Proof by Induction

• Feb 23rd 2011, 07:16 PM
zebra2147
Proof by Induction
I need some help figuring out the following proof that my professor put in my notes...

Let $f(x)=e^{-1/x^2}, x\not=0$. Then, $f'(x)=\frac{2}{x^3}f(x), f''(x)=(\frac{-6}{x^4}+\frac{2}{x^3})f(x)$. Prove by induction that $f^n(x)=q_n\frac{1}{x}f(x)$ where $q_n$ is a polynomial.

This is pretty easy to see why it works as we can continue the product rule forever here and continue to get f(x) times a polynomial. However, I'm not sure how to prove it by induction. Any help would be appreciated.
• Feb 23rd 2011, 07:23 PM
Prove It
Tis is clearly not true, as in the base step, $\displaystyle f'(x) = \frac{2}{x^3}f(x) = \frac{2}{x^2}\cdot \frac{1}{x}f(x)$, and $\displaystyle \frac{2}{x^2}$ is not a polynomial...
• Feb 23rd 2011, 07:32 PM
zebra2147
Is it true for any n>1?
• Feb 23rd 2011, 08:11 PM
DrSteve
It's false as written for all n>0, but it looks like the following is true:

$f^n(x)=q_n\frac{1}{x^{n+2}}f(x)$ where $q_n$ is a polynomial.

You should be able to prove this by induction by using the product and quotient rules, and then some algebraic simplification (getting a common denominator). I think you also need to use the fact that the derivative of a polynomial is a polynomial.
• Feb 24th 2011, 02:32 AM
Defunkt
I guess there is no real use in trying to guess the original meaning, but it seems to me that the most reasonable one is that $f^{(n)}(x) = q_n(\frac{1}{x})f(x)$, so that $q_1(x) = 2x^3, q_2(x) = -6x^4+2x^3$, etc.