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If $y=e^{-x^2}$, prove with induction that $$\frac{d^ny}{dx^n}=(-1)^nH_n(x)e^{-x^2}$$ with $H_n(x)$ a polynomial function with degree $n$.

This is my shot

Basic step: $\frac{d^0y}{dx^0}=(-1)^0H_0(x)e^{-x^2}$ which leads to the function itself, namely $e^{-x^2} = e^{-x^2}$.

Induction step: We assume that $$\frac{d^ky}{dx^k}=(-1)^kH_k(x)e^{-x^2}$$ is correct, so we need to prove that $$\frac{d^{k+1}y}{dx^{k+1}}=(-1)^{k+1}H_{k+1}(x)e^{-x^2}$$

We can write $$(-1)^{k+1}H_{k+1}(x)e^{-x^2} = \underbrace{(-1)^{k}H_{k}(x)e^{-x^2}}_{\text{equal to $\frac{d^ky}{dx^k}$ by induction}}(-1)^1 H_{1}(x)$$

I don't know how to get any further and I also don't know if this step was possible: $H_{k+1} = H_{k}H_{1}$.

NimaJan
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3 Answers3

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You're looking at this from somewhat the wrong angle. The $(-1)^n$ is actually irrelevant, since that can be absorbed into any polynomial. Instead, think of it this way. The 'base step' shows you that $H_0(x)=1$, which is a polynomial of degree zero. Now, suppose we have $f(x)=\dfrac{d^ny}{dx^n}=H_n(x)e^{-x^2}$. Take the derivative of $f(x)$ with respect to $x$, using the product rule and chain rule. What do you get? Can you express the result in the form $g(x)e^{-x^2}$, and if so, what can you say about $g(x)$?

Steven Stadnicki
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  • I know that $g(x)$ will eventually be equal to $H_k'(x)-2xH_k(x)$, but how can I verify this is actually $H_{k+1}(x)$? I started doing it in this way but I couldn't verify if these were equal in the end so I thought it was better to begin my proof from the right side of the equation. – NimaJan Oct 17 '20 at 15:28
  • @NimaJan What you've got is a definition of $H_{k+1}(x)$; the thing that you need to verify is that it's a polynomial of the appropriate degree. To do that, you need to know some properties of derivatives of polynomials, and how degrees behave under multiplication and addition/subtraction. – Steven Stadnicki Oct 17 '20 at 15:36
  • Alright, got it. Thanks Steven! – NimaJan Oct 17 '20 at 15:41
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It's just simple differentiation.

If $\frac{d^ky}{dx^k}=(-1)^kH_k(x)e^{-x^2} $ then

$\frac{d^{k+1}y}{dx^{k+1}} =(-1)^k(H_k(x)e^{-x^2})'\\ =(-1)^k(H_k'(x)e^{-x^2}+H_k(x)(e^{-x^2})')\\ =(-1)^k(H_k'(x)e^{-x^2}-2xH_k(x)e^{-x^2})\\ =(-1)^ke^{-x^2}(H_k'(x)-2xH_k(x))\\ $

so $H_{k+1}(x) =-(H_k'(x)-2xH_k(x)) =2xH_k(x)-H_k'(x) $.

marty cohen
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1

You have,

$\dfrac{d^ky}{dx^k} = (-1)^kH_k(x)e^{-x^2}$

So

$\begin{align}\dfrac{d^{k+1}y}{dx^{k+1}} &= \dfrac{d}{dx}\dfrac{d^ky}{dx^k} = \dfrac{d}{dx}(-1)^kH_k(x)e^{-x^2} = (-1)^k\left[H'_k(x)e^{-x^2}+H_k(x) (-2x)e^{-x^2}\right]\\&=(-1)^{k+1}\left[2xH_k(x)-H'_k(x)\right]e^{-x^2}\end{align}$.

If $H_k(x)$ is a polynomial of degree $k$, $2xH_k(x)$ is a polynomial of degree $k+1$

19aksh
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