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I am struggling to use the following equation:

$$ \int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = 0 $$

where $a > 1$, to deduce that $a = \text{cosec}(\frac{\pi}{4} - \frac{\alpha}{2})$, where $\alpha$ satisfies $\alpha = \cos(\alpha)$.

I integrate the integrand, via

$$ \int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = -\int_0^1 \sqrt{a^2-x^2}\, dx + \int_1^a \sqrt{a^2-x^2}\, dx $$

But once I calculate those integrals I cannot seem to get any closer to the answer.

Any help would be great.

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    $\begingroup$ What do you get when you calculate the integrals? $\endgroup$ – Jason Nov 27 '14 at 1:56
  • $\begingroup$ Using $\int \sqrt{a^2-x^2}\, dx = \frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}(x/a) + c$, one can calculate the two integrals to be $-\int_0^1 \sqrt{a^2-x^2}\, dx = -\frac{\sqrt{a^2-1}}{2}-\frac{a^2}{2}\sin^{-1}(1/a)$ and $\int_1^a \sqrt{a^2-x^2}\, dx = \frac{\pi a^2}{4} -\frac{\sqrt{a^2-1}}{2}-\frac{a^2}{2}\sin^{-1}(1/a)$. This leads to $\frac{\pi a^2}{4} -\sqrt{a^2-1}- a^2\sin^{-1}(1/a) = 0$. From there, I'm not sure what else can be done. $\endgroup$ – John Sweeney Nov 27 '14 at 2:22
  • $\begingroup$ I suppose I can divide by a^2 and rearrange, to get $1/a = \sin(\pi/4 - \sqrt{1-\frac{1}{a^2}})$, in which case my $\alpha$ would be $\alpha = 2\sqrt{1-1/a^2}$, but then how can I show this satisfies \alpha = cos(\alpha)? $\endgroup$ – John Sweeney Nov 27 '14 at 2:36
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First, $a>1$ and $x\in[0,a]$, then $|x|=x\ge0$

$$ \renewcommand\sgn{\operatorname{sgn}} \renewcommand\arcsec{\operatorname{arcsec}} \begin{array}{ll} 0\!\!\!&=\int_0^a\sqrt{a^2-x^2}\sgn(x-1)\,\mathrm dx\\ &=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx-\int_0^1\sqrt{a^2-x^2}\,\mathrm dx \end{array} $$ Then $$\int_1^a\sqrt{a^2-x^2}\,\mathrm dx=\int_0^1\sqrt{a^2-x^2}\,\mathrm dx\tag{1}$$ But $$\int_1^a\sqrt{a^2-x^2}\,\mathrm dx=\text{Area of }DAB\tag{2}$$ $$\int_0^1\sqrt{a^2-x^2}\,\mathrm dx=\text{Area of }ODBC\tag{3}$$ diagram of quarter circle

and area of $DAB+$ area of $ODBC=\dfrac14$area disc$(a)=\dfrac{\pi a^2}4$

Then $(1),(2),(3)$ yield

$$2\text{area of }DAB=\dfrac{\pi a^2}4\implies\text{ area of }DAB=\dfrac{\pi a^2}8=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx\tag{★}$$

Moreover: Area of sector $OAB=\pi\theta= \pi\arccos\left(\frac1a\right)= \pi\arcsec a$
On the other hand, Area of sector $OAB=$ Area of triangle $ODB+$ Area of $DAB$.

Then, $$\boxed{\displaystyle\pi\arcsec a=\dfrac12\sqrt{a^2-1}+\int_1^a\sqrt{a^2-x^2}\,\mathrm dx}\tag{★★}$$

From $($★$)$ and $($★★$)$ it follows that: $a$ is a solution of the equation

$$\boxed{\displaystyle\pi\arcsec a=\dfrac12\sqrt{a^2-1}+\frac{\pi a^2}{8}}$$

I transformed the initial problem into a simpler one (I hope!)

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    $\begingroup$ You don't know how to use $\LaTeX$ here? $\endgroup$ – Aditya Hase Nov 27 '14 at 2:46
  • $\begingroup$ @Itegrator, Someone else asked me some same question about LATEX. I know wrinting with LATEX but it takes much more time than the other method. Is it opposit to the rules of this forum to scan and then post scanned figures instead of LATEX? $\endgroup$ – Idris Addou Nov 27 '14 at 2:59
  • $\begingroup$ Yeah, we pretty much use $\LaTeX$ here. We are sort of looking for really nice posts that stand the test of time and serve many users who come back to benefit from your math, and so considering our grand cause, that being great questions and great answers, you will not find many people complaining that it takes a bit more time to typeset the math. Most of our better contributors typeset, and sure it takes more time, but the time taken to do this benefits the searchability of the site, the scalability of the post on various devices, and the users who read this in the future. $\endgroup$ – J. W. Perry Nov 27 '14 at 5:01
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    $\begingroup$ I note that in the 1st integral, you wrote sng instead of sgn. Had you typed this in, I could have edited to correct the typo. The way you have posted, it's impossible to correct the typo. One more reason to do it the right way! $\endgroup$ – Gerry Myerson Nov 27 '14 at 5:11
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    $\begingroup$ I have transcriped your answer into $\LaTeX$, please check that I haven't changed the meaning of your post. $\endgroup$ – Alice Ryhl Dec 2 '14 at 20:57
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Your question is now: If $\alpha =2\sqrt{1-\frac{1}{a^{2}}}$ how can one prove that $\cos \alpha =\alpha ?$

Your question in the title is :'' solve for a variable $a$ '' it means that: the problem is: solve for the variable $a$ the equation $\cos \alpha =\alpha ,$ where $a>1.$ So, lets go:

From $\frac{\alpha }{2}=\sqrt{1-\frac{1}{a^{2}}}$ then $\cos (\frac{\alpha }{% 2})=\frac{1}{a}$ (figure may help i will attach it below). But $\cos ^{2}(% \frac{\alpha }{2})=\frac{1+\cos \alpha }{2}=\frac{1}{a^{2}}$ hence $\cos \alpha =\frac{2}{a^{2}}-1.$ Therefore $\cos \alpha =\alpha $ if and only if $% \frac{2}{a^{2}}-1=2\sqrt{1-\frac{1}{a^{2}}},$ or $\sqrt{1-\frac{1}{a^{2}}}=% \frac{1}{a^{2}}-\frac{1}{2}.$ So solving this equation and taking only solution $>1$ one obtains $a=\frac{\sqrt{2}}{\sqrt[4]{3}}\simeq 1.07>1.$

enter image description here

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