmendez-rivera351 I told you this before, and I'm telling you this again, this IS NOT A PROOF, it's something that I found interesting and something that is quite intuitive and shows the beauty of Mathematics, I NEVER SAID THIS WAS A PROOF, and I don't know about you but I don't memorize stuff for what it is, math isn't about memorizing, I don't know how you came to the conclusion of MEMORIZATION by an example of a frickin Graph. And Math isn't only about proving stuff either, it's much more than that. Again, because you probably don't get this statement, THIS IS NOT A PROOF And I also said in my previous reply that THIS IS A CONSEQUENCE OF e^x being it's own derivative, in no way is this a proof, but I'm not talking about a proof in this comment, I already got that from the video, I was talking about an interesting mathematical fact, which you clearly didn't get.
@@lightspeed2014 *I told you this before, and I'm telling you this again, this IS NOT A PROOF, it's something that I found quite interesting and something that is quite intuitive,...* Breaking news: no one cares. This is a proof video. *I NEVER SAID THIS WAS A PROOF,...* No, I never said that you said it is a proof, and I have no idea why you think I said this. However, the fact that you posted this at all on a PROOF video, and you start it with your stupid pretentious "ClEaRlY, tHe LiMiT iS 1..." suggests that you want it to be treated as proof. If you did not want it to be treated as a proof, then you should have never posted your comment. *I don't know you came to the conclusion of MEMORIZATION by an example of a fricking Graph.* For someone who complains that I said that you said something you never said something, you actually accuse me of saying things I never said too much. I never arrived to "memorization" as a conclusion. I mentioned "memorization" as part of a side fact that was relevant to the point I was actually making. *And Math isn't only about proving stuff either, it's much more than that.* Technically, you are correct, but really, 90% of maths is semantics, and 9% is proof. *Again, because you probable don't get this statement, THIS IS NOT A PROOF* I am glad I proved you wrong, then. *And I also said in my previous reply that THIS IS A CONSEQUENCE OF e^x being it's own derivative,...* No, it is not. This is backwards. It is not the case that the limit is 1 _because_ exp is its own derivative. Rather, it is that exp is its own derivative _because_ the limit is 1. The fact that you do not understand this, despite me having stated this, is what makes both your comment and your attitude so irritating. *...in no way this is a proof...* Yet the fact that you say this is contradicted by your statement that "this is a consequence of...". Your argument is not even internally consistent. *I was talking about an interesting mathematical fact,...* There is nothing interesting about it. Everyone who thinks about it for 3 seconds knows that the properties of the function are made visually evident in the graph of the function. Nothing special there. Anyway, I will mute you. You use CAPS LOCK too much, despite being the one not understanding the objection being presented. It feels like I am talking to a toddler. Bye.
I'm just doing the same thing, wondering why derivative of e^x is just the same function, so I use the definition of Derivative but in little different from the video d/dx (e^x) = lim a->0 [e^(x+a) - e^x]/a d/dx (e^x) = lim a->0 [e^x*e^a - e^x]/a d/dx (e^x) = lim a->0 e^x*[e^a - 1]/a d/dx (e^x) = e^x*lim a->0 [e^a - 1]/a Subtitute [e^a - 1] = t then a = ln(t+1) and when *a* approach 0, *t* is also approach 0 d/dx (e^x) = e^x*lim t->0 t/ln(t+1) ln(t+1) is similar with t when t is approach 0 so lim t->0 t/ln(t+1) = 1 d/dx (e^x) = e^x*1 d/dx (e^x) = e^x That's it.
well, i always thought of it like this... (b^x)' = b^x lnb Doing the same for e^x, we have e^x lne, or simply e^x That doesn't prove anything, but that's how I've never forgotten the derivative of b^x (the same goes for logb(x) being 1/xlnb and lnx simply being 1/x.)
@ 1:36 "...we get zero over zero; that's indeterminate" Steve, you are the first person (other than myself) I have heard say that's indeterminate, rather than "undefined". Thank you!
It's only in the context of limits, that it's called indeterminate. In general, division by zero is undefined. And without further information about how the top and bottom both get to zero, it is undefined. He has a video called "5 levels of no solution" that explains this. Special cases of limits that involve either division by zero, or multiplication by infinity, or subtraction between two infinities, are considered indeterminate forms, which means that it is possible to do more work to resolve them as a finite number. It has to do with how fast each component of the indeterminate form, approach either the zero or the infinity, that allows you to do this. For instance, consider (x^2 - 4)/(x - 2) as x approaches 2. In this case, the numerator both approach zero, but due to the fact that they both approach zero at the same rate. The (x^2 - 4) can be factored as (x + 2)*(x - 2), and we end up with a hole, or a removable discontinuity at this problem point, due to the fact that both the top and bottom approach zero at the same rate. By contrast, had we been given (x^2 - 4)/(x - 2)^2, we'd still have a division by zero after removing the hole. This is because the bottom approaches the zero at a quadratic rate, while the top approaches zero at a linear rate, and the bottom ends up winning, and making the limit not exist, instead of being a finite value.
Maybe the best explanation of the derivation of e on youtube. The only thing missing is that you should've algebraically derived the formula for e from the difference quotient. That way viewers can see exactly how the formula for e is algebraically derived from trying to isolate an exponent base whose derivative is identical to that exponential function.
@John Spence I didn't find that to be true. It's just an extension of the chain rule, and critical for related rates. I taught differential Calc from the perspective that it's just Algebra with limits. The chain rule is easily compared to solving an equation using "opposite operations". Most of my students seemed to get that.
@@spirome28 ya I’ve since changed my position and you are correct. The only problem I find with it is it feels like a cheat of the long form proof methods.
I see what you're saying and I respect that. Keep in mind my comment was coming from my experiences working with teenagers who generally didn't plan on majoring in theoretical math at university. relating it back to Alg2 gave me the most success with the most students. personally, I barely remember "Advanced Calc" from when I was in college where we just went through Calc 1 to Diff Eq and proved the all the theorems. That's where my hatred for delta-epsilon bands comes from lol.
This is fine, and probably the best definition there is, since it is not hopelessly circular like almost every other definition there is, all while being simple.
Ça dépend peut etre des profs mais moi on m'avait bien défini e telle que e= lim(1+1/x)^x avec x-->+∞. Pour ce qui est de la fonction exponentielle, on nous la définit comme l'unique fonction f telle que f=f' et f(0)=1. On note cette fonction exp
@@ablasphemite1940 There is the additional requirement that exp(0) = 1. This immediately excludes exp(x) = 0 from being possible. The equation dy/dx = y has exactly one solution satisfying y(0) = 1. This solution is exp. Then, we define e := exp(1). Well, in theory, this comes about more naturally as [exp^(-1)][lim (1 + ε)^(1/ε) (ε -> 0)] = 1, but since e = lim (1 + ε)^(1/ε) (ε -> 0) and the above equation implies lim (1 + ε)^(1/ε) (ε -> 0) = exp(1), we have e = exp(1). Anyway, every property of the exponential function can be derived extremely easily from this definition. Deriving the Taylor series is literally trivial. The functional equation can be derived from the Taylor series using the binomial theorem and the Cauchy product theorem. Also, exp(z) = lim (1 + z·ε)^(1/ε) (ε -> 0) can be proven using Tannery's theorem, hence further justifying the connections. The properties of the natural logarithm can also be derived very easily from this.
There is for me a problem in this proof : At the beginning, you try to calculate the derivative of 2^x but at this point if you don't have the exponential function you cannot (I can be wrong) define what is 2^x and moreover you cannot compute 2^0.0001 for your limit.
I edit : In fact, you can define 2^(p/q) with p and q two integers with a q-th root therefore you can define 2^x on Q. Finally It's likely that you can use the density of Q to define 2^x for all real values but again, it feels complicated to me.
@@kabsantoor3251 No, that is incorrect. You cannot define 2^x as being the power series you mentioned, because that power series is in turn defined in terms of ln(2), which you approximated as 0.693, but ln(2) is itself defined in terms 2^x. You have thus given a circular definition, which is to say, not a definition at all.
@@valentinenprepa1379 You make a fair argument, though: 2^x is not typically regarded as well-defined by peer-reviewed publications by mathematicians for arbitrary real x. Of course, there are different constructions for which using the notation 2^x to denote a well-defined expression in certain specifiv contexts is very practical, but this notation would rarely be used otherwise. As for the exponential function, it is just defined as a power series, most of the time, and then 2^x is just taken to be as an abbreviation for exp[ln(2)·x], although in practice, such abbreviations are actually rarely used.
@@angelmendez-rivera351 @Angel Mendez-Rivera No circular reasoning. See if I define 2^x as a shorthand for lim (1+ x/n)^k*n for some(yet not defined) k and n-->infinity and expand the binomial via binomial theorem, it's not hard to show that the derivative of 2^x is this number k times 2^x. We fix k by returning to the original binomial and looking at the value of the infinite power series for k=1. This way we're led to the number e. Clearly e is the number such that e^x (important to understand in this context as the infinite polynomial rather than e multiplied a certain number of times) is its own derivative. Changing to any other number say 2 changes k as defined by the binomial - that's the proportionality constant between the rate of change and the function itself.
You can actually take the limit directly and show that it equals the log of b. \lim_{x ightarrow0}\frac{b^x-1}{x} Substitute y=b^x \lim_{x ightarrow0}\frac{b^x-1}{x}=\lim_{y ightarrow1}\frac{y-1}{\log_b(y)} Change base \log_b(y)=\frac{\log(y)}{\log(b)} \lim_{x ightarrow0}\frac{b^x-1}{x}=\log(b)\lim_{y ightarrow1}\frac{y-1}{\log(y)} Hospital rule \lim_{y ightarrow1}\frac{y-1}{\log(y)}=1 \lim_{x ightarrow0}\frac{b^x-1}{x}=\log(b)
We can also prove it by using the infinite series of sum from n = 0 to infinity of x^n/n! When we take the derivative of the whole series, then we notice that the derivative is equal to the original series Q.E.D
I’m confused at 8:07. How are you able to differentiate that without the assumption that the derivative of e to x is e to x? It seems like a circular argument. Not saying it is, but I’m not seeing how you can do the chain rule there without having proven the derivative of e to x.
Love to see some of the basic (but very important stuff). Many calculus students will find this extremely helpful. I always hated the definition of ln(x) as an integral of 1/x. It's WEIRD and backwards and comes out of nowhere.
Historically, that is the way we defined ln(x), since this property of natural log was discovered before the number e. To learn it more intuitively, I think it makes more sense to define e as the value of base b, such that d/dx b^x = b^x. Then, use this as the starting point to prove the historical ways that ln(x) and e as a limit, were previously defined.
When I was in fifth grade, I had started to really love math. It was my passion, but then my parents found out I was getting good at math so they kinda forced me to study more. This made me not feel as interested in math again, and I didn't listen to them, so I didn't study. But then, 1 year later, I decided that I wanna do math again. This is gonna be the start of my journey, hopefully my parents don't make me stop again...
Hmm... Pretty sure in my math class exp(x) was defined as an infinite series (alongside sin(x) and cos(x)). From there you prove all of its properties. One of the properties is that the infinite series happens to differentiate to itself. Another property is that exp(x) = e^x where e is a constant. So starting from "why is d/dx(e^x)=e^x?" is just doing it all backwards, which makes things conceptually more complicated...
Surely, the easiest way is to define e^x as a power series. Then, differentiate that. The first term, 1, disappears. The second term becomes 1. The third becomes the second, the fourth the third etc so you end up with the same power series you started with.
@@ProCoderIO normally yes. But you can construct the taylor series with simple observation if you start by df/dx=f(x) and then excluding zero as a solution. It's not at all rigorous and just a handwavy method I developed for my students but at least to me it's the easiest way to show e^x is it's own derivative. Ignore everything you know so far and just take any old number (do yourself a favor and take 1. Otherwise you have to divide in the end) and write two colums. df/dx and f(x) Place 1 in both. Yeah the derivative of 1 is 0. 0+1=1 that's kind of the step that almost legitimizes the method. It means we can get an infinite sum where each term is the integral of the last. But to really show that we just integrate 1, get x and fill that in the f(x) column. Because both colums have to be equal you put x in the df/dx column as well, integrate again, get x^2/2 Rinse and repeat until you figured out the pattern. The thing is you kinda need to conjure the fact this is the taylor expansion of e^x from thin air. What I do with my students is to not do that but some theatralics: I act like this infinite function gets too annoying and just plot it. After 20 terms or so it should look exponential. Then I ask my students if they think it looks like an exponential function. Most will agree after getting 2 or 3 examples (like 2^x and 3^x). Then we can simply use the fact that a^1=a and plug in 1 in our truncated polynomial. However we have to loosely make sure we got an exponential there. My kids ALWAYS are skeptical about that. After all it's a polynomial they see on the board be it an entirely useless one. So I take another exponent rule. (a^b)^c=a^bc=(a^c)^b or (e^1)^2=e^2 Meaning f(2) should be reasonably similar to f(1)^2. Typically I run through a couple more powers (like 3 and 4) before explaining that I truncated the function after 20 terms meaning, my polynomial is never perfectly it's on derivative. But e^x which is equal to the full infinite polynomial is it's own derivative. And then I show the natural log and define the derivative of exponential functions from there. I don't like my kids having to just get a derivative from the calculator and see "every exponential function has a weird constant in the derivative.". Don't know why. I just don't like that method. I prefer if they can do every step of the way themselves. Even if we have to take a detour for them to do that.
I think the point of this is to do a method that students can do when they are first learning derivatives. The traditional way to get the derivatives of these functions is to do them after learning the integral and defining the integral of 1/t as the log function, then taking an inverse to get e^x. In your way you have to wait until you go through what is typically taught in 2 years of calculus. BPRP’s way is the typical method for those that want to teach e^x right at the beginning
@@stephenbeck7222 Typical method? Almost no schools where I live define ln(x) in terms of an integral of 1/t in an introductory calculus course. In fact, integrals themselves are typically taught in calculus 2.
@@ProCoderIO "Don’t you need the derivative first to derive a Taylor series?" He's not deriving anything. It's a definition. It's a very confusing one since you already know what "e", "^", and "x" are so you shouldn't need a new definition for e^x. But it turns out that the old definition of "^" doesn't generalize very well (remember how confused you were when you first saw e^(pi i) = -1) so mathematicians like to just start over and define e^x as a power series.
There was a slight missed opportunity in the first part when trying to find the derivative of 2^x. It was all cool up until the end of the column when you decided you couldn't use L'Hopital's Rule because you'd need the derivative of 2^h. It's actually okay and gets you a little farther though obviously not all the way. The 2^h limit we get is *related* to the derivative of 2^x, but it isn't the same thing. So let's continue where we can: d/dx(2^x) =2^x * lim{h to 0}[(2^h -1)/h] Dubiously using L'Hopital's Rule turns that limit into lim{h to 0}[d/dx(2^h) / 1] which is lim{h to 0} d/dx(2^h) We don't know what that is, but call that f'(0), since that is the slope of the line tangent to 2^x when x=0. So letting f(x)=2^x We get that f'(x) = 2^x * f'(0) = f(x)*f'(0) And that's a pretty big deal: the derivative of an exponential function *anywhere* is equal to the *value* of the function at that point, scaled by the *derivative* of the function at x=*zero*. Honestly, that's a weird result in my opinion. :) Then the rest of what you said is great and follows right away like you said, but I figured I'd point out the f'(0) bit. Great video as always :)
*It was all cool up until the end of the column when you decided you couldn't use L'Hôpital's rule because you'd need the derivative of 2^h.* Correct. We need to evaluate lim (2^h - 1)/h (h -> 0) in order to prove that the derivative of x |-> 2^x exists and find its value. If you use L'Hôpital's rule to evaluate lim (2^h - 1)/h (h -> 0), then you need to take the derivative of h |-> 2^h, which is a problem, because the limit is being evaluated to prove that such a derivative exists in the first place. In other words, you are starting your proof by starting with your conclusion. *It's actually okay and gets you a little farther though obviously not all the way.* No, it is not okay. Starting the proof by assuming that the thing you are trying to prove is true is never okay. *The 2^h limit we get is **_related_** to the derivative of 2^x but it isn't the same thing.* No, you are super wrong. Of course it is the same thing: x |-> 2^x and h |-> 2^h are literally the same function. Yes, we are using different symbols to denote the input parameter, but this does not make the functions different. Your argument is like saying "1/2 and 2/4 are not the same rational number since they are different symbols".
@@angelmendez-rivera351 I think you misunderstood their post. They don't assume anything they are trying to prove, as they aren't trying to prove the value of the derivative, nor whether it exists. They simply assume the derivative exists in order to explore what properties that derivative (if it did exist) would need to have. And they come up with one: if f(x) = b^x, then f'(x) = b^x * f'(0). Or, as they say it in words: "the derivative of an exponential function *anywhere* is equal to the *value* of the function at that point, scaled by the *derivative* of the function at x=*zero.*" And, when we do eventually find the derivative, this is clearly correct. f'(x) = b^x * ln b, plug in x = 0, and you get f'(0) = ln b. Thus f'(x) = b^x * f'(0) I think it's neat that they discovered something like that.
@@turkeypedal *They don't assume anything they are trying to prove, as they aren't trying to prove the value of the derivative, nor whether it exists.* Yes, they are. They are, because the comment is in direct response to a portion of the video in which BPRP is trying to prove a specific value for the derivative, and the post is complaining that the proof did not allow for L'Hs rule to be used. Also, the post is very much trying to prove something about the derivative: that if f(x) = b^x, then f'(x) = f'(0)·b^x. *They simply assume the derivative exists in order to explore what properties that derivative (if it did exists) would need to have.* They pretended that this is what they were doing at the end of the post, but this is completely undermined by the fact that, at the beginning of the post, they specifically were complaining about the fact that L'Hs rule was not allowed in a formal proof of the statement f'(x) = ln(b)·b^x, and they even tried to argue that using L'Hs rule is not actually circular. You may not have noticed it, but this is precisely what the post did in its first few paragraphs. *I think it's neat that they discovered something like that.* "Discovered" is a strong word, but I agree with you. However, I have no issue with the discovery they made. My issue is with the first few paragraphs of the comment, where they are trying to argue that the video should have allowed L'Hs rule in the evaluation of the limit, and that BPRP "missed an opportunity" by not using L'Hs rule. The comment is objectively wrong: L'Hs rule is not permissible in a formal proof, and no opportunities "were missed", because the point of that segment of the video was to formally prove a statement, which itself was done for the sake of a grander purpose, and because the discovery in this comment is in no way relevant to the grander purpose of the video.
Thanks blackpenredpen! This makes sense. Since the derivative of a^x=a^x[ln(a)]. If the base is replaced by e then the derivative of e^x=e^x[ln(e)] and by definition ln(e)=1 hence the derivative of e^x=e^x. Can you do a explanation of why d/dx[a^x]=a^x[ln(a)]? Thanks.
The existence of e is strong evidence that we live in a computer simulation where someone was just too lazy to come up with different constants so they just plugged one universal constant in for a whole bunch of different uses.
The common way to solve lim x->0 (a^x - 1)/x, is to substitute a^x = 1 + 1/p x = loga(1 + 1/p) so that as x->0 threfore p->∞. lim x->0 (a^x - 1)/x = lim p->∞ (1 + 1/p - 1)/loga(1 + 1/p) = lim p->∞ (1/p)/loga(1 + 1/p) = lim p->∞ 1/(p.loga(1 + 1/p)) by using log's property: a log b = log(b^a) = lim p->∞ 1/(loga((1 + 1/p)^p)) as we know that: lim n->∞ (1 + 1/n)ⁿ = e = 1/loga(lim p->∞ (1 + 1/p)^p) = 1/loga(e) and by using another log's property: loga(b) = 1/logb(a) = loge(a) = ln a
@@aaaa8130 That's kinda like saying "you would have to define how to multiply first. And for that you would have to define how to add first. And for that you would have to define what a number is first. And for that you would have to define the axioms of math first. It's already defined and thats not the point so yeah.
So e was originally an undiscovered number that existed for many mathematical things, like continuous growth and stuff, and that we didn’t know could be what it is now…but it was originally defined as some number with exponent x that would produce the answer which is itself (proven in the video). But which came first…the self-generating cyclical derivative or the irrational number existing in nature what will help evolve many subjects within mathematics?
Proving that the value of the h-limit is a continuous function of the base might be tricky, assuming that we don't know about e or ln for the discussion. There might be a short way but it is necessary to know here.
Ah, you were right that the existence of the h-limit itself is plausible-but-unproven here. I suppose further technicalities can be treated the same way.
Summary: by definition!
yes sir
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🦒: I think mathematician
@蛋已经破仍爱三角函数微积分 Ash Ketchum, if he was a mathematician.
@蛋已经破仍爱三角函数微积分 🦒
Love from India 🇮🇳🇮🇳❤️❤️❤️
Fun fact, based on this definition you can actually rearrange the equation to arrive at the definition of e based on compound interest:
lim(h-->0)((e^h - 1)/h) = 1
Let h_0 be a small number such that
(e^h_0 - 1)/h_0 = 1
e^h_0 - 1 = h_0
e^h_0 = 1 + h_0
e = (1 + h_0)^(1/h_0)
Convert back to limit
e = lim(h-->0)(1+h)^(1/h)
i.e. compound interest recursion formula with infinitesimally small interest rate for an infinitely large number of compounding periods.
Damn I was just finished typing this up in my comment when I saw your comment 👌
Oh that's actually really cool
When you write "let h_0 be a small number such that: (e^(h_0)-1)/h_0=1"
How can you tell that such h_0 exist? And in particular if you rearrange the equation one may get e^(h_0)=h_0+1, and if I'm not mistaken there exist an inequality which tells that e^x>=x+1, whit equality at x=0
@@Evan-ne5bu I suppose it is equivalent of using the property that limit of f(g(x)) is equal to f(limit of g(x)) where f=(e^h - 1)/h and g=h. So h_0 is just a simpler way of writing lim(h-->0)(h).
Also, didn't know about that inequality, nice! It is true for x>0 if you graph it. In this case, the equation e^(h_0)=1+h_0 is true if we are talking about limits since h_0 is a small number approaching 0
@@Evan-ne5bu Also fun fact #2, using base 'a' in the exponential instead of base 'e' you can show that lim(x-->0)((a^x - 1)/x) = ln(a). If anyone knows a way to directly approach this limit let me know, I've been curious haha
It's really important to understand that it's not that the derivate of e^x is just e^x, but that's WHY the number e exists. Videos like yours really inspire me to share my own videos as well!
Nope.
The reason e exists is because when multiples I think it was 1.0001 the values of whole numbers were interesting to Euler and he wanted to calculate what the base was
@@Asiago9 what?
There are many definitions of e it is not the only reason e exists 😂
For example lim h->infinite (1+1/h)^h.
You can come up with this definition without calculus for example compound interest where you have an infinite amount of compound interest and every interest is 1+lim h->infinity(1/h) times higher than the previous interest.
As a veteran calculus teacher, I enjoy learning something from almost every BPRP video. Thank you!
I am glad to hear this. Thank you.
As a veteran, one can conclude this is circular reasoning which meant euler's number has NOTHING to do with reality. Which meant the actual value of pi is completely different. It is also ignorant reasoning to presuppose the perimeter of circumscribed polygon is always longer than the circumference of the unit circle at different number of sides. There must be ways to prove that and yet nobody even attempted it. Clown world this is.
@@lukiepoole9254 this dude thinks that he's more knowledgeable about math than the collective efforts of the smartest mathematicians over thousands of years
@@lukiepoole9254 what
@@Maple_MK Why are there not a single absolutely undeniably irrefutable rigorous proof for circumference of circle? The pi based on circumference of circle isn't 3.14159265; It is literally 4sqrt(1/golden ratio).
I love how this shows the process of discovering the derivatives and defining things along the way; it makes math much more interesting when you see how these things might’ve been originally figured out.
You can actually evaluate the limit of (2^h-1)/h by using the variable substitution u=2^h-1, turning it into the limit as u->0 u/log2(u+1) = 1/((1/u)log2(u+1)) =1/log2((1+u)^(1/u)).
Change of variables to t=1/u gives lim t->inf 1/log2((1+1/t)^t). This includes the limit definition of e which can be shown to exist by the monotone sequence theorem.
Then you have 1/log2(e) =log2(2)/log2(e) = ln(2) by the change of base formula.
I found out about this from Khan Academy.
Justo así es como se hace, él lo hice de una manera muy estilo ingeniero.
Jajaja
2:15 Here's how the ancient people probably did it:
By using u substitution and letting 2^h=u, we can change the limit as h approaches 0 to the limit as u approaches 1
We can say that h is log base 2 of u from this statement.
From there we can take the derivative with respect to u. By using the definition of the derivative, we can find the generalized derivative for any function log base n of x, using logarithm properties and the definition of e as a limit. From there we can find that d/dx log n of x= 1/(x*ln(n)). using that derivative we can plug in ln(2) as the solution of that limit and get 2^x*ln(2). Sorry if the text formatting on this is confusing, i could post a video about it
But before defining e in such a way, you need to prove that there exists a unique number with the described property and I can't say that that's quite easier then proving that the derivative of e^x is e^x using some other definition of e.
That was also my reaction when he did it. The existence and uniqueness of such a number are not guaranteed.
You could perhaps build a fonction that for all input a outputs the limit of (a^h-1)/h as h tends to 0, then prove it's continuous and strictly increasing, and finally apply the intermediate value theorem. That could work but that sounds more complicated than using an other definition.
Yeah it's a flawed but intuitive way of defining it.
I guess you could look at the function f(t) = d/dx (t^x) and see that f is continuous, and apply the Intermediate Value Theorem to t=2 and t=3
Excellent question! Good discussion, would like to see what more people say.
Let’s take a hypothetical number
(N^h - 1)
________. h-->0
h
Where this equals 1.
How do we know if it doesn’t settle in a value smaller than 1? Maybe it tends to a value like .85327 or something?
I’m assuming people can just say
If we plug in 2 for N, it’s lower than 1
If we plug in 3 for N, it’s more than 1
-> Therefore it’s somewhere between the two.
But plugging in number isn’t that satisfying, it’s not really a proof :/
What we can do is compare the outputs as the inputs get bigger
(a^h-1)/h Vs ((2a)^h-1)/h. The input is twice as big, but what happens to the output?
Both have a common denominator so let’s take that out.
Now what is bigger?
(a^h-1) or ((2a^h)-1)
Clearly ((2a^h)-1) is always gonna be larger than the previous value i.e (a^h - 1) as “a” gets larger.
Example:
(5^a)-1 < [(2*5)^a]-1
But the question is, how much? There’s probably a way that proves it isn’t asymptotic and extends to infinity,
I’ll try and think of a proof, but this is an excellent discussion and a reminder to always question our intuition. Good job guys!
@@skylardeslypere9909 The problem with that is that this relies on t^x being well-defined, which itself relies on the exponential function and the logarithmic function also being well-defined a priori.
We can use the definition given here to prove that e = 1/0! + 1/1! + 1/2! +1/3! + ... then we can prove that the sum converges hence proving the existence of e and it's uniqueness
We can even evaluate that limit by using the expansion
2^x = e^(xln2)
So e^(xln2) = 1+ xln2 +....
We plug that in
And get lim h->0 1+hln2...-1/h
Lim h->0 hln2+.../h
We get ln2 as limit as all other terms would give zero
for the exponential function b^x, the derivative by first principles becomes (b^x) lim h→0 ((b^h) - 1)/h = (b^x) lim h→0 ((b^h) - (b^0))/h, which is the instantaneous gradient at x = 0, and b = e is the only case where the gradient at x = 0 is 1
idk, I think it's a fun visualization of what that weird limit is actually representing
That's not a visualization.
I think when I learned it at school (in the math power lessons at grammar school with a teacher who had a math PhD), we started by pointing out that f(x)=1/x must have an antiderivative F (I'm avoiding to use the terms ln, e etc.). This cannot be calculated by using the (x^n)' = n*x^(n-1) formula because it would involve a division by 0, respectively it doesn't work for n=0. We chose to calculate the one with the constant leading to F(1)=0 and then made deductions about its inverse function G. We showed that G'=G, that G is of the form G(x) = Z^x by proving that for every x1 and x2, G(x1+x2)= G(x1)*G(x2). Finally we calculated G(1) = Z^1 = Z (which ie 2.71.....) and voila, we were done.
Thank you! This video helped me realize the relationship to define ln(a) as the limit of (a^h-1)/h as h goes to 0.
I think one intuitive way to define e is by defining e^x generally:
What we want is that e^0=1, and that the derivative is the same as the original function.
Lets set f_0 (x)=1 to achieve f(0)=1. Now to get closer to f(x)=f ' (x), we set f_1 (x)=1+x, and f_2 (x)=1+x+x^2 /2 and so on. In the limit, we get the formula that is the way to define e^x where c is a matrix or a complex number. Now with enough analytical skills, you can prove that this function f behaves like the following: f(x)=(f(1))^x. So if we set f(1)=e, then f(x)=e^x. This even works in the limit version of the derivative: ( f(x+h)-f(x) )/h=(e^x e^h -e^x)/h=(e^x (1+h+ O(h^2))-e^x)/h=(e^x +x e^x -e^x)/h=e^x h/h=e^x
Now to prove that this definition of e is the same as the others (for example ((n+1)/n)^n for n = infinity) is not simple, but not noteworthy hard.
For a few minutes I thought you were intentionally breaking up the white board into golden ratio rectangles!
That’s a great idea! I should do that one day.
Or by writing the value of it
Like:1.618........................
😅
how is this comment 19 hours ago? u dont seem to be a member....
@@blackpenredpen How the comment was done 11 hours ago?🤔
@@avishek_paul The video premiered a day ago which allowed people to comment early, but was only uploaded recently. I know, it's weird.
I'm preparing for math exams in may and this is the kind of content I need right now, I'll pray for more blackpenredpen content suggested for me by the youtube algorithm 🙏
My preferred definition of e^x is it's the solution to the differential equation y'(x)=y(x) with the initial condition y(0)=1. I like this definition because it's simplest sounding definition, you get the differentiability of e^x for free and I think it's an easy starting point to show the other 3 common definitions on wikipedia are equivalent without making circular arguments. It also makes defining the inverse, ln(x), in terms of the integral of 1/x relatively straight forward. All you have to do is start with y^-1(y(x))=x, differentiate both sides, use chain rule and solve for dy^-1/dx. You can also easily use the initial condition to deduce y^-1(1)=0 or ln(1)=0.
Using seperation of variables, we can define y=exp(x) to satisfy the integral equation from 1 to y of dt/t=x. Dividing the interval [1, x] into n equal parts and making successive linear approximations you can deduce y(x) is approximately (1+x/n)^n. You can then make this exact by taking the limit as n approaches infinity, which is the compound interest definition of exp(x). Assuming you can interchange the limit and derivative it's easy to check this limit satisfies the differential equation. After that you can apply the binomial theorem and evaluate the limit as n approaches infinity to get the power series representation of exp(x).
To prove that exp(x) is in fact an exponential function, you can use the power series representation to show that exp(x+y)=exp(x)exp(y), a property that is only applies to exponential functions, assuming the function is continuous and maybe meets a few other criteria. You could then define the limit of (1+1/n)^n as n approaches infinity to be e and use that as your base. As you can probably tell it takes a significant amount of work to rigorously justify the e^x notation using my preferred definition and BPRP's definition is arguably more intuitive. However some other people in the comments pointed out some problems BPRP's definition, so I think I prefer this one.
Theorem: There exists one unique function verifying f'=f and f(0)=1.
We define this to be the exponential function exp(x).
Therefore [exp(x)]'=exp(x) by definition
That's a bit of a cop out reasoning though.
@@xinpingdonohoe3978 sure but it's probably the easiest way to define the exponential function. We work with this one in real analysis. Even for the bprp case you need to prove unicity which he doesn't.
@@SimsHacks it's the easiest but it's hardly doing much. You're just listing criteria, naming it e^x and claiming it's correct by definition.
I do admit that sometimes definitions can be useful, like proving the integral from 0 to ∞ of e^(-x^2) - (√x)e^(-x) dx = 0, but there has to be some work going into it otherwise it's almost arbitrary.
I’d be suspicious of any mathematician who accepted this without understanding the motivation, or necessity, for e.
@@SingaporeSkaterSam Wow, you are seriously arrogant. Do you think people with Ph.D's in mathematics do not know the history behind e? I am baffled by this comments section. No respect for mathematicians in here, I guess.
e is actually e-rational
I see what you d-e-d here
@@RunningOnEmpty-f7ri +C what you did there
Finally! Someone explained this in a way that didn't seem like circular logic. I asked my professor about this when I first took calculus my freshman year of college. The answer I got never explained how we knew that d/dx(e^x) = e^x. It was frustrating because I felt like I was supposed to take it on faith. Your first demonstration opened the door to understanding why all the e and ln related rules exist. Thank you.
your teacher did not answer since this eq diff is the one defining the exponential function
Answer to Q1 & Q2:
.
.
.
.
.
Q1: What is log base b of x?
Let's name that value a, then a is log base b of x b^a=x by definition of the logarithm.
Let's write that with base e instead (i.e. b= e^(ln b) ) : x=b^a=(e^ln(b))^a=e^(a*ln(b))
Take the natural logarithm of this: ln(x)=ln(b^a)=a*ln(b) a=ln(x)/ln(b)
Q2: What is the derivative of log base b of x?
Just differentiate the answer of Q1: d/dx(a)=d/dx(ln(x)/ln(b))=(1/x)/ln(b)=1/(x*ln(b))
This is how I learned this at school...e was the value of a such that d/dx(a^x) = a^x. Very good and clear explanation from BPRP
Finally after 2 short videos one long video
Really innovative and cool 😎
I really thought you were gonna say d/dx(2^x)=e^x and 0.2 is even
Finally someone who explained it properly. Really helpful video.
wow! what a superb video. crazy to think there are nearly half a million or so people in the same situation: their teachers just gave them some generic compound interest problem to find how exponential growth works (if lucky), told e is 2.718……… and then told that e^x differentiates to itself, very few show the why.
3B1B made also video about this in his Calculus Series.
You could also find e^x sum definition and d/dx the sum aka taking the derivative of each fraction
We cant really use that to prove the derivative of e^x, because we need the derivative the get the taylor series expansion, which would result in circular reasoning.
@@Ninja20704 Ye the taylor Series expansion is also e^x’s definition in summation form right? Cause then when you take the derivative it ends up the same, since infinity-1 is simply infinity
@@Ninja20704 in fact, you can define exp as the sum of the serie without saying that it s a Taylor serie
@@Ninja20704 i was bored so i also did it with d/dx of sin x using it’s sum form/Taylor series and ended up with the sum form of cos x
@@Ninja20704 nope. Taylor series is a way to find the expansion but you don't need it to prove that sum from 0 to infinity of (x^n)/n! converges. We can just define e^x as that series
Let y=e^x
Taking log both the sides,
Ln(y)=xln(e)
Ln(y)=x
Differentiating both the sides with respect to x,
1/y(dy/dx)=1
Dy/dx=y
But y=e^x
Hence dy/dx= e^x
Thanks!
Funny enough if you arrange the limit (e^h-1)/h =1 as h->0, you get e=(1+h)^(1/h) as h->0.
e=(1+h)^(1/h) as h->0 is a generally used interest formula, that is continuous growth.
Perfectly executed! I love how you managed to connect everything so wonderfully and I could see the beauty in it.
i am think if we need to show the uniqueness of e when define as the number st lim (e^h-1/h)=1, i.e. e is the unique solution to lim (x^h-1)/h=1
I worked out that the area under the curve of 1/x (or any value b/x) must be logarithmic because you can set up proportional intervals where the trapezoids or rectangles that you use to approximate the area have constant areas. (Example: If you take trapezoids over [1, 2] that go 1, 1.1, 1.342, 1.7, 1.901, 2, then you can take trapezoids over [2, 4] that go 2, 2.2, 2.684, etc., and they'll have the same areas.) That means that the area from 1 to x^n is n times the area for 1 to x. We can approximate the areas from 1 to 2 and 1 to 3 and find them significantly less than 1 and somewhat greater than 1, respectively, so there must be some number I'll call ê that's a little less than 3 where the area is exactly 1. I haven't worked the rest of it out, but it involves the fact that dy/dx and dx/dy are reciprocals and the usual derivation of e.
Great stuff! Maths students today are so lucky to have the internet with great teachers like this man.
After defining e it's easy to differentiate a^x just take log of both sides. ln(y) = xln(a) and now differentiate. (dy/dx)1/y= ln(a) so dy/dx=yln(a) = a^x ln(a)
Alternative way to find e:
1. Graph f(x)=2^x
2. Stretch horizontally by scale factor f’(0)
3. g(x)=2^[x/f’(0)]=[2^(1/f’{0})]^x
4. Deduce e=2^[1/f^(0)]
Blackpenredpen the way to evaluate the limit of 2^h -1/h surely would be, by doing a substitution where u = 2^h - 1
Therefore evaluating lim u-> 0 of u/log2(1+u), then reicprocating and and doing 1 over allows you to take 1/u to be the coefficient of the logarithim to use as an exponential. 😊
very clear explanation, I was confused about the meaning of e for a long time!
Oh! I wanted to know about this for a long time. Thank you so much bprp for this tutorial!
If we look at the first principle i.e.
f(x+h)-f(x)/h,
when we take e^x as our function the taylor series expansion is basically
e^x = (1 + x + x^2/2! ....... x^n/n!) - (1)
when we apply this in the first principle and take tend h to 0
we find
lim h->0 e^x(e^h-1)/h
using (1) and replacing x -> h
lim h->0 e^x(h + h^2/2! .... h^n/n!)
dividing by h we get a 1+h/2! + h^2/3! .... h^n-1/n!
applying our tentative value, d(e^x)/dx = e^x
From this definition of e we can directly get another one:
We need x^h -1 to be h soo x^h is 1+h and one way is to simply put the exponent in the definition of e to be 1/h and the base 1+h soo we get limit as h goes to 0 of (1+h)^(1/h)
As illustrative as that was, the only way one really knows to start at 2^x is either using calculator to determine e is approximately 2.718 OR using the lim n approaches infinity (1+1/n)^n definition of e which really more of a result that follows from some not so obvious calculus
6:38 Oh. you wanted to know whether to use "log x" as default base e or "ln x" for this video? haha
lim e^x-1
_____
x->0 x
approaches 1, this becomes quite clear if you think about it graphically, the instantaneous slope of e^x at x=0, is it's derivative at x=0 or just 1, so when you increase x by some tiny dx, the change that takes place in e^x or y(direction) is also that same value or dx due to the slope being 1at that instant, the difference is that e^x changes from an initial value of 1 where as x changes from an initial value of 0, so when we subtract that '1' we get the same value in the numerator and in the denominator both being dx in this case, and the ratio is 1
Yes, graphically, this may be obvious, but it being obvious is not what we care about. You still need to be able to rigorously prove it.
@@angelmendez-rivera351 This is in no way a proof of any sort, it literally comes from the fact that e^x is it's own derivative, it's much more of an interesting consequence of the property of e, people generally( at least what I've seen) are focused on finding a proof for this in limits, to know why this limit approaches 1 when x approaches 0, but I think it's much easier and makes a lot of sense graphically why this limit approaches 1.
@@lightspeed2014 The fact that you are not grasping why people focus on finding proof for this limit equation tells me that you are missing the point of the entire problem. You cannot use the fact that the derivative of the exponential function is itself to prove the limit is 1, because you need to prove the limit is 1 to prove the function is its own derivative. This is why you can either start by defining e such that lim (e^h - 1)/h (h -> 0) = 1, in which case the equality is true by definition, or you can start with another definition of e, and use this definition to prove the equality. You cannot prove it without using the definition of e. Appealing to graphs does nothing, because the graphs are based on the fact that those proofs exist in the first place. Of course, if your course is not interested in explaining the proofs, but instead wants you take these statements as facts on as a matter of faith and memorize those facts, and merely wants you to be convinced that accepting these facts is not counterintuitive, then there is no issue with using invalid logic, and in that context, there is nothing wrong with using graphs for visual aid. However, this video is not such a context. This video is literally about a formal proof of why the exponential function is its own derivative, and about why the constant e relates to this.
@@angelmendez-rivera351 I told you this before, and I'm telling you this again, this IS NOT A PROOF, it's something that I found interesting and something that is quite intuitive and shows the beauty of Mathematics, I NEVER SAID THIS WAS A PROOF, and I don't know about you but I don't memorize stuff for what it is, math isn't about memorizing, I don't know how you came to the conclusion of MEMORIZATION by an example of a frickin Graph. And Math isn't only about proving stuff either, it's much more than that.
Again, because you probably don't get this statement, THIS IS NOT A PROOF
And I also said in my previous reply that THIS IS A CONSEQUENCE OF e^x being it's own derivative, in no way is this a proof, but I'm not talking about a proof in this comment, I already got that from the video, I was talking about an interesting mathematical fact, which you clearly didn't get.
@@lightspeed2014 *I told you this before, and I'm telling you this again, this IS NOT A PROOF, it's something that I found quite interesting and something that is quite intuitive,...*
Breaking news: no one cares. This is a proof video.
*I NEVER SAID THIS WAS A PROOF,...*
No, I never said that you said it is a proof, and I have no idea why you think I said this. However, the fact that you posted this at all on a PROOF video, and you start it with your stupid pretentious "ClEaRlY, tHe LiMiT iS 1..." suggests that you want it to be treated as proof. If you did not want it to be treated as a proof, then you should have never posted your comment.
*I don't know you came to the conclusion of MEMORIZATION by an example of a fricking Graph.*
For someone who complains that I said that you said something you never said something, you actually accuse me of saying things I never said too much. I never arrived to "memorization" as a conclusion. I mentioned "memorization" as part of a side fact that was relevant to the point I was actually making.
*And Math isn't only about proving stuff either, it's much more than that.*
Technically, you are correct, but really, 90% of maths is semantics, and 9% is proof.
*Again, because you probable don't get this statement, THIS IS NOT A PROOF*
I am glad I proved you wrong, then.
*And I also said in my previous reply that THIS IS A CONSEQUENCE OF e^x being it's own derivative,...*
No, it is not. This is backwards. It is not the case that the limit is 1 _because_ exp is its own derivative. Rather, it is that exp is its own derivative _because_ the limit is 1. The fact that you do not understand this, despite me having stated this, is what makes both your comment and your attitude so irritating.
*...in no way this is a proof...*
Yet the fact that you say this is contradicted by your statement that "this is a consequence of...". Your argument is not even internally consistent.
*I was talking about an interesting mathematical fact,...*
There is nothing interesting about it. Everyone who thinks about it for 3 seconds knows that the properties of the function are made visually evident in the graph of the function. Nothing special there.
Anyway, I will mute you. You use CAPS LOCK too much, despite being the one not understanding the objection being presented. It feels like I am talking to a toddler. Bye.
I'm just doing the same thing, wondering why derivative of e^x is just the same function, so I use the definition of Derivative but in little different from the video
d/dx (e^x) = lim a->0 [e^(x+a) - e^x]/a
d/dx (e^x) = lim a->0 [e^x*e^a - e^x]/a
d/dx (e^x) = lim a->0 e^x*[e^a - 1]/a
d/dx (e^x) = e^x*lim a->0 [e^a - 1]/a
Subtitute [e^a - 1] = t then a = ln(t+1) and when *a* approach 0, *t* is also approach 0
d/dx (e^x) = e^x*lim t->0 t/ln(t+1)
ln(t+1) is similar with t when t is approach 0 so lim t->0 t/ln(t+1) = 1
d/dx (e^x) = e^x*1
d/dx (e^x) = e^x
That's it.
well, i always thought of it like this...
(b^x)' = b^x lnb
Doing the same for e^x, we have
e^x lne, or simply e^x
That doesn't prove anything, but that's how I've never forgotten the derivative of b^x
(the same goes for logb(x) being 1/xlnb and lnx simply being 1/x.)
@ 1:36
"...we get zero over zero; that's indeterminate"
Steve, you are the first person (other than myself) I have heard say that's indeterminate, rather than "undefined". Thank you!
It's only in the context of limits, that it's called indeterminate. In general, division by zero is undefined. And without further information about how the top and bottom both get to zero, it is undefined. He has a video called "5 levels of no solution" that explains this.
Special cases of limits that involve either division by zero, or multiplication by infinity, or subtraction between two infinities, are considered indeterminate forms, which means that it is possible to do more work to resolve them as a finite number. It has to do with how fast each component of the indeterminate form, approach either the zero or the infinity, that allows you to do this.
For instance, consider (x^2 - 4)/(x - 2) as x approaches 2. In this case, the numerator both approach zero, but due to the fact that they both approach zero at the same rate. The (x^2 - 4) can be factored as (x + 2)*(x - 2), and we end up with a hole, or a removable discontinuity at this problem point, due to the fact that both the top and bottom approach zero at the same rate. By contrast, had we been given (x^2 - 4)/(x - 2)^2, we'd still have a division by zero after removing the hole. This is because the bottom approaches the zero at a quadratic rate, while the top approaches zero at a linear rate, and the bottom ends up winning, and making the limit not exist, instead of being a finite value.
# "division by zero is undefined."
Division by anything is perfectly well-defined; it is just not performable with zero, therefore indeterminate.
@@moebadderman227 Indeterminate and undefined mean completely different things. 0/0 is indeterminate in the right context, 1/0 is undefined.
# "Indeterminate and undefined mean completely different things."
Yes.
# "1/0 is undefined."
No.
Father of Calculus
9:26 wow thats a new way of doing a^x that I didn't know of. Mind blown 🤯
man i am addicted to your videos .It contains many information
Maybe the best explanation of the derivation of e on youtube. The only thing missing is that you should've algebraically derived the formula for e from the difference quotient. That way viewers can see exactly how the formula for e is algebraically derived from trying to isolate an exponent base whose derivative is identical to that exponential function.
Haven't visited your channel in ages. Aside from the mathematics which is always entertaining, you deserve an Excellence In Bearding Award (EIBA).
When I taught calculus to high schoolers I used the unit on implicit/logarithm differentiation to prove that e^x is it's own derivative.
ya but that blows their minds, most high schoolers I know cant handle implicit differentiation.
@John Spence I didn't find that to be true. It's just an extension of the chain rule, and critical for related rates. I taught differential Calc from the perspective that it's just Algebra with limits. The chain rule is easily compared to solving an equation using "opposite operations". Most of my students seemed to get that.
@@spirome28 ya I’ve since changed my position and you are correct. The only problem I find with it is it feels like a cheat of the long form proof methods.
I see what you're saying and I respect that. Keep in mind my comment was coming from my experiences working with teenagers who generally didn't plan on majoring in theoretical math at university. relating it back to Alg2 gave me the most success with the most students. personally, I barely remember "Advanced Calc" from when I was in college where we just went through Calc 1 to Diff Eq and proved the all the theorems. That's where my hatred for delta-epsilon bands comes from lol.
@@spirome28 lol
Just finished a level maths exams but even though I’m not doing maths at uni, I watch these vids every day, they’re so interesting
In France, e is defined as exp(1) and exp(x) is defined as exp(x)'=exp(x). No other explanation on what is e.
This is fine, and probably the best definition there is, since it is not hopelessly circular like almost every other definition there is, all while being simple.
Ça dépend peut etre des profs mais moi on m'avait bien défini e telle que e= lim(1+1/x)^x avec x-->+∞. Pour ce qui est de la fonction exponentielle, on nous la définit comme l'unique fonction f telle que f=f' et f(0)=1. On note cette fonction exp
d/dx[0]=0
Therefore e^x=0 by that definition
@@ablasphemite1940 There is the additional requirement that exp(0) = 1. This immediately excludes exp(x) = 0 from being possible. The equation dy/dx = y has exactly one solution satisfying y(0) = 1. This solution is exp. Then, we define e := exp(1). Well, in theory, this comes about more naturally as [exp^(-1)][lim (1 + ε)^(1/ε) (ε -> 0)] = 1, but since e = lim (1 + ε)^(1/ε) (ε -> 0) and the above equation implies lim (1 + ε)^(1/ε) (ε -> 0) = exp(1), we have e = exp(1). Anyway, every property of the exponential function can be derived extremely easily from this definition. Deriving the Taylor series is literally trivial. The functional equation can be derived from the Taylor series using the binomial theorem and the Cauchy product theorem. Also, exp(z) = lim (1 + z·ε)^(1/ε) (ε -> 0) can be proven using Tannery's theorem, hence further justifying the connections. The properties of the natural logarithm can also be derived very easily from this.
@@ablasphemite1940 f(x)=f'(x) has an infinite amount of solutions generalized by C*exp(x) where C is a constant. 0 is a constant
There is for me a problem in this proof :
At the beginning, you try to calculate the derivative of 2^x but at this point if you don't have the exponential function you cannot (I can be wrong) define what is 2^x and moreover you cannot compute 2^0.0001 for your limit.
I edit :
In fact, you can define 2^(p/q) with p and q two integers with a q-th root therefore you can define 2^x on Q. Finally It's likely that you can use the density of Q to define 2^x for all real values but again, it feels complicated to me.
2^x=( 2*2*2.....*2*2*2) " X" no. Of times ......
@@kabsantoor3251 No, that is incorrect. You cannot define 2^x as being the power series you mentioned, because that power series is in turn defined in terms of ln(2), which you approximated as 0.693, but ln(2) is itself defined in terms 2^x. You have thus given a circular definition, which is to say, not a definition at all.
@@valentinenprepa1379 You make a fair argument, though: 2^x is not typically regarded as well-defined by peer-reviewed publications by mathematicians for arbitrary real x. Of course, there are different constructions for which using the notation 2^x to denote a well-defined expression in certain specifiv contexts is very practical, but this notation would rarely be used otherwise. As for the exponential function, it is just defined as a power series, most of the time, and then 2^x is just taken to be as an abbreviation for exp[ln(2)·x], although in practice, such abbreviations are actually rarely used.
@@angelmendez-rivera351 @Angel Mendez-Rivera No circular reasoning. See if I define 2^x as a shorthand for lim (1+ x/n)^k*n for some(yet not defined) k and n-->infinity and expand the binomial via binomial theorem, it's not hard to show that the derivative of 2^x is this number k times 2^x. We fix k by returning to the original binomial and looking at the value of the infinite power series for k=1. This way we're led to the number e. Clearly e is the number such that e^x (important to understand in this context as the infinite polynomial rather than e multiplied a certain number of times) is its own derivative. Changing to any other number say 2 changes k as defined by the binomial - that's the proportionality constant between the rate of change and the function itself.
You could set y=eˣ so that ln y=x. Differentiating both sides with respect to x yields y'/y=1 or y'=y=eˣ.
Best video on the subject :)
1:40 "JuST usE LhOpITalS rULe"
You can actually take the limit directly and show that it equals the log of b.
\lim_{x
ightarrow0}\frac{b^x-1}{x}
Substitute y=b^x
\lim_{x
ightarrow0}\frac{b^x-1}{x}=\lim_{y
ightarrow1}\frac{y-1}{\log_b(y)}
Change base \log_b(y)=\frac{\log(y)}{\log(b)}
\lim_{x
ightarrow0}\frac{b^x-1}{x}=\log(b)\lim_{y
ightarrow1}\frac{y-1}{\log(y)}
Hospital rule \lim_{y
ightarrow1}\frac{y-1}{\log(y)}=1
\lim_{x
ightarrow0}\frac{b^x-1}{x}=\log(b)
Very good.
That was a good and needed (for me) return to the definitions.
Very nice and helpful concept of video sir
Something really cool you should have pointed out is that the limit as h goes to 0 of (x^h-1)/h could be a way of approximating ln x
We can also prove it by using the infinite series of sum from n = 0 to infinity of x^n/n!
When we take the derivative of the whole series, then we notice that the derivative is equal to the original series
Q.E.D
Very cool again! I always wondered in high school what the heck e was and now I know how they “discovered” it. Thanks again!
Who likes the video of bprp and his way of teaching me by myself he is super math man👏👏👏
This reminded me of your 2017 video "what is e, and the derivative of exponential functions"
How are you able to just plug in b for x at 8:35? What allows for that?
Summary: same reason Pikachu is Pikachu
I’m confused at 8:07. How are you able to differentiate that without the assumption that the derivative of e to x is e to x? It seems like a circular argument. Not saying it is, but I’m not seeing how you can do the chain rule there without having proven the derivative of e to x.
❤️...love u sir...love from India...I like all your video...You always motivates me to learn something new in maths
Love to see some of the basic (but very important stuff). Many calculus students will find this extremely helpful. I always hated the definition of ln(x) as an integral of 1/x. It's WEIRD and backwards and comes out of nowhere.
Historically, that is the way we defined ln(x), since this property of natural log was discovered before the number e.
To learn it more intuitively, I think it makes more sense to define e as the value of base b, such that d/dx b^x = b^x. Then, use this as the starting point to prove the historical ways that ln(x) and e as a limit, were previously defined.
So interesting. It's beautiful how you explained the deep truths behind these common derivatives. I understand the reason so much better now
That's the basics of derivatives. If you didn't know that, you never really learned derivatives.
The basics of derivatives is slope of a curve, not e. Proofs for the derivative of e^x that are not circular are also notoriously hard to find.
You re brillant ! You re far away the best maths teacher i ever seen thank you very much
I love watching your videos. The Werefrog do miss the "black pen red pen yay" at the start of the video, though.
You can also define ln(x) as the limit of (x^h-1)/h as h -> 0
Thanks, you really explained it well!
When I was in fifth grade, I had started to really love math. It was my passion, but then my parents found out I was getting good at math so they kinda forced me to study more. This made me not feel as interested in math again, and I didn't listen to them, so I didn't study. But then, 1 year later, I decided that I wanna do math again. This is gonna be the start of my journey, hopefully my parents don't make me stop again...
You’re in 6th grade? Start at algebra
5:50
e is e-rational
yup!
Hmm... Pretty sure in my math class exp(x) was defined as an infinite series (alongside sin(x) and cos(x)). From there you prove all of its properties. One of the properties is that the infinite series happens to differentiate to itself. Another property is that exp(x) = e^x where e is a constant.
So starting from "why is d/dx(e^x)=e^x?" is just doing it all backwards, which makes things conceptually more complicated...
Black pen, red pen, blue pen...😁
Nice video! 🤩
I love how his blue yeti is just a pokeball now that’s amazing
Surely, the easiest way is to define e^x as a power series.
Then, differentiate that.
The first term, 1, disappears. The second term becomes 1. The third becomes the second, the fourth the third etc so you end up with the same power series you started with.
Don’t you need the derivative first to derive a Tailor series?
@@ProCoderIO normally yes. But you can construct the taylor series with simple observation if you start by df/dx=f(x) and then excluding zero as a solution. It's not at all rigorous and just a handwavy method I developed for my students but at least to me it's the easiest way to show e^x is it's own derivative.
Ignore everything you know so far and just take any old number (do yourself a favor and take 1. Otherwise you have to divide in the end) and write two colums. df/dx and f(x)
Place 1 in both. Yeah the derivative of 1 is 0. 0+1=1 that's kind of the step that almost legitimizes the method. It means we can get an infinite sum where each term is the integral of the last.
But to really show that we just integrate 1, get x and fill that in the f(x) column.
Because both colums have to be equal you put x in the df/dx column as well, integrate again, get x^2/2
Rinse and repeat until you figured out the pattern.
The thing is you kinda need to conjure the fact this is the taylor expansion of e^x from thin air.
What I do with my students is to not do that but some theatralics: I act like this infinite function gets too annoying and just plot it. After 20 terms or so it should look exponential.
Then I ask my students if they think it looks like an exponential function. Most will agree after getting 2 or 3 examples (like 2^x and 3^x). Then we can simply use the fact that a^1=a and plug in 1 in our truncated polynomial.
However we have to loosely make sure we got an exponential there.
My kids ALWAYS are skeptical about that. After all it's a polynomial they see on the board be it an entirely useless one.
So I take another exponent rule.
(a^b)^c=a^bc=(a^c)^b or (e^1)^2=e^2
Meaning f(2) should be reasonably similar to f(1)^2.
Typically I run through a couple more powers (like 3 and 4) before explaining that I truncated the function after 20 terms meaning, my polynomial is never perfectly it's on derivative. But e^x which is equal to the full infinite polynomial is it's own derivative.
And then I show the natural log and define the derivative of exponential functions from there.
I don't like my kids having to just get a derivative from the calculator and see "every exponential function has a weird constant in the derivative.". Don't know why. I just don't like that method. I prefer if they can do every step of the way themselves. Even if we have to take a detour for them to do that.
I think the point of this is to do a method that students can do when they are first learning derivatives. The traditional way to get the derivatives of these functions is to do them after learning the integral and defining the integral of 1/t as the log function, then taking an inverse to get e^x. In your way you have to wait until you go through what is typically taught in 2 years of calculus. BPRP’s way is the typical method for those that want to teach e^x right at the beginning
@@stephenbeck7222 Typical method? Almost no schools where I live define ln(x) in terms of an integral of 1/t in an introductory calculus course. In fact, integrals themselves are typically taught in calculus 2.
@@ProCoderIO "Don’t you need the derivative first to derive a Taylor series?"
He's not deriving anything. It's a definition. It's a very confusing one since you already know what "e", "^", and "x" are so you shouldn't need a new definition for e^x. But it turns out that the old definition of "^" doesn't generalize very well (remember how confused you were when you first saw e^(pi i) = -1) so mathematicians like to just start over and define e^x as a power series.
i saw an ig reel by you some time ago and now I'm learning.. lol thanks! the pokeball really ties it all together tbh
can someone explain why this is a proove? Like he used the fact that d/dx(e^x)=e^x to "proove" it at 9:03. Or am I wrong
Love from India🤗🤗
There was a slight missed opportunity in the first part when trying to find the derivative of 2^x.
It was all cool up until the end of the column when you decided you couldn't use L'Hopital's Rule because you'd need the derivative of 2^h.
It's actually okay and gets you a little farther though obviously not all the way. The 2^h limit we get is *related* to the derivative of 2^x, but it isn't the same thing. So let's continue where we can:
d/dx(2^x)
=2^x * lim{h to 0}[(2^h -1)/h]
Dubiously using L'Hopital's Rule turns that limit into
lim{h to 0}[d/dx(2^h) / 1]
which is
lim{h to 0} d/dx(2^h)
We don't know what that is, but call that f'(0), since that is the slope of the line tangent to 2^x when x=0.
So letting f(x)=2^x
We get that
f'(x) = 2^x * f'(0) = f(x)*f'(0)
And that's a pretty big deal: the derivative of an exponential function *anywhere* is equal to the *value* of the function at that point, scaled by the *derivative* of the function at x=*zero*. Honestly, that's a weird result in my opinion. :)
Then the rest of what you said is great and follows right away like you said, but I figured I'd point out the f'(0) bit.
Great video as always :)
*It was all cool up until the end of the column when you decided you couldn't use L'Hôpital's rule because you'd need the derivative of 2^h.*
Correct. We need to evaluate lim (2^h - 1)/h (h -> 0) in order to prove that the derivative of x |-> 2^x exists and find its value. If you use L'Hôpital's rule to evaluate lim (2^h - 1)/h (h -> 0), then you need to take the derivative of h |-> 2^h, which is a problem, because the limit is being evaluated to prove that such a derivative exists in the first place. In other words, you are starting your proof by starting with your conclusion.
*It's actually okay and gets you a little farther though obviously not all the way.*
No, it is not okay. Starting the proof by assuming that the thing you are trying to prove is true is never okay.
*The 2^h limit we get is **_related_** to the derivative of 2^x but it isn't the same thing.*
No, you are super wrong. Of course it is the same thing: x |-> 2^x and h |-> 2^h are literally the same function. Yes, we are using different symbols to denote the input parameter, but this does not make the functions different. Your argument is like saying "1/2 and 2/4 are not the same rational number since they are different symbols".
@@angelmendez-rivera351 I think you misunderstood their post. They don't assume anything they are trying to prove, as they aren't trying to prove the value of the derivative, nor whether it exists. They simply assume the derivative exists in order to explore what properties that derivative (if it did exist) would need to have.
And they come up with one:
if f(x) = b^x, then f'(x) = b^x * f'(0). Or, as they say it in words: "the derivative of an exponential function *anywhere* is equal to the *value* of the function at that point, scaled by the *derivative* of the function at x=*zero.*"
And, when we do eventually find the derivative, this is clearly correct. f'(x) = b^x * ln b, plug in x = 0, and you get f'(0) = ln b. Thus f'(x) = b^x * f'(0)
I think it's neat that they discovered something like that.
@@turkeypedal *They don't assume anything they are trying to prove, as they aren't trying to prove the value of the derivative, nor whether it exists.*
Yes, they are. They are, because the comment is in direct response to a portion of the video in which BPRP is trying to prove a specific value for the derivative, and the post is complaining that the proof did not allow for L'Hs rule to be used. Also, the post is very much trying to prove something about the derivative: that if f(x) = b^x, then f'(x) = f'(0)·b^x.
*They simply assume the derivative exists in order to explore what properties that derivative (if it did exists) would need to have.*
They pretended that this is what they were doing at the end of the post, but this is completely undermined by the fact that, at the beginning of the post, they specifically were complaining about the fact that L'Hs rule was not allowed in a formal proof of the statement f'(x) = ln(b)·b^x, and they even tried to argue that using L'Hs rule is not actually circular. You may not have noticed it, but this is precisely what the post did in its first few paragraphs.
*I think it's neat that they discovered something like that.*
"Discovered" is a strong word, but I agree with you. However, I have no issue with the discovery they made. My issue is with the first few paragraphs of the comment, where they are trying to argue that the video should have allowed L'Hs rule in the evaluation of the limit, and that BPRP "missed an opportunity" by not using L'Hs rule. The comment is objectively wrong: L'Hs rule is not permissible in a formal proof, and no opportunities "were missed", because the point of that segment of the video was to formally prove a statement, which itself was done for the sake of a grander purpose, and because the discovery in this comment is in no way relevant to the grander purpose of the video.
Thanks blackpenredpen! This makes sense. Since the derivative of a^x=a^x[ln(a)]. If the base is replaced by e then the derivative of e^x=e^x[ln(e)] and by definition ln(e)=1 hence the derivative of e^x=e^x. Can you do a explanation of why d/dx[a^x]=a^x[ln(a)]? Thanks.
You basically did show why the derivative of a exponential is the same exponential times the logarithm of the base. I see it.
thank you so much im searching this for like 2 months
The existence of e is strong evidence that we live in a computer simulation where someone was just too lazy to come up with different constants so they just plugged one universal constant in for a whole bunch of different uses.
Lmao
The common way to solve
lim x->0 (a^x - 1)/x,
is to substitute
a^x = 1 + 1/p
x = loga(1 + 1/p)
so that as x->0 threfore p->∞.
lim x->0 (a^x - 1)/x
= lim p->∞ (1 + 1/p - 1)/loga(1 + 1/p)
= lim p->∞ (1/p)/loga(1 + 1/p)
= lim p->∞ 1/(p.loga(1 + 1/p))
by using log's property:
a log b = log(b^a)
= lim p->∞ 1/(loga((1 + 1/p)^p))
as we know that:
lim n->∞ (1 + 1/n)ⁿ = e
= 1/loga(lim p->∞ (1 + 1/p)^p)
= 1/loga(e)
and by using another log's property:
loga(b) = 1/logb(a)
= loge(a)
= ln a
What do you think about hyperreal numbers and nonstandard calculus?
Therefore, can we defind
ln(x) = lim (x^h - 1)/h when h -> 0 ?
I love this definition, it's more formal than "the inverse function of exp(x)".
Yes
@@aaaa8130 That's kinda like saying "you would have to define how to multiply first. And for that you would have to define how to add first. And for that you would have to define what a number is first. And for that you would have to define the axioms of math first. It's already defined and thats not the point so yeah.
Excellent video! Thank you for sharing
This man should be in the list of the top mathematicians ever.
This is how I was taught about e, and logarithms didn't come in til later. I'm in NSW, australia.
So e was originally an undiscovered number that existed for many mathematical things, like continuous growth and stuff, and that we didn’t know could be what it is now…but it was originally defined as some number with exponent x that would produce the answer which is itself (proven in the video). But which came first…the self-generating cyclical derivative or the irrational number existing in nature what will help evolve many subjects within mathematics?
Proving that the value of the h-limit is a continuous function of the base might be tricky, assuming that we don't know about e or ln for the discussion. There might be a short way but it is necessary to know here.
Ah, you were right that the existence of the h-limit itself is plausible-but-unproven here. I suppose further technicalities can be treated the same way.
I like your videos, but, I wish in this one you could have found a way to differentiate b^x without calculator or the ln(b)*e^x rule.
Simple and lovely.
The limit as h approaches zero (n^h -1)/h is equal to the natural log of n