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Question Number 26403 by abdo imad last updated on 25/Dec/17

d e v e l o p p t h e f u n c t i o n f (x) = / x / 2 π

p e r i o d i c i n f o u r i e r s e r i e . (f e v e n)

Commented by prakash jain last updated on 25/Dec/17

can you please clarify the question? Are you asking for Fourier series expansion of f(x)=|x|

Commented by abdo imad last updated on 25/Dec/17

y e s f o u r i e s s e r i e s n o t t r a n s f o r m o f f o u r i e r

Commented by abdo imad last updated on 28/Dec/17

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\propto} 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

a_{n} = \frac{2}{π} \int_{0}^{π} x c o s (n x) d x a n d t h e i n t e g r a t i o n b y p a r t s g i v e s

a_{n} = \frac{2 ({(- 1)}^{n} - 1)}{π n^{2}}

\frac{π}{2} a_{0} = \int_{0}^{π} x d x = \frac{π^{2}}{2} \Rightarrow a_{0} = π

/ x / = \frac{π}{2} + \frac{2}{π} \sum_{n = 1}^{\propto} \frac{{(- 1)}^{n} - 1}{n^{2}} c o s (n x) \Rightarrow

/ x / = \frac{π}{2} - \frac{4}{π} \sum_{n = 1}^{\propto} \frac{c o s ((2 n + 1) x)}{{(2 n + 1)}^{2}} .

Commented by prakash jain last updated on 28/Dec/17

Thanks .

when do use the notation / x /

you seem to mean ∣ x ∣ = absolute value

of x .

Commented by abdo imad last updated on 28/Dec/17

y e s / x / m e a n s a b s o l u t e v a l u e .