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Fourier serie for f(x)=3x^2-2x in -pi

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i hope u know the formulas ?
wow... anyways, could u find a0 =... ?

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Other answers:

an and bn exist before cosx and sinx
yea with oyler formula!
\(\large a_0=(1/\pi)\int \limits_{-\pi}^{\pi}(3x^2-2x)dx=....?\)
euler formula! bn and an are determined with euler formula too.
∫−ππ(3x2)d-∫−ππ(2x)d=...? yea?
i don't know how we use euler here, i would integrate by parts for an and bn
u go on,and tell what u finally get for a0=....?
yes, its 1/2 pi i thought he wrote a0/2 in the formula, which i generally do.
for find the a0 , u should separate ∫(3x2−2x)d to two ∫(3x2)dx-∫2x)dx.
yeah, that was correct...what u got a0 finally ?
a0 = 2(pi)^2
\(\large a_0=(1/2\pi)\int \limits_{-\pi}^{\pi}(3x^2-2x)dx=(1/2\pi)[x^3-x^2]^{\pi}_{-\pi}\)
3x^2 is an even function. and other one is Individual function.
we should use this state i think!
@hartnn what about other parts?!
to find an and bn, you can use integration by parts. try it...if u get stuck, i'll help.
Some good questions being asked today :)
@abb0t good solution is need for these lol
why don't u try by yourself ?
we'll correct u if u get it wrong..
i tried on paper, it is so long to type here. :D
final step u got ?
i separated that integration to two integration and find a0 an and bn for each of them. so sum them. and after that put it in initial formula!
what u got a0=...?an=...?bn=...?
an for first integration is this: an=\[\int\limits_{0}^{\pi}3x^{2}cosnxdx \rightarrow......?\]
did u miss 1/pi in the beginning ?
2/pi i missed in beginning! sry
so u having trouble with integration by parts ?
Fractional integrals is needed to solve this?
fractional integers ? no...
so how can u find the that integration?!
didn't u try integration by parts ? u know what that is, right ? \(\int uv = ... ?\)
here u=x^2 , v = cos nx
uv-∫vdu u mean this?
yes. u = x^2 what will u take dv =... ?
actually dv=cosnx and v=-sinnx
yes, yes, go ahead and solve it...
you'll again need product rule when u solve for sin nx (2x)dx
yea thnx... i'll try and finally put the answer here to prove;)

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