Solving the Fourier cosine series

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Showing the highlighted step in the Fourier cosine series solution is deemed necessary for clarity, as it aids in understanding the transition between steps. Omitting such steps can lead to confusion and make the work harder to follow. Regarding notation, while it is technically possible to use f(x) instead of x^2, doing so may obscure the specific function being analyzed. The consensus is that maintaining clarity by explicitly stating the function is preferable. Clear communication in mathematical solutions enhances comprehension and reduces potential misunderstandings.
chwala
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Homework Statement
Kindly see attached (just need clarification on highlighted part).
Relevant Equations
Fourier cosine series
1672350334369.png


My question is; is showing the highlighted step necessary? given the fact that ##\sin (nπ)=0##? My question is in general i.e when solving such questions do i have to bother with showing the highlighted part...

secondly,

1672350483726.png


Can i have ##f(x)## in place of ##x^2##? Generally, on problems to do with Fourier series- what is usually indicated is ##f(x)##... or it does not matter? yes, i know that ##f(x)=x^2##. My question is in reference to the general widely used notation on such problems.

Cheers.
 
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chwala said:
My question is; is showing the highlighted step necessary? given the fact that sin⁡(nπ)=0? My question is in general i.e when solving such questions do i have to bother with showing the highlighted part...
Yes, it's necessary, IMO. Otherwise, the transition from the line above the highlighted line to the one below it would be harder to follow. Several people have made comments in some of your threads that it was difficult to follow your work because of omitted steps.
chwala said:
Can i have f(x) in place of ##x^2##?
Why would you want to? Since you're finding the Fourier series of ##x^2##, why hide this fact by calling it f(x)?
 
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Thread 'Does this series converge uniformly?'

First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...
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