4.2 Binomial expansion for rational n and approximation — Practise

1. The expansion and the range

Answer: With $n = -1$, every coefficient $\binom{-1}{r} = (-1)^r$. So $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots$, which is valid for $|x| < 1$. The signs alternate, and the series matches the geometric series with first term $1$ and ratio $-x$.

2. The expansion and the range

Answer: Replace $x$ with $3x$ in the expansion of $(1 + x)^{-1}$: $(1 + 3x)^{-1} = 1 - 3x + 9x^2 - 27x^3 + \dots$. The expansion is valid when $|3x| < 1$, which is $|x| < \frac{1}{3}$. The coefficients are powers of $3$, because each term brings another factor of $3x$.

3. The expansion

Answer: Use $n = \frac{1}{2}$ with $-x$ in place of $x$: $(1 - x)^{\frac{1}{2}} = 1 + \frac{1}{2}(-x) - \frac{1}{8}(-x)^2 + \frac{1}{16}(-x)^3 + \dots = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} + \dots$, valid for $|x| < 1$. The odd powers stay negative, and the even power $(x)^2$ carries the negative coefficient $-\frac{1}{8}$.

4. The expansion and the range

Answer: Take out the factor $9^{\frac{1}{2}} = 3$: $\sqrt{9 + x} = 3\left(1 + \frac{x}{9}\right)^{\frac{1}{2}} = 3\left[1 + \frac{x}{18} - \frac{x^2}{648} + \frac{x^3}{11664} + \dots\right] = 3 + \frac{x}{6} - \frac{x^2}{216} + \frac{x^3}{3888} + \dots$. The expansion is valid when $\left|\frac{x}{9}\right| < 1$, which is $|x| < 9$. The factor $3$ is why the answer starts at $3$, not $1$.

5. The first three terms

Answer: Write the expression as $(1 + 4x)^{-\frac{1}{2}}$, so $n = -\frac{1}{2}$. The coefficients are $\binom{-\frac{1}{2}}{1} = -\frac{1}{2}$ and $\binom{-\frac{1}{2}}{2} = \frac{3}{8}$. So $(1 + 4x)^{-\frac{1}{2}} = 1 - \frac{1}{2}(4x) + \frac{3}{8}(4x)^2 + \dots = 1 - 2x + 6x^2 + \dots$ for $|x| < \frac{1}{4}$. The range is $|4x| < 1$.

6. The expansion

Answer: With $n = -3$: $\binom{-3}{1} = -3$, $\binom{-3}{2} = \frac{(-3)(-4)}{2} = 6$, and $\binom{-3}{3} = \frac{(-3)(-4)(-5)}{6} = -10$. So $(1 + x)^{-3} = 1 - 3x + 6x^2 - 10x^3 + \dots$ for $|x| < 1$. The coefficients grow in size, because each new factor on the top is negative and larger than the one before.

7. The first three terms

Answer: Take out the factor $2^{\frac{1}{2}} = \sqrt{2}$: $(2 + 3x)^{\frac{1}{2}} = \sqrt{2}\left(1 + \frac{3x}{2}\right)^{\frac{1}{2}} = \sqrt{2}\left[1 + \frac{3x}{4} - \frac{9x^2}{32} + \dots\right]$. So the first three terms are $\sqrt{2} + \frac{3\sqrt{2}}{4}x - \frac{9\sqrt{2}}{32}x^2$. The expansion is valid when $\left|\frac{3x}{2}\right| < 1$, which is $|x| < \frac{2}{3}$. Each coefficient carries the factor $\sqrt{2}$.

8. The coefficient

Answer: The term is $\binom{\frac{1}{2}}{2}(-2x)^2 = -\frac{1}{8} \times 4x^2 = -\frac{1}{2}x^2$. The coefficient of $x^2$ is $-\frac{1}{2}$. The minus sign comes from the coefficient, not from the bracket, because the square $(-2x)^2$ is positive.

9. The estimate

Answer: Write $1.02 = 1 + 0.02$, so $x = 0.02$. The first three terms are $1 + \frac{1}{2}x - \frac{1}{8}x^2 = 1 + 0.01 - 0.00005 = 1.00995$. The estimate is within $5 \times 10^{-7}$ of the true value, which the check confirms against a tolerance of $10^{-5}$. The term in $x^3$ is smaller than the rounding, which is why three terms are enough.

10. The estimate

Answer: With $n = -\frac{1}{2}$ and $x = 0.02$: the first three terms are $1 - \frac{1}{2}x + \frac{3}{8}x^2 = 1 - 0.01 + 0.00015 = 0.99015$. The true value is $0.9901475\dots$, so the estimate is within $3 \times 10^{-6}$ of it. The coefficient $\frac{3}{8}$ is $\binom{-\frac{1}{2}}{2}$.

11. The range

Answer: The general expansion is valid when the size of the replaced variable is less than $1$. Here the replacement is $7x$, so the condition is $|7x| < 1$. Dividing by $7$ gives $|x| < \frac{1}{7}$, which is $-\frac{1}{7} < x < \frac{1}{7}$. At the endpoints the series does not settle, which is why the interval is open.

12. The first two terms

Answer: Take out the factor $4^{\frac{1}{2}} = 2$: $\sqrt{4 + 2x} = 2\left(1 + \frac{x}{2}\right)^{\frac{1}{2}}$. The first two terms of the bracket are $1 + \frac{1}{2} \times \frac{x}{2} = 1 + \frac{x}{4}$. Multiplying by $2$ gives $2 + \frac{x}{2}$, as required. The expansion is valid when $\left|\frac{x}{2}\right| < 1$, which is $|x| < 2$.