MHB Is there a way to compute the given area exactly?

[MarkFL]

Recently on Y!Answers the following question was posed:

Compute the area of the region bounded by:

$\displaystyle y=\cos(x)$

$\displaystyle y = x$

$\displaystyle x = 0$

I puzzled for a bit, did some calculations, but could not get away from using a numeric root-finding method for:

$\displaystyle f(x)=\cos(x)-x=0$

to determine the upper limit of integration.

I was curious if someone here might have an insight I missed. By the way, both people that responded also approximated the root.

On a side note, I recall seeing once that a simple method for approximating this root is as follows:

Make sure your calculator is in radian mode.

Enter any number on your calculator.

Take the cosine of this result.

Keep successively taking the cosine of the results, and your calculator will converge (slowly) to the desired root.

I can see why this works. If r is the root, then we have both:

$\displaystyle r=\cos(r)$

$\displaystyle r=\cos^{-1}(r)$

And from this we have:

$\displaystyle r=\cos(\cos(\cos\cdots\cos(r)))$

 

[chisigma]

Recently on Y!Answers the following question was posed:

Compute the area of the region bounded by:

$\displaystyle y=\cos(x)$

$\displaystyle y = x$

$\displaystyle x = 0$

I puzzled for a bit, did some calculations, but could not get away from using a numeric root-finding method for:

$\displaystyle f(x)=\cos(x)-x=0$

to determine the upper limit of integration.

I was curious if someone here might have an insight I missed. By the way, both people that responded also approximated the root.

On a side note, I recall seeing once that a simple method for approximating this root is as follows:

Make sure your calculator is in radian mode.

Enter any number on your calculator.

Take the cosine of this result.

Keep successively taking the cosine of the results, and your calculator will converge (slowly) to the desired root.

I can see why this works. If r is the root, then we have both:

$\displaystyle r=\cos(r)$

$\displaystyle r=\cos^{-1}(r)$

And from this we have:

$\displaystyle r=\cos(\cos(\cos\cdots\cos(r)))$

Finding the [real] root of the equation $\displaystyle x= \cos x$ is equivalent to find the limit of the solution of the difference equation...

$\displaystyle a_{n+1}= \cos a_{n}$ (1)

... starting from some initial value $a_{0}$. The (1) can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n} = \cos a_{n} - a_{n} = f(a_{n})$ (2)

... and the procedure to follow is illustrated in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/#post2492

The f(x) is illustrated here... https://www.physicsforums.com/attachments/516._xfImport

There is only 'attractive fixed point' in $x_{0} \sim .739085...$ and in that point is $f^{\ '} (x_{0}) \sim -1.67361...$ so that we are in the conditions of the Threorem 4.2 and the convergence is 'oscillating'...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\chi$ $\sigma$