# Question Chebyshev`s equation may be written d dx [(1 − x dy dx] + ```Question
Chebyshev’s equation may be written
&quot;
#
1
1 dy
d
(1 − x2 ) 2
+ n2 (1 − x2 )− 2 y = 0
dx
dx
where n is a positive integer. The solutions which satisfy the boundary
conditions Tn (−1) = (−1)n and Tn (1) = 1 are polynomials of degree n. Find
the first three Chebyshev polynomials. Prove that if m 6= n then
Z
1
1
(1 − x2 )− 2 Tm (x)Tn (x)dx = 0
−1
To find the Chebyshev polynomials Tn (x) one substitutes a general polynomial into the differential equation and uses the boundary conditions. We
illustrate this with T2 (x).
The general form of T2 (x) is a quadratic so that T2 (x) = a0 + a1 x + a2 x2 .
Using T2 (1) = 1 gives a0 +a1 +a2 = 1. Using T2 (−1) = 1 gives a0 −a1 +a2 = 1.
Subtracting the equations gives a1 = 0 and hence a0 = 1 − a2 . Writing a2 as
a we see that T2 (x) = ax2 + (1 − a).
We now substitute this into Chebyshev’s equation (with n = 2) and obtain
{2a(1 − 2x2 ) + 4ax2 + 4(1 − a)}(1 − x2 )−1/2 = 0, ⇒ 4 − 2a = 0. So that
a = 2 and T2 (x) = 2x2 − 1.
Now to show orthogonality. since Tn and Tm are solutions of Chebeyshev’s
equation we have:
&quot;
#
d
dTn
(1 − x2 )
+ n2 (1 − x2 )−1/2 Tn = 0
dt
dx
&quot;
#
d
2 dTm
(1 − x )
+ m2 (1 − x2 )−1/2 Tm = 0
dt
dx
(1)
(2)
Multiplying equation (1) by Tm and equation (2) by Tn , subtracting and
integratingZ by parts between −1 and 1 gives:
1
(n2 − m2 )
(
(1 − x2 )−1/2 Tn Tm dx
&quot;−1
#
&quot;
d
dTm
d
dTn
Tn
=
(1 − x2 )
− Tm
(1 − x2 )
dx
dx
dx
dx
−1
&quot;
&Atilde;
!#1
dTm
dTn
= (1 − x2 ) Tn
− Tm
dx
dx −1
Z
1
1
#)
dx
−
Z
1
−1
(
)
dTm dTn
dTn dTm
− (1 − x2 )
dx
(1 − x )
dx dx
dx dx
2
=0
Since the first term vanishes at &plusmn;1, and the terms and the integral cancel.
2
```