Fourier less series
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- Опубліковано 20 сер 2019
- In this video, I calculate the sum of cos(2nx)/n without using Fourier series! (well, almost :P)
Integral (ln(cos(x)))^2 from 0 to pi/2: • Integral ln cos square...
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Well e^ix for some real x is always going to be on the unit circle, and seeing as ln(1-z) expressed as an infinite sum converges when |z|
Nice one, Dr. Peyam.
I loved this video.....
Answer looks like the square wave sum_{i=1}^{inf} sin((2i-1)x)/(2i-1) but with cosine and even numbers!
complex trig definitions ftw!
When using the principal branch of the complex logarithm, ln(z)+ln(w)=ln(zw) is true if, and only if, the sum of the arguments of z and w (where the argument is considered between -pi and pi, including -pi) is between -pi and pi, i.e., -pi
Cool video Dr peyam!
a powerful demonstration
fantastic video!!
Very interesting
You should've worn a cool tunic for this "2nix" problem
Let me guess, are you gonna use the complex definition of cosine and then use ln(1+x) power series?
Bingo
Will you talk about the Clausen function(s) more :) ?
Do something on hyperbolic functuons
is it possible at all to convert that original sum to a Riemann integral?
Because there isn't a factor 1/N in the series, you can't do it directly. The factor 1/n is different, and it's not related to how many terms the series has, it's just a variable within the series.
There also isn't any clear way of getting a factor 1/N, but nobody'll stop you from trying! You might try taking the derivative of the result and writing out the Riemann Integral of that, then try to prove a direct connection to the original series.
Hey Dr. Peyam! I found a really cool thing you can do with your exponential derivative! You can use it to make the Gamma(Derivative).
First, we'll look at a more general trick for getting functions of derivatives:
Consider a function F(x) = integral f(x)dx.
We'd like to be able to obtain F(D) from f(D), however, we can't do this in a useful way. If we try
integral f(D) dD, we aren't doing anything helpful, and if we try
integral f(D) dx, we aren't solving for F(D) anymore.
But, this does give us an idea.
Consider F(x) = integral f(x, t) dt.
Now, we can do some crazy integration stuff.
Γ(x) = int.0_inf t^(x-1) e^-t dt
with minimal effort, we can obtain
[Γ(D) f](x) = int.0_inf f(x+ln(t)) / te^t dt
Immediately this is cool.
I won't spoil your fun, but try taking Γ(D_x) (x^n * e^bx). It's really crazy.
Thank you! I’ll try it out :)
My only issue is this: If f is a function of one variable x, how can it suddenly become a function of two variables f(x,t)? For example, if f(x) = x^2, what is integral of f(x,t) dt?
Well, it's more that you just can't do it when you've only got an f(x).
You could have
F(x)=x^3/3, then
f(x,t) = 2t * x^3/3
F(x) = int.0_1 f(x,t) dt
But that isn't helpful. It's a method that is best applied like a method of integration: it's 'always' consistent, but it doesn't always give you an answer.
You'll want to use it when you already have some function f(x,t), like with the Gamma Function.
In the case of F(x)=x^3/3, you can sort of use f(x)=x^2, but only because, once you 'integrate' D^2 dD, you know what D^3 means.
Thinking about it now, it's not so much a method as it is the only way to do functions like Γ(D), and is just going into the explicit definition of the function and plugging in D for x. Still cool that it works, though.
Oh ok! So just to clarify, in your example, f(x,t) = t^x-1 e^-t ?
Is this a reupload? I remeber seeing this much earlier.....
Nah, it’s original. It was unlisted
@@drpeyam Phew I thought the reality glitched a liitle. Why was it unlisted?
Sir give proof of ramanujans master theorem please
What’s that?