Proving piecewise function is k-differentiable

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

My solution is,

$F(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.$

The we differentiate both sub-function of the piecewise function. Note I assume differentiable since we are proving a result that the function is differentiable, so I assume that $F^{k}(0)$ exists, that is the function is k-differentiable at zero assuming the limit exists. I will prove for first derivative k = 1 below.

$F'(x)=\left\{\begin{array}{ll} -\frac{1}{x}e^{-\frac{1}{x}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.$

$F''(x)=\left\{\begin{array}{ll} \frac{1}{x^2}e^{-\frac{1}{x}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.$

One can continue considering $k$ number of cases

Thus from our proof, the polynomial $P_k$ can be written explicitly as $P_k = \frac{1}{x^{k - 1}(-1)^k}$ for $k > 0$

Thus we consider two cases for $k$,

$F^{(k)}(x)= \begin{cases}\frac{1}{x^{k - 1}(-1)^k}\left(x^{-1}\right) \exp \left(-x^{-1}\right) & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{cases}$

for $k > 0$

And for $k = 0$, we have the original piece wise function,

$F(x)=\left\{\begin{array}{ll} \exp \left(-x^{-1}\right) & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.$

QED

So as part of our proof, we also consider a sub-proof of the reason why the function is $F^{k}$ at $x = 0$. We just consider the trival case for $k = 1$,

In order to be differentiable at zero then $lim_{x \to 0} F(x)$ must exist. This is equivalent to both the Left hand and right hand limits existing.

$\lim_{x \to 0^-} F(x) = \lim_{x \to 0^-} 0 = 0$. Therefore, left hand limit exists.

Now for right hand limit, we have,

$\lim_{x \to 0^+} e^{-\frac{1}{x}}$ DNE as I don't know the limit as $\frac{1}{x}$ goes to zero. However, for some reason, wolfram alpha says that $\lim_{x \to 0^+} e^{-\frac{1}{x}} = 0$. Does someone please know why this is true?

I will assume that that is true, and I think the next part of the sub-proof is to generalize to higher order derivative

In order to find $F^{k}(0)$ then this limit $\lim_{x \to 0} F^{k}$ must exist.

That is, $\lim_{x \to 0^-} F^{k}(x) = \lim_{x \to 0^+} F^{k}(x)$

$\lim_{x \to 0^-} F^{k}(x) = \lim_{x \to 0^-} 0 = 0$

$\lim_{x \to 0^+} F^{k}(x) = \lim_{x \to 0^+} \frac{1}{x^k(-1)^k}e^{-\frac{1}{x}} = \frac{1}{(-1)^k}\lim_{x \to 0^+} \frac{1}{x^k}e^{-\frac{1}{x}} = 0$

Does someone also please know whether this a proof by induction (or what sort of proof it is)? Or can we just use a informal generalization proof?

Thanks!

 

[docnet]

Please remind yourself of the correct definition of the chain rule $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$ and re-check your work for $$\frac{d}{dx}e^{-1/x}.$$

And you haven't yet shown that $P_k = \frac{1}{x^{k - 1}(-1)^k}$ is true for $k>2$. There are specific requirements for a valid proof by induction, and you should try to follow the steps (even if it's boring) until you understand how it works.

In order to be differentiable at zero, you need the function to be continuous at 0 and have $$\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h} =\lim_{h\to 0^-}\frac{f(x+h)-f(x)}{h}.$$
By using the hospital rule, you should be able to show that both are equal to 0 for any $k$ (this would be another proof by induction).

 

[docnet]

$\lim_{x \to 0^+} e^{-\frac{1}{x}}$ DNE as I don't know the limit as $\frac{1}{x}$ goes to zero. However, for some reason, wolfram alpha says that $\lim_{x \to 0^+} e^{-\frac{1}{x}} = 0$. Does someone please know why this is true?

$$\lim_{x\to 0^+}e^{-\frac{1}{x}}=\lim_{y\to \infty}e^{-y}=0.$$

 

[member 731016]

Thank you for your reply @docnet! I have some question about this solution,

Do you please know how they go from the $f^{n}(x)$ to $f^{n + 1}(x)$ step (I assume they differentiate $f^{n}(x)$)? I think they made a typo there is something because they did not seem to differentiate the $\frac{1}{x}$ since it is three functions multiplied together so you have to use the product rule on $p_nu$ where $u = \frac{1}{x}e^{-\frac{1}{x}}$.

I also don't understand how they got $p_1(x) = x^2$. Should it not be $p_1(x) = \frac{1}{x}$?

Thanks!

 

[SammyS]

Thank you for your reply @docnet! I have some question about this solution,

View attachment 346210

Do you please know how they go from the $f^{n}(x)$ to $f^{n + 1}(x)$ step (I assume they differentiate $f^{n}(x)$)? I think they made a typo there is something because they did not seem to differentiate the $\frac{1}{x}$ since it is three functions multiplied together so you have to use the product rule on $p_nu$ where $u = \frac{1}{x}e^{-\frac{1}{x}}$.

I also don't understand how they got $p_1(x) = x^2$. Should it not be $p_1(x) = \frac{1}{x}$?

Thanks!

Let's take your last question first.

There is no way that $\displaystyle \ p_1(x) = \frac{1}{x} \$. Fir one thing, that's not a polynomial at all.

If as they say, $\displaystyle \ p_1(x) = x^2 \ , \$ that then says that $\displaystyle \ p_1\left( \frac 1 x \right) = \left( \frac 1 x \right)^2 \ , \$

Perhaps you're still having difficulties with $\displaystyle \ F\,'(x) \ .\$ As @docnet already told you:

Please remind yourself of the correct definition of the chain rule $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$ and re-check your work for $$\frac{d}{dx}e^{-1/x}.$$

So, no, $\displaystyle \ p_1\left( \dfrac 1 x \right) \ne \frac{1}{x} \$ because $\displaystyle \ \frac{d}{dx}e^{-1/x}\ne \frac{1}{x}e^{-1/x} \ .$

As for differentiating $\displaystyle F^{(n)}(x)\,, \$ it is the product of only two functions, not three. Note that the chain rule is involved in differentiating each of those two component functions.

 

[docnet]

Thank you for your reply @docnet! I have some question about this solution,

View attachment 346210

Do you please know how they go from the $f^{n}(x)$ to $f^{n + 1}(x)$ step (I assume they differentiate $f^{n}(x)$)?

Yes, differentiating $f^n(x)$ is a crucial step of the induction procedure.

I think they made a typo there is something because they did not seem to differentiate the $\frac{1}{x}$ since it is three functions multiplied together so you have to use the product rule on $p_nu$ where $u = \frac{1}{x}e^{-\frac{1}{x}}$.

I also don't understand how they got $p_1(x) = x^2$. Should it not be $p_1(x) = \frac{1}{x}$?

Thanks!

$p_n\left(\frac{1}{x}\right)$ indicates $p_n$ is a function of $\frac{1}{x}$, like how $f(x)$ indicates $f$ is a function of $x$.

I also don't understand how they got $p_1(x) = x^2$. Should it not be $p_1(x) = \frac{1}{x}$?

You should have $p_1=\frac{1}{x^2}$ because $$\frac{d}{dx}e^{-x}=\frac{1}{x^2}e^{-x}.$$

$\frac{1}{x^2}$ isn't a polynomial but $x^2$ is one. It's true that
$$p_1\left(\frac{1}{x}\right)=\frac{1}{x^2}\Longleftrightarrow p_1(x)=x^2.$$