let give the function f(x)=x^4 2π periodic and even developp f atfourier series.


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Question Number 28072 by abdo imad last updated on 20/Jan/18

$l e t g i v e t h e f u n c t i o n f (x) = x^{4} 2 π p e r i o d i c a n d e v e n$ $d e v e l o p p f a t f o u r i e r s e r i e s .$
Commented by abdo imad last updated on 26/Jan/18

$f (- x) = f (x) a n d f 2 π p e r i o d i c s o$ $f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{+ \infty} a_{n} c o s (n x) w i t h$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x (T = 2 π)$ $a_{n} = \frac{1}{π} \int_{- π}^{π} x^{4} c o s (n x) d x = \frac{2}{π} \int_{0}^{π} x^{4} c o s (n x) d x l e t p u t$ $A_{p} = \int_{0}^{π} x^{p} c o s (n x) d x w e k n o w t h a t$ $A_{2 p} = \frac{1}{n^{2}} (2 p π^{2 p - 1} {(- 1)}^{n} - 2 p (2 p - 1) A_{2 p - 2}) s o$ $A_{4} = \frac{1}{n^{2}} (4 π^{3} {(- 1)}^{n} - 12 A_{2}) b u t$ $A_{2} = \frac{1}{n^{2}} (2 π {(- 1)}^{n} - 2 A_{0}) = \frac{2 π}{n^{2}} {(- 1)}^{n} (A_{0} = 0)$ $A_{4} = \frac{1}{n^{2}} (4 π^{3} {(- 1)}^{n} - \frac{24 π}{n^{2}} {(- 1)}^{n}) s o$ $a_{4} = \frac{2}{π} A_{4} = \frac{2}{π n^{2}} (4 π^{3} {(- 1)}^{n} - \frac{24 π}{n^{2}} {(- 1)}^{n})$ $= \frac{8 π^{2}}{n^{2}} {(- 1)}^{n} - \frac{48}{n^{4}} {(- 1)}^{n} l e t f i n d a_{0} ?$ $a_{0} = \frac{2}{π} \int_{0}^{π} x^{4} d x = \frac{2}{π} \frac{1}{5} π^{5} = \frac{2 π^{4}}{5} a n d \frac{a_{0}}{2} = \frac{π^{4}}{5} a n d$ $x^{4} = \frac{π^{5}}{5} + 8 π^{2} \sum_{n = 1}^{+ \infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x) - 48 \sum_{n = 1}^{+ \infty} \frac{{(- 1)}^{n}}{n^{4}} c o s (n x) .$
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