Question Number 26403 by abdo imad last updated on 25/Dec/17
$${developp}\:{the}\:{function}\:{f}\left({x}\right)=/{x}/\:\mathrm{2}\pi \\ $$$${periodic}\:{in}\:{fourier}\:{serie}\:.\left({f}\:{even}\right) \\ $$
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
$${yes}\:{fouries}\:{series}\:{not}\:{transform}\:{of}\:{fourier} \\ $$
Commented by abdo imad last updated on 28/Dec/17
![f(x)= (a_0 /2) + Σ_(n=1) ^∝ a_n cos(nx) with a_n =(2/T) ∫_([T]) f(x)cos(nx)dx a_n = (2/π) ∫_0 ^π x cos(nx)dx and the integration by parts gives a_n =((2( (−1)^n −1))/(π n^2 )) (π/2) a_0 = ∫_0 ^π xdx = (π^2 /2) ⇒a_0 =π /x/= (π/2) + (2/π) Σ_(n=1) ^∝ (((−1)^n −1)/n^2 ) cos(nx)⇒ /x/= (π/2) − (4/π) Σ_(n=1) ^∝ ((cos((2n+1)x))/((2n+1)^2 )) .](Q26746.png)
$${f}\left({x}\right)=\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:\:+\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:{a}_{{n}} \:{cos}\left({nx}\right)\:{with}\:{a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:{f}\left({x}\right){cos}\left({nx}\right){dx} \\ $$$${a}_{{n}} =\:\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} {x}\:{cos}\left({nx}\right){dx}\:{and}\:{the}\:{integration}\:{by}\:{parts}\:{gives} \\ $$$${a}_{{n}} \:=\frac{\mathrm{2}\left(\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right)}{\pi\:{n}^{\mathrm{2}} } \\ $$$$\frac{\pi}{\mathrm{2}}\:{a}_{\mathrm{0}} =\:\int_{\mathrm{0}} ^{\pi} \:{xdx}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\:\Rightarrow{a}_{\mathrm{0}} =\pi \\ $$$$/{x}/=\:\frac{\pi}{\mathrm{2}}\:+\:\frac{\mathrm{2}}{\pi}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{2}} }\:{cos}\left({nx}\right)\Rightarrow \\ $$$$/{x}/=\:\frac{\pi}{\mathrm{2}}\:\:−\:\frac{\mathrm{4}}{\pi}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{{cos}\left(\left(\mathrm{2}{n}+\mathrm{1}\right){x}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$
Commented by prakash jain last updated on 28/Dec/17
$$\mathrm{Thanks}.\: \\ $$$$\mathrm{when}\:\mathrm{do}\:\mathrm{use}\:\mathrm{the}\:\mathrm{notation}\:/{x}/ \\ $$$$\mathrm{you}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{mean}\:\mid{x}\mid\:=\:\mathrm{absolute}\:\mathrm{value} \\ $$$$\mathrm{of}\:{x}. \\ $$
Commented by abdo imad last updated on 28/Dec/17
$${yes}\:/{x}/\:{means}\:{absolute}\:{value}. \\ $$