Why should you care about power series? Because they allow us to approximate functions to any desired accuracy.

For instance, consider the function $h(x)=-1-2x+x^2$. Draw its graph.

Figure 1: The graph of the polynomial $h(x)=-1-2x+x^2$.

(Notice that we have chosen a particularly simple funciton for this example, a polynomial. Its power series, expanded around the origin, is just the algebraic expression for the function itself. Just by looking at it; you can see that it has only three nonzero terms. Obviously, there is no reason for you to approximate a function this simple. We are using this simple example so that you can think about the separate terms quickly and easily in your head. In § {Finding Power Series Coefficients} and § {Visualizing Power Series Approximations} you have the opportunity to work with the power series for $\sin\theta$, which has an infinite number of terms. You will have many chances to see why power series approximations are useful for more complicated functions later in this text.)

Figure 2: The graph of the polynomial $h(x)=-1-2x+x^2$ (black) together with its zeroth order approximation at the origin (blue) and its first order (linear) approximation at the origin (red).

At the origin, The value of the function is $-1$. So, the zeroth order approximation is just the constant function $-1$. It agrees with the value of the function exactly at $x=0$, but otherwise is a pretty lousy approximation. The slope of this graph at $x=0$ is given by the derivative, namely $2$. The linear approximation at the origin to this function is therefore the straight line $y=-1-2x$. Where is the linear approximation of this function a “good” approximation?

Figure 3: The graph of the polynomial $h(x)=-1-2x+x^2$ (black) together with its zeroth order (and linear) approximation at $x=1$ (blue).

The approximation that we found above, which was a (terminated) series expansion around $x=0$ isn't a very good approximation near $x=1$. We can find a better approximation if we choose to expand around another point, say $x=1$. The expansion of $h$ around $1$ has just two terms, namely $h(x)=-2+(x-1)^2$. The absence of a linear term in this expansion tells us that the graph is horizontal at $x=1$.