- #1

twoflower

- 368

- 0

Hi,

I don't know how to analyse uniform convergence/local uniform convergence for this series of functions:

[tex]

\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}

[/tex]

Then

[tex]

f_{n}^{'} = -\left(\frac{2}{3}\right)^{n}\frac{\cos \frac{1}{3^{n}x}}{x^2}

[/tex]

[tex]

f_{n}^{'} = 0 \Leftrightarrow x = \frac{1}{3^{n}\left(\frac{\pi}{2} + k\pi\right)}

[/tex]

I tried to prove the convergence using the theorem about change of sum and derivation - if I showed that

[tex]

\sum_{n=1}^{\infty} f_{n}^{'} \rightrightarrows^{loc} on (a,b)

[/tex]

I could tell that also

[tex]

\sum_{n=1}^{\infty} f_{n} \rightrightarrows^{loc} on (a,b)

[/tex]

Maybe I could use Dirichlet/Abel to show the convergence, but I'm not sure I have met the preconditions. Anyway, if I wanted to show directly convergence of the original series using the most straightforward criterion - Weierstrass, could I write it this way: ?

Maximum for [itex]f_{n}[/itex] is in [itex]x = \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi\right)}[/itex]. Then

[tex]

S_{n} := \sup_{x \in (0,\infty)} \left|2^{n}\sin \frac{1}{3^{n}x}\right| = 2^{n}

[/tex]

But

[tex]

\sum_{n=1}^{\infty} S_{n} = \infty

[/tex]

so I haven't shown the convergence. Anyway, as we can see,

[tex]

\exists n_0 \in \mathbb{N} \mbox{ such that } \forall n \ge n_0\ \ \ \ \ \ \

\frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi}\right)} < \epsilon

[/tex]

I feel this is maybe good idea, but don't know how to pull it into finish or how to use this idea...

Thank you for your help.

I don't know how to analyse uniform convergence/local uniform convergence for this series of functions:

[tex]

\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}

[/tex]

Then

[tex]

f_{n}^{'} = -\left(\frac{2}{3}\right)^{n}\frac{\cos \frac{1}{3^{n}x}}{x^2}

[/tex]

[tex]

f_{n}^{'} = 0 \Leftrightarrow x = \frac{1}{3^{n}\left(\frac{\pi}{2} + k\pi\right)}

[/tex]

I tried to prove the convergence using the theorem about change of sum and derivation - if I showed that

[tex]

\sum_{n=1}^{\infty} f_{n}^{'} \rightrightarrows^{loc} on (a,b)

[/tex]

I could tell that also

[tex]

\sum_{n=1}^{\infty} f_{n} \rightrightarrows^{loc} on (a,b)

[/tex]

Maybe I could use Dirichlet/Abel to show the convergence, but I'm not sure I have met the preconditions. Anyway, if I wanted to show directly convergence of the original series using the most straightforward criterion - Weierstrass, could I write it this way: ?

Maximum for [itex]f_{n}[/itex] is in [itex]x = \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi\right)}[/itex]. Then

[tex]

S_{n} := \sup_{x \in (0,\infty)} \left|2^{n}\sin \frac{1}{3^{n}x}\right| = 2^{n}

[/tex]

But

[tex]

\sum_{n=1}^{\infty} S_{n} = \infty

[/tex]

so I haven't shown the convergence. Anyway, as we can see,

**x**at which our function has the maximum is moving to zero as**n**grows. So, what if I chose another interval for x, let's say [itex]x \in (\epsilon, \infty), \epsilon > 0[/itex]? Then[tex]

\exists n_0 \in \mathbb{N} \mbox{ such that } \forall n \ge n_0\ \ \ \ \ \ \

\frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi}\right)} < \epsilon

[/tex]

I feel this is maybe good idea, but don't know how to pull it into finish or how to use this idea...

Thank you for your help.

Last edited: