- #1

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!

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!