Taylor's Theorem problem 1

**Pinkk** · May 1st 2010, 08:46 PM

Let $g(x) = e^{\frac{-1}{x^{2}}}$ for $x \ne 0$ and $g(0) = 0$ .

Show that $g^{n}(0) = 0$ for all $n\ge 0$ and show that the Taylor series for g about 0 agrees with $g$ only at $x=0$ .

So yeah, I am completely lost now in my real analysis class and really don't understand much of anything going on now. It says to use induction to prove that there exist polynomials $p_{kn}$ , $1\le k \le n$ so that $g^{(n)}(x) = \sum_{k=1}^{n}f^{(k)}(x^{2})p_{kn}(x)$ for $x\in \mathbb{R}, n\ge 1$ , where $f(x)=e^{\frac{-1}{x}}$ . Don't really now where to begin, thanks.

**maddas** · May 1st 2010, 09:58 PM

Faà di Bruno's formula - Wikipedia, the free encyclopedia

(to the composition $g(x) = f(x^2)$ )

**Pinkk** · May 3rd 2010, 04:16 PM

Sorry, I don't understand. We have not learned anything that advanced yet. I don't understand what Bell polynomials are and how that relates to getting a polynomial of one variable. I don't understand how two superscripts $k,n$ come into play or what it means to have a polynomial called $p_{kn}$ .

**Drexel28** · May 3rd 2010, 04:50 PM

Originally Posted by maddas

Faà di Bruno's formula - Wikipedia, the free encyclopedia

(to the composition $g(x) = f(x^2)$ )

Originally Posted by Pinkk

Sorry, I don't understand. We have not learned anything that advanced yet. I don't understand what Bell polynomials are and how that relates to getting a polynomial of one variable. I don't understand how two superscripts $k,n$ come into play or what it means to have a polynomial called $p_{kn}$ .

Faa di Bruno's formula is absolute overkill for this, I have only tangentially even heard of it by browsing MathWorld.com. Isn't only used in Umbral Calculus pretty much?

So, which part are you having trouble with?

So, we can prove by induction that $g^{(n)}(0)=0$ . To do this, we prove as was suggested that $g^{(n)}(x)=p\left(\tfrac{1}{x}\right)e^{\frac{-1}{x^2}},\text{ }x\ne 0$ .

To, see this we first note that $g^{(1)}(x)=\frac{2}{x^3}e^{\frac{-1}{x^2}}$ which clearly satisfies the conditions.

Next, we suppose that $g^{(n)}(x)=p\left(\frac{1}{x}\right)e^{\frac{-1}{x^2}}$ and so $g^{(n+1)}(x)=\frac{-1}{x^2}p'\left(\frac{1}{x}\right)e^{\frac{-1}{x^2}}+\frac{2}{x^3}p\left(\frac{1}{x}\right)e^{ \frac{-1}{x^2}}$ , but a little factoring shows this is just a polynomial in $\frac{1}{x}$ times $e^{\frac{-1}{x^2}}$

**Pinkk** · May 3rd 2010, 04:58 PM

I had an idea like that but I guess I got too caught up with the suggested hint because it says to consider that $g(x)=f(x^{2})$ where $f(x) = e^{-1/x}$ and then show that $g^{(n)}$ is actually a series of products, where each product is $f^{(k)}(x^{2})p_{kn}$ . That notation of $p_{kn}$ completely through me off, never seen that before and not sure where to begin, but if it can be done more simply as a single polynomial of $1/x$ times $e^{-1/x^{2}}$ , I much rather do that.

**Drexel28** · May 3rd 2010, 04:59 PM

Originally Posted by Pinkk

I had an idea like that but I guess I got too caught up with the suggested hint because it says to consider that $g(x)=f(x^{2})$ where $f(x) = e^{-1/x}$ and then show that $g^{(n)}$ is actually a series of products, where the each product is $f^{(k)}(x^{2})p_{kn}$ . That notation of $p_{kn}$ completely through me off, never seen that before and not sure where to begin, but if it can be done more simply as a single polynomial of $1/x$ times $e^{-1/x^{2}}$ , I much rather do that.

Yeah, of course. Because you should know that $\lim_{x\to 0}p\left(\frac{1}{x}\right)e^{\frac{-1}{x^2}}=\lim_{z\to\infty}\frac{p(z)}{e^z}=0=f(0)$

**Pinkk** · May 3rd 2010, 05:09 PM

Well, technically it should be $\frac{p(z)}{e^{z^{2}}}$ but that really doesn't make a difference since $e^{z^{2}}$ will still dominate any polynomial as $z$ tends to infinity, correct?

**Drexel28** · May 3rd 2010, 05:09 PM

Originally Posted by Pinkk

Well, technically it should be $\frac{p(z)}{e^{z^{2}}}$ but that really doesn't make a difference since [tex]e^{z^{2}} will still dominate any polynomial as $z$ tends to infinity, correct?

Mhmm.

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