# Approximating Functions with Polynomials in Taylor Series

I'm having difficulty with a series of problems I've been working on and I can't seem to find a straightforward explanation of how to solve them. The problem is approximating functions.

$f(x)=x^{1/3}$, $a=8$, approximate $7.5^{1/3}$ I was able to calculate $p_1(x)=2+(1/12)(x-8)$, $p_2(x)=2+(1/12)(x-8)-(1/288)(x-8)^2$

I know need to use the linear approximating polynomial and the quadratic approximating polynomial to come up with an answer. I'm unsure of how to use the $7.5^{1/3}$ to come up with an answer. Any help would be greatly appreciated.

• Did you think of setting $x=7.5$ in the approximating polynomials? – hardmath Jul 7 '16 at 22:54

Hint : Just plug $x=7.5$ into the approximation-polynomials.

The exact result and the approximations are (calculation with PARI/GP) :

? 7.5^(1/3)
%4 = 1.957433820584431797712468030
? 2+1/12*(7.5-8)
%5 = 1.958333333333333333333333333
? 2+1/12*(7.5-8)-1/288*(7.5-8)^2
%6 = 1.957465277777777777777777778
?

• You see : The polynomial with degree $2$ gives already a very good approximation. – Peter Jul 7 '16 at 22:51
• So just plug the 7.5 into the p1(x) to solve it, or should I plug 7.5^(1/3) into p1x? – Lumberjacked216 Jul 7 '16 at 22:55
• Only $7.5$ , not $7.5^{1/3}$. You calculated the taylor sequence around $a=8$. You want to approximate $f(7.5)$, and $f(x)$ is approximated by the taylor-polynomials. – Peter Jul 7 '16 at 22:56
• That worked! Thank you so much! – Lumberjacked216 Jul 7 '16 at 22:58


Relative Error $\ds{\sim 10^{-6}\ \%}$.