Category: Mathematics

Uniqueness of power series coefficients

The coefficients of the Taylor and Maclaurin series are unique. We are going to demonstrate this assertion because this fact will be very important in the development of the approximation techniques that we want to illustrate in this course.

Consider the power series of the form:

$\displaystyle \sum_{m=0}^\infty a_m (z - z_0)^m$

z and $z_0$ being complex numbers, we will study the convergence of the above power series in the complex plane. If R is the largest radius such that the series converges for all $|z - z_0| < R$ , then R is called the 'radius of convergence'. In its radius of convergence the power series converges to $f(z)$ :

$\displaystyle f(z) = \sum_{m=0}^\infty a_m (z - z_0)^m$

Given function $g(z)$ that is continuous on $C_r(z_0)$ (a circle in the complex plane centred on $z_0$ with radius $r < R$ and oriented counterclockwise):

$\displaystyle g(z)f(z) = g(z) \sum_{m=0}^\infty a_m (z - z_0)^m$

Using the theorem on integration of power series:

$\displaystyle \int_{C_r(z_0)} g(z)f(z) \, dz = \int_{C_r(z_0)} g(z) \sum_{m=0}^\infty a_m (z - z_0)^m \, dz$

$\displaystyle = \sum_{m=0}^\infty a_m \int_{C_r(z_0)} g(z) (z - z_0)^m \, dz$

Defining $g(z)$ :

$\displaystyle g(z) = \frac{1}{2\pi i} \frac{1}{(z - z_0)^{n+1}}$

the integral becomes:

$\displaystyle \int_{C_r(z_0)} g(z)f(z) \, dz = \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{f(z)}{(z - z_0)^{n+1}} \, dz$

Then:

$\displaystyle = \sum_{m=0}^\infty a_m \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{(z - z_0)^m}{(z - z_0)^{n+1}} \, dz$

$\displaystyle = \sum_{m=0}^\infty a_m \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{1}{(z - z_0)^{n-m+1}} \, dz$

Observing that:

$\displaystyle \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{1}{(z - z_0)^{n-m+1}} \, dz = \begin{cases} 1 & \text{for } m = n \\ 0 & \text{for } m \neq n \end{cases}$

the integral becomes:

$\displaystyle \sum_{m=0}^\infty a_m \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{(z - z_0)^m}{(z - z_0)^{n+1}} \, dz = a_n$

Using the Cauchy formula for a holomorphic function:

$\displaystyle \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{f(z)}{(z - z_0)^{n+1}} \, dz$

We have:

$\displaystyle \int_{C_r(z_0)} g(z)f(z) \, dz = \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{f(z)}{(z - z_0)^{n+1}} \, dz$

$\displaystyle \int_{C_r(z_0)} g(z)f(z) \, dz = a_n$

From Cauchy formula:

$\displaystyle \boxed{a_n = \frac{f^{(n)}(z_0)}{n!}}$

Therefore, the given power series is exactly equal to the Taylor series for $f(z)$ about the point $z_0$ and the coefficients are unique. The uniqueness of coefficients of Taylor and Maclaurin series will be used for solving very hard or even impossible-to-solve-exactly-problems with perturbation theory.

June 20, 2025
Some observations on functions
After looking at numbers, we will continue with a few practical observations about functions. The Taylor series of a real or complex-valued function $f(x)$ that is infinitely differentiable at a real or complex number $a$ is the power series:

$\displaystyle f(x) = f(a) + \frac{f'(a)}{1!} (x-a) + \frac{f''(a)}{2!} (x-a)^2 + \frac{f'''(a)}{3!} (x-a)^3 + \cdots$

$\displaystyle = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^{n}$

Using $a = 0$ , we obtain the so-called Maclaurin series:

$\displaystyle f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \cdots$

$\displaystyle = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^{n}$

The series above are clearly “power series” since we could write them as:

$\displaystyle a_0 + a_1 (x-a) + a_2 (x-a)^2 + a_3 (x-a)^3 + \cdots$

and

$\displaystyle a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$

respectively, using the definition $a_n = \frac{f^{(n)}(a)}{n!}$ for the Taylor series or $a_n = \frac{f^{(n)}(0)}{n!}$ for the Maclaurin series.

The Taylor series of a polynomial is the polynomial itself.

Taylor and Maclaurin series can be used in a practical way to program functions in a calculating machine or computer. A function such as $\texttt{cos(x)}$ or $\texttt{sin(x)}$ etc. has no meaning for a computer. For a computer, $\texttt{cos(x)}$ and $\texttt{sin(x)}$ are just symbols with no explicit algorithm defined to compute numerical results. Here are some Maclaurin series of important functions:

$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad x \in (-1,1)$

$\displaystyle \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}, \quad x \in (-1,1)$

$\displaystyle \frac{1}{(1-x)^3} = \sum_{n=2}^{\infty} \frac{(n-1)n}{2} x^{n-2}, \quad x \in (-1,1)$

$\displaystyle e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

$\displaystyle = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots$

$\displaystyle = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots, \quad x \in \mathbb{R}$

$\displaystyle \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^{n}$

$\displaystyle = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots, \quad x \in (-1,1]$

$\displaystyle \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$

$\displaystyle = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \cdots, \quad x \in \mathbb{R}$

$\displaystyle \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$

$\displaystyle = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \cdots, \quad x \in \mathbb{R}$

For programming the sine function we could use, for example, the series above as follows (in Python):
```
def sin(x):
    epsilon = 1e-16
    term = x
    sine = x
    sign = -1
    n = 1
    while abs(term) > epsilon:
        term *= x * x / ((2 * n) * (2 * n + 1))
        sine += sign * term
        sign *= -1
        n += 1
    return sine
```
June 15, 2025
Some observations on numbers
The aim of this section is to get us used to using geometric series or continued fractions to represent numbers. We will first have a look at some geometric series:

$\displaystyle 2 = 1 + 1$

$\displaystyle = 1 + \frac{1}{2} + \frac{1}{2}$

$\displaystyle = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4}$

$\displaystyle = \cdots$

$\displaystyle = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \cdots$

$\displaystyle = \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$

More generally, we can write a geometric series as:

$\displaystyle S(x) = x^0 + x^1 + x^2 + x^3 + \cdots$

$\displaystyle = 1 + x + x^2 + x^3 + \cdots$

$\displaystyle = 1 + x (1 + x + x^2 + \cdots)$

$\displaystyle = 1 + x S(x)$

Therefore:

$\displaystyle S(x) = \frac{1}{1-x}$

We have to be careful using this formula, since the radius of convergence of $S(x)$ is $(-1, 1)$ and $\frac{1}{1-x}$ is defined on $\mathbb{R} \setminus \{1\}$ . We can consider $\frac{1}{1-x}$ as some “representation” of $S(x)$ on $(-1, 1)$ .

This formula allows us, for example, to find that the argument of the geometric series of 3 is $\frac{2}{3}$ ( $\frac{2}{3} \in (-1, 1)$ ensures convergence of the series):

$\displaystyle 3 = \frac{1}{1-x} \implies x = \frac{2}{3}$

Now we can write $3$ as:

$\displaystyle 3 = \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n$

$\displaystyle = 1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \frac{16}{81} + \frac{32}{243} + \cdots$

Instead of writing numbers as a geometric series, we could also decide to write a number as a continued fraction with the form:

$\displaystyle a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \frac{1}{a_4 + \cdots}}}}$

To calculate the coefficients $a_0, a_1, a_2, \ldots$ , we can use the continued fraction algorithm, which iteratively takes the integer part of the number and inverts its fractional part. In the case of the number $\sqrt{2} \approx 1.41421356$ , we proceed as follows:

$\displaystyle \sqrt{2} \approx 1.41421356, \quad a_0 = \lfloor \sqrt{2} \rfloor = 1$

$\displaystyle \sqrt{2} - 1 \approx 0.41421, \quad \frac{1}{\sqrt{2} - 1} \approx 2.41421, \quad a_1 = \lfloor 2.41421 \rfloor = 2$

$\displaystyle \frac{1}{\sqrt{2} - 1} - 2 = \sqrt{2} - 1 \approx 0.41421, \quad \frac{1}{\sqrt{2} - 1} \approx 2.41421, \quad a_2 = \lfloor 2.41421 \rfloor = 2$

$\displaystyle \frac{1}{\sqrt{2} - 1} - 2 = \sqrt{2} - 1 \approx 0.41421, \quad \frac{1}{\sqrt{2} - 1} \approx 2.41421, \quad a_3 = \lfloor 2.41421 \rfloor = 2$

$\displaystyle \cdots$

The process repeats, leading to the coefficients $a_0 = 1$ , $a_1 = 2$ , $a_2 = 2$ , $a_3 = 2, \ldots$ . Thus, we can write $\sqrt{2}$ as a continued fraction:

$\displaystyle \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \cdots}}}$

The convergents of this continued fraction are:

$\displaystyle c_0 = 1$

$\displaystyle c_1 = 1 + \frac{1}{2} = \frac{3}{2} = 1.5$

$\displaystyle c_2 = 1 + \frac{1}{2 + \frac{1}{2}} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4$

$\displaystyle c_3 = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} = 1 + \frac{5}{12} = \frac{17}{12} \approx 1.4167$

$\displaystyle \cdots$

These convergents approach $\sqrt{2} \approx 1.41421356$ . Note that an alternative method, such as associating the partial sums of a geometric series to a continued fraction, often leads to non-standard coefficients (e.g., negative or non-integer values like $-\frac{3}{2}$ for $2$ or $-\frac{4}{3}$ for $\frac{3}{2}$ ), which may cause convergence issues. The continued fraction algorithm ensures integer coefficients and reliable convergence.

To find a continued fraction for $\pi$ , we can use its decimal expansion or a series approximation. For example, we can use the following series:

$\displaystyle \pi = \sum_{n=0}^{\infty} \frac{2^{n+1} n!^2}{(2n+1)!} \approx 3.1415926535\ldots$

Then we proceed as follows:
1. Let $a_0$ be the largest integer that does not exceed $\pi$ , namely $3$ :
  $\displaystyle a_0 = 3$
2. Compute:
  $\displaystyle \pi \approx 3.141592$
  $\pi - 3 \approx 0.141592, \quad \frac{1}{\pi - 3} \approx 7.062513, \quad a_1 = \lfloor 7.062513 \rfloor = 7$
3. Continue iteratively:
  $\displaystyle \frac{1}{\pi - 3} - 7 \approx 0.062513, \quad \frac{1}{0.062513} \approx 15.99659, \quad a_2 = \lfloor 15.99659 \rfloor = 15$
  
  $\displaystyle \frac{1}{0.06251} - 15 \approx 0.99659, \quad \frac{1}{0.996594} \approx 1.003417, \quad a_3 = \lfloor 1.003417\rfloor = 1$
  
  $\displaystyle \cdots$
Using this technique, the continued fraction for $\pi$ is:

$\displaystyle \pi = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \cdots}}}$

The number $\pi$ itself does not appear as a coefficient in this continued fraction. We could also represent $\pi$ (or any real number) using a generalized continued fraction:

$\displaystyle a_0 + \frac{b_1}{a_1 + \frac{b_2}{a_2 + \frac{b_3}{a_3 + \cdots}}}$
June 12, 2025