Soooooo cool. Never thought this integral is even solvable with elementary calculus. This integral is such a classic to introduce multivariable calculus!
I don’t know if I would call the Wallis integrals part of ‘elementary calculus’, but there is definitely a lot here that a Calc 1 or 2 student can follow.
@@bttfish How is it dirty though. Imo the usual method is dirtier because you have to know way multivariable calculus to integrate a single variable function. Also Wallis integral is just integration by parts which is elementary calculus.
It's one of those solutions that can only be found be reverse engineering. You wouldn't start with that inequality. But if you have a couple of integrals which you know give the same answer as the Gaussian, and you can show that they are the results of a limit that never exceed the Gaussian (or is never lower) the rest falls into place. Very neat.
I'm still confused. I understand you can replace with max(1-x^2,0). But in the same spirit you can also replace with f(x) = (1-x^2) for x1. I.e. for x>1 we use the gaussian function itself. f(x) then is 1 is the same). Then the integral 0-inf of this f(x) on the lhs is sqrt(pi)/2 (for the part x1 part of f(x). And in the end the integral 0-inf of the lhs will be bigger than what we want. what am i missing?
I think there's a simpler way to look at it. Integral(0 to 1) (left) is less than integral(0 to 1) (middle) which is less than integral(0 to oo) (middle). QED.
Even if we consider the left integral = middle integral it can only happen when both integral are from 0 to infinity. Since the left integral goes from 0 to 1 instead of taking 0 to infinity the lhs will become less than middle integral and the equality will break down. If we assumed from the start that left integral < middle integral then even integrating left from 0 to infinity will make left less than middle
Very cool. That's my favorite integral (solved the "usual" way: square it, convert to polar coordinates, use tricks, integrate, then square root). I've never seen this approach. Thank you.
4:00 I think we can tackle the two-sided inequality much faster by considering f(x) = exp(x) - 1 - x which (a) has critical point at x = 0 and (b) is convex everywhere, which proves f(x) >= 0 or exp(x) >= 1 + x (the derivatives are much easier than using x^2). Then, x --> -x^2 provides the lower bound and x --> x^2 and then inverting both sides provides the upper bound.
It's amazing what Spivak is able to prove in that book with relatively simple tools. The only proofs of the fundamental theorem of algebra I ever hear about are based on complex analysis or topology, but Spivak proves it with high school algebra and a smidge of real analysis.
as u know The function e^x² or e^-x² has no explicit formula for the indefinite integral. However, we can always get its integral through Taylor series. Since the function e^x is analytic. And then we replace x by x² (or-x²) and integrate ... This ideal can be used to get an approximation to the definite integral since the Taylor series is convergent we can minimize the perturbation as much as we want by just increasing the terms We take of the series.
You could also simply square the Integral, use Fubini and Substitute y=x*s, dy=x*ds. Remember 1/(1+s^2) is the Derivative of arctan(s) you'll get the result.
sir the video was very nice, but it will be very helpful for me if you come up with a video explaining the walis integral formula or share a document about it
I'm not sure I agree with the trick you used to integrate two terms in your inequality between 0 and infinity, and limit the left term to integral between 0 and 1. The function you integrate is positive only for -1 < x < 1, and clearly diverges towards -inf, so if you take a larger part of the integral, you'll end up with a smaller value. In fact the integral is maximal between 0 and 1 (if you stay on positive values of x). I think you need to explain this step a lot more. I'm not saying it's invalid, but the way it's presented it feels rather bizarre.
I'm a bit confused about what you did around 13:08. The lhs diverges to +inf for even n so the inequality doesn't hold anymore. Also by only integrating from 0 to 1 you're basically dropping the integral from 1 to inf, which is divergent. For the inequality, instead of 1-x² you should have defined a function h(x)=1-x² for |x|≤1 and h(x)=0 for all other x. Then from h(x)≤e^(-x²) you can safely arrive at the last inequality and integrate everything from 0 to inf.
The inequality holds, if you consider the up-to-one integral over all of the three and then add a (positive) one-to-infty integral to the second and third alone
"The lhs diverges to +inf for even n so the inequality doesn't hold anymore" No it doesn't. At least, not over the domain of integration (the domain that matters).
That's pretty neat! But still, IMHO, nothing beats that rectangular-to-polar conversion method for this integral. Nevertheless, this is a priceless 'arrow' in our mathematical 'quiver.' It could work some integrals the other method can't. Fred
Hi, Fine! I found the proof by double integral and polar cordinates a little bit tricky. For fun: There are bunches of places other than YT, 1 "and so on and so forth".
If you consider I=int_{0}^{infty} (e^(-x^2))dx Then, I^2=int_{0}^{infty}int_{0}^{infty} (e^(-(x^2+y^2)))dxdy Thus, changing the coordinates from the cartesian coord system to the polar coord system (taking into account that an infinite square is equal to a circle of infinite radius), and considering the determinant of the Jacobian matrix, hence: I^2=2pi*int_{0}^{infty}(r·e^(-r^2)) dr Which is an immediate integral. Once you have I^2, you get the initial I with the sqrt function!
Around 4:00, couldn't you have just substituted -x^2 for all the x^2 in the power series formula and immediately seen that e^-x^2>=1-x^2? Or would you need to do more case work after that?
The rest, in this case, is an alternating series: R(x) = x^4/2! - x^6/3! + x^8/4! - x^10/5! + ... It's not obvious that R(x) >= 0 for all x. It requires some work to prove that.
@@ricardocavalcanti3343 Exactly. But it does irk me that he proved all of these equalities for x^2 when he could have just proved them for t >= 0 and then substituted t = x^2.
At 13:07 You said the integral between 0 and 1 is less than the integral between 0 and infinity. But if n be 1 the integral is evaluated as x-x^3/3,which is 2/3 between 0 and 1 and negative infinity between 0 and infinity! I don't get it. Can you explain please?
If 1-x^2 is less than or equal to exp(-x^2), wouldn't the integral of (1-x^2)^n from 0 to infinity still be less than or equal to the Gaussian integral, i. e. converge also to sqrt(π)/2?
He did it because the result of the tool he used was from the integral from 0 to 1, and he was allowed to use that because it could maintain the inequality (if it had been an equation he couldn't have done it, but as long as it is an inequality and the leftmost thing is less than the others, it's ok to do that).
This puzzled me at first. I thought of it like this: The 1-x^2 can actually be replaced with max(1-x^2, 0) and the inequality still holds, as e^whatever >= 0. So now if we integrate from 0, inf the left hand integral is now integral from 0 to inf of max(1-x^2, 0) = integral from 0 to 1 of 1-x^2, as max(0, 1-x^2) = 0 for x>= 1.
Any Stats students know by heart that the evaluation of said integrand can be attached directly to standard normal distribution (mean = 0, s.d =1) a.k.a Gaussian pdf (probability density function).. Problem only lies within reconciling the pdf and the integrand presented here.. Integral (from -~ to ~) of [exp {(-x^2)/2} dx] equals to sqrt (2*Pi).. If you transform, exp {(-x^2)/2} as {(-x/(sqrt 2))}, and solve it, it becomes sqrt (2*Pi)/ sqrt (2) = sqrt (Pi).. Since Gaussian ranges from -~ to +~ then, the definite integral of 0 to ~ is HALF of the said value (because it's symmetrical on x=0).. id est, equals to [sqrt (Pi)]/ 2..
14:28 Homework... Imagine an election between two candidates. A receives m votes, B receives n votes, and A wins (m>n). If the ballots are cast one at a time, what is the probability that A will lead all the way throughout the voting process?
Bit confused by this myself. If n is even then the bit of the integral from 1 to infinity is positive and so dropping it is not a problem as the inequality stays the same. But it does look a bit odd.
Thinking about it a bit more the point is that the inequality which is used to construct the left hand side is only really meant to apply for x between 0 and 1. For x greater than 1 the we need 0 as 1-x^2 is negative and clearly less than exp(-x^2)
@@martinepstein9826 If you just formally integrate between zero and infinity then the integrate is diverges . It's diverged to plus infinity for n even and minus infinity for n odd. The result is none the less fine but needs a little more explanation.
It is easy to show that ∫ exp(-x^2)dx, x=0 to ∞ is convergent because we can split the integral at x=1 so.. ∫ exp(-x^2)dx, x=0 to 1 + ∫ exp(-x^2)dx, x=1 to ∞ since exp(-x^2)>0, the first definite integral is finite some number. and since exp(-x^2)⋜exp(-x) for x⋝1 the second integral must also converge because it is less by comparison ∫ exp(-x)dx, x=1 to ∞ which is 1/e.
Interesting, but not satisfying. How about some discussion of what was done? And why (motivation)? And why integrating the leftmost expression 0to1 gives the same result as 0toInf?
TL;DR: It follows from the inequality at 12:58 that their integrals over (0,1) follow the same inequality. Similarly, the integrals over (1,∞) follow the inequality. And integrals over (0,∞) are greater than integrals over (0,1) as all functions are greater than 0 for all x. Combining them together, we get what we wanted.
@@blazedinfernape886 Thanks. Then the answer is zero: the minimum of 3^((sec x)^2 - 1) in the interval [1,10] is 1, at x = pi, 2pi and 3pi, and the minimum of sqrt(9y^2-6y+2) in the same interval is sqrt(5), at y = 1. Therefore, 3^((sec x)^2 - 1)*sqrt(9y^2-6y+2) >= sqrt(5) for x, y in the interval [1,10]. (Note that y=1/3 lies outside that interval.)
Totally too complicated. Let I denote the integral. Write I^2 as the product of two integrals, one in x and the other in y. Express I^2 as a 2-dimensional integral, switch to polar coordinates, and I^2 can be easily determined. I=sqrt(I^2).
@@jean-claudearbaut7322 what about convergence of sequences of real valued functions? Uniform and pointwise. Is it on first year calculus in your country? I see sequences of real valued functions and limits... Anyway my preferred way to find that integrally is using Fourier transform
The presentation I sketched was given by a TA in a calculus section at the University of Wisconsin 55 years ago. I always kept it on mental file because it was so clever. I don't see advanced real analysis, measure theory, or anything like that being an issue. Thanks 😊 for remarks.
Soooooo cool. Never thought this integral is even solvable with elementary calculus. This integral is such a classic to introduce multivariable calculus!
I don’t know if I would call the Wallis integrals part of ‘elementary calculus’, but there is definitely a lot here that a Calc 1 or 2 student can follow.
But this method is more dirty，and you should know the Wallis integral
@@bttfish How is it dirty though. Imo the usual method is dirtier because you have to know way multivariable calculus to integrate a single variable function. Also Wallis integral is just integration by parts which is elementary calculus.
It's one of those solutions that can only be found be reverse engineering. You wouldn't start with that inequality. But if you have a couple of integrals which you know give the same answer as the Gaussian, and you can show that they are the results of a limit that never exceed the Gaussian (or is never lower) the rest falls into place.
Very neat.
For those confused about the integral from 0 to 1 at 13:20, Martin Epstein came up with this:
the integral (0,1) left
I'm still confused. I understand you can replace with max(1-x^2,0). But in the same spirit you can also replace with f(x) = (1-x^2) for x1. I.e. for x>1 we use the gaussian function itself. f(x) then is 1 is the same). Then the integral 0-inf of this f(x) on the lhs is sqrt(pi)/2 (for the part x1 part of f(x). And in the end the integral 0-inf of the lhs will be bigger than what we want. what am i missing?
@@scipionedelferro you should apply the limit when n goes to infinity, the integration of the x>1 part of f(x) will most certainly go to 0.
I think there's a simpler way to look at it. Integral(0 to 1) (left) is less than integral(0 to 1) (middle) which is less than integral(0 to oo) (middle). QED.
@@martinepstein9826 yes nice one. simple.
Even if we consider the left integral = middle integral it can only happen when both integral are from 0 to infinity. Since the left integral goes from 0 to 1 instead of taking 0 to infinity the lhs will become less than middle integral and the equality will break down. If we assumed from the start that left integral < middle integral then even integrating left from 0 to infinity will make left less than middle
Very cool. That's my favorite integral (solved the "usual" way: square it, convert to polar coordinates, use tricks, integrate, then square root). I've never seen this approach. Thank you.
4:00 I think we can tackle the two-sided inequality much faster by considering f(x) = exp(x) - 1 - x which (a) has critical point at x = 0 and (b) is convex everywhere, which proves f(x) >= 0 or exp(x) >= 1 + x (the derivatives are much easier than using x^2). Then, x --> -x^2 provides the lower bound and x --> x^2 and then inverting both sides provides the upper bound.
Thanks. In Russia this integral usually are counting in polar coordinates. Your variant of the solve is unfamous. It is great
Well that's not only in Russia, everywhere it is mostly commonly solved by polar coordinates
This method also appears in Calculus by Spivak.
I love spivak's texts. We used his single variable Calculus book at the University of Michigan.
It's amazing what Spivak is able to prove in that book with relatively simple tools. The only proofs of the fundamental theorem of algebra I ever hear about are based on complex analysis or topology, but Spivak proves it with high school algebra and a smidge of real analysis.
as u know The function e^x² or e^-x² has no explicit formula for the indefinite integral. However, we can always get its integral through Taylor series. Since the function e^x is analytic. And then we replace x by x² (or-x²) and integrate ... This ideal can be used to get an approximation to the definite integral since the Taylor series is convergent we can minimize the perturbation as much as we want by just increasing the terms We take of the series.
this is actually the way we had to solve it in our analysis 1 homework! great seeing you give this approach some respect ;)
The squeeze theorem! Never thought it would show up here!
Glad to see an approach different than I^2. 👍
You could also simply square the Integral, use Fubini and Substitute y=x*s, dy=x*ds. Remember 1/(1+s^2) is the Derivative of arctan(s) you'll get the result.
sir the video was very nice, but it will be very helpful for me if you come up with a video explaining the walis integral formula or share a document about it
using wallis product and squeeze theorem, super cool way of finding the gaussian integral
Gamma function and reflection formula - with sine
Double integral and polar coordinates
I'm not sure I agree with the trick you used to integrate two terms in your inequality between 0 and infinity, and limit the left term to integral between 0 and 1. The function you integrate is positive only for -1 < x < 1, and clearly diverges towards -inf, so if you take a larger part of the integral, you'll end up with a smaller value. In fact the integral is maximal between 0 and 1 (if you stay on positive values of x).
I think you need to explain this step a lot more. I'm not saying it's invalid, but the way it's presented it feels rather bizarre.
Ur are Great Professor
So, basically you chased the integral until it was trapped between 2 walls and couldn't escape anymore...
You are the math professor i wish i had
Haven’t seen this derivation of the result before, it’s definitely one that wouldn’t be a 1st solution lol! Great though
I'm a bit confused about what you did around 13:08. The lhs diverges to +inf for even n so the inequality doesn't hold anymore. Also by only integrating from 0 to 1 you're basically dropping the integral from 1 to inf, which is divergent. For the inequality, instead of 1-x² you should have defined a function h(x)=1-x² for |x|≤1 and h(x)=0 for all other x. Then from h(x)≤e^(-x²) you can safely arrive at the last inequality and integrate everything from 0 to inf.
The inequality holds, if you consider the up-to-one integral over all of the three and then add a (positive) one-to-infty integral to the second and third alone
"The lhs diverges to +inf for even n so the inequality doesn't hold anymore"
No it doesn't. At least, not over the domain of integration (the domain that matters).
That's pretty neat!
But still, IMHO, nothing beats that rectangular-to-polar conversion method for this integral.
Nevertheless, this is a priceless 'arrow' in our mathematical 'quiver.' It could work some integrals the other method can't.
Fred
12:25 Taking the n power can cause problems if the first term is negative and n even.
Isn't it?
Hi,
Fine! I found the proof by double integral and polar cordinates a little bit tricky.
For fun:
There are bunches of places other than YT,
1 "and so on and so forth".
If you consider I=int_{0}^{infty} (e^(-x^2))dx
Then,
I^2=int_{0}^{infty}int_{0}^{infty} (e^(-(x^2+y^2)))dxdy
Thus, changing the coordinates from the cartesian coord system to the polar coord system (taking into account that an infinite square is equal to a circle of infinite radius), and considering the determinant of the Jacobian matrix, hence:
I^2=2pi*int_{0}^{infty}(r·e^(-r^2)) dr
Which is an immediate integral.
Once you have I^2, you get the initial I with the sqrt function!
Around 4:00, couldn't you have just substituted -x^2 for all the x^2 in the power series formula and immediately seen that e^-x^2>=1-x^2? Or would you need to do more case work after that?
The rest, in this case, is an alternating series: R(x) = x^4/2! - x^6/3! + x^8/4! - x^10/5! + ... It's not obvious that R(x) >= 0 for all x. It requires some work to prove that.
@@ricardocavalcanti3343 Exactly. But it does irk me that he proved all of these equalities for x^2 when he could have just proved them for t >= 0 and then substituted t = x^2.
At 13:07
You said the integral between 0 and 1 is less than the integral between 0 and infinity.
But if n be 1 the integral is evaluated as x-x^3/3,which is 2/3 between 0 and 1 and negative infinity between 0 and infinity!
I don't get it. Can you explain please?
in which video you show this part? 9:32
캬 이거지
At 9:36 , I could not get why limit n tends to inf ((n)^(1/2) Wallis product) converges to (pi^(1/2))/2
If 1-x^2 is less than or equal to exp(-x^2), wouldn't the integral of (1-x^2)^n from 0 to infinity still be less than or equal to the Gaussian integral, i. e. converge also to sqrt(π)/2?
3:29
You should point out that you can take the reciprocal of both sides of the inequality like the way you did because both sides are non-negative.
13:18 sir please tell why did you integrate it from 0 to 1??? Please help I can't figure it out
Squeeze theorem
He did it because the result of the tool he used was from the integral from 0 to 1, and he was allowed to use that because it could maintain the inequality (if it had been an equation he couldn't have done it, but as long as it is an inequality and the leftmost thing is less than the others, it's ok to do that).
@@Catilu thank you sir
@@jackchung 👍 thanks sir
This puzzled me at first. I thought of it like this: The 1-x^2 can actually be replaced with max(1-x^2, 0) and the inequality still holds, as e^whatever >= 0. So now if we integrate from 0, inf the left hand integral is now integral from 0 to inf of max(1-x^2, 0) = integral from 0 to 1 of 1-x^2, as max(0, 1-x^2) = 0 for x>= 1.
Now do the Gaussian integral using the gamma function, it’s a cool derivation
@@angelmendez-rivera351 You can find it by the reflection formula, no?
Why the limits of integration from the left [0;1[
And that's a good place to sto
can someone give me a reference to this Wallis variant product.
Did you assume Stirling formula for those limits? If yes, it's kinda roundabout... expressing one hard theorem by another
13:06 What if we integrate from 0 to infinity, like the other two integrals? It would be bigger than √π/2, right?
@@danielelaudaz8551 Oh, that makes sense now. I was confused on why you intergrate this one only from 0 to 1.
Look up the sandwich theorem. He wants to show that the integral of the gaussian is \sqrt(pi)/2.
@@danielelaudaz8551 It diverges to +inf for even n
Any Stats students know by heart that the evaluation of said integrand can be attached directly to standard normal distribution (mean = 0, s.d =1) a.k.a Gaussian pdf (probability density function)..
Problem only lies within reconciling the pdf and the integrand presented here..
Integral
(from -~ to ~) of
[exp {(-x^2)/2} dx] equals to sqrt (2*Pi)..
If you transform, exp {(-x^2)/2} as
{(-x/(sqrt 2))}, and solve it, it becomes
sqrt (2*Pi)/ sqrt (2) = sqrt (Pi)..
Since Gaussian ranges from -~ to +~ then, the definite integral of 0 to ~ is HALF of the said value (because it's symmetrical on x=0)..
id est, equals to
[sqrt (Pi)]/ 2..
It needs to be part of the overkill playlist
מחזק
When you have the answer hand delivered to you by ups
14:28 Homework...
Imagine an election between two candidates. A receives m votes, B receives n votes, and A wins (m>n). If the ballots are cast one at a time, what is the probability that A will lead all the way throughout the voting process?
Solution and source
p = (m-n)/(m+n)
See futilitycloset.com/2013/04/29/the-ballot-box-problem for more details
How do you find this?
huh...
@@Fun_maths You mean, find the homework or find the solution?
@@goodplacetostop2973 the homework
Why can you drop the left integral into 0 to 1?
Bit confused by this myself. If n is even then the bit of the integral from 1 to infinity is positive and so dropping it is not a problem as the inequality stays the same. But it does look a bit odd.
Thinking about it a bit more the point is that the inequality which is used to construct the left hand side is only really meant to apply for x between 0 and 1. For x greater than 1 the we need 0 as 1-x^2 is negative and clearly less than exp(-x^2)
I believe we are looking for the sup and 0 to 1 gives us the sup
Why wouldn't you be able to only integrate from 0 to 1? You just need it to be less than the middle integral.
@@martinepstein9826 If you just formally integrate between zero and infinity then the integrate is diverges . It's diverged to plus infinity for n even and minus infinity for n odd. The result is none the less fine but needs a little more explanation.
It is easy to show that ∫ exp(-x^2)dx, x=0 to ∞ is convergent because we can split the integral at x=1 so..
∫ exp(-x^2)dx, x=0 to 1 + ∫ exp(-x^2)dx, x=1 to ∞
since exp(-x^2)>0, the first definite integral is finite some number.
and since exp(-x^2)⋜exp(-x) for x⋝1 the second integral must also converge because it is less by comparison ∫ exp(-x)dx, x=1 to ∞ which is 1/e.
I remember we had that as an exam problem in my analysis 2 course in first yr
At 9:15 shouldn’t it be (2n-1)? Because the way it’s written, at n=1 the first term would be 3.
Love from Pakistan ❤️
Interesting, but not satisfying. How about some discussion of what was done? And why (motivation)? And why integrating the leftmost expression 0to1 gives the same result as 0toInf?
I too was left thinking: I want to know the story behind the discovery. I am pretty certain it didn't start with that inequality.
I think I have a solution. Basically we write everything upto 12:58
Also for sake of simplicity, I will write the terms as A
TL;DR: It follows from the inequality at 12:58 that their integrals over (0,1) follow the same inequality. Similarly, the integrals over (1,∞) follow the inequality. And integrals over (0,∞) are greater than integrals over (0,1) as all functions are greater than 0 for all x. Combining them together, we get what we wanted.
You can get your second inequality much easier by using 1 + x^2 + 1/2! x^4 + ... = 1.
Natural and Beautiful..
When x = sin(t), why is dx = cos(t)*t and not dx = dsin(t) ? (7:40)
is infinity minus TREE(3) still infinity? (yes)
Fedex
behenrj
A question :-
If x,y € [1,10]
3^(sec²x-1) * √(9y²-6y+2)
This looks like a classic old jee advanced problem.
9y^2-6y+2 >=1 for all y
But 3^tan^2x * sqrt(9y^2-6y+2) =
I haven't understood the question. What are we supposed to calculate?
@@ricardocavalcanti3343 Number of pairs of solutions (x,y). It is given in the question
@@blazedinfernape886 Thanks. Then the answer is zero: the minimum of 3^((sec x)^2 - 1) in the interval [1,10] is 1, at x = pi, 2pi and 3pi, and the minimum of sqrt(9y^2-6y+2) in the same interval is sqrt(5), at y = 1. Therefore, 3^((sec x)^2 - 1)*sqrt(9y^2-6y+2) >= sqrt(5) for x, y in the interval [1,10]. (Note that y=1/3 lies outside that interval.)
Totally too complicated. Let I denote the integral. Write I^2 as the product of two integrals, one in x and the other in y. Express I^2 as a 2-dimensional integral, switch to polar coordinates, and I^2 can be easily determined. I=sqrt(I^2).
@@jean-claudearbaut7322 what about convergence of sequences of real valued functions? Uniform and pointwise. Is it on first year calculus in your country? I see sequences of real valued functions and limits... Anyway my preferred way to find that integrally is using Fourier transform
The presentation I sketched was given by a TA in a calculus section at the University of Wisconsin 55 years ago. I always kept it on mental file because it was so clever.
I don't see advanced real analysis, measure theory, or anything like that being an issue. Thanks 😊 for remarks.
Ok, fair enough. Math literate Americas generally have extremely high regard for French math training.