Hints and answers

23.9 HINTS AND ANSWERS
23.9 Hints and answers
23.1
23.3
23.5
23.7
23.9
23.11
23.13
23.15
Define y(−x) = y(x) and use the cosine Fourier transform inversion theorem;
y(x) = (2/π)1/2 exp(−x2 /2).
f (x) − f(x) = exp x; α = 3/4, β = 1/2, γ = 1/4.
(a) φ(x) = f(x) − (1 + 2n)Fn xn − (1 − 2n)F−n x−n . (b) There are no solutions for
λ = [1 ± (1 − 4n2 )−1/2 ]−1 unless F±n = 0 or Fn /F−n = ∓[(1 − 2n)/(1 + 2n)]1/2 .
b
√
√
(i)
π) sin nx; M is diago(a) a(i)
n = a hn (x)ψ (x) dx; (b) use (1/ π) cos nx and (1/ √
(k)
nal; eigenvalues λk = k/π with eigenfunctions ψ (x) = (1/ π) cos kx.
2
df̃/dω = −ω f̃, leading to f̃(ω) = Ae−ω /2 . Rearrange the integral as a convolution
2
−t2 /6
and deduce that h̃(ω) = Be−3ω /2 ; h(t)
, where resubstitution and
= Ce
Gaussian normalisation show that C = 2/(3π).
p = k0 H/(1 − 2πk0 ), q = k1 Hc /(1 − πk1 ) and r = k1 Hs /(1 − πk1 ),
2π
2π
2π
where H = 0 h(z) dz, Hc = 0 h(z) cos z dz, and Hs = 0 h(z) sin z dz. Positive
−1
values of k1 (≈ π ) are most likely to cause a conference breakdown.
a
For eigenvalue 0 : f(x) = 0 for |x| < a or f(x) is such that −a f(y)dy = 0.
For eigenvalue 2a : f(x) = µS(x, a) with µ a constant and S(x, a) ≡ [H(a + x) −
H(x − a)], where H(z) is the Heaviside step
a function.
Take f(x) = g(x) + cGS(x, a), where G = −a g(z) dz. Show that c = λ/(1 − 2aλ).
y(x) = x2 − (3I3 x + I2 ) exp x.
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