Chebyshev functions

by taratuta

on
Category: Documents
58

views

Report

Transcript

Chebyshev functions
18.4 CHEBYSHEV FUNCTIONS
Since δ(Ω − Ω ) can depend only on the angle γ between the two directions Ω and Ω ,
we may also expand it in terms of a series of Legendre polynomials of the form
δ(Ω − Ω ) =
b P (cos γ).
(18.52)
From (18.14), the coeﬃcients in this expansion are given by
2 + 1 1
δ(Ω − Ω )P (cos γ) d(cos γ)
b =
2
−1
2 + 1 2π 1
δ(Ω − Ω )P (cos γ) d(cos γ) dψ,
=
4π
0
−1
where, in the second equality, we have introduced an additional integration over an
azimuthal angle ψ about the direction Ω (and γ is now the polar angle measured from
Ω to Ω). Since the rest of the integrand does not depend upon ψ, this is equivalent
to multiplying it by 2π/2π. However, the resulting double integral now has the form of
a solid-angle integration over the whole sphere. Moreover, when Ω = Ω , the angle γ
separating the two directions is zero, and so cos γ = 1. Thus, we ﬁnd
2 + 1
2 + 1
P (1) =
,
4π
4π
and combining this expression with (18.51) and (18.52) gives
2 + 1
Ym (Ω)Ym∗ (Ω ) =
P (cos γ).
4π
m
b =
(18.53)
Comparing this result with (18.49), we see that, to complete the proof of the addition
theorem, we now only need to show that the summations in on either side of (18.53) can
be equated term by term.
That such a procedure is valid may be shown by considering an arbitrary rigid rotation
of the coordinate axes, thereby deﬁning new spherical polar coordinates Ω̄ on the sphere.
Any given spherical harmonic Ym (Ω̄) in the new coordinates can be written as a linear
combination of the spherical harmonics Ym (Ω) of the old coordinates, all having the same
value of . Thus,
Dmm Ym (Ω),
Ym (Ω̄) =
m =−
where the coeﬃcients Dmm depend on the rotation; note that in this expression Ω and Ω̄
refer to the same direction, but expressed in the two diﬀerent coordinate systems. If we
choose the polar axis of the new coordinate system to lie along the Ω direction, then from
(18.45), with m in that equation set equal to zero, we may write
P (cos γ) =
4π
C0m Ym (Ω)
Y 0 (Ω̄) =
2 + 1 m =−
C0m
that depend on Ω . Thus, we see that the equality (18.53)
for some set of coeﬃcients
does indeed hold term by term in , thus proving the addition theorem (18.49). 18.4 Chebyshev functions
Chebyshev’s equation has the form
(1 − x2 )y − xy + ν 2 y = 0,
595
(18.54)
SPECIAL FUNCTIONS
and has three regular singular points, at x = −1, 1, ∞. By comparing it with
(18.1), we see that the Chebyshev equation is very similar in form to Legendre’s
equation. Despite this similarity, equation (18.54) does not occur very often
in physical problems, though its solutions are of considerable importance in
numerical analysis. The parameter ν is a given real number, but in nearly all
practical applications it takes an integer value. From here on we thus assume
that ν = n, where n is a non-negative integer. As was the case for Legendre’s
equation, in normal usage the variable x is the cosine of an angle, and so
−1 ≤ x ≤ 1. Any solution of (18.54) is called a Chebyshev function.
The point x = 0 is an ordinary point of (18.54), and so we expect to ﬁnd
m
two linearly independent solutions of the form y = ∞
m=0 am x . One could ﬁnd
the recurrence relations for the coeﬃcients am in a similar manner to that used
for Legendre’s equation in section 18.1 (see exercise 16.15). For Chebyshev’s
equation, however, it is easier and more illuminating to take a diﬀerent approach.
In particular, we note that, on making the substitution x = cos θ, and consequently
d/dx = (−1/ sin θ) d/dθ, Chebyshev’s equation becomes (with ν = n)
d2 y
+ n2 y = 0,
dθ2
which is the simple harmonic equation with solutions cos nθ and sin nθ. The
corresponding linearly independent solutions of Chebyshev’s equation are thus
given by
Tn (x) = cos(n cos−1 x) and Vn (x) = sin(n cos−1 x).
(18.55)
It is straightforward to show that the Tn (x) are polynomials of order n, whereas
the Vn (x) are not polynomials
Find explicit forms for the series expansions of Tn (x) and Vn (x).
Writing x = cos θ, it is convenient ﬁrst to form the complex superposition
Tn (x) + iVn (x) = cos nθ + i sin nθ
= (cos θ + i sin θ)n
n
√
= x + i 1 − x2
for |x| ≤ 1.
Then, on expanding out the last expression using the binomial theorem, we obtain
Tn (x) = xn − n C2 xn−2 (1 − x2 ) + n C4 xn−4 (1 − x2 )2 − · · · ,
√
Vn (x) = 1 − x2 n C1 xn−1 − n C3 xn−3 (1 − x2 ) + n C5 xn−5 (1 − x2 )2 − · · · ,
(18.56)
(18.57)
where n Cr = n!/[r!(n − r)!] is a binomial coeﬃcient. We thus see that Tn (x) is a polynomial
of order n, but Vn (x) is not a polynomial. It is conventional to deﬁne the additional functions
Wn (x) = (1 − x2 )−1/2 Tn+1 (x)
and Un (x) = (1 − x2 )−1/2 Vn+1 (x).
(18.58)
596
18.4 CHEBYSHEV FUNCTIONS
T0
1
T2
0.5
−1
T1
−0.5
1
0.5
−0.5
T3
−1
Figure 18.3 The ﬁrst four Chebyshev polynomials of the ﬁrst kind.
From (18.56) and (18.57), we see immediately that Un (x) is a polynomial of order
n, but that Wn (x) is not a polynomial. In practice, it is usual to work entirely in
terms of Tn (x) and Un (x), which are known, respectively, as Chebyshev polynomials
of the ﬁrst and second kind. In particular, we note that the general solution to
Chebyshev’s equation can be written in terms of these polynomials as

√
c1 Tn (x) + c2 1 − x2 Un−1 (x) for n = 1, 2, 3, . . . ,
y(x) =
c + c sin−1 x
for n = 0.
1
2
The n = 0 solution could also be written as d1 + c2 cos−1 x with d1 = c1 + 12 πc2 .
The ﬁrst few Chebyshev polynomials of the ﬁrst kind are easily constructed
and are given by
T1 (x) = x,
T0 (x) = 1,
T2 (x) = 2x − 1,
T3 (x) = 4x3 − 3x,
T4 (x) = 8x4 − 8x2 + 1,
T5 (x) = 16x5 − 20x3 + 5x.
2
The functions T0 (x), T1 (x), T2 (x) and T3 (x) are plotted in ﬁgure 18.3. In general,
the Chebyshev polynomials Tn (x) satisfy Tn (−x) = (−1)n Tn (x), which is easily
deduced from (18.56). Similarly, it is straightforward to deduce the following
597
SPECIAL FUNCTIONS
4
U2
2
U1
U0
−1
−0.5
0.5
1
−2
U3
−4
Figure 18.4 The ﬁrst four Chebyshev polynomials of the second kind.
special values:
Tn (1) = 1,
Tn (−1) = (−1)n ,
T2n (0) = (−1)n ,
T2n+1 (0) = 0.
The ﬁrst few Chebyshev polynomials of the second kind are also easily found
U1 (x) = 2x,
U0 (x) = 1,
U2 (x) = 4x − 1,
U3 (x) = 8x3 − 4x,
U4 (x) = 16x4 − 12x2 + 1,
U5 (x) = 32x5 − 32x3 + 6x.
2
The functions U0 (x), U1 (x), U2 (x) and U3 (x) are plotted in ﬁgure 18.4. The
Chebyshev polynomials Un (x) also satisfy Un (−x) = (−1)n Un (x), which may be
deduced from (18.57) and (18.58), and have the special values:
Un (1) = n + 1,
Un (−1) = (−1)n (n + 1),
U2n (0) = (−1)n ,
U2n+1 (0) = 0.
Show that the Chebyshev polynomials Un (x) satisfy the diﬀerential equation
(1 − x2 )Un (x) − 3xUn (x) + n(n + 2)Un (x) = 0.
(18.59)
From (18.58), we have Vn+1 = (1 − x2 )1/2 Un and these functions satisfy the Chebyshev
equation (18.54) with ν = n + 1, namely
(1 − x2 )Vn+1
− xVn+1
+ (n + 1)2 Vn+1 = 0.
598
(18.60)
18.4 CHEBYSHEV FUNCTIONS
Evaluating the ﬁrst and second derivatives of Vn+1 , we obtain
= (1 − x2 )1/2 Un − x(1 − x2 )−1/2 Un
Vn+1
Vn+1
= (1 − x2 )1/2 Un − 2x(1 − x2 )−1/2 Un − (1 − x2 )−1/2 Un − x2 (1 − x2 )−3/2 Un .
Substituting these expressions into (18.60) and dividing through by (1 − x2 )1/2 , we ﬁnd
(1 − x2 )Un − 3xUn − Un + (n + 1)2 Un = 0,
which immediately simpliﬁes to give the required result (18.59). 18.4.1 Properties of Chebyshev polynomials
The Chebyshev polynomials Tn (x) and Un (x) have their principal applications
in numerical analysis. Their use in representing other functions over the range
|x| < 1 plays an important role in numerical integration; Gauss–Chebyshev
integration is of particular value for the accurate evaluation of integrals whose
integrands contain factors (1 − x2 )±1/2 . It is therefore worthwhile outlining some
of their main properties.
Rodrigues’ formula
The Chebyshev polynomials Tn (x) and Un (x) may be expressed in terms of a
Rodrigues’ formula, in a similar way to that used for the Legendre polynomials
discussed in section 18.1.2. For the Chebyshev polynomials, we have
√
1
(−1)n π(1 − x2 )1/2 dn
(1 − x2 )n− 2 ,
Tn (x) =
dxn
2n (n − 12 )!
√
1
dn
(−1)n π(n + 1)
(1 − x2 )n+ 2 .
Un (x) = n+1
1
n
2
1/2
dx
2 (n + 2 )!(1 − x )
These Rodrigues’ formulae may be proved in an analogous manner to that used
in section 18.1.2 when establishing the corresponding expression for the Legendre
polynomials.
Mutual orthogonality
In section 17.4, we noted that Chebyshev’s equation could be put into Sturm–
Liouville form with p = (1 − x2 )1/2 , q = 0, λ = n2 and ρ = (1 − x2 )−1/2 , and its
natural interval is thus [−1, 1]. Since the Chebyshev polynomials of the ﬁrst kind,
Tn (x), are solutions of the Chebyshev equation and are regular at the end-points
x = ±1, they must be mutually orthogonal over this interval with respect to the
weight function ρ = (1 − x2 )−1/2 , i.e.
1
Tn (x)Tm (x)(1 − x2 )−1/2 dx = 0
if n = m.
(18.61)
−1
599
SPECIAL FUNCTIONS
The normalisation, when m = n, is easily found by making the substitution
x = cos θ and using (18.55). We immediately obtain
#
1
π
for n = 0,
2 −1/2
Tn (x)Tn (x)(1 − x )
dx =
π/2 for n = 1, 2, 3, . . . .
−1
(18.62)
The orthogonality and normalisation conditions mean that any (reasonable)
function f(x) can be expanded over the interval |x| < 1 in a series of the form
f(x) = 12 a0 +
∞
an Tn (x),
n=1
where the coeﬃcients in the expansion are given by
2 1
f(x)Tn (x)(1 − x2 )−1/2 dx.
an =
π −1
For the Chebyshev polynomials of the second kind, Un (x), we see from (18.58)
that (1 − x2 )1/2 Un (x) = Vn+1 (x) satisﬁes Chebyshev’s equation (18.54) with ν =
n + 1. Thus, the orthogonality relation for the Un (x), obtained by replacing Ti (x)
by Vi+1 (x) in equation (18.61), reads
1
Un (x)Um (x)(1 − x2 )1/2 dx = 0
if n = m.
−1
The corresponding normalisation condition, when n = m, can again be found by
making the substitution x = cos θ, as illustrated in the following example.
Show that
I≡
1
−1
Un (x)Un (x)(1 − x2 )1/2 dx =
π
.
2
From (18.58), we see that
1
I=
−1
Vn+1 (x)Vn+1 (x)(1 − x2 )−1/2 dx,
which, on substituting x = cos θ, gives
0
I=
sin(n + 1)θ sin(n + 1)θ
π
1
π
(− sin θ) dθ = . sin θ
2
The above orthogonality and normalisation conditions allow one to expand
any (reasonable) function in the interval |x| < 1 in a series of the form
f(x) =
∞
an Un (x),
n=0
600
18.4 CHEBYSHEV FUNCTIONS
in which the coeﬃcients an are given by
2 1
an =
f(x)Un (x)(1 − x2 )1/2 dx.
π −1
Generating functions
The generating functions for the Chebyshev polynomials of the ﬁrst and second
kinds are given, respectively, by
∞
GI (x, h) =
GII (x, h) =
1 − xh
=
Tn (x)hn ,
1 − 2xh + h2
(18.63)
1
=
1 − 2xh + h2
(18.64)
n=0
∞
Un (x)hn .
n=0
These prescriptions may be proved in a manner similar to that used in section 18.1.2 for the generating function of the Legendre polynomials. For the
Chebyshev polynomials, however, the generating functions are of less practical
use, since most of the useful results can be obtained more easily by taking
advantage of the trigonometric forms (18.55), as illustrated below.
Recurrence relations
There exist many useful recurrence relationships for the Chebyshev polynomials
Tn (x) and Un (x). They are most easily derived by setting x = cos θ and using
(18.55) and (18.58) to write
Tn (x) = Tn (cos θ) = cos nθ,
sin(n + 1)θ
.
Un (x) = Un (cos θ) =
sin θ
(18.65)
(18.66)
One may then use standard formulae for the trigonometric functions to derive
a wide variety of recurrence relations. Of particular use are the trigonometric
identities
cos(n ± 1)θ = cos nθ cos θ ∓ sin nθ sin θ,
(18.67)
sin(n ± 1)θ = sin nθ cos θ ± cos nθ sin θ.
(18.68)
Show that the Chebyshev polynomials satisfy the recurrence relations
Tn+1 (x) − 2xTn (x) + Tn−1 (x) = 0,
Un+1 (x) − 2xUn (x) + Un−1 (x) = 0.
(18.69)
(18.70)
Adding the result (18.67) with the plus sign to the corresponding result with a minus sign
gives
cos(n + 1)θ + cos(n − 1)θ = 2 cos nθ cos θ.
601
Fly UP