1. Problem 1: Given $f(x) = \frac{(1 + 2x)^2}{(1 - 2x)^2}$
(i) Find the first 4 terms in the power series expansion.
- We can write $f(x)$ as $f(x) = (1 + 2x)^2 \cdot (1 - 2x)^{-2}$.
- Use the binomial series expansion formula for $(1 + u)^n = \sum_{k=0}^\infty \binom{n}{k} u^k$ where $\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$.
- Expand $(1 + 2x)^2 = 1 + 4x + 4x^2$ (exact, finite).
- Expand $(1 - 2x)^{-2}$ using binomial series with $n = -2$ and $u = -2x$:
$$ (1 - 2x)^{-2} = \sum_{k=0}^\infty \binom{-2}{k} (-2x)^k $$
- Calculate first 4 terms:
- $k=0$: $\binom{-2}{0}(-2x)^0 = 1$
- $k=1$: $\binom{-2}{1}(-2x)^1 = (-2)(-2x) = 4x$
- $k=2$: $\binom{-2}{2}(-2x)^2 = \frac{-2(-3)}{2} (4x^2) = 3 \cdot 4x^2 = 12x^2$
- $k=3$: $\binom{-2}{3}(-2x)^3 = \frac{-2(-3)(-4)}{6} (-8x^3) = -4 \cdot (-8x^3) = 32x^3$
- So, $(1 - 2x)^{-2} \approx 1 + 4x + 12x^2 + 32x^3$.
- Multiply $(1 + 4x + 4x^2)$ by $(1 + 4x + 12x^2 + 32x^3)$ and keep terms up to $x^3$:
$$
\begin{aligned}
f(x) &\approx (1)(1 + 4x + 12x^2 + 32x^3) + (4x)(1 + 4x + 12x^2) + (4x^2)(1 + 4x) \\
&= 1 + 4x + 12x^2 + 32x^3 + 4x + 16x^2 + 48x^3 + 4x^2 + 16x^3 \\
&= 1 + (4x + 4x) + (12x^2 + 16x^2 + 4x^2) + (32x^3 + 48x^3 + 16x^3) \\
&= 1 + 8x + 32x^2 + 96x^3
\end{aligned}
$$
(ii) The expansion is valid when the binomial series converges, which requires $|2x| < 1$ or $|x| < \frac{1}{2}$.
2. Problem 2: Given $f(x) = (1 + 2x)^{-\frac{1}{3}}$
- Use binomial expansion for fractional powers:
$$ (1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \cdots $$
- Here, $n = -\frac{1}{3}$ and $u = 2x$.
- Calculate terms up to $x^3$:
$$
\begin{aligned}
f(x) &\approx 1 + \left(-\frac{1}{3}\right)(2x) + \frac{-\frac{1}{3}(-\frac{4}{3})}{2} (2x)^2 + \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} (2x)^3 \\
&= 1 - \frac{2}{3}x + \frac{4}{9}x^2 - \frac{56}{81}x^3
\end{aligned}
$$
3. Problem 3: Approximate $(0.98)^{10}$ using binomial expansion up to $x^4$.
- Write $0.98 = 1 - 0.02$, so $x = -0.02$.
- Use binomial expansion for $(1 + x)^{10}$:
$$
(1 + x)^{10} = \sum_{k=0}^4 \binom{10}{k} x^k + \cdots
$$
- Calculate terms:
$$
\begin{aligned}
&1 + 10(-0.02) + \frac{10 \cdot 9}{2} (-0.02)^2 + \frac{10 \cdot 9 \cdot 8}{6} (-0.02)^3 + \frac{10 \cdot 9 \cdot 8 \cdot 7}{24} (-0.02)^4 \\
&= 1 - 0.2 + 0.018 - 0.00192 + 0.0005376 \\
&= 0.8166 \text{ (rounded to 4 decimal places)}
\end{aligned}
$$
4. Problem 4: Given $f(x) = \frac{15}{\sqrt{1 - x}} = 15 (1 - x)^{-\frac{1}{2}}$
- Use binomial expansion for $(1 - x)^{-\frac{1}{2}}$:
$$
(1 - x)^{-\frac{1}{2}} = \sum_{k=0}^4 \binom{-\frac{1}{2}}{k} (-x)^k
$$
- Calculate first 5 terms:
$$
\begin{aligned}
&1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 + \frac{35}{128}x^4
\end{aligned}
$$
- Multiply by 15:
$$
f(x) \approx 15 + \frac{15}{2}x + \frac{45}{8}x^2 + \frac{75}{16}x^3 + \frac{525}{128}x^4
$$
Final answers:
(i) $f(x) \approx 1 + 8x + 32x^2 + 96x^3$, valid for $|x| < \frac{1}{2}$.
(ii) $f(x) \approx 1 - \frac{2}{3}x + \frac{4}{9}x^2 - \frac{56}{81}x^3$.
(iii) $(0.98)^{10} \approx 0.8166$.
(iv) $f(x) \approx 15 + \frac{15}{2}x + \frac{45}{8}x^2 + \frac{75}{16}x^3 + \frac{525}{128}x^4$.
Binomial Expansions A5Afc5
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