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Let $P_n$ be the statement that $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$

Base case: n = 0, $\dfrac{d^0}{dx^0}(x^2-1)^0 = 1 = 0!$

Assume $P_m = \dfrac{d^m}{dx^m}(x^2-1)^m = m!$ is true.

Prove $P_{m+1} = \dfrac{d^{2(m+1)}}{dx^{2(m+1)}}(x^2-1)^{m+1} = [2(m+1)]!$

$\dfrac{d^{2(m+1)}}{dx^{2(m+1)}}(x^2-1)^{m+1}$

= $\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d^2}{dx^2}(x^2-1)^{m+1}\right)$

= $\dfrac{d^{2m}}{dx^{2m}}\left(2x(m)(m+1)(x^2-1)^{m-1}\right)$

= $[\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m][2x(m)(m+1)(x^2-1)^{-1}]$

From the inductive hypothesis,

= $(2m)! [2x(m)(m+1)(x^2-1)^{-1}]$

I got stuck here, and not sure if I have done correctly thus far? I did not know how to get to $[2(m+1)]!$. Please advise. Thank you.

Joe
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3 Answers3

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You've made several mistakes.

Hint: What does the product rule say $\frac{d}{dx} f(x) g(x)$ is equal to?
Now set $ f(x) = x^2 -1, g(x) = (x^2 -1)^m$.

Calvin Lin
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Try the binomial identity

$$ \frac{d^{2n}}{dx^{2n}}(x^{2}+1)(x^{2}+1)^{n-1}=\sum_{i=0}^{2n}{\binom{2n}{i}\frac{d^{i}}{dx^{i}}(x^{2}+1)\frac{d^{2n-i}}{dx^{2n-i}}(x^{2}+1)^{n-1}} $$

acat3
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Assume: $$P_m = \dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m = (2m)!$$

Then: $$P_{m+1}=\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d^2}{dx^2}(x^2-1)^{m+1}\right)= \dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d}{dx}\left[2x(m+1)(x^2-1)^m\right]\right)=\\ 2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2\overbrace{x^2}^{x^2-1+1}m(x^2-1)^{m-1}\right)=\\ \color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1+1)(x^2-1)^{m-1}\right)}=\\ \color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1)^{m}+2m(x^2-1)^{m-1}\right)}=\\ 2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((2m+1)(x^2-1)^m+2m(x^2-1)^{m-1}\right)=\\ 2(m+1)(2m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m+\overbrace{4m(m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^{m-1}}^{0}=\\ \color{blue}{2(m+1)(2m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m+\overbrace{4m(m+1)\dfrac{d^{2m}}{dx^{2m}}P_{2m-2}}^0}=\\ (2m+2)(2m+1)(2m)!=(2(m+1))!$$ $\color{blue}{\text{where $P_{2m-2}$ is a polynomial of degree $2m-2$, whose $2m$-th order derivative is zero.}}$

farruhota
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  • From $2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((2m+1)(x^2-1)^m+2m(x^2-1)^{m-1}\right)$ , how do you get (2m+1) from the previous step? Next, how do you get 4m(m+1) and how do you know all that terms are zero? – Joe Apr 02 '20 at 07:16
  • I added the blue steps to show details – farruhota Apr 02 '20 at 07:44
  • for $\color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1+1)(x^2-1)^{m-1}\right)}=\$ to the next step, how did you manage to get 2m(x^2-1)^m with power m? If I were to open bracket, I should get 2m(x^2-1) + 2m(x^2-1)^{m-1}. – Joe Apr 02 '20 at 12:51
  • it is algebra: $=(x^2-1)^m+2m(x^2-1)\cdot (x^2-1)^{m-1}+2m\cdot (x^2-1)^{m-1}=\cdots$ – farruhota Apr 02 '20 at 13:58
  • I see it now. thank you so much. – Joe Apr 02 '20 at 14:36
  • You are welcome. Good luck! – farruhota Apr 02 '20 at 14:37