fourier series 1/2 x^2

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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fourier series 12 x², [−L, L]

Solution

L²6 +∑_n=1^∞2(−1)ⁿL²cos(πnxL )π²n²

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Solution steps

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12 x², [−L, L]

Fourier Series

Fourier series of function f(x) on interval −L≤ x≤ L is defined as:
     f(x)=A₀+∑_n=1^∞A_n· cos(nπxL )+∑_n=1^∞B_n· sin(nπxL )
Where
     A₀=12L · ∫_−L^Lf(x)dx
     A_n=1L · ∫_−L^Lf(x)cos(nπxL )dx,      n>0
     B_n=1L · ∫_−L^Lf(x)sin(nπxL )dx,      n>0

Apply the Fourier Formula

=12L · ∫_−L^L12 x²dx+∑_n=1^∞1L · ∫_−L^L12 x²cos(nπxL )dx· cos(nπxL )+∑_n=1^∞1L · ∫_−L^L12 x²sin(nπxL )dx· sin(nπxL )

Solve the integrals