1. ## Induction Problem

having problems with the following problem:

Consider,

$\displaystyle f(x)=\frac{1}{\sqrt{1-4x}}$

Use the principle of induction to prove that for n>=1 the $\displaystyle n^{th}$ derivative $\displaystyle f^{n}(x)$ of $\displaystyle f(x)$ is:

$\displaystyle f^{n}(x)=\frac{(2n)!}{n!}(1-4x)^{-(n+\frac{1}{2})}$

if anyone could show me how to solve this problem it would be much appreciated.

2. Originally Posted by Jonny-123
having problems with the following problem:

Consider,

$\displaystyle f(x)=\frac{1}{\sqrt{1-4x}}$

Use the principle of induction to prove that for n>=1 the $\displaystyle n^{th}$ derivative $\displaystyle f^{n}(x)$ of $\displaystyle f(x)$ is:

$\displaystyle f^{n}(x)=\frac{(2n)!}{n!}(1-4x)^{-(n+\frac{1}{2})}$

if anyone could show me how to solve this problem it would be much appreciated.
note that $\displaystyle f'(x)=\frac{2}{(1-4x)^{3/2}}$

First verify the base case when n=1

$\displaystyle f^{1}(x)=\frac{2!}{1!}(1-4x)^{-(1+\frac{1}{2}}=\frac{2}{(1-4x)^{3/2}}$

So the base case checks

Now we assume it is true for n and show n+1

$\displaystyle f^{n}(x)=\frac{(2n)!}{n!}(1-4x)^{-(n+\frac{1}{2})}$

Now take the derivative of the induction hypothsis to get

$\displaystyle f^{n+1}(x)=\frac{(2n)!}{n!}[-(n+\frac{1}{2})](1-4x)^{-(n+1+\frac{1}{2})}(-4)$

lets simplify a bit

$\displaystyle f^{n+1}(x)=\frac{(2n)!}{n!}(4n+2)(1-4x)^{-(n+1+\frac{1}{2})}$

and now factor out a 2 to get

$\displaystyle f^{n+1}(x)=2\frac{(2n)!}{n!}(2n+1)(1-4x)^{-(n+1+\frac{1}{2})}$

$\displaystyle f^{n+1}(x)=2\frac{(2n+1)!}{n!}(1-4x)^{-(n+1+\frac{1}{2})}$

but this isn't quite what we want so lets build the fraction by multiplying by (n+1)

$\displaystyle f^{n+1}(x)=2\frac{(2n+1)!}{n!}\frac{n+1}{n+1}(1-4x)^{-(n+1+\frac{1}{2})}$

$\displaystyle f^{n+1}(x)=2(n+1)\frac{(2n+1)!}{(n+1)!}(1-4x)^{-(n+1+\frac{1}{2})}$

$\displaystyle f^{n+1}(x)=\frac{(2n+2)!}{(n+1)!}(1-4x)^{-(n+1+\frac{1}{2})}$

$\displaystyle f^{n+1}(x)=\frac{[2(n+1)]!}{(n+1)!}(1-4x)^{-[(n+1)+\frac{1}{2}]}$

and we are done.