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arildno said:Correct! That series have you seen before in this thread; what does it sum up to?
Hi it sums up to
[itex](-1)^{n} x^{2n}[/itex]
/Fred
arildno said:Correct! That series have you seen before in this thread; what does it sum up to?
OOPS! This is wrong!Mathman23 said:[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}[/itex]
/Fred
arildno said:No it does not; that is a TERM in the series.
What does the whole series sum up to?
arildno said:OOPS! This is wrong!
When j is even, [itex](-1)^{j}=1[/itex] wheras if j is odd, we have [itex](-1)^{j}=-1[/itex]
Use this info.
No, j alternates being even and odd.Mathman23 said:Hi
Then
[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex]
Because j is even.
/Fred
No, j is not only even! If that would be the case, your expression would be [itex]x^{4} + x^{8} + x^{12}+...[/itex]. Go back and check again.Mathman23 said:Hi
Then
[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex]
Because j is even.
/Fred
arildno said:No, it sums up to 1/(1+x^{2}).
arildno said:No, it sums up to 1/(1+x^{2}).
arildno said:No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.
We have:
[tex](-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}[/tex]
Sorry arildno, but that doesn't make sense.arildno said:We have:
[tex](-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}[/tex]
You're right, it doesn't I'll fix it right away.assyrian_77 said:Sorry arildno, but that doesn't make sense.