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. We start with $f(x) = \frac{(1+2x)^2}{1-x^2} = (1+2x)^2 \cdot \frac{1}{1-x^2}$. 2. Expand numerator: $(1+2x)^2 = 1 + 4x + 4x^2$. 3. Recall geometric series expansion for $|x|<1$: $$\frac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n} = 1 + x^2 + x^4 + \cdots$$ 4. Multiply expansions up to $x^3$ terms: $$ (1 + 4x + 4x^2)(1 + x^2) = 1 + 4x + 4x^2 + x^2 + 4x^3 + \text{higher terms} $$ Simplify: $$ 1 + 4x + 5x^2 + 4x^3 $$ 5. So, first 4 terms are: $$ 1 + 4x + 5x^2 + 4x^3 $$ 6. The expansion is valid for $|x| < 1$ because the geometric series $\frac{1}{1-x^2}$ converges when $|x^2|<1$. --- ### Problem 2 1. Given $f(x) = (1+2x)^{-1/3}$, use binomial expansion for $(1+u)^n$ where $n = -\frac{1}{3}$ and $u=2x$. 2. Binomial expansion formula for any real $n$: $$ (1+u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \cdots $$ 3. Calculate coefficients: - $n = -\frac{1}{3}$ - $n(n-1) = -\frac{1}{3} \times \left(-\frac{4}{3}\right) = \frac{4}{9}$ - $n(n-1)(n-2) = -\frac{1}{3} \times \left(-\frac{4}{3}\right) \times \left(-\frac{7}{3}\right) = -\frac{28}{27}$ 4. Substitute and expand: $$ f(x) = 1 - \frac{1}{3} (2x) + \frac{4}{9} \frac{(2x)^2}{2} - \frac{28}{27} \frac{(2x)^3}{6} $$ Simplify each term: - $-\frac{1}{3} \times 2x = -\frac{2}{3} x$ - $\frac{4}{9} \times \frac{4x^2}{2} = \frac{4}{9} \times 2x^2 = \frac{8}{9} x^2$ - $-\frac{28}{27} \times \frac{8x^3}{6} = -\frac{28}{27} \times \frac{4}{3} x^3 = -\frac{112}{81} x^3$ 5. Final expansion up to $x^3$: $$ 1 - \frac{2}{3} x + \frac{8}{9} x^2 - \frac{112}{81} x^3 $$ --- ### Problem 3 1. Approximate $(0.98)^{10}$ using binomial expansion. 2. Write $0.98 = 1 - 0.02$, so: $$ (0.98)^{10} = (1 - 0.02)^{10} $$ 3. Use binomial expansion: $$ (1 - x)^n = \sum_{k=0}^n \binom{n}{k} (-x)^k $$ with $n=10$, $x=0.02$. 4. Expand up to $x^4$: $$ \sum_{k=0}^4 \binom{10}{k} (-0.02)^k $$ Calculate terms: - $k=0$: $1$ - $k=1$: $10 \times (-0.02) = -0.2$ - $k=2$: $\binom{10}{2} ( -0.02)^2 = 45 \times 0.0004 = 0.018$ - $k=3$: $\binom{10}{3} (-0.02)^3 = 120 \times (-0.000008) = -0.00096$ - $k=4$: $\binom{10}{4} ( -0.02)^4 = 210 \times 0.00000016 = 0.0000336$ 5. Sum terms: $$ 1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8170736 $$ 6. 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 $(1-x)^n$ with $n = -\frac{1}{2}$: $$ (1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2!} (-x)^2 + \frac{n(n-1)(n-2)}{3!} (-x)^3 + \frac{n(n-1)(n-2)(n-3)}{4!} (-x)^4 + \cdots $$ 3. Calculate coefficients: - $n = -\frac{1}{2}$ - $n(n-1) = -\frac{1}{2} \times \left(-\frac{3}{2}\right) = \frac{3}{4}$ - $n(n-1)(n-2) = \frac{3}{4} \times \left(-\frac{5}{2}\right) = -\frac{15}{8}$ - $n(n-1)(n-2)(n-3) = -\frac{15}{8} \times \left(-\frac{7}{2}\right) = \frac{105}{16}$ 4. Substitute and simplify terms: $$ f(x) = 15 \left[ 1 + \left(-\frac{1}{2}\right)(-x) + \frac{3}{4} \frac{x^2}{2} + \left(-\frac{15}{8}\right) \frac{(-x)^3}{6} + \frac{105}{16} \frac{x^4}{24} \right] $$ Calculate each term: - $1$ - $+ \frac{1}{2} x$ - $+ \frac{3}{8} x^2$ - $+ \frac{15}{48} x^3 = \frac{5}{16} x^3$ - $+ \frac{105}{384} x^4 = \frac{35}{128} x^4$ 5. Multiply all terms by 15: $$ 15 + \frac{15}{2} x + \frac{45}{8} x^2 + \frac{75}{16} x^3 + \frac{525}{128} x^4 $$ 6. Final first 5 terms: $$ 15 + 7.5 x + 5.625 x^2 + 4.6875 x^3 + 4.1015625 x^4 $$