5.6 The Power Series Method
A power series about \(x_0\) is an infinite series \begin {equation} \sum ^{\infty }_{n = 0} a_n (x - x_0)^n = a_0 + a_1(x -x_0) + a_2(x -x_0)^2 + \cdots . \label {eq:powerseries} \end {equation} If (8) converges for all \(x\) with \(|x - x_0| <R\) and diverges for all \(x\) with \(|x -x_0| > R\), then \(R\) is the radius of convergence. It can be found from either \[R = \frac {1}{\lim _{n \to \infty } \sqrt [n]{|a_n|}}\qquad \text {or}\qquad R = \frac {1}{\lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right |},\] provided the limit exists, with \(R = \infty \) when the limit is \(0\), in which case the series converges for all \(x\). Within the interval of convergence a power series may be differentiated term by term.
Now consider a linear equation \[p(x) y'' + q(x)y' + r(x)y = 0 ,\] where \(p\), \(q\) and \(r\) are polynomials. The point \(x_0\) is an ordinary point if \(p(x_0) \neq 0\), so that \(\frac {q}{p}\) and \(\frac {r}{p}\) are defined near \(x_0\); otherwise it is a singular point. Near an ordinary point we try a solution \[y(x) = \sum ^{\infty }_{k=0} a_k(x -x_0)^k ,\] substitute, and solve for the coefficients \(a_k\). The series converges at least in the largest interval about \(x_0\) containing no singular point.
Solution. Let \(y = \sum ^{\infty } _{k = 0} a_k x^k\), so that \[y' = \sum ^{\infty } _{k = 1}k a_k x^{k-1},\qquad y'' = \sum ^{\infty } _{k = 2}k(k-1)a_k x^{k-2} .\] Substituting, \[\sum ^{\infty } _{k = 2}k(k-1)a_k x^{k-2} -\sum ^{\infty } _{k = 0} a_k x^k = 0 .\] Re-index the first series by replacing \(k\) by \(k + 2\), so that both series run over the same powers of \(x\): \[\sum ^{\infty }_{k = 0} (k+ 2 ) (k + 1) \, a_{k + 2} \, x^k - \sum ^{\infty }_{k = 0} a_k\, x^k = \sum ^{\infty }_{k=0}\big [(k+2)(k+1)a_{k+2} - a_k\big ] x^k = 0 .\] A power series is zero only if every coefficient is zero, so \[a_{k+2} = \frac {a_k}{(k+2)(k+1)} ,\qquad k = 0,1,2,\dots .\] This recurrence relation determines all the coefficients from \(a_0\) and \(a_1\), which remain arbitrary: \[a_2 = \frac {a_0}{2!},\quad a_3 = \frac {a_1}{3!},\quad a_4 = \frac {a_2}{3\cdot 4} = \frac {a_0}{4!},\quad a_5 = \frac {a_1}{5!},\quad a_6 = \frac {a_0}{6!},\ \dots \] So \[y(x) = a_0\left ( 1 + \frac {x^2}{2!}+ \frac {x^4}{4!} +\cdots \right ) + a_1 \left (x + \frac {x^3}{3!} + \frac {x^5}{5!}+ \cdots \right ) = a_0 \sum ^{\infty }_{n = 0} \frac {x^{2n}}{(2n)!} + a_1 \sum ^{\infty }_{n=0} \frac {x^{2n+1}}{(2n+1)!},\] that is, \(y(x) = a_0 \cosh x + a_1 \sinh x\), valid for all \(x\).
Example 5.6.2. Find the first few terms of the power series solutions of Airy’s equation \(y'' - xy = 0\) about \(x = 0\).
Solution. With \(y = \sum a_kx^k\), \(xy = \sum _{k\geq 1}a_{k-1}x^k\), and \(y'' = \sum _{k\geq 0}(k+2)(k+1)a_{k+2}x^k\). The constant term gives \(2a_2 = 0\), and for \(k\geq 1\), \[a_{k+2} = \frac {a_{k-1}}{(k+2)(k+1)} .\] So \(a_2 = a_5 = a_8 = \cdots = 0\), \(a_3 = \frac {a_0}{6}\), \(a_6 = \frac {a_3}{30} = \frac {a_0}{180}\), \(a_4 = \frac {a_1}{12}\), \(a_7 = \frac {a_4}{42} = \frac {a_1}{504}\), and \[y = a_0\left (1 + \frac {x^3}{6} + \frac {x^6}{180} + \cdots \right ) + a_1\left (x + \frac {x^4}{12} + \frac {x^7}{504} + \cdots \right ),\] convergent for all \(x\). These functions are not elementary; the series is their definition.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.