#### Dr.Peterson

##### Elite Member
These can both be proved by mathematical induction. Is that what you are learning? Have you tried yet?

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.

• topsquark

#### muiz205

##### New member
If anyone can help me break it down step-by-step i would truly appreciate it. here's the problem

#### Dr.Peterson

##### Elite Member
Did that get cut off?

Please do as I asked, and show where your difficulty is. We don't just do the work for you, and we can't teach you a whole lesson. Your textbook or notes will have "broken down the steps" in teaching you induction. Your task now is to make an attempt at applying what you were taught, so we can see where you might be missing something, and help out. The ball is in your court.

#### Otis

##### Senior Member
This thread is marked as 'Solved'. #### muiz205

##### New member
I have entered n = 1 but not the same.

#### Jomo

##### Elite Member
OK, so can you please show us your work so we can see where you went wrong?

#### MarkFL

##### Super Moderator
Staff member
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?

#### Jomo

##### Elite Member
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?
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 errors

#### Dr.Peterson

##### Elite Member
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?

#### muiz205

##### New member
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?
is there still the next step?

#### muiz205

##### New member
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 solved this part because I asked here

#### Jomo

##### Elite Member
You 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?

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#### muiz205

##### New member
You 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?
I haven't tried it and I'll try what you tell me

#### MarkFL

##### Super Moderator
Staff member
is there still the next step?
Yes, take the induction hypothesis, and differentiate both sides with respect to $$x$$, and see if that leads to $$P_{n+1}$$. #### Attachments

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#### MarkFL

##### Super Moderator
Staff member
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.

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#### muiz205

##### New member
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.
why can it change (1 + x) to (1-x) ?