- #1
Poirot1
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express $\frac{1}{z+1}$ as a power series about z=1. My working: I know $\frac{1}{z+1}=1-z+z^2-z^3+...$ but this is macluarin series.
To express $\frac{1}{z+1}$ as a power series about $z=1$, we can use the geometric series formula: $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$. Substituting $x=z-1$, we get $\frac{1}{z+1} = \frac{1}{1+(z-1)} = \frac{1}{1-x}$, which can be expressed as a power series about $z=1$.
The general form of the power series for $\frac{1}{z+1}$ about $z=1$ is $\frac{1}{z+1} = \sum_{n=0}^{\infty} (-1)^n (z-1)^n$.
The number of terms needed in the power series for a given level of accuracy depends on the value of $z$ and the desired level of precision. In general, the more terms included, the more accurate the approximation will be.
The power series for $\frac{1}{z+1}$ can only be used to evaluate the function for values of $z$ within the radius of convergence, which in this case is $|z-1| < 1$. For values of $z$ outside this range, the series will not converge and cannot be used to evaluate the function.
The power series for $\frac{1}{z+1}$ is a shifted version of the Taylor series for $\frac{1}{z}$, with a center at $z=1$ instead of $z=0$. This can be seen by substituting $x=z-1$ in the Taylor series for $\frac{1}{z}$, which results in the power series for $\frac{1}{z+1}$.