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The question asks

Represent the function $f(z) = \frac{z+1}{z-1}$ by its Maclaurin series for $|z|<1$ and its Laurent series for $1<|z|<\infty$.

The answers the book gives are:

Maclaurin series: $-1-2\sum_{n=1}^{\infty}z^n$

Laurent series: $1+2\sum_{n=1}^{\infty}\frac{1}{z^n}$

The answers I got were:

Maclaurin series: $-(z+1)\sum_{n=0}^{\infty}z^n$

Laurent series: $\frac{z+1}{z}\sum_{n=0}^{\infty}\frac{1}{z^n}$, for the same intervals.

Where do the $1$ and $2$ constant/coefficient come from? Are my answers correct but just not in the given form? What am I missing?

Kevin Sheng
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1 Answers1

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Hint. Your answer for the Maclaurin series can be expanded: $$\eqalign{-(z+1)\sum_{n=0}^\infty z^n &=-(z+1)(1+z+z^2+\cdots)\cr &=-z-1-z^2-z-z^3-z^2-\cdots\ .\cr}$$ As you can see, there is often more than one $z$ term with the same exponent. If you collect terms you will get the simplified series.

David
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  • Dang, so my answer was correct and just not in the form of the answer? I spent too much time on this haha. – Kevin Sheng Nov 10 '14 at 02:16
  • Debatable whether or not your answer was correct. It was "algebraically correct" but it was not exactly a power series. If it was an exam and I was marking it I would give part marks, but not full marks. – David Nov 10 '14 at 02:22