Maclaurin’s series 71
Substituting 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 7. Expand ln( 1 −x)to five terms.Replacing x by −x in the series for ln( 1 +x)in
Problem 6 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 8. Determine the power series for
ln(
1 +x
1 −x)
.ln
(
1 +x
1 −x)
=ln( 1 +x)−ln( 1 −x)by the laws of log-arithms, and from Problems 6 and 7,
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 9. Use Maclaurin’s series to find the
expansion of( 2 +x)^4.f(x)=( 2 +x)^4 f( 0 )= 24 = 16
f′(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 applying
the binomial theorem.)Problem 10. Expand ex 2
as far as the term inx^4.f(x)=ex(^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 +···