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Let function $f(t)$ is represented by Fourier series, $$\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}}),$$ where $a$ and $b$ are lower and upper boundary, $$a_0=\frac{2}{b-a}\int_{a}^{b}f(t)dt,$$ $$a_n=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt,$$ $$b_n=\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt.$$

My question is, what conditions must be met so I can find derivative as (term by term) $$\frac{d}{dt}\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}})=\frac{d}{dt}(\sum_{n=1}^{\infty}a_n\cos{\frac{2n\pi t}{b-a}}+\sum_{n=1}^{\infty}b_n\sin{\frac{2n\pi t}{b-a}})=\sum_{n=1}^{\infty}\frac{d}{dt}(a_n\cos{\frac{2n\pi t}{b-a}})+\sum_{n=1}^{\infty}\frac{d}{dt}(b_n\sin{\frac{2n\pi t}{b-a}})?$$

Yiorgos S. Smyrlis
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etf
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2 Answers2

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A sufficient condition for differentiability of the series and commuting of sum and derivative is that $$ \sum_{n=1}^\infty n\big(\lvert a_n\rvert+\lvert b_n\rvert\big)<\infty. $$ See: Rudin, Principles of Mathematical Analysis, Theorem 7.17, p. 152.

Yiorgos S. Smyrlis
  • 83,933
1

We need the Fourier Series to converge uniformly on the interval $ [A,B] $ (which may be open or closed). Then we can integrate and differentiate the Fourier Series term by term. To check for uniform convergence, we can use the Weierstrass M Test:

If $$ |x| \leq M_n \forall \space t \in I $$

Where $M_n$ is independent of t, and $$ \sum_{n=0}^\infty M_n$$ converges, then $$ \sum_{k=0}^\infty u_k(t)$$ converges uniformly on the interval $I$

Eweler
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