tuto8

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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) X 1 (n!) 2 (2n)! ; (ii) X 1 n! n n ; (iii) X n=10 1 ln n ; (iv) X n=10 1 n ln n ; (v) X n=10 1 n p ln n ; (vi) X n=100 1 ln(ln n) ; (vii) X n=1 n 1000 n! ; (viii) X 10 1 n(ln n) p . 2. For what values of x, the following series are convergent? (i) X 1 x n n! ; (ii) X 1 x n n ; (iii) n X 1 (x n + x -n ); (iv) X 1 1 x n + x -n ; (v) X n=r n(n - 1) ... (n - r + 1)x n-r ; (vi) X n=1 x n+r n(n + 1)(n + 2) ··· (n + r) . 3. If X 1 a n is a convergent series of positive terms, prove that each of the following series (i) X 1 a 2 n ; (ii) X 1 a n 1+ a n ; (iii) X 1 a n n ; (iv) X 1 ln(1 + a n ) is convergent. 4. If X na n is convergent, prove that a n is convergent. 5. Find the radius of convergence of the following power series: (i) X x n (ii) X x n n! ; (iii) X n!x n (iv) X n=k n(n - 1) ··· (n - k + 1)x n ; (v) X (2n)! 2 2n (n!) 2 x n ; 1

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Page 1: Tuto8

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

Page 2: Tuto8

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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