1. Problem: For $f(x) = \frac{(1+2x)^2}{(1-x)^2}$, find (i) the first 4 terms in its power series expansion, and (ii) state when the expansion is valid. (i) First, rewrite the function as: $$f(x) = (1+2x)^2 \cdot (1-x)^{-2}$$ Expand each term separately: - Expand $(1+2x)^2$ using the binomial formula: $$ (1+2x)^2 = 1 + 2\cdot2x + (2x)^2 = 1 + 4x + 4x^2 $$ - Expand $(1-x)^{-2}$ using the generalized binomial series for $(1 - x)^{-n}$: $$ (1-x)^{-2} = \sum_{k=0}^{\infty} \binom{2 + k -1}{k} x^k = 1 + 2x + 3x^2 + 4x^3 + ... $$ Here the first four terms are $1, 2x, 3x^2, 4x^3$. Now multiply the two series and keep terms up to $x^3$: $$ (1 + 4x + 4x^2)(1 + 2x + 3x^2 + 4x^3) $$ Calculating term by term: - Constant term: $1 \times 1 = 1$ - $x$-term: $1 \times 2x + 4x \times 1 = 2x + 4x = 6x$ - $x^2$-term: $1 \times 3x^2 + 4x \times 2x + 4x^2 \times 1 = 3x^2 + 8x^2 + 4x^2 = 15x^2$ - $x^3$-term: $1 \times 4x^3 + 4x \times 3x^2 + 4x^2 \times 2x = 4x^3 + 12x^3 + 8x^3 = 24x^3$ So the first four terms are: $$ f(x) = 1 + 6x + 15x^2 + 24x^3 + ... $$ (ii) The series expansion for $(1 - x)^{-2}$ converges for $|x| < 1$, and $(1+2x)^2$ is a polynomial, so no convergence restriction there. Therefore, the power series expansion of $f(x)$ is valid for: $$ |x| < 1 $$ --- 2. Problem: Given $f(x) = (1+2x)^{-1/3}$, find the binomial expansion up to $x^3$. Use the binomial expansion formula for any real exponent $a$: $$ (1 + u)^a = 1 + au + \frac{a(a-1)}{2!}u^2 + \frac{a(a-1)(a-2)}{3!}u^3 + ... $$ Here, $a = -\frac{1}{3}$ and $u = 2x$. Calculate each term: - First term: $1$ - Second term: $$ a u = - \frac{1}{3} \times 2x = - \frac{2}{3}x $$ - Third term: $$ \frac{a(a-1)}{2} u^2 = \frac{-\frac{1}{3}(-\frac{4}{3})}{2} (2x)^2 = \frac{\frac{4}{9}}{2} \times 4x^2 = \frac{4}{9} \times 2 x^2 = \frac{8}{9} x^2 $$ Explanation: $a(a-1) = -\frac{1}{3} \times (-\frac{4}{3}) = \frac{4}{9}$ - Fourth term: $$ \frac{a(a-1)(a-2)}{6} u^3 = \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} (2x)^3 $$ Calculate numerator: $$ -\frac{1}{3} \times -\frac{4}{3} = \frac{4}{9} $$ $$ \frac{4}{9} \times -\frac{7}{3} = -\frac{28}{27} $$ Divide by 6: $$ \frac{-28/27}{6} = -\frac{28}{162} = -\frac{14}{81} $$ Now multiply by $(2x)^3 = 8x^3$: $$ -\frac{14}{81} \times 8x^3 = -\frac{112}{81} x^3 $$ So the expansion up to $x^3$ is: $$ f(x) = 1 - \frac{2}{3}x + \frac{8}{9}x^2 - \frac{112}{81}x^3 + ... $$ --- 3. Problem: Approximate $(0.98)^{10}$ using binomial expansion up to $x^4$, to 4 decimal places. Rewrite with $x$: $$ (1 - 0.02)^{10} $$ Use the binomial theorem: $$ (1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k $$ Here $n=10$, $x = -0.02$. Calculate terms up to $x^4$: - $k=0$: $\binom{10}{0} (-0.02)^0 = 1$ - $k=1$: $\binom{10}{1} (-0.02)^1 = 10 \times (-0.02) = -0.2$ - $k=2$: $\binom{10}{2} (-0.02)^2 = 45 \times 0.0004 = 0.018$ - $k=3$: $\binom{10}{3} (-0.02)^3 = 120 \times (-0.000008) = -0.00096$ - $k=4$: $\binom{10}{4} (-0.02)^4 = 210 \times 0.00000016 = 0.0000336$ Sum these: $$ 1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8179736 $$ Rounded to 4 decimal places: $$ 0.8180 $$ --- 4. Problem: For $f(x) = \frac{15}{\sqrt{1-x}}$, find the first 5 terms of the expansion. Rewrite $f(x)$ as: $$ f(x) = 15(1-x)^{-1/2} $$ Use binomial expansion for $(1 - x)^a$ with $a = -\frac{1}{2}$: $$ (1 - x)^a = \sum_{k=0}^\infty \binom{a}{k} (-x)^k $$ Calculate terms: - $k=0$: $\binom{-1/2}{0}(-x)^0 = 1$ - $k=1$: $\binom{-1/2}{1}(-x)^1 = a(-x) = -\frac{1}{2} (-x) = \frac{1}{2}x$ - $k=2$: $$ \binom{-1/2}{2} = \frac{a(a-1)}{2} = \frac{-\frac{1}{2}(-\frac{3}{2})}{2} = \frac{\frac{3}{4}}{2} = \frac{3}{8} $$ Term: $$ \binom{-1/2}{2} (-x)^2 = \frac{3}{8} x^2 $$ - $k=3$: $$ \binom{-1/2}{3} = \frac{a(a-1)(a-2)}{6} = \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{6} $$ Calculate numerator: $$ -\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4} $$ $$ \frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8} $$ Divide by 6: $$ -\frac{15}{8} \times \frac{1}{6} = -\frac{15}{48} = -\frac{5}{16} $$ Term: $$ - \frac{5}{16} x^3 $$ - $k=4$: $$ \binom{-1/2}{4} = \frac{a(a-1)(a-2)(a-3)}{24} $$ Calculate numerator: $$ -\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4} $$ $$ \frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8} $$ $$ -\frac{15}{8} \times -\frac{7}{2} = \frac{105}{16} $$ Divide by 24: $$ \frac{105}{16} \times \frac{1}{24} = \frac{105}{384} $$ Term: $$ \frac{105}{384} x^4 $$ Now multiply all terms by 15: $$ f(x) = 15 \left(1 + \frac{1}{2} x + \frac{3}{8} x^2 - \frac{5}{16} x^3 + \frac{105}{384} x^4 + ... \right) $$ Simplify: $$ = 15 + \frac{15}{2} x + \frac{45}{8} x^2 - \frac{75}{16} x^3 + \frac{1575}{384} x^4 + ... $$ --- Final answers: 1. $f(x) = 1 + 6x + 15x^2 + 24x^3 + ...$, valid for $|x| < 1$ 2. $f(x) = 1 - \frac{2}{3}x + \frac{8}{9}x^2 - \frac{112}{81}x^3 + ...$ 3. Approximate $(0.98)^{10} \approx 0.8180$ 4. $f(x) = 15 + \frac{15}{2}x + \frac{45}{8}x^2 - \frac{75}{16}x^3 + \frac{1575}{384}x^4 + ...$