Use a linear approximation to estimate the number $64.07^{2/3}$

Question

That's all they give you. I tried putting it into the linear approximation equation of:

$$ f(a)+f'(a)(x-a) $$

but I get almost the same value as $64.07^{2/3}$, which is around $16.0116$. Just not sure how else to go about doing this problem. Thanks.

Answer 1

Consider function $f(x)=x^{2/3}$ and establish, from definitions, Taylor series at $x=64$ since $64^{2/3}=16$.

You will get for first order (linear approximation) $$f(x)=16+\frac{x-64}{6}+O\left((x-64)^2\right)$$ from which $x=\frac{9607}{600}\approx 16.01166667 $. This is exactly what you did.

For more accuracy, you could use the second order expansion (quadratic approximation) would be $$f(x)=16+\frac{x-64}{6}-\frac{(x-64)^2}{2304}+O\left((x-64)^3\right)$$ from which $x=\frac{368908751}{23040000}\approx 16.01166454$

For ten significant digits, the exact solution is $\approx 16.01166454$.

Answer 2

An equivalent formula for $x^a$ is $(1+x)^a \approx 1+ax $ for small $x$.

Therefore, for small $v$, $(u^{a}+v)^{1/a} =u(1+\frac{v}{u^a})^{1/a} \approx u(1+\frac{v}{au^a}) = u+\frac{uv}{au^a} = u+\frac{u^{1-a}v}{a} $.

In this case, $a = \frac32$, so $(u^{3/2}+v)^{2/3} \approx u+\frac{u^{-1/2}v}{\frac32} =u+\frac{2v}{3\sqrt{u}} $.

If $u=16$ and $v = .07$, this gives $16+\frac{.14}{12} =16.01166... $.