30 NUMBER AND ALGEBRA
for ex. Thus
e^0.^05 = 1 + 0. 05 +(0.05)^2
2!+(0.05)^3
3!+(0.05)^4
4!+(0.05)^5
5!+···= 1 + 0. 05 + 0. 00125 + 0. 000020833+ 0. 000000260 + 0. 000000003and by adding,
e^0.^05 = 1 .0512711, correct to 8 significant figuresIn this example, successive terms in the series grow
smaller very rapidly and it is relatively easy to deter-
mine the value of e^0.^05 to a high degree of accuracy.
However, whenxis nearer to unity or larger than
unity, a very large number of terms are required for
an accurate result.
If in the series of equation (1),xis replaced by−x,
then,
e−x= 1 +(−x)+(−x)^2
2!+(−x)^3
3!+···i.e. e−x= 1 −x+
x^2
2!−x^3
3!+···In a similar manner the power series for exmay be
used to evaluate any exponential function of the form
aekx, whereaandkare constants. In the series of
equation (1), letxbe replaced bykx. Then,
aekx=a{
1 +(kx)+(kx)^2
2!+(kx)^3
3!+···}Thus 5 e^2 x= 5
{
1 +(2x)+(2x)^2
2!+(2x)^3
3!+···}= 5{
1 + 2 x+4 x^2
2+8 x^3
6+···}i.e. 5 e^2 x= 5
{
1 + 2 x+ 2 x^2 +4
3x^3 +···}Problem 14. Determine the value of 5 e^0.^5 , cor-
rect to 5 significant figures by using the power
series for ex.ex= 1 +x+x^2
2!+x^3
3!+x^4
4!+···Hence e^0.^5 = 1 + 0. 5 +(0.5)^2
(2)(1)+(0.5)^3
(3)(2)(1)+(0.5)^4
(4)(3)(2)(1)+(0.5)^5
(5)(4)(3)(2)(1)+(0.5)^6
(6)(5)(4)(3)(2)(1)
= 1 + 0. 5 + 0. 125 + 0. 020833+ 0. 0026042 + 0. 0002604+ 0. 0000217i.e. e^0.^5 = 1 .64872,
correct to 6 significant figuresHence 5e0.5=5(1.64872)=8.2436,
correct to 5 significant figuresProblem 15. Expand ex(x^2 −1) as far as the
term inx^5.The power series for exis,ex= 1 +x+x^2
2!+x^3
3!+x^4
4!+x^5
5!+···Hence ex(x^2 −1)=(1 +x+x^2
2!+x^3
3!+x^4
4!+x^5
5!+···)(x^2 −1)=(x^2 +x^3 +x^4
2!+x^5
3!+···)−(1 +x+x^2
2!+x^3
3!+x^4
4!+x^5
5!+···)Grouping like terms gives:ex(x^2 −1)=− 1 −x+(
x^2 −x^2
2!)
+(
x^3 −x^3
3!)+(
x^4
2!−x^4
4!)
+(
x^5
3!−x^5
5!)+···=− 1 −x+1
2x^2 +5
6x^3 +11
24x^4 +19
120x^5when expanded as far as the term inx^5.