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Let's say I have Fourier series of some function, [tex]f(t)[/tex], [tex]f(t)=\frac{a0}{2}+\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})[/tex], where [tex]a[/tex] and [tex]b[/tex] are lower and upper boundary of function, [tex]a0=\frac{2}{b-a}\int_{a}^{b}f(t)dt[/tex], [tex]an=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt[/tex], bn=[tex]\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt[/tex]. My question is, can I find derivative of Fourier series on this way:
[tex]\frac{d}{dt}\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})=\frac{d}{dt}(\sum_{n=1}^{\infty}an\cos{\frac{2n\pi t}{b-a}}+\sum_{n=1}^{\infty}bn\sin{\frac{2n\pi t}{b-a}})=[/tex]
[tex]\frac{d}{dt}\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})=\frac{d}{dt}(\sum_{n=1}^{\infty}an\cos{\frac{2n\pi t}{b-a}}+\sum_{n=1}^{\infty}bn\sin{\frac{2n\pi t}{b-a}})=[/tex]
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