
Proof by Induction
I need some help figuring out the following proof that my professor put in my notes...
Let $\displaystyle f(x)=e^{1/x^2}, x\not=0$. Then, $\displaystyle f'(x)=\frac{2}{x^3}f(x), f''(x)=(\frac{6}{x^4}+\frac{2}{x^3})f(x)$. Prove by induction that $\displaystyle f^n(x)=q_n\frac{1}{x}f(x)$ where $\displaystyle q_n$ 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.

Tis is clearly not true, as in the base step, $\displaystyle \displaystyle f'(x) = \frac{2}{x^3}f(x) = \frac{2}{x^2}\cdot \frac{1}{x}f(x)$, and $\displaystyle \displaystyle \frac{2}{x^2}$ is not a polynomial...


It's false as written for all n>0, but it looks like the following is true:
$\displaystyle f^n(x)=q_n\frac{1}{x^{n+2}}f(x)$ where $\displaystyle q_n$ 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.

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 $\displaystyle f^{(n)}(x) = q_n(\frac{1}{x})f(x)$, so that $\displaystyle q_1(x) = 2x^3, q_2(x) = 6x^4+2x^3$, etc.