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2007
P2 Q3
Topical
Maclaurin
07 P2 Q3
2007 H2 Mathematics Paper 2 Question 3
Maclaurin Series
Answers
(i)
1
+
n
x
+
n
(
n
−
1
)
2
x
2
+
n
(
n
−
1
)
(
n
−
2
)
6
x
3
+
…
1+nx+\frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3 + \ldots
1
+
n
x
+
2
n
(
n
−
1
)
x
2
+
6
n
(
n
−
1
)
(
n
−
2
)
x
3
+
…
(ii)
8
−
3
x
+
387
16
x
2
−
1151
128
x
3
+
…
8 - 3 x + \frac{387}{16} x^2 - \frac{1151}{128} x^3 + \ldots
8
−
3
x
+
16
387
x
2
−
128
1151
x
3
+
…
(iii)
{
x
∈
R
:
−
1
2
<
x
<
1
2
}
{\{ x\in\mathbb{R}: -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \}}
{
x
∈
R
:
−
2
1
<
x
<
2
1
}
Full solutions
(i)
Let
f
(
x
)
=
(
1
+
x
)
n
f
′
(
x
)
=
n
(
1
+
x
)
n
−
1
f
′
′
(
x
)
=
n
(
n
−
1
)
(
1
+
x
)
n
−
2
f
′
′
′
(
x
)
=
n
(
n
−
1
)
(
n
−
2
)
(
1
+
x
)
n
−
3
\begin{align*} \textrm{Let } f(x) &= (1+x)^n \\ f'(x) &= n(1+x)^{n-1} \\ f''(x) &= n(n-1)(1+x)^{n-2} \\ f'''(x) &= n(n-1)(n-2)(1+x)^{n-3} \\ \end{align*}
Let
f
(
x
)
f
′
(
x
)
f
′′
(
x
)
f
′′′
(
x
)
=
(
1
+
x
)
n
=
n
(
1
+
x
)
n
−
1
=
n
(
n
−
1
)
(
1
+
x
)
n
−
2
=
n
(
n
−
1
)
(
n
−
2
)
(
1
+
x
)
n
−
3
f
(
0
)
=
1
f
′
(
0
)
=
n
f
′
′
(
0
)
=
n
(
n
−
1
)
f
′
′
′
(
0
)
=
n
(
n
−
1
)
(
n
−
2
)
\begin{align*} f(0) &= 1 \\ f'(0) &= n \\ f''(0) &= n(n-1) \\ f'''(0) &= n(n-1)(n-2) \\ \end{align*}
f
(
0
)
f
′
(
0
)
f
′′
(
0
)
f
′′′
(
0
)
=
1
=
n
=
n
(
n
−
1
)
=
n
(
n
−
1
)
(
n
−
2
)
Hence the Maclaurin's series for
(
1
+
x
)
n
:
{(1+x)^n:}
(
1
+
x
)
n
:
(
1
+
x
)
n
=
1
+
n
x
+
n
(
n
−
1
)
2
x
2
+
n
(
n
−
1
)
(
n
−
2
)
6
x
3
+
…
■
\begin{align*} & (1+x)^n \\ & = 1+nx+\frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3 + \ldots \; \blacksquare \end{align*}
(
1
+
x
)
n
=
1
+
n
x
+
2
n
(
n
−
1
)
x
2
+
6
n
(
n
−
1
)
(
n
−
2
)
x
3
+
…
■
(ii)
(
4
−
x
)
3
2
(
1
+
2
x
2
)
3
2
=
4
3
2
(
1
−
x
4
)
3
2
(
1
+
3
2
(
2
x
2
)
+
…
)
=
8
(
1
−
3
x
8
+
3
2
(
1
2
)
2
x
2
=
8
(
1
+
3
2
(
3
2
−
1
)
(
3
2
−
2
)
6
x
3
+
…
)
(
1
+
3
x
2
+
…
)
=
8
(
1
−
3
8
x
+
3
128
x
2
+
1
1024
x
3
)
(
1
+
3
x
2
)
+
…
=
8
(
1
−
3
8
x
+
(
3
+
3
128
)
x
2
+
(
1
1024
−
9
8
)
x
3
)
+
…
=
8
−
3
x
+
387
16
x
2
−
1151
128
x
3
+
…
■
\begin{align*} & (4-x)^\frac{3}{2}(1+2x^2)^\frac{3}{2} \\ & = 4^{\frac{3}{2}}\left(1-\frac{x}{4}\right)^{\frac{3}{2}}\left( 1 + \frac{3}{2}(2x^2) + \ldots \right) \\ & = 8 \Bigg( 1- \frac{3x}{8} + \frac{\frac{3}{2}\left(\frac{1}{2}\right)}{2} x^2 \\ & \phantom{= 8 \Bigg( 1} + \frac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{6} x^3 + \ldots \Bigg) \left( 1 + 3 x^2 + \ldots \right) \\ & = 8 \left( 1 - \frac{3}{8} x + \frac{3}{128} x^2 + \frac{1}{1024} x^3 \right) \left(1 + 3 x^2 \right) + \ldots \\ &= 8 \Bigg( 1 -{\textstyle \frac{3}{8}}x + \left( {\textstyle 3 + \frac{3}{128}} \right)x^2 + \left( {\textstyle \frac{1}{1024} - \frac{9}{8}} \right)x^3 \Bigg) + \ldots \\ &= 8 - 3 x + \frac{387}{16} x^2 - \frac{1151}{128} x^3 + \ldots \; \blacksquare \end{align*}
(
4
−
x
)
2
3
(
1
+
2
x
2
)
2
3
=
4
2
3
(
1
−
4
x
)
2
3
(
1
+
2
3
(
2
x
2
)
+
…
)
=
8
(
1
−
8
3
x
+
2
2
3
(
2
1
)
x
2
=
8
(
1
+
6
2
3
(
2
3
−
1
)
(
2
3
−
2
)
x
3
+
…
)
(
1
+
3
x
2
+
…
)
=
8
(
1
−
8
3
x
+
128
3
x
2
+
1024
1
x
3
)
(
1
+
3
x
2
)
+
…
=
8
(
1
−
8
3
x
+
(
3
+
128
3
)
x
2
+
(
1024
1
−
8
9
)
x
3
)
+
…
=
8
−
3
x
+
16
387
x
2
−
128
1151
x
3
+
…
■
(iii)
The expansion for
(
4
−
x
)
3
2
=
4
3
2
(
1
−
x
4
)
3
2
{(4-x)^{\frac{3}{2}}=4^{\frac{3}{2}}\left(1-\frac{x}{4}\right)^{\frac{3}{2}}}
(
4
−
x
)
2
3
=
4
2
3
(
1
−
4
x
)
2
3
is valid for:
∣
−
x
4
∣
<
1
−
4
<
x
<
4
\begin{gather*} \left| -\frac{x}{4} \right| < 1 \\ -4 < x < 4 \end{gather*}
−
4
x
<
1
−
4
<
x
<
4
The expansion for
(
1
+
2
x
2
)
3
2
{(1+2x^2)^{\frac{3}{2}}}
(
1
+
2
x
2
)
2
3
is valid for:
∣
2
x
2
∣
<
1
−
1
2
<
x
<
1
2
\begin{gather*} \left| 2x^2 \right| < 1 \\ -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \end{gather*}
2
x
2
<
1
−
2
1
<
x
<
2
1
Hence the set of values of
x
{x}
x
for which the expansion in part (ii) is valid:
{
x
∈
R
:
−
1
2
<
x
<
1
2
}
■
\{ x\in\mathbb{R}: -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \} \; \blacksquare
{
x
∈
R
:
−
2
1
<
x
<
2
1
}
■
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