(1-x2){d2y\overdx2}-x{dy\overdx}+p2y=0
where p is a real (or complex) constant. The equation is named after Russian mathematician Pafnuty Chebyshev.
The solutions can be obtained by power series:
y=
| infty | |
| \sum | |
| n=0 |
| n | |
| a | |
| nx |
where the coefficients obey the recurrence relation
an+2={(n-p)(n+p)\over(n+1)(n+2)}an.
The series converges for
|x|<1
The recurrence may be started with arbitrary values of a0 and a1,leading to the two-dimensional space of solutions that arises from second orderdifferential equations. The standard choices are:
a0 = 1 ; a1 = 0, leading to the solution
F(x)=1-
| p2 | |
| 2! |
x2+
| (p-2)p2(p+2) | |
| 4! |
x4-
| (p-4)(p-2)p2(p+2)(p+4) | |
| 6! |
x6+ …
a0 = 0 ; a1 = 1, leading to the solution
G(x)=x-
| (p-1)(p+1) | |
| 3! |
x3+
| (p-3)(p-1)(p+1)(p+3) | |
| 5! |
x5- … .
The general solution is any linear combination of these two.
When p is a non-negative integer, one or the other of the two functions has its series terminateafter a finite number of terms: F terminates if p is even, and G terminates if p is odd.In this case, that function is a polynomial of degree p and it is proportional to theChebyshev polynomial of the first kind
Tp(x)=(-1)p/2 F(x)
Tp(x)=(-1)(p-1)/2 p G(x)