Calculus: Analytic Geometry and Calculus, with Vectors

498 Exponential and logarithmic functions

With or without more attention to details, jump to the conclusion that, for each

n=1, 2, 3, ... ,

n xk xn+1 Jn+1

(6) O S e= - I -k!< e"(n + l)!<eA (n + 1)!

k=O

While this may not be a suitable time to worry about the details, we can be sure

that if 4 = 400, then the quantity

,4n+1 14 .4 14 A A .4 A .4

(n+1)!- 1 2 3 4 5 n-1 nn+l

will be large when n is about J. The quantity will nevertheless be near 0 when

n is sufficiently great. Therefore,

n

(7) ex = lim I = i + x +

xz

+

x3

+

x4+ ....

n-iw k-0 2! 3!^41

While (7) is more spectacular than (6), it is not always as useful. Show that

putting x = B = 1 in (6) gives the formula

(8)^11

(n+ 151*^3

Verify that

1=1.

1 = 1.

1/2! = 0.5

1/3! = 0.16666 66666

1/4! = 0.04166 66666

1/5! = 0.00833 33333

1/6! = 0.00138 88888

and that the next term is obtained by dividing by 7. Continue the work to

obtain a decimal approximation to e that is correct to 6D (5 decimal places after

the decimal point).

9 Supposing that x < 0, observe that

(1) OSel51.

Replace x by i and integrate over the interval from x to 0 to obtain

(2) 051-e=5-x.

Replace x by i and integrate over the interval from x to 0 to obtain

?z

(3) 05e 1 -x5

Repeat the process to obtain

(4)

(5)

051+x+2a=<3

xz xa x4

O T!,