# mental math: approximating fractional exponents

Does anyone have any good tricks for estimating expressions with fractional exponents (besides guess and check)?

For example, I want to easily calculate $9.1^{1/3}$. Currently, the best I've got is trial and error, but besides that $2^3 = 8$ and $3^3 = 27$, it gets pretty ugly... Could I approximately interpolate between these two?

Using calculus, specifically Taylor approximation, centered at 8 (which is the closest value to 9.1 whose cube root we know), we get $$(8+x)^{1/3}=2+\frac{1}{12}x-\frac{1}{288}x^2+\frac{5}{20736}x^3\cdots$$ Hence, in your example, we have $x=1.1$, or $$(8+1.1)^{1/3}=2+\frac{1}{12}1.1-\frac{1}{288}1.1^2+\frac{5}{20736}1.1^3\cdots$$ You can compute as many terms as you need. One term givs you 2. Two terms gives you $2.091\overline{6}$. Three terms gives you $2.0874652\overline{7}$.
Use one step Newton formula for the zero of $f(x) = x^3-a\;$ with $a=9.1\;$ and the starting value $x_0 = 2$ $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}= 2 - \frac{2^3-9.1}{2\times2^2} = 2 + \frac{1.1}{12} = 2 + \frac{1}{10}\frac{110}{120} \\ x_1\approx 2 + \frac{1}{10}\frac{108}{120} = 2 + \frac{9}{100} = 2.09$$
• Thanks! Can you explain briefly how you're constructing $f(x)$? – DilithiumMatrix Apr 30 '14 at 14:22
• This is the function for cube roots $x^3=a\;$ and you want to find a zero of $x^3-a=0\;$ thus $f(x)=x^3-a$ – gammatester Apr 30 '14 at 14:28