let f(x) =x∣x∣ 2π periodic odd developp f at fourier series


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 65287 by mathmax by abdo last updated on 27/Jul/19

$l e t f (x) = x ∣ x ∣ 2 π p e r i o d i c o d d$ $d e v e l o p p f a t f o u r i e r s e r i e s$
Commented by mathmax by abdo last updated on 28/Jul/19
$f(x) =Σ_(n=1) ^∞ a_n sin(nx) with a_n =(2/T) ∫_([T]) f(x)sin(nx)dx =(2/(2π)) ∫_(−π) ^π x∣x∣ sin(nx)dx =(2/π) ∫_0 ^π x^2 sin(nx)dx ⇒ (π/2)a_n =∫_0 ^π x^2 sin(nx)dx by parts u =x^2 and v^′ =sin(nx) ⇒ (π/2)a_n =[−(x^2 /n)cos(nx)]_0 ^π −∫_0 ^π (2x)(−(1/n))cos(nx)dx =−(π^2 /n)(−1)^n +(2/n) ∫_0 ^π x cos(nx)dx again by parts ∫_0 ^π x cos(nx)dx =[(x/n)sin(nx)]_0 ^π −∫_0 ^π (1/n)sin(nx)dx =−(1/n) ∫_0 ^π sin(nx)dx =(1/n^2 )[cos(nx)]_0 ^π =(1/n^2 )((−1)^n −1) ⇒ (π/2)a_n =−(π^2 /n)(−1)^n +(2/n^3 ) {(−1)^n −1} a_n =(2/π){−(π^2 /n)(−1)^n +(2/n^3 ){ (−1)^n −1)} =−2π (((−1)^n )/n) +(4/(πn^3 )){ (−1)^n −1} ⇒ x∣x∣ =Σ_(n=1) ^∞ (−2π (((−1)^n )/n) +(4/(πn^3 )){(−1)^n −1})sin(nx) =−2π Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx) +(4/π) Σ_(n=1) ^∞ (((−1)^n −1)/n^3 )sin(nx)$
$f (x) = \sum_{n = 1}^{\infty} a_{n} s i n (n x) w i t h a_{n} = \frac{2}{T} \int_{[T]} f (x) s i n (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} x ∣ x ∣ s i n (n x) d x = \frac{2}{π} \int_{0}^{π} x^{2} s i n (n x) d x \Rightarrow$ $\frac{π}{2} a_{n} = \int_{0}^{π} x^{2} s i n (n x) d x b y p a r t s u = x^{2} a n d v^{^{'}} = s i n (n x) \Rightarrow$ $\frac{π}{2} a_{n} = {[- \frac{x^{2}}{n} c o s (n x)]}_{0}^{π} - \int_{0}^{π} (2 x) (- \frac{1}{n}) c o s (n x) d x$ $= - \frac{π^{2}}{n} {(- 1)}^{n} + \frac{2}{n} \int_{0}^{π} x c o s (n x) d x a g a i n b y p a r t s$ $\int_{0}^{π} x c o s (n x) d x = {[\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x$ $= - \frac{1}{n} \int_{0}^{π} s i n (n x) d x = \frac{1}{n^{2}} {[c o s (n x)]}_{0}^{π} = \frac{1}{n^{2}} ({(- 1)}^{n} - 1) \Rightarrow$ $\frac{π}{2} a_{n} = - \frac{π^{2}}{n} {(- 1)}^{n} + \frac{2}{n^{3}} {{(- 1)}^{n} - 1}$ $a_{n} = \frac{2}{π} {- \frac{π^{2}}{n} {(- 1)}^{n} + \frac{2}{n^{3}} {{(- 1)}^{n} - 1)}$ $= - 2 π \frac{{(- 1)}^{n}}{n} + \frac{4}{π n^{3}} {{(- 1)}^{n} - 1} \Rightarrow$ $x ∣ x ∣ = \sum_{n = 1}^{\infty} (- 2 π \frac{{(- 1)}^{n}}{n} + \frac{4}{π n^{3}} {{(- 1)}^{n} - 1}) s i n (n x)$ $= - 2 π \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} s i n (n x) + \frac{4}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} - 1}{n^{3}} s i n (n x)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum