Answers & Hints for Selected Exercises 695- cos x = cos[c + (x - c)] =cos ccos(x - c) - sin csin(x - c)
- oo (-l)k(x-c)2k. oo (-1t(x-c)2k+4 - oo (x-c)k
- cosc I: ( 2 k)! - smc I: ( 2 k+l)! - I: ak~
k=O k=O k=O
where ak = (-l)kf^2 cosc if k is even; ak = (-l)(k+l)/^2 sinc if k is odd.
( )
x3 xs x7 xg C
ll. a x + 3-TI + 5-2! + 7-3! + 9 -4! + · · · +(b) (^1) n I I x - x + 2-2! x2 - 3.31 x3 + 4-4! x4 - 5·5! xs + · · · + C
00
- By Exercise 6.6.16, the Maclaurin series off is I: 0 · xk.
k=O - f(x) = (1 + x^2 )-1, f'(x) = (1:;:;1),, /' (x) = (l^6 ~;;lp. In general, j(n)(x) is
a rational function whose denominator is an integral power of 1 + x^2 and so is
never 0. Thus, f is infinitely differentiable everywhere. But its Maclaurin series
00
is I: (-l)kx^2 k, which converges only in (-1, 1).
k=O
00 - Ri = -1, but for i 2: 2, Ri = 0. :. I: Ri = -1;
i=l
00
C1 = 1, but for j 2: 2, Cj = o .... L cj = 1.
J=l
n m - (a) I: I: aij is the sum of the entries in a rectangle in the upper left
i=l j=l
corner.
00
Suppose the sum by rows of I: laijl converges. Then, by Thm. 8.7.17,
i,j=l
00
all the row sums Ri = I: aij converge, and the sum by rows converges, say
i,j=l
i~ ~ = s. Let E: > 0. Then :3no E N 3 n 2: no =} Ii~ Ri - si < c/2,
and Vi EN, :3mi EN 3 m 2: mi=} If aij - ~1 < t::/2i+^1. Then m , n 2:
J=l
max{ no, m1, · · · , mn 0 } =} If, ( f aij) - si =If, ( f aij - Ri) + f, Ri - si
i=l J=l i=l J=l i=l:=::If, (f aij - ~) l+I f, ~ - si:::; f, If aij - Ril+~ < f, 2 ,~1 +~ = t::.
i=l J=l i=l i=l J=l i=l
n n n+l n
(b) For the double series of Ex. 17 , Vi EN, I: I:%= 0, I: I:%= 1,
i=lj=l i=lj=land i~~t: aij = -1. Thus,~ S,no 3 m,n 2: no=} li~j~l aij -SI< 2.