# Proof by induction

Ok so I have found a formula for d^n/dx^n 1/x^2
= (-1)^n * (1+n)! * x^-(n+2)

So I have to do d/dx [(-1)^n * (1+n)! * x^-(n+2)] and see what I end up with. But how do I do that.

My book gives an example: (from d/dx (1+x)^-1)
d/dx [(-1)^k * k!(1+x)^(-k-1)] = (-1)^k * k!(-k-1)(1+x)^(-k-2)=...

What on earth is going on?! My book just drops explaning _how_ . Where does (-k-1) come from? I'm stuck...

Thanks

sony said:
Ok so I have found a formula for d^n/dx^n 1/x^2
= (-1)^n * (1+n)! * x^-(n+2)

So I have to do d/dx [(-1)^n * (1+n)! * x^-(n+2)] and see what I end up with. But how do I do that.

My book gives an example: (from d/dx (1+x)^-1)
d/dx [(-1)^k * k!(1+x)^(-k-1)] = (-1)^k * k!(-k-1)(1+x)^(-k-2)=...

What on earth is going on?! My book just drops explaning _how_ . Where does (-k-1) come from? I'm stuck...

Thanks

the essence of proof by induction is to show that

a) there is a minimum case where what you want to prove is true
b) that if n=k is true, it follows automatically (after some manipulation) that n=k+1 is true.

there is a theorem that says that if these conditions are satisfied, the statement in question is true.

so the authors are trying to show that n=k implies that n=k+1.

the essence of proof by induction is to show that

a) there is a minimum case where what you want to prove is true
b) that if n=k is true, it follows automatically (after some manipulation) that n=k+1 is true.

there is a theorem that says that if these conditions are satisfied, the statement in question is true.

so the authors are trying to show that n=k implies that n=k+1.
Yes I get _that_ :)

But how do I do the derivative of that expression???

sony said:
Ok so I have found a formula for d^n/dx^n 1/x^2
= (-1)^n * (1+n)! * x^-(n+2)

So I have to do d/dx [(-1)^n * (1+n)! * x^-(n+2)] and see what I end up with. But how do I do that.

Reread the chapter on derivatives. (d/dx)(x^n) = n*x^(n-1).

Nevermind, it was how to get from (-1)^k to (-1)^k+1 and the faculty thing I didnt get.