Question #224578

For the Fourier series in Problem 1, deduce a

series for

π

4

at the point where x =

π

2


1
Expert's answer
2021-08-11T12:53:47-0400

For the Fourier series in Problem 1, deduce a series for π4\dfrac{\pi}4 at the point where x=π2x= \dfrac{\pi}2


f(x)=a02+Σansinnπx2+Σbncosnπx2f(x)= \dfrac{a_0}2+ \Sigma a_n \sin \dfrac{nπx}{2}+ \Sigma b_n\cos\dfrac{nπx}{2}

but bn = 0

Hence, f(x)=a02+Σansinnπx2f(x)= \dfrac{a_0}2+ \Sigma a_n \sin \dfrac{nπx}{2}

When π2\dfrac{\pi}2, f(x) = 2, hence;


2=8πsinπ2+83πsin3π2+85πsin5π2+...2 = \dfrac8{\pi}\sin\dfrac{\pi}2+\dfrac8{3\pi}\sin\dfrac{3\pi}2+ \dfrac8{5\pi}\sin\dfrac{5\pi}2+...


i.e

2=8π(113+1517+...)2 =\dfrac8{\pi}(1-\dfrac13+\dfrac15-\dfrac17+...)


π4=113+1517+...\dfracπ4 = 1-\dfrac13+\dfrac15-\dfrac17+...



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