Calculus: Analytic Geometry and Calculus, with Vectors

636

3 Write two more terms of each series appearing in the calculations

rr/2

K - 1 dB

J 0 1/1 - k2 sin2 B

+ (-x) .+.

1

2-)

=1+Zx+2-4x2+ ...

K foA/2[ 1+2k2sin26+2-4k4sin20+ .ld6

in which 0 < k < 1. Use of the formula

Series

Ix/2 1-3-5 (2p - 1)a

sin.p 0 dO = 2.4.6 (2p) 2 = ,2,3, ...)1,2,3,

then gives

111 *3' z 2-5 2

C1 +2

22

k2+22.42 k4+22.42.62k6+

.. .l.

4 Supposing that 0 < k < 1, let

E= r./21/1 -k2sin2BdO

Jo

and modify the work of the preceding problem to obtain

• 1 - 12.3 -12.32.5 6-

E=

(1

22

k2

22.42

k'

22.42.62k

It is not easy to know everything, and we may be unable to say whether our

formula for E is valid when k = 1. Show that if it is, then

1 12.3 12.32.5 12.32.52.7^2

22 + 22-4z + 22.42.62 22.42.62.82+ = 1 -

5 While this may not be an appropriate time to enter into details of proofs,

the function f for which f(0) = 0 and

f(x) = e u== (x 0)

has continuous derivatives of all orders over the whole infinite interval - m <

x < oo. Moreover, f«>(0) = 0 for each k = 1, 2, 3,. In this case the

Taylor formula (12.53) with a = 0 becomes

f(x) = 0+0+

The Taylor expansion of f in powers of x is therefore the series

0+0+0+ ...

which converges for each x but converges to f (x) only when x = 0.