Prove through induction that if $f(x) := \frac{2}{1-x^2}$ then $f^{(n)}(x) = n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$

Question

The original formula is $f(x) := \frac{2}{1-x^2}$, for $f\colon\mathbb{R}- ({-1,1}) \to \mathbb{R}$

Through Induction prove that $f^{(n)}(x)=n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$.

So I start of with $n=0$ as the base case and see that: $$f^{(0)}(x)=0!(\frac{1}{(1-x)^{0+1}}+\frac{(-1)^0}{(1+x)^{0+1}})$$ $$f^{(0)}(x)=(\frac{1}{(1-x)}+\frac{1}{(1+x)})$$ $$f^{(0)}(x)=\frac{1(1+x)+1(1-x)}{(1-x)(1+x)}$$ $$f^{(0)}(x)=\frac{2}{(1-x)(1+x)}$$ $$f^{(0)}(x)=\frac{2}{1-x^2}$$

Then I presumed that because it works for $n=0$ then it must also work for all $n∈\mathbb N$. Then I proceeded to state that it must also work for $n=n+1$.

$$f^{(n+1)}(x)=(n+1)!(\frac{1}{(1-x)^{n+1+1}}+\frac{(-1)^{n+1}}{(1+x)^{n+1+1}})$$ $$f^{(n+1)}(x) = f^{(n)'}(x)\\$$ $$=[(n!)((\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}}))]´$$ $$=(n!)[(\frac{1}{(1-x)^{n+1}}]´ + [\frac{(-1)^n}{(1+x)^{n+1}}]´$$ $$=n!(-(-n-1)(1-x)^{-n-2})+((-1)^n(-n-1)(1+x)^{-n-2})$$

But I'm kinda stuck on how to continue on from there ...

Answer 1

You are almost done. $-n-1=-(n+1)$ so your last line becomes $$f^{(n+1)}(x)=n![(n+1)(1-x)^{-(n+2)}+(-1)^n(-1)(n+1)(1+x)^{-(n+2)}]=(n+1)!\left[\frac{1}{(1-x)^{n+2}}+\frac{(-1)^{n+1}}{(1+x)^{n+2}}\right]$$

Answer 2

Hint: $-(-n-1)=n+1$ and $(-1)^n(-n-1) = (-1)^{n+1}(n+1)$.

Also, you should be careful about how you state the process of induction. You showed it works for $n=0$. Then you presume it works for some value $n \in \mathbb{N}$ (not all $n \in \mathbb{N}$), and then you show it works for $n+1$.

Answer 3

you did all right until the last line

$$....=(n!)\left[(\frac{1}{(1-x)^{n+1}}\right]' + \left[\frac{(-1)^n}{(1+x)^{n+1}}\right]' =(n!)(\frac{n+1}{(1-x)^{n+2}} - \frac{(n+1)(-1)^n}{(1+x)^{n+2}} \\= (n+1)!\left(\frac{1}{(1-x)^{n+2}} + \frac{((-1)^{n+1}}{(1+x)^{n+2}}\right)$$