How can I prove that I can put the differentiation operation

>>8381200
>sen
is this the retarded version of sine?

>>8381211
It is the spanish version of sine.

>>8381215
so yes?

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20KB, 490x278px

I found this. Can anyone help me prove that I can do that with bn*sin(npix/L)?

>>8381200
You can simply rely on the fact that the derivative is a linear operator. That is to say that it has the property that given a sum of two functions f and g, the derivative of their sum D(f+g) is equal to Df + Dg, where D is the derivative operator. Therefore, if you have D(sum(dicks)), you can rewrite it as sum(D(dicks)) where dicks represents some function that OP sucks

>>8381226
I know that, but it doesn't apply on INFINITE sums.

>>8381200
I'm really, really fucking drunk, but you justify this with a uniform convergence argument like >>8381223
Consult any analysis text for a proof
That sine converges uniformly is more or less immediate from the definition as a power series

>>8381231
Sure it does. What you're confused about is whether or not the resulting series converges, which the theorem you just posted proves for you. Linear operators have that property I mentioned regardless of the amount of terms in the sum.

>>8381235
Could you recommend me a text?

>>8381236
Linear operators can always be pulled inside of finite sums, but not always infinite sums.

Example:

Define [math] g_n(x) [/math] so that
[math] \sum_{k=1}^n g_k(x)=sin(nx)/n [/math] for each n

(for example, let g_1=sin(x) and [math] g_n=sin(nx)/n-\sum_{k=1}^{n-1}g_{k}[/math])

On the one hand,

[math] (d/dx) \sum_{k=1}^{\infty} g_k(x)=(d/dx)\lim_{n\to\infty} sin(nx)/n=(d/dx)0=0 [/math]

But

[math] \sum_{k=1}^{\infty} g_k'(x)=\lim_{n\to\infty}\sum_{k=1}^n g_k'(x)=\lim_{n\to\infty} (d/dx) \sum_{k=1}^n g_k(x)=\lim_{n\to\infty} (d/dx) sin(nx)/n=\lim_{n\to\infty} cos(nx) [/math]

which is not even a well-defined function

>>8381305
Naw just do it faggot

>>8381247
Baby Rudin, chapter 7

>>8381315
>shown explicit example where it gives you the wrong answer
>"just do it bro"

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