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Calculus confusion

Anonymous Tue 15 Feb 03:54:35 2022 No.14230740 View Reply Original Report

Quoted By: >>14230780 >>14230785 >>14231912 >>14232209 >>14232227 >>14232232

Just how....

Anonymous

Anonymous Tue 15 Feb 2022 04:22:44 No.14230780 Report

Quoted By:

>>14230740
I did a class on ODEs last year. I got through it by just memorizing the algorithm for doing the fourier bullshit.

Anonymous

Anonymous Tue 15 Feb 2022 04:25:35 No.14230785 Report

Quoted By: >>14230809

>>14230740
Where are you confused?

Anonymous

Anonymous Tue 15 Feb 2022 04:34:18 No.14230809 Report

Quoted By: >>14230850 >>14233619

>>14230785
I'm confused on how they made the fourier series into that integral. Tbh I don't even know how they got those integrations, it just had me confused

Anonymous

Anonymous Tue 15 Feb 2022 04:51:24 No.14230850 Report

Quoted By: >>14230854 >>14230922 >>14231881

>>14230809
The fourier series is basically defined by the values of each $c_n$ and $d_n$ , right? So the goal is to figure out those values. The way to find the values is kind of a trick. Lets say you want the value $d_1$ . In this case $n=1$ . What you do is multiply $f(x)$ by $/cos(/pix/b)$ (note again n =1) and integrate over the interval $[-b,b]$ . What happens is most of the terms of the integral are zero. This is what it is talking about when they say for $m /neq n$ the cosine functions are orthogonal. The only term that isn't zero has $d_1$ in it. so you can solve for $d_1$ . What they've done here is solved in general for $d_m$ doing the method I just described.

Anonymous

Anonymous Tue 15 Feb 2022 04:53:17 No.14230854 Report

Quoted By: >>14231917

>>14230850
eh, hopefully you can read that and it kinda makes sense. if you don't know what it means for functions to be orthogonal, review that with your book or youtube or google.

Anonymous

Anonymous Tue 15 Feb 2022 05:37:25 No.14230922 Report

Quoted By: >>14233561

>>14230850
>What you do is multiply f(x) by /cos(/pix/b)
I followed about the rest, but this part kind of confuses me still. I really appreciate it for the rest though it seems to make sense, especially subbing n=1

Anonymous

Anonymous Tue 15 Feb 2022 15:50:02 No.14231881 Report

Quoted By:

>>14230850
based

Anonymous

Anonymous Tue 15 Feb 2022 15:59:03 No.14231912 Report

Quoted By: >>14233009

>>14230740
how what? It was explained in the fucking picture
are there some terms you are not familiar with? If not, then look them up and understand them

Anonymous

Anonymous Tue 15 Feb 2022 16:00:13 No.14231917 Report

Quoted By:

>>14230854
it means that if you multiply them with eachother and integrate over them, the result will be zero (If I remember correctly)

Anonymous

Anonymous Tue 15 Feb 2022 17:38:09 No.14232209 Report

Quoted By:

>>14230740
Just think of it in terms of linear algebra, the integral is a projection in the direction of that basis function. For instance, if x = ax1 + bx2 + cx3, you take the inner product of x and x1 and get (x, x1) = a if x1, x2, x3 is an orthonormal basis.

Anonymous

Anonymous Tue 15 Feb 2022 17:43:11 No.14232227 Report

Quoted By:

>>14230740
exponential notation is nicer than cos+sin, tbqhwyfamalam

Anonymous

Anonymous Tue 15 Feb 2022 17:45:50 No.14232232 Report

Quoted By:

>>14230740
Math

Anonymous

Anonymous Tue 15 Feb 2022 21:36:59 No.14233009 Report

Quoted By:

>>14231912
>find x
>it's right there bro
thx prof

Anonymous

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Anonymous Wed 16 Feb 2022 00:53:25 No.14233561 Report

Quoted By:

>>14230922
>>What you do is multiply f(x) by /cos(/pix/b)
Multiply f(x) by pic related. This is the n = 1 case. In general, you do n = m. I think you should try writing out the steps on a whiteboard or paper, if you haven't already.

Anonymous

Anonymous Wed 16 Feb 2022 01:15:23 No.14233619 Report

Quoted By:

>>14230809
First order of taylor extention

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