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

If it does, then you can conclude that the series converges locally uniformly on the interval (ε,∞). However, you cannot use this method to directly show the convergence of the original series, as the supremum of f_n does not go to 0 as n grows. In summary, the Weierstrass criterion can be used to analyse uniform convergence/local uniform convergence, but it may not hold for the entire domain and may need to be applied to subsets of the domain.
  • #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, 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:
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  • #2
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.
 
  • #3




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
 

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

1. What is uniform convergence in mathematical analysis?

In mathematical analysis, uniform convergence is a type of convergence that occurs when the terms of a sequence of functions approach their limit function at the same rate at every point in the domain. This means that for any given value of x, the difference between the sequence of functions and the limit function becomes smaller and smaller as the sequence progresses.

2. How is uniform convergence different from pointwise convergence?

Pointwise convergence occurs when the terms of a sequence of functions approach their limit function at each individual point in the domain. This means that for each x value, the difference between the sequence of functions and the limit function approaches zero. Uniform convergence, on the other hand, requires that the difference between the sequence of functions and the limit function approaches zero at every point in the domain simultaneously.

3. Why is uniform convergence important in the analysis of infinite series?

Uniform convergence is important in the analysis of infinite series because it guarantees that the limit function of the series is continuous. This allows for more accurate approximations of the function and ensures that the series will converge to the correct value regardless of the order in which the terms are summed.

4. How is uniform convergence of a series tested?

Uniform convergence of a series is typically tested using the Weierstrass M-test. This test involves finding an upper bound for the absolute value of each term in the series and determining whether the series converges using this bound. If the series converges, then the series is said to be uniformly convergent.

5. What is the significance of the series $\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}$ in mathematics?

The series $\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}$ is a well-known example of a uniformly convergent series. It is often used to illustrate the concept of uniform convergence and its importance in infinite series analysis. Furthermore, this series has applications in Fourier analysis and signal processing.

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