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Math Analysis 2

Series of functions, power series and Taylor series

Pointwise and uniform convergence, the Weierstrass M-test, radius of convergence, term-by-term differentiation and integration, Taylor series and computing integrals with series.

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In Analysis 1 series were infinite sums of numbers. In Analysis 2 we add up infinitely many functions, and the result is a new function. This is how calculators compute sines and exponentials, how differential equations with no elementary solution are solved, and how signals are approximated. In this lesson we cover sequences and series of functions, uniform convergence, power series and Taylor series.

Pointwise and uniform convergence

A sequence of functions f_n converges pointwise to f on a set I if f_n(x)\to f(x) for every x\in I. It is the most natural notion, but it is weak: the functions f_n(x) = x^n on [0, 1] are all continuous, yet the pointwise limit is 0 for x \lt 1 and 1 for x = 1, which is discontinuous.

We say f_n converges uniformly to f if

    \[ \sup_{x\in I}|f_n(x) - f(x)| \to 0 \]

that is, if from some n on the whole graph of f_n lies in a “strip” of width \varepsilon around the graph of f, for all x at once. In the x^n example, for every n there is an x close to 1 where x^n \gt 1/2: the convergence is not uniform.

Why it matters

  • The uniform limit of continuous functions is continuous. Given \varepsilon, choose n with |f_n - f| \lt \varepsilon/3 everywhere, then \delta from the continuity of f_n; then |f(x) - f(x_0)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| \lt \varepsilon.
  • You can take the limit under the integral sign on a bounded interval: \int_a^b f_n \to \int_a^b f, because |\int (f_n - f)| \le (b - a)\sup|f_n - f|.
  • Derivatives need more: uniform convergence of the derivatives f_n'.

For a series of functions \sum f_k(x) one studies the partial sums. The most used test is the Weierstrass M-test: if |f_k(x)| \le M_k for every x and \sum M_k converges, the series converges uniformly. Example: \sum \frac{\sin kx}{k^2} converges uniformly on the whole of \mathbb R, because |\sin kx/k^2| \le 1/k^2.

Power series

A power series centred at x_0 is

    \[ \sum_{k=0}^{\infty} a_k\,(x - x_0)^k \]

Theorem. There is R\in[0, +\infty], the radius of convergence, such that the series converges absolutely for |x - x_0| \lt R and does not converge for |x - x_0| \gt R. The end points must be checked separately. The radius is found with the ratio or root test:

    \[ \frac1R = \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| \quad\text{or}\quad \frac1R = \limsup_{k\to\infty}\sqrt[k]{|a_k|} \]

Examples.

  • \sum x^k: R = 1, sum \frac{1}{1 - x}. It does not converge at the end points.
  • \sum \frac{x^k}{k}: R = 1; at x = -1 it converges (alternating series), at x = 1 it diverges (harmonic series). The sum is -\ln(1 - x).
  • \sum \frac{x^k}{k!}: R = +\infty, sum e^x.

Inside the interval of convergence the series converges uniformly on every smaller closed interval, and it can be differentiated and integrated term by term without changing the radius. For example, integrating \frac{1}{1 + t^2} = \sum (-1)^k t^{2k} from 0 to x gives

    \[ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \]

and at x = 1 the Leibniz series: \frac\pi4 = 1 - \frac13 + \frac15 - \dots

Taylor series

If f has derivatives of every order at x_0, its Taylor series is

    \[ \sum_{k=0}^{\infty}\frac{f^{(k)}(x_0)}{k!}\,(x - x_0)^k \]

The partial sums are the Taylor polynomials of Analysis 1. The series converges to f if and only if the remainder R_n(x) = f(x) - T_n(x) tends to zero. With Lagrange’s remainder, R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - x_0)^{n+1} for some \xi between x_0 and x.

The sine function and its Taylor polynomials of degree 1, 3, 5 and 7
The sine (blue) and its Taylor polynomials centred at 0 of degree 1, 3, 5 and 7 (dashed). Each new term widens the interval where the approximation is good; the full series converges to the sine on the whole line.

Proof: the sine series converges everywhere

Every derivative of the sine is \pm\sin x or \pm\cos x, so |f^{(n+1)}(\xi)| \le 1 and

    \[ |R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!} \to 0 \quad\text{for every } x \]

because the factorial grows faster than any power. Hence

    \[ \sin x = \sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{(2k+1)!},\qquad \cos x = \sum_{k=0}^{\infty}\frac{(-1)^k x^{2k}}{(2k)!},\qquad e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!} \]

Substituting x = i\theta in the exponential series and separating even and odd terms gives Euler’s formula, e^{i\theta} = \cos\theta + i\sin\theta.

Careful: not every function is the sum of its Taylor series. The function f(x) = e^{-1/x^2} (with f(0) = 0) has all derivatives zero at 0: its Taylor series is the zero function, which equals f only at the origin. Functions that equal their Taylor series near every point are called analytic.

An application: computing with series

To compute \int_0^1 e^{-x^2}dx, which has no elementary antiderivative, integrate the series term by term:

    \[ \int_0^1 e^{-x^2}dx = \sum_{k=0}^{\infty}\frac{(-1)^k}{k!\,(2k+1)} = 1 - \frac13 + \frac1{10} - \frac1{42} + \frac1{216} - \dots \approx 0.7468 \]

Since it is an alternating series with decreasing terms, the error is less than the first term left out: with five terms the error is below 1/1320. This integral is proportional to the error function erf, which describes the bit error probability of a digital transmission with Gaussian noise.

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