- Thread starter muiz205
- Start date

- Joined
- Nov 12, 2017

- Messages
- 3,634

Please read our submission guidelines as you have been instructed. Then tell us the context of the problems, what you have done so far, and where you are stuck. We need that in order to help you most effectively.

- Joined
- Nov 12, 2017

- Messages
- 3,634

Please do as I asked, and

- Joined
- Nov 24, 2012

- Messages
- 1,587

I believe that the induction hypothesis \(P_n\) should actually be written:1. Prove that if View attachment 11349 then View attachment 11348 ?

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

We're dealing with differentiation, rather than exponentiation.

I would begin by writing:

\(\displaystyle f(x)=\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\)

Next, let's check if the base case \(P_0\) is true:

\(\displaystyle f(x)=f^{(0)}(x)=\frac{0!}{2}\left(\frac{1}{(1+x)^{0+1}}+\frac{(-1)^0}{(1-x)^{0+1}}\right)=f(x)\quad\checkmark\)

Now, as our induction step, we should differentiate \(P_n\) with respect to \(x\)...what do you get when you do so?

- Joined
- Dec 30, 2014

- Messages
- 3,526

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:

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

We're dealing with differentiation, rather than exponentiation.

I would begin by writing:

\(\displaystyle f(x)=\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\)

Next, let's check if the base case \(P_0\) is true:

\(\displaystyle f(x)=f^{(0)}(x)=\frac{0!}{2}\left(\frac{1}{(1+x)^{0+1}}+\frac{(-1)^0}{(1-x)^{0+1}}\right)=f(x)\quad\checkmark\)

Now, as our induction step, we should differentiate \(P_n\) with respect to \(x\)...what do you get when you do so?

- Joined
- Nov 12, 2017

- Messages
- 3,634

Do you think the others are right, that is it meant to be about the

- Joined
- Mar 13, 2019

- Messages
- 12

is there still the next step?I believe that the induction hypothesis \(P_n\) should actually be written:

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

We're dealing with differentiation, rather than exponentiation.

I would begin by writing:

\(\displaystyle f(x)=\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\)

Next, let's check if the base case \(P_0\) is true:

\(\displaystyle f(x)=f^{(0)}(x)=\frac{0!}{2}\left(\frac{1}{(1+x)^{0+1}}+\frac{(-1)^0}{(1-x)^{0+1}}\right)=f(x)\quad\checkmark\)

Now, as our induction step, we should differentiate \(P_n\) with respect to \(x\)...what do you get when you do so?

- Joined
- Mar 13, 2019

- Messages
- 12

I haven't solved this part because I asked hereas written. (I didn't bother expanding the denominator, though.)

Do you think the others are right, that is it meant to be about thenth derivativerather than thenth power? Can you show us the original?

- Joined
- Mar 13, 2019

- Messages
- 12

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?

- Joined
- Nov 24, 2012

- Messages
- 1,587

Yes, take the induction hypothesis, and differentiate both sides with respect to \(x\), and see if that leads to \(P_{n+1}\).is there still the next step?

- Joined
- Mar 13, 2019

- Messages
- 12

- Joined
- Nov 24, 2012

- Messages
- 1,587

Let's start with the induction hypothesis \(P_n\) (which I incorrectly state before):

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

Differentiating both sides, we obtain (after some minor simplification):

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

We have derived \(P_{n+1}\) from \(P_n\), thereby completing the proof by induction.

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

Differentiating both sides, we obtain (after some minor simplification):

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

We have derived \(P_{n+1}\) from \(P_n\), thereby completing the proof by induction.

Last edited:

- Joined
- Mar 13, 2019

- Messages
- 12

why can it change (1 + x) to (1-x) ?Let's start with the induction hypothesis \(P_n\) (which I incorrectly state before):

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

Differentiating both sides, we obtain (after some minor simplification):

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

We have derived \(P_{n+1}\) from \(P_n\), thereby completing the proof by induction.