Loser66
  • Loser66
How to change the exponent of z to -n? \(\sum_{n=0}^\infty 2^{n-1}((-1)^n +1)z^{-(n+1)}\) Please, help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Loser66
  • Loser66
it is right? \(\sum_{n=1}^\infty 2^{n-2}((-1)^{n-1}+1)z^{-n}\)
thomas5267
  • thomas5267
\[ \sum_{n=1}^\infty 2^{n-1}((-1^n+1)z^{-(n+1)}=z^{-1}\sum_{n=1}^\infty 2^{n-1}((-1^n+1)z^{-n} \]You look like you just had a massive brain fart.
thomas5267
  • thomas5267
Actually me too. \[ \sum_{n=1}^\infty 2^{n-1}((-1)^n+1)z^{-(n+1)}=z^{-1}\sum_{n=1}^\infty 2^{n-1}((-1)^n+1)z^{-n} \]

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Loser66
  • Loser66
in the first sum, the lower limit is n =0
Loser66
  • Loser66
Hence, if we use k is a new letter in second limit, then we change n+1 =k so that \(z ^{-(n+1)}= z^{-k}\) for lower limit, if n+1 =k and n =0, then k =1, the limit of the sum is \(\sum_{k=1}^\infty\) For \(2^{n-1}\), n +1=k, then n = k -1, then n-1 = k-1-1=k-2. So that \(2^{n-1}=2^{k-2}\) for (-1) ^n, just replace n = k-1, hence it becomes \((-1)^{k-1}\) combine all: \(\sum_{k=1}^\infty 2^{k-1}((-1)^{k-1}+1)z^-k\)
Loser66
  • Loser66
sorry for the last one of 2^. It is \(2^{k-2}\)
thomas5267
  • thomas5267
You sneaky index-changing bastard!!!! jk \[ \sum_{n=0}^\infty 2^{n-1}((-1)^n+1)z^{-(n+1)}=(2z)^{-1}\sum_{n=0}^\infty 2^n((-1)^n+1)z^{-n} \]I would prefer not to do any index shift. This some is also equal to \[z^{-1}\sum_{n=0}^\infty 2^{2n}z^{-2n}\] since \((-1)^n+1=0\) for odd n and 2 for even n.
Loser66
  • Loser66
|dw:1450037781759:dw|
Loser66
  • Loser66
right?
Loser66
  • Loser66
oh, I got that.
thomas5267
  • thomas5267
\[ \begin{align*} (-1)^n+1&=0\text{ if n odd}\\ &=2\text{ if n even} \end{align*} \]That's why I hate numbers.
Loser66
  • Loser66
Thank you so much. I think I got it.
thomas5267
  • thomas5267
A sufficent condition for convergence would be \(|4z^{-2}|<1\) I think.

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