552 Real Analysis
=
∑∞
k= 0(− 1 )k1 · 3 ···( 2 k− 1 )
2 k·k!
xk=∑∞
k= 0(− 1 )k( 2 k)!
22 k·k!·k!
xk=
∑∞
k= 0(− 1 )k1
22 k(
2 k
k)
xk.Replacingxby−x^2 then taking antiderivatives, we obtain
arcsinx=∫x0( 1 −t^2 )−(^12)
dt=
∑∞
k= 01
22 k(
2 k
k)∫x0t^2 kdt=
∑∞
k= 01
22 k( 2 k+ 1 )(
2 k
k)
x^2 k+^1 ,as desired.
492.(a) Differentiating the identity from the second example from the introduction, we
obtain
2 arcsinx
√
1 −x^2=
∑
k≥ 11
k( 2 k
k) 22 kx^2 k−^1 ,whence
xarcsinx
√
1 −x^2=
∑
k≥ 11
k( 2 k
k) 22 k−^1 x^2 k.Differentiating both sides and multiplying byx, we obtain
xarcsinx+x√
1 −x^2
( 1 −x^2 )^3 /^2=
∑
k≥ 01
( 2 k
k) 22 kx^2 k.Substitutingx 2 forx, we obtain the desired identity.
Part (b) follows from (a) if we letx=1.
(S. Radulescu, M. R ̆ adulescu, ̆ Teoreme ̧si Probleme de Analiza Matematic ̆ ̆a(Theorems
and Problems in Mathematical Analysis), Editura Didactica ̧ ̆si Pedagogic ̆a, Bucharest,
1982).
493.Consider the functionfof period 2πdefined byf(x)=xif 0≤x< 2 π. This
function is continuous on( 0 , 2 π), so its Fourier series converges (pointwise) on this
interval. We compute
a 0 =1
2 π∫ 2 π0xdx=π, am= 0 , form≥ 1 ,