1. Prove that if View attachment 11349 then View attachment 11348 ?
I agree with what you wrote. I am just concerned that the OP did not get equality with n=1 so I think that we should look at that work to correct the errorsI believe that the induction hypothesis \(P_n\) should actually be written:
[MATH]f^{(n)}(x)=\frac{n!}{2}\left(\frac{1}{(1+x)^{n+1}}+\frac{(-1)^n}{(1-x)^{n+1}}\right)[/MATH]
We're dealing with differentiation, rather than exponentiation.
I would begin by writing:
[MATH]f(x)=\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)[/MATH]
Next, let's check if the base case \(P_0\) is true:
[MATH]f(x)=f^{(0)}(x)=\frac{0!}{2}\left(\frac{1}{(1+x)^{0+1}}+\frac{(-1)^0}{(1-x)^{0+1}}\right)=f(x)\quad\checkmark[/MATH]
Now, as our induction step, we should differentiate \(P_n\) with respect to \(x\)...what do you get when you do so?
I believe that the induction hypothesis \(P_n\) should actually be written:
[MATH]f^{(n)}(x)=\frac{n!}{2}\left(\frac{1}{(1+x)^{n+1}}+\frac{(-1)^n}{(1-x)^{n+1}}\right)[/MATH]
We're dealing with differentiation, rather than exponentiation.
I would begin by writing:
[MATH]f(x)=\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)[/MATH]
Next, let's check if the base case \(P_0\) is true:
[MATH]f(x)=f^{(0)}(x)=\frac{0!}{2}\left(\frac{1}{(1+x)^{0+1}}+\frac{(-1)^0}{(1-x)^{0+1}}\right)=f(x)\quad\checkmark[/MATH]
Now, as our induction step, we should differentiate \(P_n\) with respect to \(x\)...what do you get when you do so?
That's what I got when I tried taking n=1 for the statement as written. (I didn't bother expanding the denominator, though.)
Do you think the others are right, that is it meant to be about the nth derivative rather than the nth power? Can you show us the original?
I haven't tried it and I'll try what you tell meYou only computed the right hand side for n=1! What about the LHS, that is what is f' (x) and does it equal what you already computed?
is there still the next step?
why can it change (1 + x) to (1-x) ?Let's start with the induction hypothesis \(P_n\) (which I incorrectly state before):
[MATH]f^{(n)}(x)=\frac{n!}{2}\left(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}}\right)[/MATH]
Differentiating both sides, we obtain (after some minor simplification):
[MATH]f^{(n+1)}(x)=\frac{(n+1)!}{2}\left(\frac{1}{(1-x)^{(n+1)+1}}+\frac{(-1)^{n+1}}{(1-x)^{(n+1)+1}}\right)[/MATH]
We have derived \(P_{n+1}\) from \(P_n\), thereby completing the proof by induction.