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

#### MarkFL

Staff member
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:

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

##### Well-known member
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:

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...

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'...

Marry Christmas from Serbia

$\chi$ $\sigma$