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How would I go about evaluating $I = \int_0^{0.1} \sqrt{1-1/2\sin^2(t)} dt$ to four decimal accuracy?

TMM
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MathPower
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3 Answers3

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You can get 4 decimal places by a simple mid-point approximation:

$$I \approx 0.1 \sqrt{1-0.5 \sin^2{(0.05)}} \approx 0.0999375$$

where a numerical integration gives $I \approx 0.0999168$.

Ron Gordon
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    How can you tell that the simple midpoint-approximation gives 4 decimal accuracy? – MathPower Apr 03 '13 at 18:50
  • It is a smooth function being integrated over a small interval. Think of it as a trapezoidal approximation having a single interval. The error is then, roughly, $(0.1)^3/12$, which is about 4 decimal places. – Ron Gordon Apr 03 '13 at 18:55
  • This argument does not show that the approximation is valid to 4 decimal places. You may have a smooth function that has a sudden jump in the middle. – Lord Soth Apr 03 '13 at 19:35
  • @LordSoth: but we don't have that, do we? The approximation, I should have mentioned, depends on the function having a second derivative in the interval. – Ron Gordon Apr 03 '13 at 19:52
  • @RonGordon I see the intuition behind your argument, after all, the function is roughly linear around such a "small" interval. You will however need to estimate the derivatives for a formal proof of "$4$ decimal accuracy." – Lord Soth Apr 03 '13 at 20:05
  • @LordSoth: as said 2nd derivative is bounded by $1$, we are OK. http://www.wolframalpha.com/input/?i=D%5BSqrt%5B1-(1%2F2)%20Sin%5Bt%5D%5E2%5D%20%2C%20%7Bt%2C2%7D%5D&t=crmtb01 That said, this was to me an exercise in intuition and how to come up with a quick and dirty way to approximate to 4 places. – Ron Gordon Apr 03 '13 at 20:10
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \int_{0}^{0.1}\root{1 - \half\sin^{2}\pars{t}}\,\dd t = {\rm E}\pars{0.1,{\root{2} \over 2}} \end{align} where $\ds{{\rm E}\pars{\phi,k}}$ is the Elliptic Integral of the Second Kind. For 'small' $\ds{\phi}$: $$ {\rm E}\pars{\phi,{\root{2} \over 2}} \approx \phi\braces{1 - {\phi^{2} \over 12}\bracks{1 - {\phi^{2} \over 8}}} $$ such that $$ \color{#00f}{\large\int_{0}^{0.1}\root{1 - \half\sin^{2}\pars{t}}\,\dd t} \approx 0.1\braces{1 - {0.01 \over 12}\bracks{1 - {0.01 \over 8}}} = \color{#00f}{\large 0.0999168} $$ with 'relative error' of the order $\ds{10^{-8}}$ %.

Felix Marin
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$I = \int_0^{0.1} \sqrt{1-1/2\sin^2(t)} dt \leq \int_0^{0.1} (1-1/4\sin^2(t)) dt = \frac{7}{80}+\frac{\sin(0.2)}{16} = 0.09991683317469133..$. The first inequality is since $\sqrt{1-x} \leq 1-\frac{x}{2}$ in your range of integration.

Also, since $\sin(x) \leq x$ in your range of integration, $I \geq \int_0^{0.1} \sqrt{1-t^2/2} dt = \frac{x\sqrt{2-x^2}-2\sin^{-1}(x/\sqrt{2})}{2\sqrt{2}} |_{x=0.1} = 0.09991660405478747...$.

So, $.09991660405478747... \leq I \leq 0.09991683317469133..$

Of course this assumes you can evaluate sines and arcsines and stuff, but perhaps you can use Taylor series, etc. for them.

Lord Soth
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