While elementary functions like $\sin x$ and $e^x$ satisfy basic differential equations, many physical phenomena—such as heat distribution or quantum states—are governed by equations that have no "closed-form" solutions. This slide introduces the Taylor series as the fundamental bridge, allowing us to represent unknown solutions as infinite power series.
By assuming a solution is analytic at a point, we transform the problem of solving a differential equation into a problem of determining a sequence of numerical coefficients.
1. The Foundation of Analyticity
A function $f$ that has a Taylor series expansion about $x = x_0$ with a radius of convergence $\rho > 0$ is said to be analytic at $x = x_0$. This property is the prerequisite for seeking series solutions to ordinary differential equations. If the coefficient functions of our ODE are analytic at $x_0$, the solution $y(x)$ is guaranteed to be analytic there as well.
2. Taylor Series Representation
The series $\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$ is called the Taylor series for the function $f$ about $x = x_0$. Here, the coefficients are defined by:
$$\displaystyle a_n = \frac{f^{(n)}(x_0)}{n!}$$
This links the global behavior of the function to its local derivatives at a single point.
3. Convergence and Validity
A power series solution is only meaningful within its radius of convergence. For example, while the exponential function $\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ converges for all $x$ ($\rho = \infty$), other series derived from differential equations may only converge within a specific distance from the expansion point $x_0$. This distance is usually determined by the singularities (where the equation's coefficients break down) of the equation.
Consider the differential equation $y' = y$ with the initial condition $y(0)=1$. Instead of guessing the solution, we assume a power series form:
$$y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1x + a_2x^2 + \dots$$
Differentiating gives $y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}$. Substituting into $y'=y$:
$$\sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} a_n x^n$$
Aligning indices, we find $(n+1)a_{n+1} = a_n$, which implies $\displaystyle a_n = \frac{a_0}{n!}$. Since $y(0)=1$, $a_0=1$. The result is the Taylor series for $e^x$.