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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.

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  • $\begingroup$ Did you think of setting $x=7.5$ in the approximating polynomials? $\endgroup$ – hardmath Jul 7 '16 at 22:54
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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
?
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  • $\begingroup$ You see : The polynomial with degree $2$ gives already a very good approximation. $\endgroup$ – Peter Jul 7 '16 at 22:51
  • $\begingroup$ So just plug the 7.5 into the p1(x) to solve it, or should I plug 7.5^(1/3) into p1x? $\endgroup$ – Lumberjacked216 Jul 7 '16 at 22:55
  • $\begingroup$ 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. $\endgroup$ – Peter Jul 7 '16 at 22:56
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    $\begingroup$ That worked! Thank you so much! $\endgroup$ – Lumberjacked216 Jul 7 '16 at 22:58
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} 7.5^{1/3} & = \pars{15 \over 2}^{1/3} = \pars{60 \over 8}^{1/3} = \pars{64 - 4 \over 8}^{1/3} = \bracks{{64 \over 8}\pars{1 - {1 \over 16}}}^{1/3} = 2\pars{1 - {1 \over 16}}^{1/3} \\[3mm] & \approx 2 - {1 \over 24} - {1 \over 1152} - {5 \over 165888} - {5 \over 3981312} \approx 1.95743\color{#f00}{388109246} \end{align}

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

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