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Q) for what values of Z is
Now im not sure if im going about this the right way, but so far ive done
can anyone tell me whether im going about this the right way?
thank you
Last edited by GemmaJ1988 (2008-12-08 04:06:36)
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I'd do it by changing it into [Z/(1-Z)]^n.
I'm guessing you know that ∑ x^n converges if x is in [-1,1), so then you can find the answer by investigating which values of Z give Z/(1-Z) within that interval.
Why did the vector cross the road?
It wanted to be normal.
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ok i understand the bit about x^n converges when x is in [-1,1) but i dont know how to appy this to Z/(1-Z)
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You need to find the range of Z that solves -1 ≤ Z/(1-Z) < 1.
Why did the vector cross the road?
It wanted to be normal.
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im still confused how to do this
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ive got the range of z that solves -1<=Z/(1-Z)<1 is when z<=0
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When I work it out, I get Z < 1/2 to be the solution.
Why did the vector cross the road?
It wanted to be normal.
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