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Binomial Series Approximations

Binomial Series Approximations 1

Determine the first three non-zero terms of the binomial series for $\sqrt{1+x}$ and state its radius of convergence.

Hence determine the value of $\sqrt{101}$ correct to 3 decimal places.

solution - press button to display

The binomial series is given by $$ (1+x)^\nu = 1 + \nu x + \frac{\nu(\nu-1)}{2!}x^2 + \dots$$

The series is convergent when $|x|<1$.

In this case ($\nu = \frac{1}{2}$) we get $$ \sqrt{1+x} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^2 $$

To evaluate $\sqrt{101}$, we need to find an appropriate value of $x$ within the radius of convergence.

Let us rewrite $\sqrt{101}$ as follows: $$\sqrt{101} = \sqrt{100\times 1.01} = 10\sqrt{1 + \frac{1}{100}}$$ So we can evaluate our series when $x = \frac{1}{100}$ and multiply our answer by 10. $$\begin{align} \sqrt{1+ \frac{1}{100}} &\approx 1 + \frac{1}{2}\frac{1}{100} - \frac{1}{8}\left(\frac{1}{100}\right)^2 \\ 1 + \frac{1}{2}\frac{1}{100} - \frac{1}{8}\left(\frac{1}{100}\right)^2 &= 1.00499 \end{align}$$ Hence $\sqrt{101} \approx 10.050$

Binomial Series Approximations 2

Determine the first three non-zero terms of the expansion of $$ ^3\!\sqrt{1+x} $$ hence determine the value of $^3\!\sqrt{480}$, accurate to 3 decimal places
solution - press button to display
To proceed, we must now rewrite $^3\!\sqrt{480}$ in the form $^3\!\sqrt{k^3(1+x)}, |x|\lt1$. Consequently, we need to find some suitable $k^3$. The obvious first choice is cube numbers around 480. So, let $k=8$, then $k^3 = 512$. We get
$$^3\!\sqrt{480} = ^3\!\sqrt{512\left(1-\frac{1}{32}\right)} $$