Sketching a periodic function and Fourier analysis

SU403RUNFAST
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Homework Statement


So i have a function f(x)=x^2 that is periodic -a<x<a and need to sketch this function from -3a<x<a. I know how to find the Fourier coefficients though.

Homework Equations


f(x)=x^2 sketch it periodically

The Attempt at a Solution


I know that a function is only periodic if it repeats its value in a given interval, so f(x)=f(x+D) where D is some distance down the x-axis where f(x) has the same value as f(x+D), and I know what a graph of x^2 looks like in my head but it does not repeat... How is this function periodic? Sketching it is the first step of my homework problem. It appears to me that the interval given -a<x<a is supposed to be the interval of which it repeats, so it would repeat -3a<x<a and a<x<3a. Is the problem given to us missing other conditions? A sketch or some ideas is helpful thanks in advance
 
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SU403RUNFAST said:
It appears to me that the interval given -a<x<a is supposed to be the interval of which it repeats,
I think that should be the case. This just means that the period is ##2a##.
 
You plot it from -a to +a and copy that picture 2a to the right, 4a etc.
 
That makes sense, thanks I can complete the question
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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