Urgend Geometric series question

[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.