Shea Stuart

Answered

2022-07-03

Is my reasoning for whether

$F(x)={\int}_{0}^{x}\sum _{0}^{\mathrm{\infty}}\frac{\mathrm{cos}(nt)}{{2}^{n}}\text{d}t$

$F(x)={\int}_{0}^{x}\sum _{0}^{\mathrm{\infty}}\frac{\mathrm{cos}(nt)}{{2}^{n}}\text{d}t$

Answer & Explanation

Tamia Padilla

Expert

2022-07-04Added 16 answers

It looks good, but some additional justifications should be made. At the start (and for the later series) you should state why the series converges uniformly (by the Weierstrass_M-test, e.g.).

Also, to justify that the sum of the series is integrable and that switching the order of summation and integration is valid, you should state that the terms of $G(x)=\sum _{n=0}^{\mathrm{\infty}}\frac{\mathrm{cos}(nt)}{{2}^{n}}$ are integrable over any interval $[0,x]$

But, can save a few steps in your argument. Since $\sum _{n=0}^{\mathrm{\infty}}\frac{\mathrm{cos}(nt)}{{2}^{n}}$ converges uniformly to $G(x)$, and since the terms of this series are continuous, $G(x)$ is a continuous function. The Fundamental Theorem of Calculus immediately gives you the continuity of $F(x)={\int}_{0}^{x}G(t)\phantom{\rule{thinmathspace}{0ex}}dt$

Also, to justify that the sum of the series is integrable and that switching the order of summation and integration is valid, you should state that the terms of $G(x)=\sum _{n=0}^{\mathrm{\infty}}\frac{\mathrm{cos}(nt)}{{2}^{n}}$ are integrable over any interval $[0,x]$

But, can save a few steps in your argument. Since $\sum _{n=0}^{\mathrm{\infty}}\frac{\mathrm{cos}(nt)}{{2}^{n}}$ converges uniformly to $G(x)$, and since the terms of this series are continuous, $G(x)$ is a continuous function. The Fundamental Theorem of Calculus immediately gives you the continuity of $F(x)={\int}_{0}^{x}G(t)\phantom{\rule{thinmathspace}{0ex}}dt$

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