MACLAURIN’S SERIES 69A
fiv(x)=
− 6
(1+x)^4fiv(0)=− 6
(1+0)^4=− 6fv(x)=24
(1+x)^5fv(0)=24
(1+0)^5= 24Substituting these values into equation (5) gives:
f(x)=ln (1+x)= 0 +x(1)+x^2
2!(−1)+x^3
3!(2)+x^4
4!(−6)+x^5
5!(24)i.e.ln( 1 +x)=x−
x^2
2+x^3
3−x^4
4+x^5
5−···Problem 5. Expand ln (1−x) to five terms.Replacingxby−xin the series for ln (1+x)in
Problem 4 gives:
ln (1−x)=(−x)−
(−x)^2
2+(−x)^3
3−(−x)^4
4+(−x)^5
5−···i.e.ln(1−x)=−x−
x^2
2−x^3
3−x^4
4−x^5
5−···Problem 6. Determine the power series forln(
1 +x
1 −x)
.ln
(
1 +x
1 −x)
=ln (1+x)−ln (1−x) by the laws oflogarithms, and from Problems 4 and 5,
ln
(
1 +x
1 −x)
=(x−x^2
2+x^3
3−x^4
4+x^5
5−···)−(−x−x^2
2−x^3
3−x^4
4−x^5
5−···)= 2 x+2
3x^3 +2
5x^5 +···i.e.ln
(
1 +x
1 −x)
= 2(
x+x^3
3+x^5
5+ ···)Problem 7. Use Maclaurin’s series to find the
expansion of (2+x)^4.f(x)=(2+x)^4 f(0)= 24 = 16f′(x)=4(2+x)^3 f′(0)=4(2)^3 = 32f′′(x)=12(2+x)^2 f′′(0)=12(2)^2 = 48f′′′(x)=24(2+x)^1 f′′′(0)=24(2)= 48fiv(x)= 24 fiv(0)= 24Substituting in equation (5) gives:( 2 +x)^4=f(0)+xf′(0)+x^2
2!f′′(0)+x^3
3!f′′′(0)+x^4
4!fiv(0)= 16 +(x)(32)+x^2
2!(48)+x^3
3!(48)+x^4
4!(24)= 16 + 32 x+ 24 x^2 + 8 x^3 +x^4(This expression could have been obtained by apply-
ing the binomial theorem.)Problem 8. Expand ex(^2) as far as the term inx^4.
f(x)=e
x
(^2) f(0)=e^0 = 1
f′(x)=
1
2
e
x
(^2) f′(0)=
1
2
e^0 =
1
2
f′′(x)=
1
4
e
x
(^2) f′′(0)=
1
4
e^0 =
1
4
f′′′(x)=
1
8
e
x
(^2) f′′′(0)=
1
8
e^0 =
1
8
fiv(x)=
1
16
e
x
(^2) fiv(0)=
1
16
e^0 =
1
16
Substituting in equation (5) gives:
e
x
(^2) =f(0)+xf′(0)+x
2
2!
f′′(0)
- x^3
3!
f′′′(0)+
x^4
4!
fiv(0)+···
= 1 +(x)
(
1
2
)
x^2
2!
(
1
4
)
x^3
3!
(
1
8
)
x^4
4!
(
1
16
)
+···
i.e. e
x
(^2) = 1 +
1
2
x+
1
8
x^2 +
1
48
x^3 +
1
384
x^4 +···