Approximaths

Uniqueness of power series coefficients

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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.

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