1
$\begingroup$

Use a power series to approximate the definite integral to six decimal places.

$\int_{0}^{0.3} \frac{x^2}{1+x^4}$

I'm not sure how to find the sum of this for solving when it has x's in the numerator, this is what I assumed however.

$\frac{1}{1-x} $ $\sum_{n=0}^\infty $ x^n

$\sum_{n=0}^\infty \frac{1}{1-x^4}$

$\sum_{n=0}^\infty {(-x^4)}^n$ ==> $\frac{x}{1+x^4}$

I'm not sure if this is right nor what the next step is to get the approximate definite integral.

$\endgroup$
  • $\begingroup$ This is definitely in the right direction. Then multiply the last fraction by $x$ to make it equal to the integrand. $\endgroup$ – Simply Beautiful Art Oct 26 '16 at 20:04
  • $\begingroup$ so it becomes x^2/1+x^4? isn't this back in the same direction? how do i find the intergrand $\endgroup$ – Megan Byers Oct 26 '16 at 20:14
  • $\begingroup$ To be more clear, $$\frac1{1+x^4}=\sum_{n=0}^\infty(-x^4)^n$$ $$\implies\frac{x^2}{1+x^4}=x^2\sum_{n=0}^\infty(-x^4)^n$$ and expand the sum as many digits as you need. $\endgroup$ – Simply Beautiful Art Oct 26 '16 at 21:10
2
$\begingroup$

$$ I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$ and the last series is a series with positive terms. Since $$ \sum_{n\geq 2}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)<\frac{1}{19}\sum_{n\geq 2}\left(\frac{3}{10}\right)^{8n+3}<10^{-11} \tag{2}$$ the first eight figures of $I$ are given by the sum appearing in the RHS of $(1)$ restricted to $n=0$ and $n=1$: $$ \sum_{n=0}^{1}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)=\frac{3453033133161387}{385000000000000000}=\color{green}{0.00896891}72289906\ldots\tag{3}$$

$\endgroup$
2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {x^{2} \over 1 + x^{4}} & = x^{2} - x^{6} + x^{10} - x^{14} + \mrm{O}\pars{x^{16}} \end{align} You must truncate the series whenever the 'last term' is smaller than the Machine Precision $\ds{\texttt{mp}}$. Namely, $$ x^{n} < \texttt{mp} \implies n < {\ln\pars{\texttt{mp}} \over \ln\pars{x}} = {\verts{\ln\pars{\texttt{mp}}} \over \verts{\ln\pars{x}}} < {\verts{\ln\pars{\texttt{mp}}} \over \verts{\ln\pars{0.3}}} $$ Typical $\ds{\texttt{mp}}$ are of order $\ds{10^{-16}}$ which truncates the series up to $\ds{x^{14}}$: $$ \left\{\begin{array}{rcl} \ds{\int_{0}^{0.3}{x^{2} \over 1 + x^{4}}\,\dd x} & \ds{=} & \ds{0.00896891723506713\ldots} \\[1mm] \ds{\int_{0}^{0.3}\pars{x^{2} - x^{6} + x^{10} - x^{14}}\,\dd x} & \ds{=} & \ds{0.0089689172\color{#f00}{289906155844}\ldots} \end{array}\right. $$

That's the best you can do when $\ds{\texttt{mp} \sim 10^{-16}}$.

$\endgroup$
  • $\begingroup$ hmm, we never learned machine precision, but thank you for this answer! $\endgroup$ – Megan Byers Oct 26 '16 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.