Uniform Convergence Analysis of $\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}$

[twoflower]

Hi,

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

<br /> \sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}<br />

Then
<br /> f_{n}^{&#039;} = -\left(\frac{2}{3}\right)^{n}\frac{\cos \frac{1}{3^{n}x}}{x^2}<br />
<br /> f_{n}^{&#039;} = 0 \Leftrightarrow x = \frac{1}{3^{n}\left(\frac{\pi}{2} + k\pi\right)}<br />

I tried to prove the convergence using the theorem about change of sum and derivation - if I showed that
<br /> \sum_{n=1}^{\infty} f_{n}^{&#039;} \rightrightarrows^{loc} on (a,b)<br />

I could tell that also
<br /> \sum_{n=1}^{\infty} f_{n} \rightrightarrows^{loc} on (a,b)<br />

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 f_{n} is in x = \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi\right)}. Then
<br /> S_{n} := \sup_{x \in (0,\infty)} \left|2^{n}\sin \frac{1}{3^{n}x}\right| = 2^{n}<br />

But
<br /> \sum_{n=1}^{\infty} S_{n} = \infty<br />
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 x \in (\epsilon, \infty), \epsilon &gt; 0? Then

<br /> \exists n_0 \in \mathbb{N} \mbox{ such that } \forall n \ge n_0\ \ \ \ \ \ \ <br /> \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi}\right)} &lt; \epsilon<br />

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.

 

[member 553083]

To analyse uniform convergence/local uniform convergence for this series of functions, you can use the Weierstrass criterion. To do this, you need to find a maximum for f_n, which is at x = 1/(3^n((π/2) + 2kπ)). Then you can calculate the supremum of f_n, which is 2^n. You then need to calculate the sum of the supremums, which is ∞. This means that the Weierstrass criterion does not hold, and so the series does not converge uniformly or locally uniformly. However, you can use the same method to analyse a subset of the domain. For example, if you choose an interval (ε,∞), ε>0, then there exists an n_0 such that for all n ≥ n_0, 1/(3^n((π/2)+2kπ)) < ε. This means that for all n ≥ n_0, the maximum of f_n is within the interval (ε,∞). Therefore, you can calculate the supremum of f_n in this interval, and check whether the Weierstrass criterion holds.

 

[thepatientmental]

Hi there,

Thank you for sharing your thoughts on how to analyze the uniform convergence of this series. It seems like you have some good ideas, but there are a few things that need to be clarified in order to fully understand the convergence of this series.

Firstly, it is important to note that this series does not converge uniformly on the entire interval (0,∞). This is because as n increases, the function 2^n sin(1/(3^n x)) becomes increasingly oscillatory and does not approach a single limiting function. Instead, it converges pointwise to a different function at each x-value. This means that the convergence is not uniform on the entire interval, but it may be uniform on certain subintervals.

One approach you could take to analyze the uniform convergence of this series is to consider it on smaller intervals, as you mentioned. For example, if you restrict the x-values to a smaller interval (a, b), then you could try to show that the series converges uniformly on that interval. As you mentioned, this could potentially be done using the Weierstrass M-test or other convergence tests such as the Dirichlet or Abel tests.

Another approach could be to use the Cauchy criterion for uniform convergence. This criterion states that a series of functions converges uniformly on a given interval if and only if for every ε > 0, there exists an N such that for all n ≥ N and for all x in the interval, the sum of the remaining terms is less than ε. In other words, we need to show that the series converges uniformly on a given interval by showing that the tail of the series (the sum of the remaining terms) gets arbitrarily small as n increases.

In order to apply the Cauchy criterion, you could try to bound the terms of the series by a convergent series. For example, you could try to show that |2^n sin(1/(3^n x))| ≤ 1/n^2 for all x in the interval (a, b) and for all n ≥ N, where N is some fixed integer. This would imply that the tail of the series is bounded by a convergent series, and therefore the series converges uniformly on the interval (a, b).

In conclusion, there are various approaches you could take to analyze the uniform convergence of this series on different intervals. It may be helpful to try different convergence tests and see which one works