tuto8
TRANSCRIPT
INDIAN INSTITUTE OF TECHNOLOGY GANDHINAGAR
DISCIPLINE OF MATHEMATICS
MA 201: Autumn, 2014-15 Tutorial Sheet No.8—————————————————————————————————
1. Examine the following series for convergence.
(i)∞∑1
(n!)2
(2n)!; (ii)
∞∑1
n!
nn;
(iii)∞∑
n=10
1
lnn; (iv)
∞∑n=10
1
n lnn;
(v)∞∑
n=10
1
np lnn; (vi)
∞∑n=100
1
ln(lnn);
(vii)∞∑n=1
n1000
n!; (viii)
∞∑10
1
n(lnn)p.
2. For what values of x, the following series are convergent?
(i)∞∑1
xn
n!; (ii)
∞∑1
xn
n;
(iii)n∑1
(xn + x−n); (iv)∞∑1
1
xn + x−n;
(v)∞∑n=r
n(n− 1) . . . (n− r + 1)xn−r; (vi)∞∑n=1
xn+r
n(n+ 1)(n+ 2) · · · (n+ r).
3. If∞∑1
an is a convergent series of positive terms, prove that each of the following series
(i)∞∑1
a2n;
(ii)∞∑1
an1 + an
;
(iii)∞∑1
ann;
(iv)∞∑1
ln(1 + an)
is convergent.
4. If∑
nan is convergent, prove that∑
an is convergent.
5. Find the radius of convergence of the following power series:
(i)∑
xn (ii)∑ xn
n!; (iii)
∑n!xn (iv)
∞∑n=k
n(n− 1) · · · (n− k + 1)xn; (v)∑ (2n)!
22n(n!)2xn;
1
6. Consider the equation (1 + x2)y′′ + 2xy′ − 2y = 0.
(i) Find its general solution y =∑
anxn in the form y = a0y2(x) + a1y1(x), where y1(x) and
y2(x) are power series.
(ii) Find the radius of convergence for y1(x) and y2(x).
(iii) Show that y1(x) is x. Use this fact to find a second independent solution [same as y2(x)in (i)].
7. Find a power series solution in powers of x for the following differential equation(1 + x2)y′′ − 4xy′ + 6y = 0
8(a). Show that the fundamental system of solutions of Legendre equation
(1− x2)y′′ − 2xy′ + p(p+ 1)y = 0
consists of y1(x) =∑∞
n=0 a2nx2n and y2(x) =
∑∞n=0 a2n+1x
2n+1, where a0 = a1 = 1 and
a2n+2 = −(p− 2n)(p+ 2n+ 1)
(2n+ 1)(2n+ 2)a2n, n = 0, 1, 2, · · ·
a2n+1 = −(p− 2n+ 1)(p+ 2n)
2n(2n+ 1)a2n−1, n = 1, 2, 3, · · · .
(b). Verify that
y1(x) = P0(x) = 1, y2(x) =1
2ln
1 + x
1− xfor p = 0
y2(x) = P1(x) = x, y1(x) = 1− x
2ln
1− x
1 + xfor p = 1.
(c). The expression, Pn(x) =1
2nn!
dn
dxn[(x2 − 1)n], is called the Rodrigues’ formula for
Legendre polynomial Pn of degree n. Assuming this, find P1, P2, P3, P4.
9. Show that(i) Pn(−x) = (−1)nPn(x) (ii) P ′
n(−x) = (−1)n+1P ′n(x)
(iii)∫ 1
−1Pn(x)Pm(x) dx =
2
2n+ 1δmn (iv)
∫ 1
−1xmPn(x) dx = 0 if m < n
10. Suppose m > n. Show that∫ 1
−1xmPn(x) dx = 0 if m− n is odd. What happens if
m− n is even?
11. The function on the left side of1√
1− 2xt+ t2=
∞∑n=0
Pn(x)tn is called the generating function
of the Legendre polynomial Pn. Using this relation, show that
(i) (n+ 1)Pn+1(x)− (2n+ 1)xPn(x) + nPn−1(x) = 0 (ii) nPn(x) = xP ′n(x)− P ′
n−1(x)
(iii)P ′n+1(x)− xP ′
n(x) = (n+ 1)Pn(x) (iv)Pn(1) = 1, Pn(−1) = (−1)n
(v) P2n+1(0) = 0, P2n(0) = (−1)n1 · 3 · 5 · · · · · · (2n− 1)
2nn!.
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