Continuous and nowhere differentiable

[saltydog]

"relatively simple matter"

. . . . . . yea, right . . . . . I'm wrong and I ain't proud. Can someone show me how to prove:

$$f(x)=\sum_{r=0}^{\infty} A^r Sin[B^r x]$$

Is nowhere differentiable. Even with all I did above I can't apply it to this equation with A<1 and B an integer: I end up getting the difference quotient is larger than minus infinity which is meaningless.

 

[saltydog]

I was incorrectly interpreting the partial sums of a geometric series. It should be:


$$\sum_{r=0}^{n-1}(\frac{1}{A})^r+(\frac{1}{A})^n+\sum_{r=n+1}^{\infty}(\frac{1}{A})^r=$$


$$\frac{1-(\frac{1}{A})^n}{1-\frac{1}{A}}+(\frac{1}{A})^n+\frac{(\frac{1}{A})^{n+1}}{1-\frac{1}{A}}$$


The important point is to devise sizes for A and B such that the following partial differences are less than 1. The most difficult is for $h_n(x)$. I rationalized that since factorials squared were being used above, then I should try to make the B term quite large with respect to A. In the following example, I used $B=A^2$.

$$|k_n(x)-k_n(x_0)|\geq(\frac{1}{A})^n$$

$$|I_n(x)-I_n(x_0)|\leq\frac{2}{A-1}(\frac{1}{A})^n$$

$$|h_n(x)-h_n(x_0)|\leq(\frac{A^n-1}{A-1})\frac{3\pi}{A^nA^n}$$

Plugging this into the differential difference quotient leads to the following expression:


$$\mathop{\lim}\limits_{n\to\infty}\frac{(\frac{1}{A})^n[1-\frac{2}{a-1}-\frac{3\pi(A^n-1)}{A^n(A-1)}]}{\frac{3\pi}{B^n}}$$

With $B=A^2$

This limit tends to infinity if $A\geq 13$

Thus,

$$f(x)=\sum_{r=0}^\infty(\frac{1}{13})^r Sin[(13)^{2r}x]$$

is nowhere differentiable.

(this is not a proof)