Definition 4.3.2.
If \(\{a_n\}_{n=1}^\infty\) is a sequence of real numbers, a generating function for the sequence is a Taylor series (or other series in some contexts) where the \(a_n\) are the series coefficients.
Section 4.3 Legendre Functions
There is a nice example of series solutions which deserves its own section, since it leads to an important class of functions. Let \(k \in \NN\) and consider the differential equation
\begin{equation*}
(1-t^2)y^{\prime \prime} - 2ty^\prime + k(k+1)y=0 \text{.}
\end{equation*}
I need to divide by \(1-t^2\) to get this DE in standard form, implying that \(t = \pm 1\) are singular points. A solution centered at \(0\) should have radius of convergence at least \(R=1\text{.}\)
\begin{align*}
(1-t^2)y^{\prime \prime} - 2ty^\prime + k(k+1) y \amp =
(1-t^2) \sum_{n=2}^\infty c_n n(n-1)t^{n-2}\\
\amp - 2t \sum_{n=1}^\infty c_n nt^{n-1} +k(k+1)
\sum_{n=0}^\infty c_n t^n = 0
\end{align*}
I’ve taken the derivatives of the series and inserted them. Then I multiply the powers of \(t\) into the sums. Then I’ll follow the standard steps: I’ll shift to make the exponents match, pull out initial terms to make the indices match, then combine the sums into one.
\begin{align*}
0 \amp = \sum_{n=2}^\infty c_n n (n-1)t^{n-2} -
\sum_{n=2}^\infty c_n n(n-1) t^n
- \sum_{n=1}^\infty 2c_n n t^n +
\sum_{n=0}^\infty k(k+1) c_n t^n\\
\amp = \sum_{n=0}^\infty c_{n+2} (n+2) (n+1)t^n -
\sum_{n=2}^\infty c_n n(n-1) t^n\\
\amp - \sum_{n=1}^\infty 2c_n n t^n +
\sum_{n=0}^\infty k(k+1) c_n t^n\\
\amp = 2c_2 + 6c_3 t + \sum_{n=2}^\infty c_{n+2} (n+2)
(n+1)t^n\\
\amp - \sum_{n=2}^\infty c_n n(n-1) t^n -2c_1 t -
\sum_{n=2}^\infty 2c_n n t^n\\
\amp + k(k+1) c_0 + k(k+1) c_1 t +
\sum_{n=2}^\infty k(k+1) c_n t^n\\
\amp = \left[ 2c_2 + k(k+1) c_0 \right] + \left[ 6c_3 - 2c_1 +
k(k+1) c_1 \right] t\\
\amp + \sum_{n=2}^\infty \left[ c_{n+2} (n+2) (n+1) - c_n
n(n-1) - 2c_n n + k(k+1) c_n \right] t^n
\end{align*}
I need all these coefficients to be zero. I leave \(c_0\) and \(c_1\) as unknown parameters, since I haven’t imposed initial conditions. The first two isolated terms (the constant and the linear term) give unique equations for their coefficients.
\begin{align*}
\left[ 2c_2 + k(k+1) c_0 \right] \amp = 0\\
c_2 \amp = \frac{-c_0 k (k+1)}{2!}\\
\left[ 6c_3 - 2c_1 + k(k+1) c_1 \right] \amp = 0\\
6c_3 + (k+2)(k-1) c_1 \amp = 0\\
c_3 \amp = \frac{-c_1 (k+2)(k-1)}{3!}
\end{align*}
The remaining terms are determined by the recurrence relation. I’ll do some algebra with the complicated term inside the sum to try to make the recurrence relation clear.
\begin{align*}
c_{n+2} (n+2) (n+1) - c_n n(n-1) - 2c_n n + k(k+1) c_n \amp =
0\\
c_{n+2} (n+2) (n+1) + (-n^2 + n - 2n + k(k+1)) c_n \amp = 0\\
c_{n+2} (n+2) (n+1) + (-n^2 - n + k^2 +k ) c_n \amp = 0\\
c_{n+2} (n+2) (n+1) + (k-n)(k+n+1) c_n \amp = 0\\
c_{n+2} \amp = \frac{-c_n (k-n)(k+n+1)}{(n+1)(n+2)}
\end{align*}
This is an order two recurrence relation. I calculate coefficients using this relation.
\begin{align*}
c_0 \amp = c_0\\
c_1 \amp = c_1\\
c_2 \amp = \frac{-c_0 k (k+1)}{2!}\\
c_3 \amp = \frac{-c_1 (k+2)(k-1)}{3!}\\
c_4 \amp = \frac{-c_2 (k-2)(k+3)}{(3)(4)} = \frac{c_0
(k)(k+1)(k-2)(k+3)}{4!}\\
c_5 \amp = \frac{-c_3 (k-3)(k+4)}{(4)(5)} = \frac{c_1
(k-1)(k+2)(k-3)(k+4)}{5!}\\
c_6 \amp = \frac{-c_4 (k-4)(k+5)}{(5)(6)} = \frac{-c_0
(k)(k+1)(k-2)(k+3)(k-4)(k+5)}{6!}\\
c_7 \amp = \frac{-c_5 (k-5)(k+6)}{(6)(7)} = \frac{-c_1
(k-1)(k+2)(k-3)(k+4)(k-5)(k+6)}{7!}
\end{align*}
The pattern continues and gives two series, one for the \(c_0\) terms and one for the \(c_1\) terms.
I assumed that \(k \in \NN\text{.}\) Now I’ll consider two cases: \(k\) even or \(k\) odd. If \(k\) is odd, then the term involving \(c_1\) eventually has \((k-k)\) in the numerator. That means the numerator will be zero: this recurrence relationship eventually stop and there is a polynomial solution. If \(k\) is even, then terms involving \(c_0\) eventually have \((k-k)\) in the numerator, and again there is a polynomial solution. So, for any \(k\text{,}\) I find a polynomial solution. (Note this extends the expected radius of convergence. I am only guaranteed a radius of \(R=1\text{,}\) but polynomials converge everywhere, so \(R=\infty\)).
These special polynomial solutions are the solutions I am interested in. They are called Legendre Polynomials for \(k \in
\NN\text{.}\) They are historically important special functions with several other important properties in addition to the property of solving the Legendre DE. I’ll calculate the first few. There is the unknown \(c_0\) or \(c_1\) to consider; I’ll always choose \(c_0\) or \(c_1\) so that the polynomial goes through the point \((1,1)\text{.}\) (Which is equivalent to the initial condition \(y(1) = 1\text{.}\)) This convention leads to one of the important properties of the Legendre polynomials.
\begin{align*}
k=0 \implies P_0(t) \amp = c_0 \amp \amp \text{ choose }
c_0 = 1\\
P_0(t) \amp = 1 \amp \amp \\
k=1 \implies P_1(t) \amp = c_1t \amp \amp \text{ choose }
c_1 = 1\\
P_1(t) \amp = t \amp \amp \\
k=2 \implies P_2(t) \amp = c_0 (1 - 3t^2) \amp \amp
\text{ choose } c_0 = \frac{-1}{2}\\
P_2(t) \amp = \frac{1}{2} (3t^2 -1) \amp \amp \\
k=3 \implies P_3(t) \amp = c_1 \left( \frac{-5}{3}t^3 + t
\right) \amp \amp \text{ choose } c_1 = \frac{-3}{2}\\
P_3(t) \amp = \frac{-3}{2} \left( \frac{-5}{3}t^3 + t \right)
\amp \amp
\end{align*}
I’ll not repeat all the details, but here are the next eight polynomials.
\begin{align*}
P_3(t) \amp = \frac{1}{2} (5t^3 -3t ) \amp \amp \\
P_4(t) \amp = \frac{1}{8} (35t^4 - 30 t^2 + 3) \amp \amp \\
P_5(t) \amp = \frac{1}{8} (63t^5 - 70t^3 + 15t) \amp \amp \\
P_6(t) \amp = \frac{1}{16} (231t^6 - 315t^4 + 105t^2 - 5) \amp
\amp \\
P_7(t) \amp = \frac{1}{16} (429t^7 - 693t^5 + 315t^3 - 35t)
\amp \amp \\
P_8(t) \amp = \frac{1}{128} (6435t^8 - 12012t^6 + 6903t^4 -
1260 + 35) \amp \amp \\
P_9(t) \amp = \frac{1}{128} (12155t^9 - 25740t^7 + 18018t^5 -
4620t^3 + 315t) \amp \amp \\
P_{10}(t) \amp = \frac{1}{256} (46189t^{10} - 109395t^8 +
90090t^6 - 30030t^4 + 3465t^2 - 63) \amp \amp
\end{align*}
I can notice a number of interesting properties. First, all the even \(P_k\) are even functions, all the odd are odd. The even one all pass through \((-1,1)\) and \((1,1)\text{.}\) All the odd ones pass through \((-1,-1)\) and \((1,1)\text{.}\) (It is conventional to only work with the Legendre polynomials on the domain \([-1,1]\text{.}\)) I can also notice interesting patterns in the coefficients. There are powers of \(2\) in the denominators, but some are skipped. Interesting patterns are also found in the prime divisors of the numerators.
Here’s a list of further interesting properties.
\begin{align*}
P_k(-t) \amp = (-1)^k P_k(t)\\
\int_{-1}^1 P_k(t) P_l(t) \amp = \frac{2}{2l+1} \delta_{kl}\\
0 \amp = (k+1)P_{k+1}(t) - (2k+1)tP_k(t) + k P_{k-1}(t) \\
(2k+1) P_k(t) \amp = \frac{d}{dt} \left( P_{k+1}(t) -
P_{k-1}(t) \right)\\
P_k(t) \amp = \sum_{j=0}^k (-1)^j \binom{k}{j}^2 \left(
\frac{1+t}{2} \right)^{k-j} \left( \frac{1-t}{2} \right)^j
\end{align*}
The \(\delta_{kl}\) in the second property is the Kronecker delta. It evalutes to \(1\) if \(k=l\) and to \(0\) otherwise. It’s a very useful piece of notation. This second property is an orthogonality property: if I define an inner product of two functions as the integral of their product over \([-1,1]\text{,}\) then the Legendre polynomials are all orthogonal to each other. In linear algebra language, they form an orthogonal basis for the infinite dimensional linear space of polynomials (with this particular inner product). Orthogonal bases are nice to work with, another fact that recommends the Legendre polynomials.
The third property is called the functional equation. This section is the start of a whole branch of mathematicals on so-called ‘special functions’. The existence of a functional equation which relates members of the family to each other is quite typical for special functions.
As an interesting aside, I’ll very briefly detour to talk about generating functions.
In some way, the generating function ‘knows’ the properties of the sequence. I will share two interesting examples.
Example 4.3.3.
In some way, the function on the left relates to and accounts for the sequences of integer squares.
\begin{align*}
\amp \frac{t(t+1)}{(1-t)^3} = \sum_{n=0}^\infty n^2 t^n
\amp \amp |t| \lt 1
\end{align*}
Example 4.3.4.
Again, this strange polynomial/exponential function somehow captures the sequence of squares divided by factorials.
\begin{align*}
\amp t(t+1)e^t = \sum_{n=0}^\infty \frac{n^2}{n!} t^n \amp
\amp t \in \RR
\end{align*}
It turns out, if I briefly consider functions of two variables, that there are generating functions for Legendre polynomials. This relevant identity is
\begin{equation*}
\frac{1}{\sqrt{1- 2tx+x^2}} = \sum_{k=0}^\infty P_k(t) x^n\text{.}
\end{equation*}
Somehome, this strange two-variables square root functions ‘knows’ all the Legendre polynomials. There are all there, encoded in the Taylor series coefficients.
