Proof by induction that $f^{(n)}(x)=p_n(x)e^{x^2}$ where $f(x)=e^{x^2}$

Question

Let $f(x)=e^{x^2}$. Show by induction that $f^{(n)}(x)=p_n(x)e^{x^2}$ where $p_n(x)$ is an $n$th degree polynomial. $f^{(n)}(x)$ is the $n$th derivative of $f(x)$.

Tip: You do not need to find a formula for $p_n(x)$.

Hopefully I don't need someone to write it out for me. Just tell me what the "trick" is.

Thanks a bunch

I don't think there is a "trick". You just do it. What have you tried? — Chris Eagle, Jan 29 '13 at 19:20
The only think that came to mind was figuring out the Nth derivative of e^(x^2) and going from there, but that is too advanced for me and is not required by the problem. — John Smith, Jan 29 '13 at 19:25

score 2 · Answer 1 · answered Jan 29 '13 at 19:33

2

The statement is clear for $n=0$. We have to show that it is true for $n=k+1$, on the assumption it is true for $n=k$.

So we're interested in $f^{(k+1)}(x)$. By definition this is the derivative of $f^{(k)}(x)$. But our induction hypothesis tells us what this is: $f^{(k)}(x)=p_k(x)e^{x^2}$, where $p_k(x)$ is a degree $k$ polynomial. Differentiating (using the product and chain rules), we get that $f^{(k+1)}(x)=p_k'(x)e^{x^2}+2xp_k(x)e^{x^2}$. So all that remains is to show that $p_k'(x)+2xp_k(x)$ is a polynomial of degree $k+1$.

answered Jan 29 '13 at 19:33

Chris Eagle

33,306

Thanks for the help, Chris! I see that $p_k'(x)+2xp_k(x)$ is a polynomial of degree $k+1$ because $p_k(x)$ is multiplied by x. Would it be sufficient to point that out? – John Smith Jan 29 '13 at 19:47
That shows that $2xp_k(x)$ has degree $k+1$. But why does adding the other term not change the degree? – Chris Eagle Jan 29 '13 at 19:49
Well, because it's been differentiated, right? So it would be of degree k-1. Maybe I'm missing something – John Smith Jan 29 '13 at 19:52
What you're missing is that you didn't say that. You now have a correct argument: one summand has degree $k+1$, one has degree $k-1$, and the sum of two such polynomials always has degree $k+1$. – Chris Eagle Jan 29 '13 at 19:55
I'm sorry! I'm probably missing something pretty obvious. My bad. Would you mind enlightening me? – John Smith Jan 29 '13 at 19:59
What? Why do you believe you are missing something? – Chris Eagle Jan 29 '13 at 19:59
Didn't see your edit in time. Still getting used to the comment section. Thanks for your help! – John Smith Jan 29 '13 at 20:01

score 1 · Answer 2 · answered Jan 29 '13 at 19:25

1

You just need to show that the form $p(n)e^{x^2}$ is closed under differentiation.

Since $$\frac{d}{dx}\left(x^ne^{x^2}\right) = nx^{n-1}e^{x^2} + 2x^{n+1}e^{x^2} = (nx^{n-1} + 2x^{n+1})e^{x^2}$$ is of the form $(\text{some polynomial in }x)e^{x^2}$, and since differentiation is linear, we can conclude that $$ \frac{d}{dx}\left((\text{some polynomial in $x$})e^{x^2}\right)=(\text{another polynomial in $x$})e^{x^2}. $$

answered Jan 29 '13 at 19:25

Tunococ

10,303

Thank you. That was very helpful! – John Smith Jan 29 '13 at 19:36

Proof by induction that $f^{(n)}(x)=p_n(x)e^{x^2}$ where $f(x)=e^{x^2}$

2 Answers2