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The Harvey Mudd College Math dept presents the Weierstrass' function:
f(x)=\sum_{n=0}^{\infty} B^nCos(A^n \pi x)
as an example of a continuous function nowhere differentiable if 0<B<1 and AB>1+\frac{3\pi}{2}. Surely it converges to a continuous function if 0<B<1 regardless of the value of A. I base this on the Weierstrass M-test for this series.
However, I can differentiate the series leading to:
f^{'}(x)=-\pi \sum_{n=0}^{\infty}(AB)^n Sin(A^n \pi x)
Doesn't this series also converge uniformly (same convergence test as above) for all x if 0<AB<1? If so, why are they requiring AB>1+\frac{3\pi}{2}? I would think you could say it's nowhere differentialbe if AB>1 although greater than 1+\frac{3\pi}{2} still qualifies.
I haven't asked them, I guess I could, maybe someone here knows though.
f(x)=\sum_{n=0}^{\infty} B^nCos(A^n \pi x)
as an example of a continuous function nowhere differentiable if 0<B<1 and AB>1+\frac{3\pi}{2}. Surely it converges to a continuous function if 0<B<1 regardless of the value of A. I base this on the Weierstrass M-test for this series.
However, I can differentiate the series leading to:
f^{'}(x)=-\pi \sum_{n=0}^{\infty}(AB)^n Sin(A^n \pi x)
Doesn't this series also converge uniformly (same convergence test as above) for all x if 0<AB<1? If so, why are they requiring AB>1+\frac{3\pi}{2}? I would think you could say it's nowhere differentialbe if AB>1 although greater than 1+\frac{3\pi}{2} still qualifies.
I haven't asked them, I guess I could, maybe someone here knows though.