Let's break down the problem step-by-step:

1. Expansion of $(1 - \frac{x}{16})^{\frac{1}{3}}$ up to $x^2$

We will use the binomial expansion for $(1 + u)^n$ where $u = -\frac{x}{16}$ and $n = \frac{1}{3}$: $(1 + u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2 + \dots$ Substitute $u = -\frac{x}{16}$ and $n = \frac{1}{3}$: $(1 - \frac{x}{16})^{\frac{1}{3}} \approx 1 + \frac{1}{3}\left(-\frac{x}{16}\right) + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2}\left(-\frac{x}{16}\right)^2$

Let's simplify this: $1 - \frac{x}{48} + \frac{\frac{1}{3} \cdot -\frac{2}{3}}{2} \cdot \frac{x^2}{256}$ $1 - \frac{x}{48} - \frac{1}{9 \cdot 512} x^2$ $1 - \frac{x}{48} - \frac{x^2}{4608}$

2. State the range of $x$ for which the expansion is valid

The binomial expansion $(1 + u)^n$ is valid when $|u| < 1$: $\left|-\frac{x}{16}\right| < 1$ $\frac{|x|}{16} < 1$ $|x| < 16$ So, the range of $x$ is $-16 < x < 16$.

3. Show that $\sqrt[3]{8 - \frac{x}{2}} = 2(1 - \frac{x}{16})^{\frac{1}{3}}$

First, express the left side: $\sqrt[3]{8 - \frac{x}{2}} = (8 - \frac{x}{2})^{\frac{1}{3}}$ Factor out 8: $= 8^{\frac{1}{3}} \left(1 - \frac{x}{16}\right)^{\frac{1}{3}}$ Since $8^{\frac{1}{3}} = 2$: $= 2(1 - \frac{x}{16})^{\frac{1}{3}}$

4. Approximate $\sqrt[3]{7}$ using $x = 2$

Set $x = 2$ in the previous expression: [ \sqrt[3]{8 - \frac{2}{2}} = \sqrt[3]{7} ] $= 2(1 - \frac{2}{16})^{\frac{1}{3}}$ Simplify: $= 2(1 - \frac{1}{8})^{\frac{1}{3}}$ $= 2(1 - 0.125)^{\frac{1}{3}}$

Use the binomial expansion found earlier: $\approx 2\left(1 - \frac{1}{48} \times 2 - \frac{2^2}{4608}\right)$ $= 2\left(1 - \frac{1}{24} - \frac{4}{4608}\right)$ $= 2\left(1 - 0.0416667 - 0.0008681\right)$ $= 2 \times 0.9574652$ $\approx 1.9149$

Thus, $\sqrt[3]{7} \approx 1.9149$ to four significant figures.

Would you like more details on any part of this process or have any questions?

Related Questions

1. How would the expansion change if the power was different, say $(1 - \frac{x}{16})^{\frac{1}{2}}$?
2. Can we approximate $\sqrt[3]{7}$ using other methods such as Newton's method?
3. How does the error of the binomial approximation change as more terms are added?
4. Can the binomial expansion be used to approximate functions other than polynomials?
5. How does the range of $x$ for convergence change if the constant inside the binomial expansion changes?

Tip

When using binomial expansions, always check the range of $x$ for which the expansion is valid to ensure accurate approximations.