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Math Help - Proof by Induction

  1. #1
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    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.
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  2. #2
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    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...
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  3. #3
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    Is it true for any n>1?
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  4. #4
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    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.
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  5. #5
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    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.
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