Two Proofs of a Useful Inequality
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- Опубліковано 5 чер 2024
- We prove the inequality 1 + x ≤ e^x in two different ways, both of which relate to proof by induction.
00:00 Intro
00:09 Proof 1: First steps
03:26 Induction 1
06:53 Proof 2: Extending from ℤ to ℝ
09:30 Induction 2
12:22 Conclusion
In proof 1...
You show that (1 + x/2)
This is a really good point, and quite subtle. Well-spotted!
As I said, it's not a problem if you just consider the case for x < -2 separately, which is trivial. (When x < -2, 1 + x < 0, and 0 < e^x, so 1 + x < e^x). So, it's not a completely lost cause.
One could also take the function f(x)=e^x-1-x. It's easy to see that f(0)=0. Moreover, f has a global minima at x=0 since f'(x)=e^x-1 is greater then zero if x is greater then zero, smaller then zero if x is smaller then zero and equal to zero if x=0. Then:
f(x) ≥ 0
e^x -1-x ≥0
e^x ≥ 1+x
Q.E.D.
Honestly, I think this is the simplest and most straightforward method. It avoids all the sign issues that pop up when trying to extend the inductive steps into the negative x values, and it doesn't require any additional steps to extend the proof from integers to reals since it just deals with the reals right from the start.
You could use Bernoulli's inequality too. Not hard to prove by induction. It states that (1+x)^n >= 1+nx for all x>-1, and non-negative integer n.
Right, so if you put in x = t/n, you get (1 + t/n)^n >= 1 + t. Taking the limit as n -> inf, you get e^t >= 1 + t.
@@chaosredefined3834 Indeed! And the requirement that x>-1 is not a problem, because even for negative values of t, eventually n will outgrow it, so in the limit it's fine.
@@barutjeh The x
i like the idea of extending from an induction over the integers to one over the reals using the floor function trick
It should also be fairly easy to show that the point in which e^x gets "closest" to (or furthest past) x+1 is the point where the tangent slope of e^x is 1 (same as the line), which is at x=0, where the equations are tangent, which means e^x never passes through the line, and must always be greater than it.
For more detail: it should be easy to show that for any strictly increasing function with monotonic derivative, the furthest point that function will get through a given positive-sloped line (measured perpendicular to the line) will be at the point where that function is tangent to the line. Anything less than that before that point, and it gains tangent distance from the line up to the tangent area. Anything more than that after that point, and it's closing in the gap. While it's at a tangent slope, it's neither gaining nor losing distance. If that tangent point is above the line, then the whole function must be.
The formalization could look like this, since e^x-x is differentiable it's minimum should occur when it's derivative is zero which happens only at x = 0 where the function is zero. Then we check it's a minimum with second derivative test and conclude the result.
The nice thing about the proofs in the video is that they can be followed by people fairly early in their mathematical lives. If you tell them the limit definition of e, the first proof can be followed perfectly fine without even knowing that calculus exists.
Of course, if you know about derivatives, this becomes a lot easier.
@@chaosredefined3834 I completely agree, a very good way to teach people slightly more advanced proofs without completely diving in.
Isn't it enough to point that the second derivative of e^x is always positive and the second derivative of the tangent is zero.
It's hard to get too excited about this analytical proof for x>==0, given the linear and exponential growth and the easy hand sketches of both graphs. We only have to know 2
I do see how to fill in the details, basically just noting that for any fixed x it's eventually weakly with respect to n, so you can generalize from n=2^m
easier way :
knowing that the tangent to a convex function is always below the function, and that e^x is a convex function then :
e^x > (e^a)'(x-a)+e^a
by taking a = 0, we get
e^x > x + 1
another way is to use the result that a convex function lies entirely in a half plane determined by any of its tangents so e^x lies entirely above 1+x, the tangent line at 0
Didn’t watch the video, but the easiest way should be to take derivatives and show that e^x - x - 1 has only a single critical point at 0 which is a local, hence global, minimum.
La fonction exponentielle est convexe donc la tangente à sa courbe en (0;1) est au dessous de la courbe : ce qui se traduit analytiquement par l'inégalité voulue.
I tried it by the Maclauren expansion of e^x and showed that e^x = 1 + x + more stuff, which proved the inequality instantly for x>=0 but I am stumped with using this approach to show that it holds for x
Can compare derivatives and show that the derivatives of e^x is less than that of 1 + x for all x < 0
Derivate f(x) = exp(×)-x-1, find there is an absolute minimum at x = 0 which is f(0) = 0, then f(x) is positive, then the wanted inequality.
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Thanks PROF 👏
No offense to Dr Barker, but I think he's only a doctor. A professor is a higher ranking. If he is a prof, he should update his name :)
@@chaosredefined3834 😂
@@chaosredefined3834 lol but hes acc a math teacher he teaches at my skl and he was my math teacher
This is a pretty long comment but I feel like its a much quicker solution (so much so that i did it in my head)
f(x)=eˣ-x-1
f'(x)=eˣ-1
eˣ-1=0
eˣ=1
x=0
We've shown that there is either a minimum, a maximum, or a stationary point at x=0
That means that the function is either strictly increasing or decreasing as |x| increases.
Now just take any 2 derivatives outside 0,
f(1)= eˣ-1 ≈ 1.718
f(-1) = (1/e)+1 =1/e ≈ 1.368
Both of these are greater than 0, therefore 0 is a global minimum
eˣ-x-1>=0
eˣ >= x+1.
The comment I was looking for!
Or f''(x) = eˣ and f''(0) = 1 > 0 and so is a global minimum
Please do a proof of |sin(x)| ≤ |x|
It's one of my favorite inequalities as it is useless almost everywhere, but quite intricate around x=0
x > 1, sorted
For x > 0 f(x) = x - sinx implies
f(0) - 0
f'(x) = 1 - cos x > 0 except when x =0 and thus f' = 0 and thus f is monotonic increasing when x>0
thus x > sin(x) for x > 0
Using symmetry result follows :-)
You can also show that geometrically, without using derivatives. Consider the definition of sin(x) on the unit circle and compare the area of the right-angled triangle with the area of the corresponding circular sector: 1/2 times 1 times sin(x) ≤ 1/2 times 1² times x.
For x < 0, the proof is similar, you just have to consider the triangle and the sector below the x-axis.
you only need consider values of x between 0 and 1. and for those values, sin is an alternating series of decreasing terms so sin x will lie between x - x^3/(3!)
Could’t you just study the function f(x) = e^x - x - 1 and simply conclude via monotony and continuity? It takes three rows at most.
can't you just use graphing i usually find it boring but this look complicated but when you use graphing you jsut can see that e^x is always above x+1 (i hope it is i didn't really graph it but it gotta be true😭)
While yes the graph does show this, it's not really a good proof method, as it assumes that your graph is infinitely accurate, especially around where the two lines touch. How can you be completely sure that the lines don't overlap a tiny tiny bit?
Regardless of whether you draw it by hand or use a tool, any graph is ultimately just an approximation with finite precision.
Your question is: Can one just use technology to prove mathematical propositions? We can't stop you from believing this is valid I guess. After all, the unproven Goldbach conjecture has been enormously strengthened by super computing....
@@MyOneFiftiethOfADollar I didn't use technology you dont need technology to graph e^x
@@phiefer3 thanks for the explanation in my country we mostly do tests so teachers don't ask explanation for your answer that's why i just use easy methods like this
And the only point they are equal is when x=0
I dont understand shit but dats cool
Write in English.