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 [tex][/tex]

 

[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.