## Section7.2Calculating Power Series Coefficients

### Derivation of the Formula for the Coefficients of a Power Series.

One way of finding the coefficients is using Taylor's theorem, derived as follows:

We evaluate both sides of equation (7.1.1) at the point $$z=a\text{,}$$ to obtain:

$$f(a)=c_0 + c_1(a-a) + c_2(a-a)^2 + c_3(a-a)^3 + \dots\tag{7.2.1}$$

Since all of the terms, except the first, on the right hand side of (7.2.1) are zero, the equation simplifies to:

$$c_0=f(a)\tag{7.2.2}$$

To find the next coefficient, $$c_1\text{,}$$ we first differentiate (7.1.1)

$$f'(z)=c_1 + 2 c_2 (z-a) + 3 c_3 (z-a)^2 +\dots\tag{7.2.3}$$

We then evaluate (7.2.3) at $$z=a$$ to obtain

$$c_1 = f'(a)\tag{7.2.4}$$

We continue to differentiate (7.2.3) and then evaluate at $$z=a\text{,}$$ reordering the equation as necessary

\begin{align*} f^{\prime\prime}(z) \amp = 2 c_2 + (3)(2)(z-a) + \dots\\ c_2 \amp = \frac{1}{2} f^{\prime\prime}(a) \end{align*}

The $$n$$th coefficient is given by

\begin{align} c_n \amp = {1\over n!}\,{d^n\over dz^n} f(z)\Big|_{z=a}\tag{7.2.5}\\ \amp = {1\over n!}\,f^{(n)}(a) \tag{7.2.6} \end{align}

where the last line is just a common notation for the previous line.

### Notation.

i.e. the symbol $$f^{(n)}$$ means to take $$n$$ derivatives of $$f\text{.}$$

Make sure to evaluate the derivatives at $$z=a$$ only after you have taken the required number of derivatives.

By plugging these values of the coefficients into Equation (7.1.1), we obtain the following form of the power series:

\begin{align} f(z) \amp = \sum_{n=0}^{\infty} {1\over n!}\,{d^n\over dz^n} f(a)\, (z-a)^n \tag{7.2.7} \end{align}

### Sensemaking7.1.

Why is there a factor of $$1/n!$$ in (7.2.6), the expression for the $$n$$th coefficient?

To check your understanding of this section you should complete the following activity.

### Activity7.2.

1. Find the first five nonzero coefficients for $$\sin(\theta)$$ expanded around the origin.

2. Write out a series approximation, correct to fourth order, for $$\sin(\theta)$$ expanded around the origin.

\begin{equation*} \sin(\theta) = \end{equation*}
3. Find the first four nonzero coefficients for $$\sin(\theta)$$ expanded around $$\theta_0={\pi\over 6}\text{.}$$

4. Write out a series approximation, correct to fourth order, for $$\sin(\theta)$$ expanded around $$\theta_0={\pi\over 6}\text{.}$$

\begin{equation*} \sin(\theta) = \end{equation*}
Hint.

Some important observations are given below:

1. Pay attention to the name of the independent variable. The equation for the coefficients is given in terms of the variable $$z\text{.}$$ What is the independent variable in $$\sin\theta\text{?}$$

2. Commonly, but not always, there are an infinite number of terms in a power series expansion. Unless you can find a general expression for the coefficients, it will take you an infinite amount of time to find all of them. In practice, one usually stops at some stage. Some language for this: “find the first four non-zero terms” means find the coefficients for the four lowest powers of the independent variable, continuing until you have four that are not zero; “find the expansion correct to fourth order” means find the coefficients for all of the low powers of the independent variable, up to and including the fourth power.

3. If you are asked to find the power series expansion around $$z=a\text{,}$$ then you must plug the number $$a$$ into all of the derivatives.

4. Your final answer for a power series expansion should be of the form

$$f(z)=c_0 + c_1(z-a) + c_2(z-a)^2 + c_3(z-a)^3 + \dots\tag{7.2.8}$$

or

$$f(z)\approx c_0 + c_1(z-a) + c_2(z-a)^2 + c_3(z-a)^3\tag{7.2.9}$$

where you have plugged in numbers for all of the $$c_n$$ and for $$a\text{.}$$ Use the first form of the equation, with an equals sign, if you include the symbol $$\dots$$ to remind yourself that there are (an infinite number of) terms that you have not written down. Use the second form of the equation, with an approximately equals sign, if you have truncated the series, leaving out (a possibly infinite number of) terms.