What's so amazing about these videos is how these results at first look (to me) well beyond the ability of a mid-level college math major (that was me in college, but I never studied stuff like this), yet they frequently use just basic first-year calculus, applied creatively.
Just an integration tip I thought I would share: when changing variables, you can multiply the integrand by the modulus of the derivative and make the bounds of the integral ordered, ie the smaller bound on the bottom. The reason for this: when doing change of basis in multivariable calculus, we use the modulus of the Jacobian. If you just use the one dimensional version, you get what I just described. I personally find it easier to work with, but it's up to you.
@@OH-pc5jx yes well change of basis theorem doesnt hold if the derivative is zero anywhere. I suppose you may be able to split the domain of the integral before substitution, but it may get messy ;)
If we want to do this in other direction 1/sqrt(1-x^2) - by binomial expansion integrate to get arc sin(x) Use Cauchy product to get series for arcsin(x)/sqrt(1-x^2) but we would have to deal with double sum
A nice convergence of summation of proper fractions.The definite integration of sine power function is mysteriously connected to the summation.Very interesting.
There are two factorial signs in the expression in the thumbnail; one inside parentheses, the other outside. Both are correct. The summand is just 1/C(2n,n). Fred
But ²F¹ is the usual "original" (Gauss's) hypergeometric function. It's only really "generalised" if the indices are different from 2 and 1. So much so that the indices 2 and 1 are usually omitted in this case and one simply write F(1,1; 1/2;1/4)
A nice convergence of summation of proper fractions plus one.The definite integration of sine power function is mysteriously connected to the summation.Very interesting.
I think you would better mentioned that formula of infinite sum which you used is only worth for convergent sums of course it is arcsin so x is between -1 and 1 however it was not really obvious
The general case is: Sum[(n!)^p/(q n)!,{n,0,Infinity}]=pF(q-1)[1,1,...,1;1/q,2/q,...,(q-1)/q;q^q] where pFq is the Hypergeometric function. The particular sum in the video is 2F1[1,1;1/2;1/4]
Write the fraction as a beta function then change the order of sum and beta integral. Then do the sum first (which will be an easy geometric series) Then do the integral
Thank you for your comment. I thought about gamma functions at first when seeing the problem but I had forgotten about beta. I tried it as you said and indeed, a bit long to compute but very easy approach
Very great idea! Using Beta function properties, I obtain that the general term of the series is B(n,n)*n/2. One has to be careful, because Beta funcion isn't defined in B(0,0), and is necessary to pull apart the first term, 1, from the rest of the series. Then, using the Beta funcion definition and Lebesgue's dominated convergence theorem, I exchange the integral symbol for the sumation symbol. Now, the series inside the integral is the derivative of a geometric series (the term is (n+1)(t(1-t))^n, from n=0 to infinity). This series is equal to 1/(1-t(1-t))^2 when t \in (0,1). So the initial series is equal now to 1+1/2 \int_0^1 1/(1-t(1-t))^2dt. This integral isn't funny at all, but can be done and gives the same result obtained by professor Penn.
How would one approach the sum of (2k choose k) * (-4)^(-k) from k=0 to infinity? I got to the sum from the integral of e^(-x) erf(sqrt(x)) dx from x=0 to infinity. That integral has a surprisingly simple closed form, and I imagine its evaluation is quite elegant. The sum I have verified is correct, and I obtained it through the taylor series for e^x and some gamma function identities.
Excellent! Michael would you do a presentation on the Dominated Convergence Theorem, also what do you recommend as a good reference for detail on it? rgds, Rod
Enough Sound explanation....but may I get the same series by the help of fourier series? May I get a periodic function which may help me?...eagerly waiting for your reply.... From India ♥️
Remarkable result! Remarkable that you can even get a result! Especially seeing what it takes to get it!! Might I ask how you came across this result? 2π/(9√3) + 4/3 = 1.736399858718715077909795168364923... Finally, kudos for your teaching style. It is really exemplary! Fred
for me it is interesting to listen to the whole demonstration one day and then (one day, two days later) go back to listening to it 1.5 times faster. complete understanding. thank you Mr. Penn!
In general Sum( (n!)^b / (b n)! ), n=0 to inf for b>=2 can be expressed as the generalized hypergeometric function p F q b F b-1 [1, ... (b times) ... ,1 ; 1/b, 2/b, ... , (b-1)/b ; 1/(b^b)]
Nice problem. Did it differently though (after looking at the video, there are some similarities) - defined sum y(x) but adding x^(2n-1) to the sequence and starting at n=1 (so our final sum = y(1)+1). - after some manipulation and derivation, I get a differential equation y'(4-x^2)-3xy=2 with y(0)=0 - Using integration factor, I get in the end y(x) = 2(4-x^2)^(-3/2) * integral(sqrt(4-t^2)dt, t=0..x). Which can be integrated with simple substitution t=sin(u). - End result y(x) = 1/(4-x^2)^(3/2) * (4*arcsin(x/2)+x(4-x^2)^(1/2). Substituting x=1, I get same result.
the way it's written on the post cover (2n!) is wrong, it leads to divergence. The right notation is (2n)! as it appears on the board in the video. 2 x n! is not (2n)!
@@angelmendez-rivera351 it definitely is, don't get me wrong. By the way Michael prefaced the video I expected the sum to be straight up pi or maybe pi ^ k or something. Having a closed form be it convoluted as it may be is always amazing starting with such weird sums
What's so amazing about these videos is how these results at first look (to me) well beyond the ability of a mid-level college math major (that was me in college, but I never studied stuff like this), yet they frequently use just basic first-year calculus, applied creatively.
What?!!! This series has the sum in a closed form?!
man where are your videos?
Hello!
DJ VALENTE DO CHP If you check his instagram you’ll see he got a surgery recently (appendicitis I think), so he needs time
@@HeyKevinYT thanks for the information 👍
Get well soon, bprp!!
Fred
13:32
🤣
27:57 ?
This video is gonna get alot of views, I can feel it
Just an integration tip I thought I would share: when changing variables, you can multiply the integrand by the modulus of the derivative and make the bounds of the integral ordered, ie the smaller bound on the bottom. The reason for this: when doing change of basis in multivariable calculus, we use the modulus of the Jacobian. If you just use the one dimensional version, you get what I just described. I personally find it easier to work with, but it's up to you.
Yes - unless the derivative is zero within the range, in which case you’re ✨ trouble ✨
Usually a sign of a bad substation tho so I don’t think it comes up too often
@@OH-pc5jx yes well change of basis theorem doesnt hold if the derivative is zero anywhere. I suppose you may be able to split the domain of the integral before substitution, but it may get messy ;)
William Churcher yeah in 1D you can get around it with a bit of thought but in nD I think I’d just use a different substitution
If we want to do this in other direction
1/sqrt(1-x^2) - by binomial expansion
integrate to get arc sin(x)
Use Cauchy product to get series for arcsin(x)/sqrt(1-x^2)
but we would have to deal with double sum
Sir can you do a series of lesbegue integration and measure theory , functional analysis?
A nice convergence of summation of proper fractions.The definite integration of sine power function is mysteriously connected to the summation.Very interesting.
Very claasic video my Michael. Excellent!
15:05 Nice clothing change!
28:41
:)
Woah I just realised what a long video that was
username checks
15:06 was not a good place to stop
Michael, you have a typo in your thumbnail.
the factorial in the thumbnail is inside the brackets
I saw that at first and then thought “infinity, how is this video so long”
There are two factorial signs in the expression in the thumbnail; one inside parentheses, the other outside. Both are correct.
The summand is just 1/C(2n,n).
Fred
ffggddss no He changed the thumbnail, previously it was (2n!) which is completely different from (2n)!
@@sauravthegreat8533 I see... Well that *does* make a difference.
Fred
As always, a wonderful video. Thank you for the great math content!
The result can be expressed as
²F¹[1,1;0.5;(1/2²)]
F is the generalized hypergeometric function
wolfram notation
Hypergeometricpfq[{1,1},{0.5},0.25]
But ²F¹ is the usual "original" (Gauss's) hypergeometric function. It's only really "generalised" if the indices are different from 2 and 1. So much so that the indices 2 and 1 are usually omitted in this case and one simply write F(1,1; 1/2;1/4)
A nice convergence of summation of proper fractions plus one.The definite integration of sine power function is mysteriously connected to the summation.Very interesting.
Hey I can't wait for the video on the dominated convergence theorem. I'm struggling with it quite a bit
Very nice! Wouldn’t know where to start without the hints ngl
I would love a video about the dominating convergence theorem
Would love to see some videos on some of the more sensitive convergence tests like Gauss's test
Yes please to the video on dominated convergeance theorem.
You are starting to remind me of Sami Hamiyyee.... my favorite professor.. Thank you.
Beautiful. How did you find these three lemmas?
27:56 was that a burp?
How did you get this? Like seriously it seems so arbitrary but boom everything works and fits in perfectly
Oh my God! It’s awesome! Good job
19:02 but by writing 1-x^2 as sqrt(1-x^2)^2 you are assuming that x =< 1, right..?
X=1/2
Even from the very beginning, the condition |x|
not really since the integral is from 0 to 1.
a lot of prep pays off
I think you would better mentioned that formula of infinite sum which you used is only worth for convergent sums
of course it is arcsin so x is between -1 and 1
however it was not really obvious
The general case is:
Sum[(n!)^p/(q n)!,{n,0,Infinity}]=pF(q-1)[1,1,...,1;1/q,2/q,...,(q-1)/q;q^q]
where pFq is the Hypergeometric function.
The particular sum in the video is 2F1[1,1;1/2;1/4]
Please a video on the dominated convergence theorem!
Factory-al.
What's the second tool called
And iis there any other way of proving it ?
Amazing!!
Shouldn't the result be pi/9/3^0.5 + 4/3 though? I can't figure where the 2 in the numerator comes from
15:02 magic
Write the fraction as a beta function then change the order of sum and beta integral.
Then do the sum first (which will be an easy geometric series)
Then do the integral
Thank you for your comment. I thought about gamma functions at first when seeing the problem but I had forgotten about beta. I tried it as you said and indeed, a bit long to compute but very easy approach
Very great idea! Using Beta function properties, I obtain that the general term of the series is B(n,n)*n/2. One has to be careful, because Beta funcion isn't defined in B(0,0), and is necessary to pull apart the first term, 1, from the rest of the series.
Then, using the Beta funcion definition and Lebesgue's dominated convergence theorem, I exchange the integral symbol for the sumation symbol. Now, the series inside the integral is the derivative of a geometric series (the term is (n+1)(t(1-t))^n, from n=0 to infinity).
This series is equal to 1/(1-t(1-t))^2 when t \in (0,1). So the initial series is equal now to 1+1/2 \int_0^1 1/(1-t(1-t))^2dt. This integral isn't funny at all, but can be done and gives the same result obtained by professor Penn.
please do a proof of the dominated convergence theorem!!!
He have also used the differentiation theorem of power series.He should also do a video of that.
How would one approach the sum of (2k choose k) * (-4)^(-k) from k=0 to infinity? I got to the sum from the integral of e^(-x) erf(sqrt(x)) dx from x=0 to infinity. That integral has a surprisingly simple closed form, and I imagine its evaluation is quite elegant. The sum I have verified is correct, and I obtained it through the taylor series for e^x and some gamma function identities.
So good solution, I like this video .
Funny, usually when factorials appear in infinite series, e usually shows up. This time it's pi.
Try the numerator without the square...
Sum[(n!)/(2 n)!,{n,0,Infinity}]
Excellent! Michael would you do a presentation on the Dominated Convergence Theorem, also what do you recommend as a good reference for detail on it?
rgds, Rod
Can you evaluate the integral log(x)/(1-x^2) from 0 to 1 by means of a contour?
Do a video about catalan number please
First definit integral can be easily derived by walleys method than induction
Could you prove that (2n)!!/(2n-1)!!~sqrt(πn) please?
Hello, I want ti know how did you find those tools? Because if I have to solve this problem from nothing how can I find those tools?
Sir its great
Did anyone else notice when he changed shirts after the second tool?
Me: doing the exercise
Michael: and thats a good place to stop
Me: okay michael senpai :3 i will stop 4 u uwu
Suggestion: evaluate the same sum but instead of (2n!) in the denominator, put (2n+1)!
Enough Sound explanation....but may I get the same series by the help of fourier series? May I get a periodic function which may help me?...eagerly waiting for your reply.... From India ♥️
In your title you forgot the rejoinder, "... so little time!!!". Seriously, you put out great content.
Dat burp doe
Remarkable result! Remarkable that you can even get a result! Especially seeing what it takes to get it!! Might I ask how you came across this result?
2π/(9√3) + 4/3 = 1.736399858718715077909795168364923...
Finally, kudos for your teaching style. It is really exemplary!
Fred
15:06 Everything changed when the red-shirt nation attacked.
15:05
for me it is interesting to listen to the whole demonstration one day and then (one day, two days later) go back to listening to it 1.5 times faster. complete understanding. thank you Mr. Penn!
Hello, I have a question. What is (319!)! ?
@Michael Penn Your answer is wrong, it's:
pi/(6×sqrt(3)) + 4/3
This remind me the beta function.
In general
Sum( (n!)^b / (b n)! ), n=0 to inf
for b>=2
can be expressed as the generalized hypergeometric function p F q
b F b-1 [1, ... (b times) ... ,1 ; 1/b, 2/b, ... , (b-1)/b ; 1/(b^b)]
The teacher has to tell where term
Wise differentiation ,integration is applicable
His answer and the way I did it numerically in Matlab gives a consistent result. I wouldn't have gotten it analytically though... This is really good!
Isn't math(s) amazing. One never knows what will turn up in series. Here we have pi again!
Nice problem.
Did it differently though (after looking at the video, there are some similarities)
- defined sum y(x) but adding x^(2n-1) to the sequence and starting at n=1 (so our final sum = y(1)+1).
- after some manipulation and derivation, I get a differential equation y'(4-x^2)-3xy=2 with y(0)=0
- Using integration factor, I get in the end y(x) = 2(4-x^2)^(-3/2) * integral(sqrt(4-t^2)dt, t=0..x). Which can be integrated with simple substitution t=sin(u).
- End result y(x) = 1/(4-x^2)^(3/2) * (4*arcsin(x/2)+x(4-x^2)^(1/2).
Substituting x=1, I get same result.
That's close to how I did it too, though your equation was slightly simpler than mine.
Amazing! Could you explain how did you found this ODE? Also, why add x^(2n-1) instead of x^n?
Hello Michael! Where is this problem from? Was it from a math competition?
13:33 burp xd
the way it's written on the post cover (2n!) is wrong, it leads to divergence. The right notation is (2n)! as it appears on the board in the video. 2 x n! is not (2n)!
It seems so strange to obtain a sum as the result ....
Nice
this one is cute, reminds me of quantum mechanics prroblems
Nice problem and thanks for sharing this video! By the way, there's a minor mistake in that the derivative of arcsin(x) should be 1/sqrt(1 - x^2).
Even though the answer is a closed form. The fact that is a sum leaves me... Unfulfilled
you're welcome to take a common denominator lol, unless you would also be unfulfilled by the resulting sum in the numerator xD
@@angelmendez-rivera351 it definitely is, don't get me wrong. By the way Michael prefaced the video I expected the sum to be straight up pi or maybe pi ^ k or something. Having a closed form be it convoluted as it may be is always amazing starting with such weird sums
7th comment
13:33 Michael penn remain silent for 2sec straight giving respect to Chadwick Boseman
@Adam Romanov are you Russian
@@parameshwarhazra2725 по-моему, нет.
@@IoT_ what did you say please tell me in English
@@parameshwarhazra2725 I said, in my opinion he's not а Russian, but allegedly he knows it a bit.
@@IoT_ oh okay