1. **Problem 1:** Given $f(x) = \frac{(1+2x)^2}{(1-x)^2}$, find the first 4 terms in the power series expansion and state when the expansion is valid. 2. **Formula and rules:** Use binomial series expansions: - For $(1+u)^n = \sum_{k=0}^\infty \binom{n}{k} u^k$ valid for $|u|<1$. - Here, $(1+2x)^2$ is a polynomial, and $(1-x)^{-2}$ can be expanded using binomial series with $n=-2$. 3. **Expand $(1+2x)^2$:** $$(1+2x)^2 = 1 + 4x + 4x^2$$ 4. **Expand $(1-x)^{-2}$:** Using binomial theorem for negative integer powers: $$ (1-x)^{-2} = \sum_{k=0}^\infty \binom{-2}{k} (-x)^k $$ But more simply, the known expansion is: $$ (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \cdots $$ 5. **Multiply the two expansions up to $x^3$ terms:** $$ f(x) = (1 + 4x + 4x^2)(1 + 2x + 3x^2 + 4x^3) $$ Calculate terms: - Constant: $1 \times 1 = 1$ - $x$: $1 \times 2x + 4x \times 1 = 2x + 4x = 6x$ - $x^2$: $1 \times 3x^2 + 4x \times 2x + 4x^2 \times 1 = 3x^2 + 8x^2 + 4x^2 = 15x^2$ - $x^3$: $1 \times 4x^3 + 4x \times 3x^2 + 4x^2 \times 2x = 4x^3 + 12x^3 + 8x^3 = 24x^3$ 6. **First 4 terms:** $$ f(x) = 1 + 6x + 15x^2 + 24x^3 + \cdots $$ 7. **Validity:** The expansion is valid where the binomial series converges, i.e., for $|x| < 1$. --- 8. **Problem 2:** Given $f(x) = (1+2x)^{1/3}$, find the binomial expansion up to $x^3$. 9. **Formula:** For fractional powers: $$ (1+u)^r = 1 + r u + \frac{r(r-1)}{2!} u^2 + \frac{r(r-1)(r-2)}{3!} u^3 + \cdots $$ with $r=\frac{1}{3}$ and $u=2x$. 10. **Calculate coefficients:** - First term: $1$ - Second term: $\frac{1}{3} \times 2x = \frac{2}{3}x$ - Third term: $\frac{\frac{1}{3}(\frac{1}{3}-1)}{2} (2x)^2 = \frac{\frac{1}{3}(-\frac{2}{3})}{2} 4x^2 = \frac{-2/9}{2} 4x^2 = -\frac{4}{9} x^2$ - Fourth term: $\frac{\frac{1}{3}(\frac{1}{3}-1)(\frac{1}{3}-2)}{6} (2x)^3$ Calculate numerator: $$ \frac{1}{3} \times -\frac{2}{3} \times -\frac{5}{3} = \frac{10}{27} $$ Divide by 6: $$ \frac{10}{27} \times \frac{1}{6} = \frac{10}{162} = \frac{5}{81} $$ Multiply by $(2x)^3 = 8x^3$: $$ \frac{5}{81} \times 8x^3 = \frac{40}{81} x^3 $$ 11. **Expansion up to $x^3$:** $$ f(x) = 1 + \frac{2}{3}x - \frac{4}{9}x^2 + \frac{40}{81}x^3 $$ --- 12. **Problem 3:** Approximate $(0.98)^{10}$ using binomial expansion up to $x^4$. 13. **Rewrite:** $$ 0.98^{10} = (1 - 0.02)^{10} $$ Use binomial expansion: $$ (1 - x)^n = \sum_{k=0}^n \binom{n}{k} (-x)^k $$ with $n=10$, $x=0.02$. 14. **Calculate terms up to $x^4$:** - $k=0$: $1$ - $k=1$: $10 \times (-0.02) = -0.2$ - $k=2$: $\binom{10}{2} ( -0.02)^2 = 45 \times 0.0004 = 0.018$ - $k=3$: $\binom{10}{3} (-0.02)^3 = 120 \times (-0.000008) = -0.00096$ - $k=4$: $\binom{10}{4} ( -0.02)^4 = 210 \times 0.00000016 = 0.0000336$ 15. **Sum terms:** $$ 1 - 0.2 + 0.018 - 0.00096 + 0.0000336 = 0.8170736 $$ Rounded to 4 decimal places: $$ 0.8171 $$ --- 16. **Problem 4:** Given $f(x) = \frac{15}{\sqrt{1-x}}$, find the first 5 terms in the expansion. 17. **Rewrite:** $$ f(x) = 15 (1-x)^{-1/2} $$ Use binomial expansion for fractional power $r = -\frac{1}{2}$: $$ (1-x)^r = 1 + r x + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \frac{r(r-1)(r-2)(r-3)}{4!} x^4 + \cdots $$ 18. **Calculate coefficients:** - $r = -\frac{1}{2}$ - $r(r-1) = -\frac{1}{2} \times -\frac{3}{2} = \frac{3}{4}$ - $r(r-1)(r-2) = \frac{3}{4} \times -\frac{5}{2} = -\frac{15}{8}$ - $r(r-1)(r-2)(r-3) = -\frac{15}{8} \times -\frac{7}{2} = \frac{105}{16}$ 19. **Terms:** - $1$ - $-\frac{1}{2} x$ - $\frac{3}{4} \frac{x^2}{2} = \frac{3}{8} x^2$ - $-\frac{15}{8} \frac{x^3}{6} = -\frac{15}{48} x^3 = -\frac{5}{16} x^3$ - $\frac{105}{16} \frac{x^4}{24} = \frac{105}{384} x^4 = \frac{35}{128} x^4$ 20. **Multiply by 15:** $$ f(x) = 15 \left(1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 \right) $$ $$ = 15 - \frac{15}{2} x + \frac{45}{8} x^2 - \frac{75}{16} x^3 + \frac{525}{128} x^4 $$ 21. **Final first 5 terms:** $$ f(x) = 15 - 7.5 x + 5.625 x^2 - 4.6875 x^3 + 4.1015625 x^4 $$