Download (direct link):
5. Use the ratio test:
| (2x+1) n+1/(n+1) 2| n2 ³ żż ³
lim ---------------------- = lim --------- | 2x+1 | = 12x+1 | .
n ^ ro | (2x+1) n/n2| n ^ ro (n+1) 2
Therefore the series converges absolutely for |2x+1| < 1, or |x+1/2| < 1/2. At x = 0 and x = -1 the series also converge absolutely. However, for |x+1/2| > 1/2 the series diverges by the ratio test. The radius of convergence is p = 1/2.
9. For this problem f(x) = sinx. Hence f'(x) = cosx, f"(x) = -sinx, f'"(x) = -cosx,... . Then f(0) = 0,
f'(0) = 1, f"(0) = 0, f'"(0) = -1,... . The even terms in the series will vanish and the odd terms will alternate
in sign. We obtain sinx = (-1)n x2n+1/(2n+1)!. From
the ratio test it follows that p = ro.
12. For this problem f(x) = x . Hence f (x) = 2x, f (x) = 2,
and f(n)(x) = 0 for n > 2. Then f(-1) = 1, f'(-1) = -2,
22 f (-1) = 2 and x = 1 - 2(x+1) + 2(x+1) /2! = 1 - 2(x+1) +
(x+1) . Since the series terminates after a finite number of terms, it converges for all x. Thus p = ro.
13. For this problem f(x) = lnx. Hence f'(x) = 1/x,
f'(x) = -1/x2, f' (x) = 1-2/x3,... , and f(n)(x) =
(-1) n+1(n-1)!/xn. Then f(1) = 0, f'(1) = 1, f''(1) = -1,
f"'(1) = 1-2,... , f(n)(1) = (-1) n+1(n-1)! The Taylor
series is lnx = (x-1) - (x-1) /2 + (x-1) /3 - ... =
(-1) n+1(x-1) n/n. It follows from the ratio test that
the series converges absolutely for |x-l| < 1. However, the series diverges at x = 0 so p = 1.
Writing the individual terms of y, we have
y = ao + axx + a2x +...+aax + ..., so
y' = a! + 2a2x + 3a3x2 + ... + (n+1)an+1xn + ..., and
y" = 2a2 + 3'2a3x + 4-3a4x2 + ... + (n+2)(n+1)an+2xn + ... .
If y" = y, we then equate coefficients of like powers of x to obtain 2a2 = a0, 3-2a3 = a1, 4-3a4 = a2, ... (n+2)(n+1)an+2 =
an, which yields the desired result for n = 0,1,2,3... .
Set m = n-1 on the right hand side of the equation. Then n = m+1 and when n = 1, m = 0. Thus the right hand side
becomes am(x-1) m+1, which is the same as the left hand
side when m is replaced by n.
Multiplying each term of the first series by x yields *x nanxn 1 = nanxn = nanxn, where the last
n=1 n=1 n=0
equality can be verified by writing out the first few terms. Changing the index from k to n (n=k) in the second series then yields
X nanxn + Xanxn = X (n+1)anxR.
nn n=0 n=0 n=0
m(m-1)amxm 2 + x^^ kakxk 1 =
X (n+2)(n+1)an+2xn + X kakxk =
X, [(n+2)(n+1)an+2 + nan]xn. In the first case we have
let n = m - 2 in the first summation and multiplied each term of the second summation by x. In the second case we have let n = k and noted that for n = 0, nan = 0.
28. If we shift the index of summation in the first sum by
letting m = n-1, we have
nanxn-1 = (m+1)am+1xm. Substituting this into the
given equation and letting m = n again, we obtain:
1 (n+1)an+1 xn + 2 1 anxn = 0, or
1 [(n+1)an+1 + 2an]xn = 0.
Hence an+]_ = -2an/(n+1) for n = 0,1,2,3,... . Thus
a1 = -2a0, a2 = -2a1/2 = 2 a0/2, a3 = -2a2/3 = -2 a0/2-3 =
-23a0/3!... and an = (-1) n2na0/n!. Notice that for n = 0 this formula reduces to a0 so we can write
1 anxn = 1 (-1) n2n a0xn/n! = a01 (-2x) n/n! = a0e-2x
n=0 n=0 n=0
Section 5.2, Page 247
2. y = 1 anxn; y' = I nanxn 1 and since we must multiply
y' by x in the D.E. we do not shift the index; and
y" = n(n-1)anxn-2 = (n+2)(n+1)an+2xn. Substituting
in the D.E., we obtain
1 (n+2)(n+1)an+2xn - nanxn - anxn = 0. In order to
n=0 n=1 n=0
have the starting point the same in all three summations, we let n = 0 in the first and third terms to obtain the following
(2-1 a2 - a0)x0 + 1 [(n+2)(n+1)an+2 - (n+1)an]xn = 0.
Thus an+2 = an/(n+2) for n = 1,2,3,... . Note that the recurrence relation is also correct for n = 0. We show how to calculate the odd a's: