THE CONFUSING DERIVATIVES
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- Опубліковано 15 жов 2024
- So did the derivatives of tanh^-1(x) and coth^-1(x) and found out they are the same! But.... are they???
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so basically, its like saying ln(-x) and ln(x) have the same derivative, but they correspond to different domains of 1/x
Yup, that's another example.
Yes, which is why although we always get taught that d Ln(x)/dx = 1/x, they still tell us without explanation that the antiderivative of 1/x is Ln|x| + C which is not the same as the expected Ln(x) + C. They always forget to cover function domains because the education of mathematics is absolute horrible and ill-executed.
@@angelmendez-rivera351 I totally agree with you
actually, ln(-x)=ln(-1)+ln(x)
ln(-1) is i*pi*(2n+1), wehere n is an integer
So,
ln(-x)=ln(x)+i*pi*(2n+1)
Just in case you are considering complex numbers
@@angelmendez-rivera351 When you study mathematics, this will most definitely be discussed. I don't know where you are doing mathematics, but I think you should change uni.
8:00
"one plus one is two......yes"
the more you know
Are u a fan of Fermi dirac
@@nazishahmad1337 I think it's safe to assume he's a fan of Paul Dirac, perhaps even Enrico Fermi, but I'm not so sure he knows Fermi Dirac
Actually I should Fermi and dirac
@@nazishahmad1337 I'm a physicist, so...
I am too a physicist
If you work with the complex logarithms then you actually can argue that those two functions differ by a constant complex number because 1/2 ln( (1+x)/(1-x) ) = 1/2 ln( (x+1)/(x-1) ) + 1/2 ln(-1).
Complex analysis fixes real yet again. XD
Ah, but watch out for the branch cuts... I don't think it's the same constant in every region.
I was about to point out the same thing.
Also, since both functions would be defined on all real numbers except -1 and 1 their 'real' graphs are missing some branches... which would be the same on both, maybe off by a constant
...to expand on my other comment: if we conventionally take the principal value of ln(z) to have a branch cut along the negative real axis, then ln(-z) would have its branch cut along the positive real axis, and ln(z) - ln(-z) is not necessarily a constant ln(-1) = i pi. It's -i pi on half of the complex plane.
(You can see there's a potential ambiguity by just doing the substitution u = -z; that turns it into ln(-u) - ln(u), which is apparently the same expression with the opposite sign. So by the same reasoning above that would be -ln(-1) = -i pi. Why would that give you a different answer, if it's the same thing? Because ln(z) is multi-valued in the complex world and you need some kind of principal-branch convention.)
Actually... tanh^-1(x) and coth^-1(x) can still be said as 'off by a constant', it is just that the constant is πi/2
actually, there is a constant of 1/2((2n+1)πi) where n is an integer
edit: oops i forgot about the 1/2
Yeah yeah yeah but didnt we learn recently that pi is an integer
:thinking:
Yep.... you can just seperate it and youll get tis. Ln(-1)/2+arctanh(x)=arccoth(x)
I don't think _(2n+1) π i_ is right? Because doesn't _π/2 i_ also work?
Edit: I think the proper form is (k + 1/2) π i, where k can be any integer.
@@markorezic3131 yeah it's 3, or sqrt(g) which is just sqrt(10), so we also learned that sqrt(10)=3
@@XanderGouws that's exactly what he said but in other words
Nice video!
Always good to hear about things that work similarly, but on different domains! 🙂
Especially that many people forget to consider differences in domains.
It would be great to see more content like this one! 🙂
Wait though, if you include complex numbers then are the graphs the exact same?
Yes and no, they would look the same, but be just off by a constant (ln(-1)/2 in this case). But those would not be the functions this video is about, the behavior of real functions actually differs a lot from their complex-valued forms. But yeah, sketching ln(-1)/2 + arctanh(x) provides the same graph as sketching arccoth(x) on the complex plane.
Very interesting indeed. I have never faced this problem because I have always worked with these functions in the complex world. But in the real world the indefinite integral of 1/(1-x^2) would be a piecewise function right?
What about when we allow things to be complex?
If we let 1/2 ln((1+x)/(1-x)) = 1/2ln((1+x)/(x-1)) + C, we get C = ln(i) = i π/2.
So technically, we can say that they are the same function + a constant, just that the constant is complex.
Right?
Edit: If we're looking for a set of all constants that work I think it's C = (k + 1/2) π i, where k is any integer.
I love that you've been covering the hyperbolic trig functions so much recently. It's been helping to demystify them. :)
The mystery will unfold if the graphs of the inverse functions of tanh and coth are are drawn in the same coordinate system.
krischan67 yup!
They are indeed off by a constant, which is pi*i/2
On the first expression, factor the -1 from the denominator:
1-x = (-1)*(x-1)
so it becomes:
1/2*ln( (-1)*( x+1/x-1))
using the ln properties, this becomes:
1/2*(ln(-1) + ln(...))
ln(-1) = pi*i
Thus the expression equals:
1/2(pi*i +ln(...)= cosh^-1(x) +pi*i/2
Amazing! You blew my mind. It is great to see these simple examples that shatter my confidence that I understand math. Thank you.
Thank you Steve! How are you? Good to see you again!! I still remember you from 3 years ago, especially I still have the note that you left me at my school.
@@blackpenredpen I'm doing well here in Beijing. Thanks for remembering me. 😀
different domain. ln can't be negative. The numerators in the ln are identical but the denominators are negatives of each other because -(1-x)= -1+x = x-1.
This means that whenever one is positive, the other is negative and thus not real.
mrBorkD Okay, but having a non-real value is fine.
@@angelmendez-rivera351 not in real analysis it's not.
The Cartesian plane only has real numbers on it.
The constant is there but its a complex constant right?
That's what I got, when you restrict it to the real numbers only it becomes a domain issue, when using complex numbers they're off by a constant.
Yup, they're off by πi/2.
It doesn't really matter if the constant is complex
@@abdula1717 yes it does, because if you only look at it from a real perspective there is no constant and that's an issue as if two functions give the same derivative they can only be off by a constant. However this problem is solved by the complex constant.
An alternative explanation is that you have to consider the domains however it is much more reassuring that there is still a constant.
no, I'm studying for my linear algebra final and I ended up here. but great video though.blessed efforts.
: )
THAT ... WAS .... AWESOME.. your channel is my new favorite! Keep up the great work
C is ln(-1)
divided by two!
@@imadhamaidi you are right
Hi, bprp!
So I was playing around with angles, and I found out that there exists an angle α between 0 and 90 degrees such that cos(α) = tan(α).
And, the neat part is, sin(α) is exactly 1/φ (or you can say, csc(α) = φ).
(φ is the golden ratio)
Pretty cool, isn't it? Maybe you should make a video about this!
Cos = tan
Cos = sin/cos
Cos^2 = sin
1 - sin^2 = sin
0 = sin^2 + sin - 1
sin = -1/2 +- sqrt(5)/2
Limiting to 0 to 90 we get only the positive.
Phi being 1/2 + sqrt(5)/2 makes our answer -phi in the negative case. One of the properties of phi is that it is tied to its inverse with the 2 quadratic solutions. One solution is phi, the other is -1/phi
At 1:15 you can just take d/dy to get dx/dy, then you just flip it over to get dy/dx. it simplifies real nicely with the hyp trig identities.
You can do it with y=arccoth(x) too obvi
Hey Bprp, I’ve just bought two shirts for a Christmas present! I hope they’ll arrive before that date, but they have to travel the globe because I’m from Italy. How long do you think will it take?
I’m really looking forward to them!
Keep pushing with your videos, you teach me a lot every single day..
Hmmm I would say about 2 weeks or so. Teespring needs to print them and then send them out. I hope they can get to you on time! Thank you for your support! : )
Request: Do a problem that requires the use of quartics! Thanks
The domain explanation is exactly what I suspected was gonna happen, I feel a little smarter lol
But what if we extend domain and range into Complex numbers? Is arctanh analytic in places arccotanh isn't and vice-verse?
Yes. If you extend the domain and range to the complex numbers, then other than at the poles, arctanh(x) and arccoth(x) are both the same function that only differ by a constant in each section.
The projection of both of their outputs onto the real numbers is:
1/2*ln(|(x+1)/(x-1)|)
arctanh(x) is real when x is a real number between the two poles, and it is complex outside the two poles.
arccoth(x) is real, when x is a real number outside the two poles, and it is complex between the two poles.
If you add up arctanh(x) and arccoth(x), you'll get the following:
ln(|(x+1)/(x-1)|) - i*pi*sgn(x)
Where sgn(x) tells us the sign of x. The complex number has a jump discontinuity from +i*pi for negative x's, to -i*pi for positive x's.
friend, really clear explanations throughout this video and other videos!!
Name of the channel is blackpenredpen but he also used blue pen
So I guess you can basically say.. (Its propably wrong the way I say it in english).. the tanh^-1(x) being the steady continuation of coth^-1(x) to define a function all over the real numbers?
It's not over all real numbers, neither function can have 1 or -1 in its domain.
So for the indefinite integral ∫1/(1-x²) dx, do we have to write (ln|x+1| - ln|x-1|)/2 +C or is picking either tanh⁻¹(x)+C or coth⁻¹(x)+C is okay too because we do not have any x values?
Actually, you can even use the x=1/t sub to get the same things if you choose the hyperbolic subs, but what you get is that coth⁻¹(x)=tanh⁻¹(1/x) and that is true for every real valued x.
If you write the answer in terms of hyperbolic functions, you have to use a pointwise function. The right answer would be
| tanh⁻¹(x) for -1 < x < 1
∫1/(1-x²) dx= |
| coth⁻¹(x) for -1 > x or x > 1
@@Koisheep Use the x=1/t sub ;)
Hannah Winter Oh ok, thanks
The key here is that the graph of 1/(1-x^2) has three disjoint pieces, on domain (-infinity to -1 ) (-1 to 1) and ( 1 to +infinity) respectively.
Different domains was my guess - nice to see it worked out.
You were right, this was super cool!
What a cool quirk in the fabric of the universe.
Great example! I loved it lol thnx for every video like this 👍
6:02 when you did calc 2 for 10 hours straight and you wonder where is the absolute value?
"so what's the matter?"
"domain."
*blows every mind in the vicinity*
Could you do a video on the integral from 0 to 1 of -ln(-ln(t))dt, aka the Euler Mascheroni constant γ? Such an interesting definition to γ.
Absolute value signs would solve this problem and would be defined everywhere but x=1 and x=-1. It would be 1/2 * log | (1+x) / (1-x) | + c. In fact, this is the result of the integral of 1 / (1-x^2) by using partial fractions. It's why the absolute value is important in the integral of 1/x.
The full statement of the theorem is a bit cumbersome, easy to forget the condition: If f is continuous on [a,b] and differentiable on (a,b) and for all x in (a,b) f'(x)=g'(x) holds, then there exists a real C such that for all x in [a,b] f(x)=g(x)+C.
Kinda interesting, the hyperbolic functions are really just pieces of the exponential function, so it's kinda fitting that the derivatives of arctanh and arccoth would be pieces of 1/(1-x^2)
I remember doing this & my Prof tested the class to figure it out re both Derivatives being the same.
Nobody had a clue to use Geometry & check the Domains, as we thought the proof required 'rigor' rather than visual insight. A similar thing occurred for an exercise with the classic limit (sin Θ/Θ=1)....🤣🤣🤣
I just factored out the negative sign once you solved the coth part. ln(-1) = i*pi (+i*2*pi*m if we want to be precise) since e^i*pi = cos(pi) + i*sin(pi) = -1 so they differ by a constant so their derivatives must be the same.
Integral of (tan(x) * sec²(x))
=Tg²(x)
And
=Sec²(x)
They have the same graph, but diverge by a constant, its very easy to see that in this case, because sec²(x) = 1+tg²(x)
What's Tg^2(x)?
7:15 Chen Lu, you will not t be forgotten!
This makes me love maths
For solving the integral 1/(1-x^2), you can do the sub x=1/u and well... The result is still interesting.
lets say that i were integrating some function that could be written in terms of tanh^-1 or coth^-1 from -inf to inf when evaluated. To evaluate it would i do...
coth^-1(x)]-inf to -1 +
tanh^-1(x)]-1 to 1 +
coth^-1(x)]1 to inf?
If you look at the physics of an object moving through the air with a thrust, you have to solve the integral of 1 / 1-x^2 , but on different domains. This actually gives you three functions, because inverse regular tangent sneaks in, too. (This is brought up to 5 functions if you want to have a different initial velocity. (It's doubled except that the coth^-1 already requires certain initial velocities.))
Here's a neat desmos thing I made for it:
www.desmos.com/calculator/lraydh20ae
interesting fact
if you take k*coth(k*x) where k is an infinitesimal number then it’s graph approaches 1/x
Opened my eyes!!! Wow!!!
uhm so you take again the derivative after getting the answer? so does that means that the 1/1+x^2 is the second derivative now? Sorry Im still confused.
This exercise is beautiful!
nicely done, ty for that content
Sorry for the question, but if I solve the integral of that function with no upper and lower limits... cosh^-1 (x) and tanh-1 (x) are both correct answers?
in their respective domains!
a very nice reminder to always identify where the functions are defined before you spend hours going back over your Algebra ...
they differ in complex constant
To get domain of tanh-1(x),
Solve (1+x)/(1-x)>0.
To get domain of coth-1(x),
Solve (x+1)/(x-1)>0.
So if you allow complex numbers as your solution and thus expand both domains to include all reals except x = 1 or -1 then tanh^-1(x) and coth^-1(x) would only differ by the constant i*pi/2.
So when I'm integrating 1/(1-x^2) dx what is correct to answer?
tanh-1(x) or coth-1(x)?
Whynotboth.jpg
Use a pointwise function because the domain of both is smaller than the domain of the integrand. Either that or write it in terms of logs
The correct answer is to use partial fraction decomposition.
The answer is 1/2 * ln|(1+x)/(1-x)| for any value of x ≠ 1,-1 (you can solve this using partial fraction decomposition). This result leads to both cases shown on the video. For -1
You are a good teacher i love
Awesome video, thanks
It is super-cool & a great job to you!
I get surprised when i get my writing on the white board to fits too. 6:50
Awesome explanation😘 but I need help....
What is the general solution of differential equation x^2.y"=2y.... 😞
Very late answer but.. this should be the answer -
+/- y = (4×sqrt(2)/3)×(lnx)^(3/2) - 2×sqrt(2)×k×x^2)
Where k is any real number.
Note: this is not the complete answer. This is for the case when first constant of integration = 0. The question was too difficult. I might go back and try to solve it in a more general manner.
In the meantime you can verify that answer by forming its DE.
In other words, these functions have the same derivative, BUT they never have the same domain (otherwise they would have to equal one another). Do these functions analytically continue one another (up to a constant) then? Branch cuts might screw that argument up though, but if not then I think they have to analytically continue one another. Still, that's really interesting. I wonder how many functions there are that have the same derivative form on the real axis, but have different, non-overlapping domains? I imagine in most cases this must have to do with analytic continuation. (e.g. look at the series expansions for 1/(1 - x) with the pole at x=1. On one side you'd say it's a divergent sum, but for x > 1, the complex function definition is obviously well-defined and there's no need to appeal to any geometric series' convergence.)
Really helpful! Thanks!
Actually, tangent and cotangent without the hyperbolic function is different. So, it should be different between the hyperbolic function of tangent and cotangent.
Wahyu Adi No, not really, because the relationships do not translate linearly.
No, because sin^2(x)+cos^2(x)=1, but cosh^2(x)-sinh^2(x)=1. This messes up most of the other trigonometric identities when translated to hyperbolas
What’s the integral of that derivative? Would you wind up in a differential equation situation where the solution is both of those functions?
I think you would get a piecewise function where it would be one of them between -1 and 1, and the other outside that range.
Is there such a derivative such that it is different for a countably infinite number of domains, yet is still the same derivative? In the example, there are only 2 domains, but is it possible that there is more than 2 or even a countably infinite set of domains? How would someone prove this?
so, what should i do if have to integrate [1 / (1-x^2)] ?
I did a Google search and found a neat antiderivative calculator.
www.symbolab.com/solver/antiderivative-calculator
It returned a result of (ln|x+1|)/2-(ln|x-1|)/2+C
I realize at once that if we want to integrate 1/(1-x^2) we will get both function with absolute value brilliant !
woah that new intro is impressive!
BPRP.
The second column in black would have been much shorter if you had used the fact that coth is the reciprocal of tanh.
Ah!! So you mean I could have just done inverse coth (x) = inverse tanh (1/x), just like the regular trig one inverse cot (x) = inverse tan (1/x)
Miss the " Black Pen Red Pen YAY" :D
Where do you find this "error" in the process of differentiation?
No error in the differentiation steps. It's the domain issue.
outstanding....plz...do any video regarding on imo questions☺️☺️
Perhaps nicer to say that tanh^-1(x) is defined when |x|1.
Omg!!!! This question takes me in 10 class and nobody can tell me why it is
I love bprp
Thank you!
Ben's debates are the best to watch!
blackpenredpen im not really ben shapiro and i actually love you
@@athanasiuscontramundum4127
I know! I just wanted to bring up Ben as well.
blackpenredpen he responded twice! I have been blessed.
Huh. Why did none of my calculus classes mention this possible gotcha?
OtherTheDave
Did you go over tanh? Anyways trig integration can lead to multiple answers differing by a constant
@@duckymomo7935 Yeah that always happens. I was thinking of a similar situation with the integral of sinxcosx dx. You can find it using 3 methods (at least to my knowledge) The answers all differing by constants due to trig identities. imgur.com/gallery/jZBcnqP
so if you were to integrate 1/(1-x²), what would the result be?
If you want a universal real numbered output for this integral, you get:
1/2*ln(|(x + 1)/(x - 1)|)
This function matches arctanh(x) when x is between the two poles of x=-1 and x=+1. And when x is outside the poles, it matches arccoth(x). The two functions differ from each other by a complex constant, of i*pi/2.
I noticed they differ by a multiple of -1 under the logarithm. Does that mean they differ by a constant, and that constant is ipi+2pin where n is an integer?
Yes, they do differ by a constant, of i*pi/2.
Love that intro
Lorenzo Barbano thank you!
question: do tanh-1(z) and coth-1(z) differ by a complex constant?
Yes
Yup, ln(-1)/2
Yes ... Super Cool
Thank you ❤️
So, at what step in the anti-differentiation of 1/(1-x^2) must one consider the domain?
I would assume it would have to be the first step, as there is no going forward without knowing which function the integral will produce. Otherwise, the answer would be both functions with the domains for each being expressed.
This is where the absolute values on the ln(x) function come in to play. This example is a simple partial fraction expansion, that we then integrate as two natural log functions that are added together. Including the absolute values, we get a complete solution in the real numbered domain. This gives:
1/2*ln(|(x + 1)/(x - 1)|)
It turns out that arctanh(x) and arccoth(x), do ultimately differ by a constant, which is i*pi/2. Both of these functions project onto the real numbers, to deliver the log function I mentioned above.
Please solve this
Integration from 0 to Π √cosx/√1+x^2
What happens if you integrate the derivatives? What do you get? The Tanh or the Coth? Both?
Nvm i did it in the integral calculator website and it turns out you get a graph that is a combination of both the coth^-1 and tanh^-1
What do you get if you integrate 1/(1-X^2)
You get:
1/2*ln(|(x + 1)/(x - 1)|)
For values of x that are between -1 and +1, arctanh(x) will equal this log expression.
For values of x that are outside of the two poles, arccoth(x), will equal this log expression.
The two functions do differ by a constant in each interval of x, but that constant is a complex number, of i*pi/2.
Dude can u do a vid about wheels theory plssss??
ZAID SALAMEH YES
Wheel see, I expect.
So is there 2 answers off by more than a constant for the integral of 1/(1 - x^2) ?
If you want a universal real numbered output for this integral, you get:
1/2*ln(|(x + 1)/(x - 1)|)
This function matches arctanh(x) when x is between the two poles of x=-1 and x=+1. And when x is outside the poles, it matches arccoth(x). The two functions differ from each other by a complex constant, of i*pi/2.
OK, so now here is what I want to see: Do the integral of 1/(1-x^2) and show how/why you get the inverse hyperbolic tangent in one domain, and the inverse hyperbolic cotangent in the other domain.
If you want a universal real numbered output for this integral, you get:
1/2*ln(|(x + 1)/(x - 1)|)
This function matches arctanh(x) when x is between the two poles of x=-1 and x=+1. And when x is outside the poles, it matches arccoth(x). The two functions differ from each other by a complex constant, of i*pi/2.
Ok but what's the right way to integrate 1/(1-x²) then?
If you want a universal real numbered output for this integral, you get:
1/2*ln(|(x + 1)/(x - 1)|)
This function matches arctanh(x) when x is between the two poles of x=-1 and x=+1. And when x is outside the poles, it matches arccoth(x). The two functions differ from each other by a complex constant, of i*pi/2.
Those two functions are actually moved by a constant ln(-1)/2 = πi/2 in the complex world. So their derivatives are the same
Not in the real world though.
What a cool intro!
Thanks!!
why is the d/dx of 1/2 just 1/2 should it be 0 because 1/2 is a constant?
The half is not a separate term; it is a coefficient. As such it is not affected by differentiation or integration. Now if there was a +/- 1/2 at the end, then that *would* differentiate down to zero, but as it's being *multiplied* by the ln function (or perhaps more precisely, the ln is being multiplied by *it*), it will remain as 1/2. You could use log rules to take it inside the natural logarithm as a square root, but that is just unnecessary work that would require the chen lu.
You could even treat the half as a function of x (exactly why you would I don't know). If you used the product rule then you'd multiply the derivative of 1/2 (0) by the ln function and then add the derivative of the ln function multiplied by 1/2. As anything multiplied by zero is zero, you'd be left with half the derivative of the first function, which is the same result as if you leave the half on place after differentiating the ln function.
niiiiiice new intro
What are other examples of "pretty" different functions with the same derivatives? Someone else said ln(-x) and ln(x), but that is a lot more obvious than this one is. Any other examples where it doesn't seem that obvious that they have the same derivative?
Almost any example you can give will be a composition of a ratio off linear functions with the natural logarithm, and the example in the video is in fact a special case of Ln(x) and Ln(-x) having the same derivative, since arcoth and artanh only differ in that the inside of one is x - 1 and the other is 1 - x, both in the denominator, and x - 1 = (-1)(1 - x).
In fact, if we look at why the derivatives are the same, the reason is simple: Ln f(x) differentiated is f’(x)/f(x). If both f(x) and f’(x) shed a negative sign, then the negatives cancel. And the sign difference in the original function is what causes different domains
PDY ! Another fact too is that the functions actually technically are different by a constant, as is also the case with Ln(x) and Ln(-x). Namely, the domains are actually the same if you allow the functions to be complex-valued, which you can allow as long as the input is always real valued. And in our case, it is true that Ln(-x) = Ln(x) + (2n + 1)πi, and the functions in the video differ by the same set of constants (although constants are just sets too technically :P )
@Angel Mendez-Rivera: That's an interesting way to see too.
In fact, this reminds me of the derivatives of arctan(x) and -arccot(x) (both equal to 1/(1 + x²)) which are off by a constant too but different constants depending on the sign of x and, moreover, you can extend arctan(x) and -arccot(x) if you consider that both of them are defined per complex logarithm which outputs multiple values too.
Sorry but ln(-1)= i*pi that is the euler formula
you cold have skipped the derivation of the alternative expression of *coth* by simply taking the reciprocate of the alt expression of the *tanh*. Basically skip the part 3:36 to 6:20
why is this not listed? D:
Just wanted to space out the uploads
"X" should NEVER be used for the hyperbolic function. "U" is more appropriate to represent the hyperbolic ANGLE. I'll never understand why everyone uses the "e" definition. I'm starting to think that next to no one really understands the hyperbolic functions. While I LOVE your videos, I feel they are all about "operations" and not the substance of the problems.