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Power Series

Most functions can be represented as a power series, whose general form is given by: \begin{eqnarray} f(z) &=& \sum_{n=0}^{\infty} c_n(z-a)^n \nonumber\\ &=& c_0 + c_1(z-a) + c_2(z-a)^2 + c_3(z-a)^3 + \dots \label{Series} \end{eqnarray} where $z$ is the independent variable of the function, $a$ represents the point “around” which the function is being expanded, each of the constants $c_n$ is called the coefficient of the $n$th term, and the entire $n$th term, i.e. $c_n(z-a)^n$, is called the $n$th order term. For further information about the geometric meaning of these new vocabulary words, see § {Approximations Using Power Series}.

For each function $f(z)$ and point $a$, the coefficients are unique. One way of finding the coefficients is using Taylor's theorem, derived as follows:

We evaluate both sides of equation ($\ref{Series}$) at the point $z=a$, to obtain: \begin{equation} f(a)=c_0 + c_1(a-a) + c_2(a-a)^2 + c_3(a-a)^3 + \dots \label{c0} \end{equation} Since all of the terms, except the first, on the right hand side of ($\ref{c0}$) are zero, the equation simplifies to: \begin{equation} c_0=f(a) \end{equation} To find the next coefficient, $c_1$, we first differentiate ($\ref{Series}$) \begin{equation} f'(z)=c_1 + 2 c_2 (z-a) + 3 c_3 (z-a)^2 +\dots \label{c1diff} \end{equation} We then evaluate ($\ref{c1diff}$) at $z=a$ to obtain \begin{equation} c_1 = f'(a) \end{equation} We continue to differentiate ($\ref{c1diff}$) and then evaluate at $z=a$, reordering the equation as necessary \begin{eqnarray} f^{\prime\prime}(z) &=& 2 c_2 + (3)(2)(z-a) + \dots\\ c_2 &=& \frac{1}{2} f^{\prime\prime}(a) \end{eqnarray} The $n$th coefficient is given by \begin{eqnarray} c_n &=& {1\over n!}\,{d^n\over dz^n} f(z)\Big|_{z=a}\label{cn}\\ &=& {1\over n!}\,f^{(n)}(a) \nonumber \end{eqnarray} where the last line is just another way to write the previous line, i.e. the symbol $f^{(n)}$ means to take $n$ derivatives of $f$. Make sure to evaluate the derivatives at $z=a$ only after you have taken the required number of derivatives. Why is there a factor of $1/n!$ in ($\ref{cn}$), the expression for the $n$th coefficient?

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