Show function series involving arctan is not differentiable at x=0

[schniefen]

Homework Statement

Show $f(x)=\sum_{k=1}^\infty \frac{\arctan (kx)}{k^2}$ is not differentiable at $x=0$.

Relevant Equations

Unsure.

I have previously shown that the function series is differentiable at $x\neq 0$. The series converges uniformly (thus pointwise) on $\mathbb R$ and the term wise differentiated series is uniformly convergent on any interval $d\leq |x|$, where $d>0$. Moreover, the terms are continuously differentiable. So from all that, we can conclude that $f(x)$ is differentiable on $\mathbb{R}\setminus \{0\}$, however, I am unsure how to show it is not differentiable at $x=0$. How would you do that?

 

[PeroK]

Homework Statement: Show $f(x)=\sum_{k=1}^\infty \frac{\arctan (kx)}{k^2}$ is not differentiable at $x=0$.
Relevant Equations: Unsure.

I have previously shown that the function series is differentiable at $x\neq 0$. The series converges uniformly (thus pointwise) on $\mathbb R$ and the term wise differentiated series is uniformly convergent on any interval $d\leq |x|$, where $d>0$. Moreover, the terms are continuously differentiable. So from all that, we can conclude that $f(x)$ is differentiable on $\mathbb{R}\setminus \{0\}$, however, I am unsure how to show it is not differentiable at $x=0$. How would you do that?

You could investigate the expression $\sum_{k=1}^\infty \frac{\arctan(kh) - \arctan(0)}{hk^2}$ for small $h$. If the function is differentiable, then the limit of that expression must be finite as $h \rightarrow 0$. Perhaps you could find a sequence of $h_n$ where the expression is unbounded? Just an idea.

 

[fresh_42]

Homework Statement: Show $f(x)=\sum_{k=1}^\infty \frac{\arctan (kx)}{k^2}$ is not differentiable at $x=0$.
Relevant Equations: Unsure.

I have previously shown that the function series is differentiable at $x\neq 0$. The series converges uniformly (thus pointwise) on $\mathbb R$ and the term wise differentiated series is uniformly convergent on any interval $d\leq |x|$, where $d>0$. Moreover, the terms are continuously differentiable. So from all that, we can conclude that $f(x)$ is differentiable on $\mathbb{R}\setminus \{0\}$, however, I am unsure how to show it is not differentiable at $x=0$. How would you do that?

Can you use the series expansion of $\arctan(kx)=\displaystyle{\sum_{j=0}^\infty (-1)^j\dfrac{(kx)^{2j+1}}{2j+1}}$?

 

[schniefen]

Can you use the series expansion of $\arctan(kx)=\displaystyle{\sum_{j=0}^\infty (-1)^j\dfrac{(kx)^{2j+1}}{2j+1}}$?

Definitely.

 

[pasmith]

Homework Statement: Show $f(x)=\sum_{k=1}^\infty \frac{\arctan (kx)}{k^2}$ is not differentiable at $x=0$.
Relevant Equations: Unsure.

I have previously shown that the function series is differentiable at $x\neq 0$. The series converges uniformly (thus pointwise) on $\mathbb R$ and the term wise differentiated series is uniformly convergent on any interval $d\leq |x|$, where $d>0$. Moreover, the terms are continuously differentiable. So from all that, we can conclude that $f(x)$ is differentiable on $\mathbb{R}\setminus \{0\}$, however, I am unsure how to show it is not differentiable at $x=0$. How would you do that?

We know that f(0) = 0, so it suffices to show that <br /> \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \lim_{K \to \infty} \sum_{k = 1}^{K} \frac{\arctan(kx)}{kx} \frac 1k does not exist.

That \lim_{y \to 0} \frac{y}{\tan(y)} = 1 might be useful.

 

[schniefen]

We know that f(0) = 0, so it suffices to show that <br /> \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \lim_{K \to \infty} \sum_{k = 1}^{K} \frac{\arctan(kx)}{kx} \frac 1k does not exist.

That \lim_{y \to 0} \frac{y}{\tan(y)} = 1 might be useful.

Can we exchange those limits? That is, does $\lim_{x \to 0} \lim_{K \to \infty}\ldots=\lim_{K \to \infty} \lim_{x \to 0}\ldots$?

 

[PeroK]

Can we exchange those limits? That is, does $\lim_{x \to 0} \lim_{K \to \infty}\ldots=\lim_{K \to \infty} \lim_{x \to 0}\ldots$?

I assume you would have to justify that!

 

[fresh_42]

Can we exchange those limits? That is, does $\lim_{x \to 0} \lim_{K \to \infty}\ldots=\lim_{K \to \infty} \lim_{x \to 0}\ldots$?

This is the point that is left to show. If you use the series expansion that I have mentioned, then you run into the same problem: can we exchange the two series? I would approach it with an indirect proof. Assume the limit exists, because if it does not, then exchanging the limits is impossible - they do not exist.

Another idea that I first had: fix a point $x=\varepsilon$ near $x=0$ and show that the series diverges but I'm not sure it does.

 

[PeroK]

I assume you would have to justify that!

I assume you need to be rigorous here. Otherwise, you could just take:
$$f'_K(x) = \sum_{k = 1}^K \frac 1 {k^2(1 + k^2x^2)}$$$$f'_K(0) = \sum_{k = 1}^K \frac 1 {k^2}$$$$f'(0) = \lim_{K \to \infty}f'_K(0) = \frac{\pi^2}{6}$$And, unless you are being rigorous, how do you know whether that's correct or not?

 

[schniefen]

The statement that a limit function $f$ of a sequence $(f_n)_1^\infty$ (of continuous functions) is continuous at a point $a$ means that $\lim_{x\to a} f(x)=f(a)$, i.e.

$\lim_{x\to a}(\lim_{n \to \infty}f_n(x))=\lim_{n \to \infty}(\lim_{x\to a} f_n(x)).$​

This exchange is permitted when the convergence is uniform. In this case, $a=0$ and

$f_n(x)= \sum_{k = 1}^{n} \frac{\arctan(kx)}{kx} \frac1k .$
​

So I guess I need to

1. Check the uniform convergence of $f_n(x)$.​
2. Check that $f_n$ and $f$ are continuous at $x=0$.​

It looks to me like $f_n(x)$ isn't even defined at $x=0$, so how can it be continuous there?

 

[PeroK]

The statement that a limit function $f$ of a sequence $(f_n)_1^\infty$ (of continuous functions) is continuous at a point $a$ means that $\lim_{x\to a} f(x)=f(a)$, i.e.

$\lim_{x\to a}(\lim_{n \to \infty}f_n(x))=\lim_{n \to \infty}(\lim_{x\to a} f_n(x)).$​

This exchange is permitted when the convergence is uniform. In this case, $a=0$ and

$f_n(x)= \sum_{k = 1}^{n} \frac{\arctan(kx)}{kx} \frac1k .$
​

So I guess I need to

1. Check the uniform convergence of $f_n(x)$.​
2. Check that $f_n$ and $f$ are continuous at $x=0$.​

It looks to me like $f_n(x)$ isn't even defined at $x=0$, so how can it be continuous there?

Where's the derivative here? For the derivative you are dealing with points that are explicity not $0$.

 

[schniefen]

Where's the derivative here? For the derivative you are dealing with points that are explicity not $0$.

You are right! I forgot. So $x\neq 0$.

That said, I'm still unsure what exactly it is I need to check in order for the exchange of limits to be valid.

 

[PeroK]

You are right! I forgot. So $x\neq 0$.

Okay, you seem all at sea here, IMO. So. We're trying to prove that a function is NOT differentiable. In this case at $x = 0$. In general, this means we have to show that the following limit does NOT exist:
$$\lim_{h \to 0}\frac{f(0 + h) - f(0)}{h}$$In this case $f(0) = 0$, so the limit becomes:$$\lim_{h \to 0}\frac{f(h)}{h}$$How do you, in general, show that a limit does not exist? That depends to some extent on what you think goes wrong: do the limiting values oscillate or become unbounded? In any case, one approach is to find a sequence (in this case $h_n \to 0$) where $\frac{f(h_n)}{h_n}$ does not converge to a single limit. This is the idea that both @pasmith and I had. Note that it's sufficient to find one such sequence.

Now, let's look at the specific function you have. The function values for our sequence will be:
$$\frac{f(h)}{h} = \sum_{k = 1}^{\infty}\frac{\arctan(kh)}{hk^2}$$We could proceed with $\arctan$ as it is or we could use the Taylor series, as suggested above. My instinct would be to try as it is first. And, try to make that infinite series become unbounded. That might not work. That would be one approach.

For example, if you could find a sequence $h_n$ where
$$\forall n: \frac{f(h_n)}{h_n} > n$$Then you would be done. The limit couldn't then be finite and the function would have been shown to be non-differentiable at $0$

 

[PeroK]

PS I have a solution using the above method, this hint and a well-known divergent series!

That \lim_{y \to 0} \frac{y}{\tan(y)} = 1 might be useful.

 

[pasmith]

Another approach is proof by contradiction: Assume that f&#039;(0) exists and see what follows.

 

[schniefen]

Ok, thanks for clarifying things @PeroK. I had to remind myself of the sequential characterization of the limit and the aim is clear now.

PS I have a solution using the above method, this hint and a well-known divergent series!

Have you been able to find a sequence such that $$\forall n: \frac{f(h_n)}{h_n} > n?$$

 

[PeroK]

Have you been able to find a sequence such that $$\forall n: \frac{f(h_n)}{h_n} > n?$$

Yes. Following the hints in post #14.

 

[schniefen]

Ok, I think I also may have a solution. Grateful for any feedback.

Take $h_k=1/k^2$, then we have $h_k\to 0$ as $n\to\infty$. Also, $$\frac{f(h_k)}{h_k}=\sum_{k=1}^\infty \arctan (1/k).$$ This series diverges according to the limit comparison test with $1/k$, i.e. we have $$0<\lim_{k\to\infty}\frac{\arctan(1/k)}{1/k}=1<\infty.$$ So, we have found a sequence that converges to zero (and is nowhere zero), but for which $\lim_{k\to\infty}\frac{f(h_k)}{h_k}$ does not exist.

 

[PeroK]

Ok, I think I also may have a solution. Grateful for any feedback.

Take $h_k=1/k^2$,

$k$ is a dummy variable within the summation. $h$ can't vary with each term in the sum.

 

[schniefen]

We'll always have uniform convergence of $$\sum_{k = 1}^{\infty}\frac{\arctan(kh)}{hk^2}$$ no matter how we fix $h$, be it as $h_n=1/n$ or some other sequence tending to $0$ as $n\to\infty$.

 

[PeroK]

We'll always have uniform convergence of $$\sum_{k = 1}^{\infty}\frac{\arctan(kh)}{hk^2}$$ no matter how we fix $h$, be it as $h_n=1/n$ or some other sequence tending to $0$ as $n\to\infty$.

This is irrelevant. If you have $f(x) = \sum_{k = 1}^{\infty} a_k(x)$, you can't set $x = k$. In that case, $f(k)$ makes no sense.

 

[schniefen]

Hmm, I'm confused. Are you claiming there is a sequence $(h_n)_1^\infty$ (independent of $k$) such that $$\sum_{k = 1}^{\infty}\frac{\arctan(kh_n)}{h_nk^2}$$ diverges? I don't see how that would be possible.

 

[PeroK]

Hmm, I'm confused.

I see that.

Are you claiming there is a sequence $(h_n)_1^\infty$ (independent of $k$)

$k$ is a dummy variable. What you are trying to do makes no sense.

such that $$\sum_{k = 1}^{\infty}\frac{\arctan(kh_n)}{h_nk^2}$$ diverges? I don't see how that would be possible.

Nevertheless, we can find $h_1 > 0$ with$$\sum_{k = 1}^{\infty}\frac{\arctan(kh_1)}{h_1k^2} > 1$$And $h_1 > h_2 > 0$ with $$\sum_{k = 1}^{\infty}\frac{\arctan(kh_2)}{h_2k^2} > 2$$Etc.

Given the nature of the function, that's the only way that $f$ can fail to be differentiable at $x = 0$.

 

[PeroK]

I have previously shown that the function series is differentiable at $x\neq 0$.

It might be worth posting this proof, at least in outline, to let us see how much rigorous maths you understand.

 

[PeroK]

PS it's getting late in the UK, so I'm signing off for today.

 

[pasmith]

Can you use the series expansion of $\arctan(kx)=\displaystyle{\sum_{j=0}^\infty (-1)^j\dfrac{(kx)^{2j+1}}{2j+1}}$?

If <br /> f_K(x) = \sum_{k=1}^K \frac{\arctan(kx)}{k^2} then the series expansion <br /> f_K(x) = \sum_{k=1}^K \sum_{n=0}^\infty \frac{(-1)^nk^{2n-1}x^{2n+1}}{2n+1} is only valid for x\in \bigcap_{k=1}^K \{t \in \mathbb{C} : k|t| &lt; 1\} = \{ t \in \mathbb{C}: |t| &lt; K^{-1}\} so that the radius of convergence tends to zero as K \to \infty.

 

[fresh_42]

If <br /> f_K(x) = \sum_{k=1}^K \frac{\arctan(kx)}{k^2} then the series expansion <br /> f_K(x) = \sum_{k=1}^K \sum_{n=0}^\infty \frac{(-1)^nk^{2n-1}x^{2n+1}}{2n+1} is only valid for x\in \bigcap_{k=1}^K \{t \in \mathbb{C} : k|t| &lt; 1\} = \{ t \in \mathbb{C}: |t| &lt; K^{-1}\} so that the radius of convergence tends to zero as K \to \infty.

I thought about fixing a point near zero and assuming differentiability so that
$$
f(\varepsilon )=L\varepsilon +o(\varepsilon )= \sum_{k=1}^\infty \sum_{n=0}^\infty \frac{(-1)^nk^{2n-1}\varepsilon ^{2n+1}}{2n+1}
$$
and then compare the coefficients at $\varepsilon^n.$ This would result in $L=\displaystyle{\sum_{k=1}^\infty \dfrac{1}{k}}$ which is divergent. But it feels a bit like cheating, i.e. I'm not sure whether this is rigorous.