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I have a function which is a convergent series:

$$f(x) = \sin(x) + \frac{1}{10}\sin(10x) + \frac{1}{100}\sin(100x) + \cdots \frac{1}{10^n}\sin(10^nx)$$

This function is convergent because for any E you care to specify, the function has a term which is smaller than E. However, the function is not differentiable, and I don't understand why.

$$\frac{d}{dx}f(x) = \cos(x)+\frac{10}{10}\cos(10x) + \frac{100}{100}\cos(100x) + \cdots \frac{10^n}{10^n}\cos(10^nx)$$

Is this a special case to A Continuous Nowhere-Differentiable Function ?

Community
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user1928764
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2 Answers2

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First note that $\sum a_k$ is absolutely convergent if $$\sum \left|a_k\right|\Rightarrow \mbox{converges}$$ Since absolute convergence implies convergence, then let's check for absolute convergence $$ \sum\limits_{k=0}^n \left|\frac{\sin\left(10^kx\right)}{10^k}\right|$$ First note that $$ \left|\sin\left(10^kx\right)\right|\leq 1 $$ $$ \frac{\left|\sin\left(10^kx\right)\right|}{\left|10^k\right|} \leq \frac{1}{\left|10^k\right|} $$ $$ \left|\frac{\sin\left(10^kx\right)}{10^k}\right| \leq \left|\frac{1}{10}\right|^k $$ Since $\left|\frac{1}{10}\right|\lt 1$, then by the geometric series test, $$ \sum\limits_{k=0}^n \left|\frac{1}{10}\right|^k \Rightarrow \mbox{converges} $$ Which implies that $$ \sum\limits_{k=0}^n \left(\frac{1}{10}\right)^k \Rightarrow \mbox{converges absolutely} $$ Therefore by the direct comparison test, we have $$ f(x)=\sum\limits_{k=0}^n \frac{\sin\left(10^kx\right)}{10^k} \Rightarrow \mbox{converges absolutely}$$ Now let's talk about the differentiability of $f(x)$. If we take the derivative of each term in the sum, we'll end up with $$ \sum\limits_{k=0}^n \cos\left(10^kx\right) $$ Now let's show that this series diverges. Note that $$ -1\leq \cos\left(10^nx\right) $$ Since $$ \lim\limits_{n\to\infty} 1=1\not=0 $$ Then by the divergence test $$ -\sum\limits_{k=0}^n 1\Rightarrow \mbox{diverges} $$ Therefore by the direct comparison test $$ \sum\limits_{k=0}^n \cos\left(10^kx\right) \Rightarrow \mbox{diverges}$$ So although each term of $f(x)$ is differentiable and the sum of each term is convergent, the sum of the derivative of each term does not converge. This fact makes $f(x)$ a non differentiable function.

k170
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Notice that if you plug $x=k\pi$, where $k$ is any integer, into the derivative of the that function, you get $\infty$, thereby making the derivative discontinuous. This, in other words, means that the function is not differentiable.

Arturo don Juan
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  • These answers are correct, but they do not answer my question. Why is my function not differentiable when it is convergent and piecewise continuous? Is there another test that a function has to have in order to be differentiable, or do I have see if the derivative exists before I can say that the function is differentiable. In other words, it isn't sufficient that if a function is continuous, then it is differentiable. – user1928764 Dec 18 '14 at 04:59
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    @user1928764, continuity does not imply differentiability. A differentiable function is a function that is differentiable at any point in its domain. – k170 Dec 18 '14 at 05:46
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    To add to @k170 comment, consider abs(x). Certainly is continuous, but not differentiable (at all points). – jameselmore Dec 18 '14 at 20:25
  • @user1928764 The first question you asked in that comment is like asking, "How can't a person walk even though they have legs?" Clearly if the person is dead or paralyzed waist down they can't walk. In the same way that a person having legs doesn't necessarily mean they can walk,a function being piece-wise continuous and convergent doesn't necessarily mean that it is differentiable, despite how tempting it is to assume so. – Arturo don Juan Dec 20 '14 at 23:07