Continuously differentiable series

[mathmari]

Hey!

I want to show that series $$f(x)=\sum_{k=1}^{\infty}2^k\sin (3^{-k}x)$$ is continuously differentiable. We have that $|2^k\sin (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$, or not?

The sum $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges as a geometric series. So by comparison test the series $f(x)$ converges absolutely.


The derivative is equal to $$f'(x)=\sum_{k=1}^{\infty}2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)$$

It holds that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$

The sum $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges as a geometric series. So by comparison test the series $f'(x)$ converges absolutely. So, we have tha tthe function and its derivative are absolutely convergent. Does this imply that the function $f(x)$ is continuousy differentiable? (Wondering)

 

[I like Serena]

We have that $|2^k\sin (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$, or not?

Hey mathmari!

Shouldn't that be $|2^k\sin (3^{-k}x)|\leq 2^k\cdot |3^{-k}x|$? (Wondering)

So, we have that the function and its derivative are absolutely convergent. Does this imply that the function $f(x)$ is continuousy differentiable? (Wondering)

If a function is well-defined, and if it's differentiable, that implies indeed that it's continuous as well. (Nerd)

 

[mathmari]

Shouldn't that be $|2^k\sin (3^{-k}x)|\leq 2^k\cdot 1$? (Wondering)

Oh yes. So here we cannot check the convergence, as I did.

If a function is well-defined, and if it's differentiable, that implies indeed that it's continuous as well. (Nerd)

So, we have shown that the derivative is absolutely convergent. That means that $f(x)$ is differentiable, right?
So, does this mean that this implies that $f(x)$ is continuously differentiable? (Wondering)

 

[I like Serena]

Oh yes. So here we cannot check the convergence, as I did.

Actually, I modified my previous post, because we do have that $|\sin y| < |y|$, so with a little tweak we do have enough to confirm convergence. (Thinking)

So, we have shown that the derivative is absolutely convergent. That means that $f(x)$ is differentiable, right?
So, does this mean that this implies that $f(x)$ is continuously differentiable?

It means that $f$ is differentiable on its domain. We also need to establish that $f(x)$ converges everywhere, so that its domain is $\mathbb R$.
If we have both, then $f$ is indeed continuously differentiable on $\mathbb R$. (Nod)

Recall that $g'(x) = \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$, where $g$ must be defined in a neighborhood around $x$.
So if $g$ is not continuous in $x$, then the numerator is non-zero, meaning the derivative does not exist.
Therefore, if the derivative does exist, then the function is also continuous, and therefore continuously differentiable. (Nerd)

 

[mathmari]

Actually, I modified my previous post, because we do have that $|\sin y| < |y|$, so with a little tweak we do have enough to confirm convergence. (Thinking)

Ah ok. So we have that $$|2^k\sin (3^{-k}x)|\leq 2^k\cdot |3^{-k}x|\leq 2^k\cdot 3^{-k}$$ for $x\in [-1,1]$, right? (Wondering)

So, we have to check the convergence also for $|x|>1$, or not? (Wondering)

Recall that $g'(x) = \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$, where $g$ must be defined in a neighborhood around $x$.
So if $g$ is not continuous in $x$, then the numerator is non-zero, meaning the derivative does not exist.
Therefore, if the derivative does exist, then the function is also continuous, and therefore continuously differentiable. (Nerd)

Does this mean that we have to check only if the derivative of $f(x)$ exists, i.e. if the series of $f'(x)$ converges?

 

[I like Serena]

Ah ok. So we have that $$|2^k\sin (3^{-k}x)|\leq 2^k\cdot |3^{-k}x|\leq 2^k\cdot 3^{-k}$$ for $x\in [-1,1]$, right? (Wondering)

So, we have to check the convergence also for $|x|>1$, or not?

It's implicit.
We can treat $x$ as a constant, so that $\sum 2^k\cdot |3^{-k}x| = |x|\sum (\frac 23)^k$ converges. (Thinking)

Does this mean that we have to check only if the derivative of $f(x)$ exists, i.e. if the series of $f'(x)$ converges?

That, together with a check that $f(x)$ exists. (Nod)

 

[mathmari]

It's implicit.
We can treat $x$ as a constant, so that $\sum 2^k\cdot |3^{-k}x| = |x|\sum (\frac 23)^k$ converges. (Thinking)
That, together with a check that $f(x)$ exists. (Nod)

So, we have the following:

It holds that $$|2^k\sin (3^{-k}x)|\leq 2^k\cdot |3^{-k}x|\leq |x| \cdot 2^k\cdot 3^{-k}=|x|\left (\frac{2}{3}\right )^k$$ Since the series $$\sum_{k=1}^{\infty}|x|\left (\frac{2}{3}\right )^k=|x|\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k=2|x|$$ converges, by the comparison test we have that the series of $f(x)$ converges absolutely, and so $f(x)$ exists. We have that $$f'(x)=\sum_{k=1}^{\infty}2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)$$ Since $|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$ and $$\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k=2$$ it follows by the comparison test that $f'(x)$ converges absolutely, and so $f'(x)$ exists. From the above it follows that $f(x)$ is continuously differentiable. Is everything correct? (Wondering)

 

[I like Serena]

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great!

Next, I want to determine the intervals where $f(x)$ is uniformly convergent. Do we apply here the Weierstrass M-test? We have that $$|2^k\sin (3^{-k}x)|\leq 2^k\cdot |3^{-k}x|\leq |x| \cdot 2^k\cdot 3^{-k}=|x|\left (\frac{2}{3}\right )^k$$ Let $x\in [-a,a]$ for any $a>0$. Then we have $$|2^k\sin (3^{-k}x)|\leq a\left (\frac{2}{3}\right )^k$$ According to Weierestrass M-Test we have that that the series converges uniformly on any bounded interval.

Is this correct? (Wondering)

 

[Opalg]

Two comments about this thread:

Frist, "continuously differentiable" usually means "has a continuous derivative". So you need to show that $f'(x)$ is continuous (the function itself is automatically continuous if it is differentiable, but in general the derivative might not be).

Second, you are assuming that the derivative of this infinite series can be obtained by differentiating the series term by term. In general, that need not hold. In other words, if $f(x) = \sum u_n(x)$, and each term $u_n(x)$ is differentiable, it need not be true that $f'(x) = \sum u_n'(x).$ But there is a theorem saying that if the differentiated series $\sum u_n'(x)$ converges uniformly, then it is true that $f'(x) = \sum u_n'(x).$ There is a careful statement of that theorem here (Theorem 2).

In the case of your function $$f(x) = \sum_{k=1}^\infty 2^k\sin(3^{-k}x),$$ that series converges pointwise (but not uniformly). But the differentiated series $\sum(\frac23)^k\cos(3^{-k}x)$ does converge uniformly. So the Theorem applies, and it follows that $f(x)$ is continuously differentiable.

Next, I want to determine the intervals where $f(x)$ is uniformly convergent. Do we apply here the Weierstrass M-test? We have that $$|2^k\sin (3^{-k}x)|\leq 2^k\cdot |3^{-k}x|\leq |x| \cdot 2^k\cdot 3^{-k}=|x|\left (\frac{2}{3}\right )^k$$ Let $x\in [-a,a]$ for any $a>0$. Then we have $$|2^k\sin (3^{-k}x)|\leq a\left (\frac{2}{3}\right )^k$$ According to Weierestrass M-Test we have that that the series converges uniformly on any bounded interval.

Is this correct? (Wondering)

Yes, that is correct: the series converges uniformly on bounded intervals, but not on the whole real line.