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Weierstrass function

Klaas van Aarsen

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Mar 5, 2012
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But at post #21 how did we get the inequality $|\gamma_n|\leq 4^n$ ? Shouldn't we get there also the equality then? I got stuck right now.
I believe you are right. For $n<k$ there won't be an integer in between either, so we get equality as well. (Nod)

Still, doesnt $|\gamma_n|=4^n$ imply that $|\gamma_n|\le 4^n$?
And doesn't it suffice for the purpose of the proof that we have the inequality? (Wondering)
 

mathmari

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Apr 14, 2013
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I believe you are right. For $n<k$ there won't be an integer in between either, so we get equality as well. (Nod)

Still, doesnt $|\gamma_n|=4^n$ imply that $|\gamma_n|\le 4^n$?
And doesn't it suffice for the purpose of the proof that we have the inequality? (Wondering)
Ok! I just wanted to understand the differnece of $|\gamma_n|$and $|\gamma_k|$, why at the one we had inequality and at the other one equality.


Since $|\gamma_k|=4^k$ we have that $$\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n=\frac{1}{2}\left (3^k+1\right )$$ so for $k\rightarrow +\infty$, $\delta_k\rightarrow 0$ and $3^k\rightarrow +\infty$, it follows that....
Do we apply here the IVT for $f$ ?

If yes, then we have for $x_0<\xi <x_0+\delta_k$ that \begin{align*}f'(\xi)&=\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ \Rightarrow |f'(\xi)| &=\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

Is it correct so far? (Wondering)

We have that $|\gamma|\leq 4^n$, so isn't the inequality symbol above at $$\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n$$ wrong? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Do we apply here the IVT for $f$ ?

If yes, then we have for $x_0<\xi <x_0+\delta_k$ that \begin{align*}f'(\xi)&=\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ \Rightarrow |f'(\xi)| &=\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}
We can't because we do not know if $f'$ exists.
Worse, we are trying to prove that $f'$ does not exist, not for any value of $x$, aren't we? (Worried)

Instead I believe we should just substitute our results so far, which amounts to the same thing, except that we can't say that it is equal to $|f'(\xi)|$.

We have that $|\gamma|\leq 4^n$, so isn't the inequality symbol above at $$\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n$$ wrong?
Don't we have for $n < k$:
$$\left |\left (\frac{3}{4}\right )^n\gamma_n\right | \le 3^n \implies \left (\frac{3}{4}\right )^n\gamma_n \ge -3^n$$
(Thinking)
 

mathmari

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Apr 14, 2013
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We can't because we do not know if $f'$ exists.
Worse, we are trying to prove that $f'$ does not exist, not for any value of $x$, aren't we? (Worried)

Instead I believe we should just substitute our results so far, which amounts to the same thing, except that we can't say that it is equal to $|f'(\xi)|$.
So do we take the right side of IVT and we show that this doesn't converge and so it cannot be equal to the derivative? (Wondering)


Don't we have for $n < k$:
$$\left |\left (\frac{3}{4}\right )^n\gamma_n\right | \le 3^n \implies \left (\frac{3}{4}\right )^n\gamma_n \ge -3^n$$
(Thinking)
We have the following:
\begin{align*}&\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ & \Rightarrow \left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

So for $0\leq n\leq k$ we have that $|\gamma_n|\leq 4^n$ so we get that \begin{align*}-4^n\leq \gamma_n\leq 4^n &\Rightarrow -\left (\frac{3}{4}\right )^n4^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq \left (\frac{3}{4}\right )^n4^n \\ & \Rightarrow -3^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq 3^n \\ & \Rightarrow \sum_{n=0}^{k}\left (-3^n\right )\leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\sum_{n=0}^{k}3^n \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\frac{3^{k+1}-1}{2} \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \frac{3^{k+1}-1}{2}\end{align*}

Doesn't this means that $$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\leq \frac{3^{k+1}-1}{2}$$ ? I got stuck how we get the other inequality symbol. (Wondering)


Or do we take it as follows?

$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\geq -\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n$$
(Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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So do we take the right side of IVT and we show that this doesn't converge and so it cannot be equal to the derivative?
More specifically, we have a sub sequence and we are showing that it diverges.
Consequently that means that the limit that we need for $f'$ does not exist, and thus $f'$ does not exist. (Thinking)

We have the following:
\begin{align*}&\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ & \Rightarrow \left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

So for $0\leq n\leq k$ we have that $|\gamma_n|\leq 4^n$ so we get that \begin{align*}-4^n\leq \gamma_n\leq 4^n &\Rightarrow -\left (\frac{3}{4}\right )^n4^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq \left (\frac{3}{4}\right )^n4^n \\ & \Rightarrow -3^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq 3^n \\ & \Rightarrow \sum_{n=0}^{k}\left (-3^n\right )\leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\sum_{n=0}^{k}3^n \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\frac{3^{k+1}-1}{2} \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \frac{3^{k+1}-1}{2}\end{align*}

Doesn't this means that $$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\leq \frac{3^{k+1}-1}{2}$$ ? I got stuck how we get the other inequality symbol. (Wondering)


Or do we take it as follows?

$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\geq -\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n$$
(Wondering)
Can't we split it up like:
$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |
\geq \left |\left (\frac{3}{4}\right )^k\gamma_k\right | -\sum_{n=0}^{k-1}\left (\frac{3}{4}\right )^n|\gamma_n|$$
(Thinking)
 

mathmari

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Apr 14, 2013
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Can't we split it up like:
$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |
\geq \left |\left (\frac{3}{4}\right )^k\gamma_k\right | -\sum_{n=0}^{k-1}\left (\frac{3}{4}\right )^n|\gamma_n|$$
(Thinking)
Ahh ok! So we split it in that way to get a lower bound that goes to infinity if $k\rightarrow \infty$ and so the limit of IVT that we need for $f'$ does not exist, and thus $f'$ does not exist and so the proof is complete, right? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Ahh ok! So we split it in that way to get a lower bound that goes to infinity if $k\rightarrow \infty$ and so the limit of IVT that we need for $f'$ does not exist, and thus $f'$ does not exist and so the proof is complete, right?
Yep. (Nod)
 

mathmari

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Apr 14, 2013
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Great!! Thank you so much!! 😊
 

mathmari

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Apr 14, 2013
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For $x\in [-1,1]$ we define $\phi (x)=|x|$ and then we extend $\phi$ to the whole $\mathbb{R}$ such that $\phi (x+2)=\phi (x)$.
How does this look graphically? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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How does this look graphically?
\begin{tikzpicture}
\draw[help lines] (-4,-2) grid (4,2);
\draw[->] (-4.4,0) -- (4.4,0);
\draw[->] (0,-2.2) -- (0,2.2);
\draw foreach \i in {-4,...,4} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
\draw foreach \i in {-2,...,2} { (0.1,\i) -- (-0.1,\i) node[ left ] {$\i$} };
\draw[domain=-4.2:4.2, variable=\x, blue, ultra thick, samples=200] plot ({\x}, {abs(\x-floor((\x-1)/2)*2-2)}) node[above right] {$\phi$};
\end{tikzpicture}
:geek:
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,028
We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$
Is it correct that we can take the limit into the sum? :unsure:
Can we do that because the series converges?
 
Last edited:

Klaas van Aarsen

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Mar 5, 2012
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Is it correct that we can take the limit into the sum?
Can we do that because the series converges?
It's because Tannery's theorem says:

Let $ S_n = \sum\limits_{k=0}^\infty a_k(n) $ and suppose that $ \lim\limits_{n\rightarrow\infty} a_k(n) = b_k $. If $ |a_k(n)| \le M_k $ and $ \sum\limits_{k=0}^\infty M_k < \infty $, then $ \lim\limits_{n\rightarrow\infty} S_n = \sum_{k=0}^{\infty} b_k $.

Can we find such $M_k$? 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,028
It's because Tannery's theorem says:

Let $ S_n = \sum\limits_{k=0}^\infty a_k(n) $ and suppose that $ \lim\limits_{n\rightarrow\infty} a_k(n) = b_k $. If $ |a_k(n)| \le M_k $ and $ \sum\limits_{k=0}^\infty M_k < \infty $, then $ \lim\limits_{n\rightarrow\infty} S_n = \sum_{k=0}^{\infty} b_k $.

Can we find such $M_k$? 🤔
So we want to find an upper bound of $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |$, right?
We have that $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |\leq \left |\left (\frac{3}{4}\right )^n\right |\cdot \left |\phi (4^nx)\right |\leq 1\cdot 1=1$ or not?
But this $M_k$ does not satisfy the condition $ \sum\limits_{k=0}^\infty M_k < \infty $.

:unsure:
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,713
So we want to find an upper bound of $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |$, right?
We have that $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |\leq \left |\left (\frac{3}{4}\right )^n\right |\cdot \left |\phi (4^nx)\right |\leq 1\cdot 1=1$ or not?
But this $M_k$ does not satisfy the condition $ \sum\limits_{k=0}^\infty M_k < \infty $.
How about $M_k=\left (\frac{3}{4}\right )^k$? 🤔