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```SPECIAL FUNCTIONS
18.24
The solutions y(x, a) of the equation
d2 y
− ( 14 x2 + a)y = 0
dx2
are known as parabolic cylinder functions.
(∗)
(a) If y(x, a) is a solution of (∗), determine which of the following are also
solutions: (i) y(a, −x), (ii) y(−a, x), (iii) y(a, ix) and (iv) y(−a, ix).
(b) Show that one solution of (∗), even in x, is
y1 (x, a) = e−x
2 /4
M( 12 a + 14 , 12 , 12 x2 ),
where M(α, c, z) is the conﬂuent hypergeometric function satisfying
dM
d2 M
+ (c − z)
− αM = 0.
dz 2
dz
You may assume (or prove) that a second solution, odd in x, is given by
2
y2 (x, a) = xe−x /4 M( 12 a + 34 , 32 , 12 x2 ).
z
2
(c) Find, as an inﬁnite series, an explicit expression for ex /4 y1 (x, a).
(d) Using the results from part (a), show that y1 (x, a) can also be written as
y1 (x, a) = ex
2 /4
M(− 12 a + 14 , 12 , − 12 x2 ).
(e) By making a suitable choice for a deduce that
∞
∞
bn x2n
(−1)n bn x2n
2
1+
= ex /2 1 +
,
(2n)!
(2n)!
n=1
n=1
5
where bn = nr=1 (2r − 32 ).
18.1
18.3
18.5
18.7
18.9
18.11
18.13
Note that taking the square of the modulus eliminates all mention of φ.
Integrate both sides of the generating function deﬁnition from x = 0 to x = 1,
and then expand the resulting term, (1 + h2 )1/2 , using a binomial expansion. Show
that 1/2 Cm can be written as [ (−1)m−1 (2m − 2)! ]/[ 22m−1 m!(m − 1)! ].
Prove the stated equation using the explicit closed form of the generating function.
Then substitute the series and require the coeﬃcient of each power of h to vanish.
(b) Diﬀerentiate result (a) and then use (a) again to replace the derivatives.
(a) Write the result of using Leibnitz’ theorem on the product of xn+m and e−x as a
ﬁnite sum, evaluate the separated derivatives, and then re-index the summation.
(b) For the ﬁrst recurrence relation, diﬀerentiate the generating function with
respect to h and then use the generating function again to replace the exponential.
Equating coeﬃcients of hn then yields the result. For the second, diﬀerentiate the
corresponding relationship for the ordinary Laguerre polynomials m times.
x2 f + xf + (λx3 − 14 )f = 0. Then, in turn, set x3/2 = u, and 23 λ1/2 u = v; then v
satisﬁes Bessel’s equation with ν = 13 .
(a) (1 − z)−a . (b) x−1 ln(1 + x). (c) Compare the calculated coeﬃcients with those
of tan−1 x. F( 12 , 1, 32 ; −x2 ) = x−1 tan−1 x. (d) x−1 sin−1 x. (e) Note that a term
containing x2n can only arise from the ﬁrst n + 1 terms of an expansion in powers
of sin2 x; make a few trials. F(−a, a, 12 ; sin2 x) = cos 2ax.
Looking for f(x) = u such that u + 1 is an inverse
√ power of
√ x with f(0) = ∞ and
f(1) = 1 leads to f(x) = 2x−1 − 1. I = B( 12 , 32 )/ 2 = π/(2 2).
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18.15
18.17
18.19
18.21
18.23
(a) Show that the ratio of two deﬁnitions based on m and n, with m > n > −Re z,
is unity, independent of the actual values of m and n.
(b) Consider the limit as z → −m of (z + m)z!, with the deﬁnition of z! based on
n where n > m.
the integrand in partial fractions and use J, as given, and J =
Express
∞
exp[ −iu(k − ia) ] du to express I as the sum of two double integral expressions.
0
Reduce them using the standard Gaussian integral, and then make a change of
variable 2v = u + 2a.
(b) Using the representation
1
Γ(b)
M(a, b; z) =
ezt ta−1 (1 − t)b−a−1 dt
Γ(b − a) Γ(a) 0
allows the equality to be established, without actual integration, by changing the
integration variable to s = 1 − t.
Calculate y (x) and y (x) and then eliminate x−1 e−x to obtain xy + (n + 1 + x)y +
ny = 0; M(n, n + 1; −x). Comparing the expansion of the hypergeometric series
with the result of term by term integration of the expansion of the integrand
shows that A = n.
(a) If the dummy variable
in the incomplete gamma function is t, make the change
√
of variable y = + t. Now choose a so that 2(a − 1) + 1 = 0; erf(x) = P ( 21 , x2 ).
(b) Change the integration
variable u in the standard representation of the RHS
√
to s, given by u = 12 π(1 − i)s, and note that (1 − i)2 = −2i. A = (1 + i)/2. From
part (a), C(x) + iS(x) = 12 (1 + i)P ( 12 , − 12 πi x2 ).
647
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