1. **Problem statement:** Expand $$\sqrt{\frac{1+m}{1-m}}$$ in ascending powers of $m$ up to and including the term in $m^3$. State the values of $m$ for which the expansion is valid. Then, by substituting $m=0.2$, find an approximate value for $$\sqrt{1.5}$$. 2. **Formula and rules:** We use the binomial expansion for $(1+x)^n$ where $|x|<1$ and $n$ is any real number: $$ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$ 3. **Step 1: Rewrite the expression** $$ \sqrt{\frac{1+m}{1-m}} = \frac{(1+m)^{1/2}}{(1-m)^{1/2}} = (1+m)^{1/2} (1-m)^{-1/2} $$ 4. **Step 2: Expand each term using binomial expansion** - For $(1+m)^{1/2}$: $$ 1 + \frac{1}{2}m - \frac{1}{8}m^2 + \frac{1}{16}m^3 $$ - For $(1-m)^{-1/2}$: $$ 1 + \frac{1}{2}m + \frac{3}{8}m^2 + \frac{5}{16}m^3 $$ 5. **Step 3: Multiply the two expansions and keep terms up to $m^3$** $$ \left(1 + \frac{1}{2}m - \frac{1}{8}m^2 + \frac{1}{16}m^3\right) \times \left(1 + \frac{1}{2}m + \frac{3}{8}m^2 + \frac{5}{16}m^3\right) $$ Multiply term by term and collect powers: - Constant term: $1$ - Coefficient of $m$: $\frac{1}{2} + \frac{1}{2} = 1$ - Coefficient of $m^2$: $\frac{1}{2} \times \frac{1}{2} + \left(-\frac{1}{8}\right) + \frac{3}{8} = \frac{1}{4} - \frac{1}{8} + \frac{3}{8} = \frac{1}{2}$ - Coefficient of $m^3$: $\frac{1}{2} \times \frac{3}{8} + \frac{1}{16} + \frac{5}{16} - \frac{1}{8} \times \frac{1}{2} = \frac{3}{16} + \frac{1}{16} + \frac{5}{16} - \frac{1}{16} = \frac{8}{16} = \frac{1}{2}$ So the expansion is: $$ 1 + m + \frac{1}{2}m^2 + \frac{1}{2}m^3 $$ 6. **Step 4: Validity of expansion** The binomial expansions are valid for $|m|<1$ because the expansions require $|x|<1$ for convergence. 7. **Step 5: Approximate $$\sqrt{1.5}$$ by substituting $m=0.2$** $$ \sqrt{1.5} = \sqrt{\frac{1+0.2}{1-0.2}} \approx 1 + 0.2 + \frac{1}{2}(0.2)^2 + \frac{1}{2}(0.2)^3 = 1 + 0.2 + 0.02 + 0.004 = 1.224 $$ **Final answer:** $$ \sqrt{\frac{1+m}{1-m}} = 1 + m + \frac{1}{2}m^2 + \frac{1}{2}m^3 \quad \text{for} \quad |m|<1 $$ Approximate value for $$\sqrt{1.5}$$ when $m=0.2$ is approximately $1.224$.