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I am looking at Abbott's second edition of Understanding Analysis, and I am unsure if my answer is accurate.

Exercise:

The function $g(x) =\sum_{n=0}^{\infty} \frac{\cos (2^{n}x)}{2^{n}}$

has been used as an example of a continuous nowhere differentiable function. What happens if we try to use the Differentiable Limit Theorem to explore whether g is differentiable?

My solution:

$g(x) =\sum_{n=0}^{\infty} \frac{\cos (2^{n}x)}{2^{n}}$

$g'(x) =\frac{d}{dx}\sum_{n=0}^{\infty} \frac{\cos (2^{n}x)}{2^{n}}=\sum_{n=0}^{\infty}\frac{d}{dx} \frac{\cos (2^{n}x)}{2^{n}}$

Consider the nth term

$y =\frac{\cos (2^{n}x)}{2^{n}}$

Let $\Delta x$ be a small increment given to $x$, and $\Delta y$ the corresponding increment in $ y.$

$\Delta y=\frac{\cos (2^{n}(x+\Delta x))}{2^{n}}-\frac{\cos (2^{n}x)}{2^{n}}$

$=\frac{1}{2^n}(-2 \sin 2^{n-1}(2x+\Delta x) \sin 2^{n-1}\Delta x)$

Divide by $\Delta x$

$\frac{\Delta y}{\Delta x}=\frac{1}{\Delta x}\frac{1}{2^n}(-2 \sin 2^{n-1}(2x+\Delta x) \sin 2^{n-1}\Delta x)$

As $\Delta x$ tends to $0$, $\frac{\sin \Delta x}{\Delta x}$ tends to $1$

$=\frac{\Delta y}{\Delta x}=\frac{1}{2^n}(-2 \sin 2^{n-1}(2x+\Delta x))2^{n-1}$

$= (-2 \sin 2^{n-1}(2x))$

Thus it gives some answer for derivative of nth term

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    $\begingroup$ I cannot imagine that the point of this question was for you to take the derivative of $\cos$ directly from the definition. I am 100% sure that you are expected to know that $\cos'(x) = -\sin(x)$, and how to use the chain rule. What you were expected to do is go look up the statement of the "Differential Limit Theorem" and attempt to apply to this series, and in particular to figure out why it fails here. I.e., why this series does not meet the requirements of the theorem. $\endgroup$ – Paul Sinclair Nov 27 '15 at 0:14
  • $\begingroup$ The point of this exercise is to show that $g_n(x)$ is differentiable but $g(x)$ is not differentiable, then $(g_n)\to g$ not uniformly. $\endgroup$ – Masacroso Jun 22 '16 at 11:54
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The Relevant Thm, "Term-by-term differentiability thm" (6.4.3), can't be used to justify interchanging the infinite sum and derivative.
In the setup of the theorem, you have $$f_n(x)=\frac{\cos (2^n x)}{2^n}, \text{ for }n=0,1,...$$ A key hypothesis of the theorem is that $\sum_{n=0}^\infty f_n '(x)=- \sum_{n=0}^\infty \sin(2^n x)$ converges uniformly on an interval [a,b] that you're interested in, but in fact it doesn't even converge pointwise on any interval.

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