How to show that $\frac{d^n}{dx^n} (x^2-1)^n = 2^n \cdot n!$ for $x=1$

Question

I am trying to show that

$$ \frac{d^n}{dx^n} (x^2-1)^n = 2^n \cdot n!, $$ for $x = 1$. I tried to prove it by induction but I failed because I lack axioms and rules for this type of derivatives.

Can someone give me a hint?

Answer 1

Since $x^2-1=(x-1)(x+1)$ one have to differentiate $(x-1)^n(x+1)^n$. It's clear that the result would have the form $\sum a_{km}(x-1)^k(x+1)^m$ with $k,m\le n$. If we insert $x=1$, then all summands with $k>0$ became zero. So it remains to consider the summands of the form $\sum_{k=0} a_{km}(x+1)^m$. There is actually only one such summand, which is $(x+1)^n\frac{d^n}{dx^n}(x-1)^n= n!(x+1)^n$, because otherwise $k$ ist $>0$ in $a_{km}(x-1)^k(x+1)^m$. So, in the point $x=1$ You have $n!(1+1)^n$.

Answer 2

$$ (x^2-1)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)(-1)^{n-k}(x^2)^k $$ or your problem becomes $$ \frac{d^n}{dx^n}\sum_{k=0}^n\left(\matrix{n\\k}\right)(-1)^{n-k}(x^2)^k = \sum_{k=0}^n\left(\matrix{n\\k}\right)(-1)^{n-k}\frac{d^n}{dx^n}x^{2k} $$ Can you take it from here?

Answer 3

Hint:

You could do this by induction, with the induction step:

$$\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^{n+1}=\frac{d^n}{dx^{n}}\left (\frac{d}{dx}\left [(x^2-1)^n (x^2-1)\right ]\right )=2x(n+1)\frac{d^n}{dx^n}(x^2-1)^n+(2n+2)\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$

Now you only have to show that $$\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n |_{x=1}=0$$

Answer 4

$$\frac{d^n}{dx^n}(x^2-1)^n=2nx\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^{n-1}+2n\frac{d^{n-2}}{dx^{n-2}}(x^2-1)^{n-1}$$

The second term is $0$ at $x=1$.

So we have the recursion $I_n=2nI_{n-1}$.

Answer 5

We have: $$y=(x^2-1)^n$$ Using the chain rule: $$y=a[f(x)]^n\to\frac{dy}{dx}=an[f'(x)][f(x)]^{n-1}$$ We get: $$\frac{dy}{dx}=2xn(x^2-1)^{n-1}$$ Then: $$\frac{d^2y}{dx^2}=2xn\cdot2x(n-1)\cdot(x^2-1)^{n-2}=n(n-1)(2x)^2(x^2-1)^{n-2}$$ $$\frac{d^3y}{dx^3}=2xn\cdot2x(n-1)\cdot2x(n-2)\cdot(x^2-1)^{n-2}=(2x)^3[n(n-1)(n-2)][(x^2-1)^{n-3}]$$ Can you show it from here?

Answer 6

Apply General Leibniz rule

$$(fg)^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x) g^{(k)}(x)$$

to $f(x) = (x+1)^n$ and $g(x) = (x-1)^n$. Notice for $k \le n$, we have $$g^{(k)}(x) = \frac{n!}{(n-k)!} (x-1)^{n-k} \implies g^{(k)}(1) = \begin{cases} 0, & k < n\\n! & k = n\end{cases}$$

We obtain

$$\left. (x^2-1)^{(n)} \right|_{x=1} = \binom{n}{n}f(1)g^{(n)}(1) = (1+1)^n n! = 2^n n!$$

Answer 7

You can use Taylor's formula and try to expland $x^2-1$ around $x=1$ and the coefficient ahead of $(x-1)^n$ would be $\frac{d^n}{dx^n} (x^2-1)^n{\Big|_{x=1}} \cdot \frac{1}{n!}$

We have $x^2 - 1 = (x-1)((x-1)+2)$ and so $(x^2-1)^n = (x-1)^n((x-1)^n+2)^n$. From here it's not hard to conclude that the coefficient in front of $(x-1)^n$ is $2^n$. You can use Binomial Formula or explicitly just write out the $n$ factors and see that you have to multiply by $2$ just to keep the exponent $n$ in the explansion. Hence we have that:

$$2^n = \frac{d^n}{dx^n} (x^2-1)^n{\Big|_{x=1}} \cdot \frac{1}{n!}$$

Hence the proof.