Urgend Geometric series question

[Mathman23]

Hi

I have the following problem:

show that

1/(1+x^2)) = 1-x^2 + x^4 + (-1)^n*(x^2n-2) + (-1)^n * (x^2n)/(1+x^2)

I that know this arctan function can be expanded as a geometric series by using:

1 + q + q^2 + q^3 + ... + = 1/(1-q)


Then by putting q = -x^2. I get:


1/(1-(-x^2) = 1 - x^2 - (-x^2) - (-x^2)^3 + ... +

My question is how do I proceed from this to get the desired result?

Sincerley and Best Regards

Fred

 

[arildno]

No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2

 

[Mathman23]

No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2

I still get first result, but has it something to do that the series converges? And then I then add a sum to the first result?

/Fred

 

[arildno]

No, you simply haven't done it correctly. Try again.
Insert (-x^2) on the q-places in the following expression:
1+q+q^2+q^3+++
What do you get?

 

[assyrian_77]

Then by putting q = -x^2. I get: 1/(1-(-x^2) = 1 - x^2 - (-x^2) - (-x^2)^3 + ... +

This is wrong. Try again.

 

[Mathman23]

No, you simply haven't done it correctly. Try again.
Insert (-x^2) on the q-places in the following expression:
1+q+q^2+q^3+++
What do you get?

Hello by inserting q = -x^2 into the above I get:


1/(1-x^2) = 1 - x^2 + x^4 - x^6 + ... +

but how do I proceed from there ?

Sincerley

Fred

 

[assyrian_77]

Now you got it right and should be able to "see" how the general term looks like. Just see how the sign changes and how the exponent increases.

 

[arildno]

Hello by inserting q = -x^2 into the above I get:


1/(1-x^2) = 1 - x^2 + x^4 - x^6 + ... +

but how do I proceed from there ?

Sincerley

Fred

Great! One flaw though:
You've been working with 1/(1-(-x^2)), not 1/(1-x^2)!

Now, find the n'th term, and scrutinize the infinite part of the series starting with the n'th term.

 

[Mathman23]

Now you got it right and should be able to "see" how the general term looks like. Just see how the sign changes and how the exponent increases.

Hello I can see that the the function changes in the following way

(-1)^n * (x^2)^n, then


1/(1-x^2) = 1/(1-x^2) = 1 - x^2 + x^4 - x^6 + ... + (-1)^n * (x^2)^n


Do I then rewrite the above result to get the result required ?

Sincerely
Fred

 

[arildno]

You have now found the general term of the series expansion of 1/(1-(-x^2))
For all N>=n, draw out the common factor (-1)^(n)x^(2n)
What are you left with then?

 

[Mathman23]

The n-1 term ?

/Fred

You have now found the general term of the series expansion of 1/(1-(-x^2))
For all N>=n, draw out the common factor (-1)^(n)x^(2n)
What are you left with then?

 

[arildno]

No.
Let's see:
We have:
(-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)
Try to deduce how the +++++ will look like..

 

[Mathman23]

The n + 1 term ?

/Fred

No.
Let's see:
We have:
(-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)
Try to deduce how the +++++ will look like..

 

[arildno]

What are you talking about?
Try to figure out the rest by yourself

 

[Mathman23]

I'v have been looking at mathworld, and then arrieved at the following idear:
Then I need to do it for infinity?

/Fred


S_n = \sum_{k=1} ^ {\infty} (-1)^k \frac{1}{1-x^k}

What are you talking about?
Try to figure out the rest by yourself

 

[arildno]

Try to tackle it this way:
The infinite series starting with the n'th term is of the form:
\sum_{i=n}^{\infty}(-1)^{i}x^{2i}
Now, introduce the new index j=i-n, that is i=j+n
Then, the series can be written in the form:
\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}
What can you do about this expression?

 

[Mathman23]

I rewrite it as a product of a two Sums ??

/Fred

Try to tackle it this way:
The infinite series starting with the n'th term is of the form:
\sum_{i=n}^{\infty}(-1)^{i}x^{2i}
Now, introduce the new index j=i-n, that is i=j+n
Then, the series can be written in the form:
\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}
What can you do about this expression?

 

[arildno]

No, draw out the common factor!
\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}
What is the j-sum equal to?

 

[Mathman23]

No, draw out the common factor!

\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1)^{n} 2^{n}
What is the j-sum equal to?

??


\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1)^{n} x^{2n}

/Fred

 

[arildno]

Please write out the first few terms of the series \sum_{j=0}^{\infty}(-1)^{j}x^{2j}

 

[Mathman23]

Please write out the first few terms of the series \sum_{j=0}^{\infty}(-1)^{n}x^{2j}

Hello

\sum_{j=0}^{\infty}(-1)^{n}x^{2j} = (-1)^{n} + (-1)^{n}x^{2} + (-1)^{n}x^{4} + (-1)^{n}x^{6} + (-1)^{n}x^{8}

/Fred

 

[assyrian_77]

Ok, what arildno meant was this: \sum_{j=0}^{\infty}(-1)^{j}x^{2j}. The index is j everywhere.

 

[arildno]

Oops; sorry about that mistake in the index.

 

[Mathman23]

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}

/Fred

 

[arildno]

Correct! That series have you seen before in this thread; what does it sum up to?

 

[Mathman23]

Correct! That series have you seen before in this thread; what does it sum up to?

Hi it sums up to

(-1)^{n} x^{2n}

/Fred

 

[arildno]

No it does not; that is a TERM in the series.
What does the whole series sum up to?

 

[arildno]

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}

/Fred

OOPS! This is wrong!
When j is even, (-1)^{j}=1 wheras if j is odd, we have (-1)^{j}=-1

Use this info.

 

[Mathman23]

Hi the series summes up to \frac{1}{1-x^2}

/Fred

No it does not; that is a TERM in the series.
What does the whole series sum up to?

 

[Mathman23]

Hi

Then

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}

Because j is even.

/Fred

OOPS! This is wrong!
When j is even, (-1)^{j}=1 wheras if j is odd, we have (-1)^{j}=-1

Use this info.

 

[arildno]

No, it sums up to 1/(1+x^{2}).

 

[arildno]

Hi

Then

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}

Because j is even.

/Fred

No, j alternates being even and odd.

 

[assyrian_77]

Hi

Then

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}

Because j is even.

/Fred

No, j is not only even! If that would be the case, your expression would be x^{4} + x^{8} + x^{12}+.... Go back and check again.

 

[Mathman23]

No, it sums up to 1/(1+x^{2}).

Then

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8} Sums up to

\frac{1}{(1+x^{2})}

 

[Mathman23]

No, it sums up to 1/(1+x^{2}).

Then

\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8} Sums up to

 

[assyrian_77]

First of all, \sum_{j=0}^{\infty}(-1)^{j}x^{2j} \neq x^{2} + x^{4} + x^{6} + x^{8}, it is \sum_{j=0}^{\infty}(-1)^{j}x^{2j}=1-x^{2} + x^{4} - x^{6} + x^{8}....

 

[arildno]

No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.

We have:
(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}

 

[assyrian_77]

And second; you know also this: \frac{1}{1-q}=1 + q + q^2 + q^3 +....

Can you see the connection now?

I strongly suggest you to over the entire problem again and - as arildno is saying - read it properly.

 

[Mathman23]

No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.

We have:
(-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}

Then
\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=1-x^{2} + x^{4} - x^{6} + x^{8} +\cdots + (-1)^{n}x^{2n} = \sum_{j=0}^{\infty}\frac{(-1)^{n}x^{2n}}{1+x^{2}}

 

[assyrian_77]

We have:
(-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}

Sorry arildno, but that doesn't make sense.

 

[arildno]

Sorry arildno, but that doesn't make sense.

You're right, it doesn't I'll fix it right away.