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Multivariable CalculusTaylor Series

Reading time: ~20 min

We can define a polynomial which approximates a smooth function in the vicinity of a point with the following idea: match as many derivatives as possible.

The utility of this simple idea emerges from the convenient simplicity of polynomials and the fact that a wide class of functions look pretty much like polynomials when you zoom in around a given point.

First, a bit of review on the exponential function x\mapsto \exp(x): we define \exp to be the function which maps 0 to 1 and which is everywhere equal to its own derivative. It follows (nontrivially) from this definition that \exp(x) = \exp(1)^x, so may define \mathrm{e} = \exp(1) and write the exponential function as x\mapsto \mathrm{e}^x. The value of \mathrm{e} is approximately 2.718.

Find the quadratic polynomial P_2 whose zeroth, first, and second derivatives at the origin match those of the exponential function.

Solution. Since P_2 is quadratic, we must have

\begin{align*}P_2(x) = a_0 + a_1x + a_2x^2\end{align*}

for some a_0, a_1, and a_2. To match the derivative, we check that P_2(0) = a_0 and f(0) = 1. So we must have a_0 =1. Similarly, P_2'(0) = a_1, so if we want P_2'(0) = f'(0) = 1, have to choose a_1 = 1 as well.

For a_2, we calculate P_2''(x) = (a_1 + 2a_2x)' = 2a_2, so to get P_2''(0) = f''(0) = 1, we have to let a_2 = \tfrac{1}{2}. So

\begin{align*}P_2(x) = 1 + x + \tfrac{1}{2}x^2\end{align*}

is the best we can do. Looking at the figure, we set that P_2 does indeed do a better job of 'hugging' the graph of f near x=0 than the best linear approximation (L(x) = 1 + x) does.

The best constant, linear, and quadratic approximations of \exp(x) = \mathrm{e}^x near the origin

We can extend this idea to higher order polynomials, and we can even include terms for all powers of x, thereby obtaining an infinite series:

Definition (Taylor Series)
The Taylor series, centered at c, of an infinitely differentiable function f is defined to be

\begin{align*}f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \cdots\end{align*}

Find the Taylor series centered at the origin for the exponential function.

Solution. We continue the pattern we discovered for the quadratic approximation of the exponential function at the origin: the $n$th derivative of a_0 + a_1x + \cdots + a_n x^n + \cdots is n!a_n, while the $n$th derivative of the exponential function is 1 at the origin. Therefore, a_n = 1/n!, and we obtain the Taylor series

\begin{align*}1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\end{align*}

It turns out that this series does in fact converge to \mathrm{e}^x, for all x \in \mathbb{R}.

Taylor series properties

It turns out that if the Taylor series for a function converges, then it does so in an interval centered around c. Furthermore, inside the interval of convergence, it is valid to perform term-by-term operations with the Taylor series as though it were a polynomial:

  • We can multiply or add Taylor series term-by-term.
  • We can integrate or differentiate a Taylor series term-by-term.
  • We can substitute one Taylor series into another to obtain a Taylor series for the composition.

All the operations described above may be applied wherever all the series in question are convergent. In other words, f and g have Taylor series P and Q converging to f and g in some open interval, then the Taylor series for fg, f+g, f', and \int f converge in that interval and are given by PQ, P+Q, P', and \int P, respectively. If P has an infinite radius of convergence, then the Taylor series for f\circ g is given by P\circ Q.

The following example shows how convenient this theorem can be for finding Taylor series.

Find the Taylor series for f(x) = \cos x + x \mathrm{e}^{x^2} centered at c = 0.

Solution. Taking many derivatives is going to be no fun, especially with that second term. What we can do, however, is just substitute x^2 into the Taylor series for the exponential function, multiply that by x, and add the Taylor series for cosine:

\begin{align*}&\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right) + x\left(1 + x^2 + \frac{(x^2)^2}{2!} + \frac{(x^2)^3}{3!} + \cdots\right) \\\ &= 1 + x - \frac{x^2}{2!} + x^3 + \frac{x^4}{4!} + \frac{x^5}{2!} + \cdots.\end{align*}

In summation notation, we could write this series as \sum_{n=0}^\infty a_n x^n where a_n is equal to (-1)^{n/2}/n! if n is even and 1/((n-1)/2)! if n is odd.

Find the Taylor series for 1/(1-x) centered at the origin, and show that it converges to 1/(1-x) for all -1 < x < 1.

Use your result to find x + 2x^2 + 3x^3 + 4x^4 + \cdots. Hint: think about differentiation.

Solution. Calculating derivatives of 1/(1-x), we find that the Taylor series centered at the origin is 1 + x + x^2 + \cdots. Furthermore, we know that

\begin{align*}\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots,\end{align*}

for -1 < x < 1, by the formula for infinite geometric series.

We can use this result to find \sum_{k = 1}^\infty k x^k by differentiating both sides and multiplying both sides by x:

\begin{align*}\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots\end{align*}

We get

\begin{align*}\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + 4x^4 + \cdots\end{align*}

Show that \lim_{n\to\infty}(1+x/n)^n is equal to \mathrm{e}^x by showing that \lim_{n\to\infty}\log (1+x/n)^n = x.

Solution. Integrating the equation

\begin{align*}\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - \cdots\end{align*}

term by term, we find that

\begin{align*}\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\end{align*}

Substituting gives

\begin{align*}n \log (1+x/n) = x - \frac{x}{2n} + \frac{x^3}{3n^2} - \cdots.\end{align*}

Each of the terms other than the converges to 0, and we can take limits term-by-term since x/n is inside of the interval of convergence for this series. Therefore, \lim_{n\to\infty}\log(1+x/n)^n = x, and since the exponential function is continuous, this implies that \lim_{n\to\infty}(1+x/n)^n = \mathrm{e}^x

Bruno Bruno