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The overpowered Laplace technique for summing series.
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- čas přidán 23. 08. 2023
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Very cool but a and B cannot be the same
even if a and b are both positive? r/b-a will blow up but Xᵃ - Xᵇ will go to zero. hmmmm
its not apparent that a=b is invalid from the series alone
you can take a limit as b approaches a and notice it is the derivative of x^(a) with respect to a
@@aliinci1874 can confirm. Evaluating for b approaches a at a=0 r=1 you find the solution to the basel problem
@@GeoffryGifari of a=b u can't use partial fraction decomposition, not in the way Michael did
a=b is a special case of decomposition.
Mind blown... We used Laplace transforms to solve diff eqs... a lot. Using it like this opens your mind. I love pure maths more and more every day!
The formula only works for a != b. Can we use limits to extend it for the a = b case?
Computer scientist spotted!!!!!!!!!
Also just take the original problem and consider the case a=b and you kind of just solve it from there
Say a == b, not a = b.
Congratulations!! A very interesting point of view linking infinite series to integrals!! Keep up the great job, Mike!!...
At 7:00, we must require both a and b to be > -1 to ensure absolute convergence for both integrals. In fact, take for example the first Laplace transform: multiplying e^(-nt) with e^(-at) under the integral sign, we have e^[-t(n+a)]. Remember that t > 0, so for the integral to converge for every natural n > 0 we need n+a > 0 for every natural n > 0. This is true if and only if a > -1. The same goes for b.
Well... He assumed a,b to be nonnegative integers right at the start, didn't he?
an interesting case analysis of the formula would be at r=1 a = 0 and b approaching 0 as a limit, the value of this should be the basel sum
A bit of a basic question, but at 13:35, when performing the u substitution, why is it okay to write sqrt(u^2)=u instead of |u|? The way Michael did it in the video, after writing dx=2udu you're left with u^2 on the numerator while I would have thought the result to be |u|*u
Thank you.
Both x and u are positive on the interval (0, 1).
The integration is within the interval which is more or equal to zero ,i.e. from 0 to 1
@@michaelguenther7105 Of course! Thanks.
@@IoT_ Thank you!
Actually if u=-sqrtx, x=1,u=-1,u can be negative
However, u=sqrtx and u=-sqrtx are the same integral
Let x=u^2 and -sqrtx=u
dx=2udu
x=0,u=0
x=1,u=-1
∫_(0,1) sqrtx/(x-1)dx
=∫_(0,-1) -u/(u^2-1)*2udu
=∫_(-1,0) 2u^2/(u^2-1)du
=∫_(0,1) 2u^2/(u^2-1)du
Because it is an even function
hmmm can the resulting integral be solved with contour integration?
At the end the case r=0 works and gives 0 not because ln(1)=0, but because the limit of all that stuff is 0 for r->0
you can't really use dominated convergence here, but monotone convergence will do since we're summing positive terms
A thing of beauty.
The calculation in first example was wrong. U should integrate them separately because when x=u^2, u=sqrtx, x=1-, u=sqrt(1-), when x=u^3, u=cbrtx, x=1-, u=cbrt(1-)
If u assumed x=1, u=sqrt1=1 and u=cbrt1=1, u will get wrong in final answer.
Very satisfying method I agree.
18:45
Sums are my favorites
7:21
Why are the sigma and integral interchangeable?
There is a theorem about this which has to do with different types of convergence, I think when you have "absolute" convergence then it is okay to make the switch. Maybe someone can expand on my answer.
The "Dominated Convergence Theorem". Basically says as long as your series is 'nicely behaved', you can swap the order. (*very* oversimplified explanation of 'nice' in this case is absolutely bounded and converges at every value of the summation index. don't use this explanation in your homework.)
Iirc the condition for dominanted convergence is that (or at least in one of the versions) there is a nonnegative function such that at each point within the integration bounds, the expression doesn’t exceed the function regardless of n, and that the function is integrable
Hi,
I used this technic, when I was about 20, to study this functional series : f(x)=1/(x+1)-1/(x+2)+1/(x+3)-... , and show that it had some thing to do with (pi x) / (sin pi x).
Knowing, now, that that is equal to gamma(1-x) gamma(1+x), may be I could retrieve my function out of the gamma function. 🤔
By the same method on the video I reduced -sum_{n=1}^{infty} (-1)^n/(n+x) to int_0^infty e^-(1+x)t/(1+e^(-t)) dt. I'm not sure if I did it right and if it is how it is related to gamma(1-x)gamma(1+x).
When plotted on Desmos the integral is a really got approximation for gamma(1-x)gamma(1+x) in the interval (-1,-1/2]
Amazing!
[Edit: This is wrong, but just in case someone else has the same brainfart] 14:06 is clearly wrong, it should be Integral u/(u²-1)du and Integral u/(u³-1)du.
I think you forgot to multiply the dx terms.
No, it is right because of the du term.
@@megauser8512 Sorry, you are right.
13.36
He said dx=2udu but he wrote du=2udu
region of convergence
Excellent video as always
I dont feel comfortable in that change of variables.
It feels like because of the limit nature of integrals, youre doing there two different infinite limits with the same variable taken at the same time, something that is not entirely rigorous. Can you explain why you are allowed to do that?
Since the original series is absolutely convergent, then any change of variables or order is valid.
Well, you can tear the former integral into two, do the change of variable for each of them as Michael has done. The variables used in the change should probably differ as they result from different changes, but we can rename the variable we integrate over as we please. Then we would be left with the addition of the two integrals over the same integral, which can be joined again.
Schematically:
int_0^1 [ f(x) + g(x) ] dx = int_0^1 f(x) dx + int_0^1 g(x) dx = int_0^1 h(u) du + int_0^1 k(v) dv = int_0^1 h(u) du + int_0^1 k(u) du = int_0^1 [ h(u) + k(u) ] du
The thing is we are able to do this because both of the changes of variables give the same interval (0, 1) "by coincidence".
0:54 Yeah, that's the problem with most of what you learn in school. It only works in school.
This is interesting
Check it out !
This is not really overpowered when you can turn the summand into an integral as already shown by yourself.
handsome
This is a pet peeve of mine that I have written about before. When you say (a lot) “something like” it assumes you are putting an approximation to the answer rather than saying “the answer is or will be”
19:21