1. Problem 1: Given $f(x) = \frac{(1+2x)^2}{(1-x)^2}$ (i) Find the first 4 terms in the power series expansion. (ii) State when the expansion of $f(x)$ is valid. 2. Problem 2: Given $f(x) = \frac{1}{(1+2x)^{1/3}}$, find the binomial expansion up to $x^3$. 3. Problem 3: Given $(0.98)^{10}$, use binomial expansion to approximate to 4 decimal places, expanding up to $x^4$. 4. Problem 4: Given $f(x) = \frac{15}{\sqrt{1-x}}$, find the first 5 terms in the expansion. --- ### Problem 1 1. Write $f(x)$ as a product of two binomial expansions: $$f(x) = (1+2x)^2 (1-x)^{-2}$$ 2. Expand $(1+2x)^2$ using binomial theorem: $$(1+2x)^2 = 1 + 2 \times 2x + (2x)^2 = 1 + 4x + 4x^2$$ 3. Expand $(1-x)^{-2}$ using binomial series for negative powers: $$(1-x)^{-2} = \sum_{n=0}^\infty \binom{-2}{n} (-x)^n$$ Using formula for negative binomial coefficients: $$\binom{-2}{n} = (-1)^n \binom{n+1}{n} = (-1)^n (n+1)$$ So, $$(1-x)^{-2} = \sum_{n=0}^\infty (n+1) x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots$$ 4. Multiply the two expansions up to $x^3$ terms: $$(1 + 4x + 4x^2)(1 + 2x + 3x^2 + 4x^3)$$ Calculate terms: - Constant: $1 \times 1 = 1$ - $x$: $1 \times 2x + 4x \times 1 = 2x + 4x = 6x$ - $x^2$: $1 \times 3x^2 + 4x \times 2x + 4x^2 \times 1 = 3x^2 + 8x^2 + 4x^2 = 15x^2$ - $x^3$: $1 \times 4x^3 + 4x \times 3x^2 + 4x^2 \times 2x = 4x^3 + 12x^3 + 8x^3 = 24x^3$ 5. So the first 4 terms are: $$f(x) = 1 + 6x + 15x^2 + 24x^3 + \cdots$$ 6. The expansion is valid where both series converge. The binomial series for $(1-x)^{-2}$ converges for $|x| < 1$, and $(1+2x)^2$ is a polynomial (valid everywhere). So the radius of convergence is $|x| < 1$. --- ### Problem 2 1. Given $f(x) = (1+2x)^{-1/3}$, use binomial expansion for fractional powers: $$ (1 + u)^k = 1 + k u + \frac{k(k-1)}{2!} u^2 + \frac{k(k-1)(k-2)}{3!} u^3 + \cdots $$ where $k = -\frac{1}{3}$ and $u = 2x$. 2. Calculate coefficients: - First term: $1$ - Second term: $k u = -\frac{1}{3} \times 2x = -\frac{2}{3} x$ - Third term: $$\frac{k(k-1)}{2} u^2 = \frac{-\frac{1}{3}(-\frac{4}{3})}{2} (2x)^2 = \frac{\frac{4}{9}}{2} 4x^2 = \frac{4}{18} \times 4 x^2 = \frac{8}{9} x^2$$ - Fourth term: $$\frac{k(k-1)(k-2)}{6} u^3 = \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} (2x)^3$$ Calculate numerator: $$-\frac{1}{3} \times -\frac{4}{3} = \frac{4}{9}, \quad \frac{4}{9} \times -\frac{7}{3} = -\frac{28}{27}$$ Divide by 6: $$-\frac{28}{27} \times \frac{1}{6} = -\frac{28}{162} = -\frac{14}{81}$$ Multiply by $(2x)^3 = 8x^3$: $$-\frac{14}{81} \times 8 x^3 = -\frac{112}{81} x^3$$ 3. So the expansion up to $x^3$ is: $$f(x) = 1 - \frac{2}{3} x + \frac{8}{9} x^2 - \frac{112}{81} x^3 + \cdots$$ --- ### Problem 3 1. Write $(0.98)^{10}$ as $(1 - 0.02)^{10}$. 2. Use binomial expansion: $$(1 - x)^n = \sum_{k=0}^n \binom{n}{k} (-x)^k$$ with $n=10$, $x=0.02$. 3. Expand up to $x^4$: $$\begin{aligned} (1 - 0.02)^{10} &= 1 - 10 \times 0.02 + \binom{10}{2} (0.02)^2 - \binom{10}{3} (0.02)^3 + \binom{10}{4} (0.02)^4 \\ &= 1 - 0.2 + 45 \times 0.0004 - 120 \times 0.000008 + 210 \times 0.00000016 \end{aligned}$$ Calculate each term: - $45 \times 0.0004 = 0.018$ - $120 \times 0.000008 = 0.00096$ - $210 \times 0.00000016 = 0.0000336$ 4. Sum terms: $$1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8170736$$ 5. Rounded to 4 decimal places: $$0.8171$$ --- ### Problem 4 1. Given $f(x) = \frac{15}{\sqrt{1-x}} = 15 (1-x)^{-1/2}$. 2. Use binomial expansion for $(1-x)^k$ with $k = -\frac{1}{2}$: $$ (1-x)^k = \sum_{n=0}^\infty \binom{k}{n} (-x)^n $$ 3. Calculate first 5 terms ($n=0$ to $4$): - $n=0$: $1$ - $n=1$: $k (-x) = -\frac{1}{2} (-x) = \frac{1}{2} x$ - $n=2$: $\frac{k(k-1)}{2!} (-x)^2 = \frac{-\frac{1}{2}(-\frac{3}{2})}{2} x^2 = \frac{3}{8} x^2$ - $n=3$: $\frac{k(k-1)(k-2)}{3!} (-x)^3 = \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{6} (-x)^3$ Calculate numerator: $$-\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4}, \quad \frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8}$$ Divide by 6: $$-\frac{15}{8} \times \frac{1}{6} = -\frac{15}{48} = -\frac{5}{16}$$ Multiply by $(-x)^3 = -x^3$: $$-\frac{5}{16} \times (-x^3) = \frac{5}{16} x^3$$ - $n=4$: $\frac{k(k-1)(k-2)(k-3)}{4!} (-x)^4$ Calculate numerator: $$k(k-1)(k-2)(k-3) = -\frac{1}{2} \times -\frac{3}{2} \times -\frac{5}{2} \times -\frac{7}{2}$$ Calculate stepwise: $$-\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4}$$ $$\frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8}$$ $$-\frac{15}{8} \times -\frac{7}{2} = \frac{105}{16}$$ Divide by $4! = 24$: $$\frac{105}{16} \times \frac{1}{24} = \frac{105}{384} = \frac{35}{128}$$ Multiply by $(-x)^4 = x^4$: $$\frac{35}{128} x^4$$ 4. So the expansion is: $$f(x) = 15 \left(1 + \frac{1}{2} x + \frac{3}{8} x^2 + \frac{5}{16} x^3 + \frac{35}{128} x^4 + \cdots \right)$$ 5. Multiply through by 15: $$f(x) = 15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4 + \cdots$$ --- Final answers: - Problem 1: $f(x) = 1 + 6x + 15x^2 + 24x^3 + \cdots$, valid for $|x| < 1$. - Problem 2: $f(x) = 1 - \frac{2}{3} x + \frac{8}{9} x^2 - \frac{112}{81} x^3 + \cdots$. - Problem 3: Approximation of $(0.98)^{10} \approx 0.8171$. - Problem 4: $f(x) = 15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4 + \cdots$