Find the regions where sin(x) is greater than or equal to X

**Pupil** · November 18th 2010, 02:43 PM

I know the graphs intersect three times, one at (0,0) and the other 2 points are reflected about the y-axis. Is there a solution for this exercise? I can't think of any of these points where a value of X is less than its sin value. And also, is there any way I can solve for the other 2 points algebraically? Because I've been approximating and it seems my approximates aren't that good...

Thanks in advance guys, I appreciate it.

**Prove It** · November 18th 2010, 03:08 PM

Have you tried graphing the functions? The answer

Spoiler:

should then be obvious.

**Archie Meade** · November 18th 2010, 03:12 PM

Originally Posted by Pupil

I know the graphs intersect three times, one at (0,0) and the other 2 points are reflected about the y-axis. Is there a solution for this exercise? I can't think of any of these points where a value of X is less than its sin value. And also, is there any way I can solve for the other 2 points algebraically? Because I've been approximating and it seems my approximates aren't that good...

Thanks in advance guys, I appreciate it.

$\frac{d}{dx}Sinx=Cosx$

The maximum value of the derivative of $Sinx$ is 1.

The derivative of $f(x)=x$ is always 1.

Hence $Sinx=x$ only when $x=0$

Therefore $Sinx>x$ if $x<0$

**Archie Meade** · November 18th 2010, 03:34 PM

modem error....
please delete

**Pupil** · November 18th 2010, 03:52 PM

Alright guys, thanks for the help. I understand the inequality now.

Just one more question: Is there any way I can get the exact values by algebraic methods? If I try to do it graphically, my approximations are not as nearly as accurate. I know the graphs intersect 3 times and one is (0,0), but how would I determine the other points without approximating? Because frankly, my graphing software sucks. ><

**Archie Meade** · November 18th 2010, 03:56 PM

Originally Posted by Pupil

Alright guys, thanks for the help. I understand the inequality now.

Just one more question: Is there any way I can get the exact values by algebraic methods? If I try to do it graphically, my approximations are not as nearly as accurate. I know the graphs intersect 3 times and one is (0,0), but how would I determine the other points without approximating? Because frankly, my graphing software sucks. ><

If the graph of the functions intersect to the right of the y-axis,
then Sinx would have had to have a gradient >1 at some stage but it doesn't!

Why are you convinced that there is more than one point of intersection?

**Pupil** · November 18th 2010, 04:06 PM

Originally Posted by Archie Meade

If the graph of the functions intersect to the right of the y-axis,
then Sinx would have had to have a gradient >1 at some stage but it doesn't!

Why are you convinced that there is more than one point of intersection?

Ah, it was just a mistake with my software then. But I have other questions where there is more than 1 intersection, and my approximation with the software is not that good.

So how would I be able to solve for the exact points of intersection? Ex: finding the 4 points where $cos(x) = 1/2^x$ Thanks for all the help, by the way.

**Archie Meade** · November 18th 2010, 04:30 PM

Originally Posted by Pupil

Ah, it was just a mistake with my software then. But I have other questions where there is more than 1 intersection, and my approximation with the software is not that good.

So how would I be able to solve for the exact points of intersection? Ex: finding the 4 points where $cos(x) = 1/2^x$ Thanks for all the help, by the way.

The first point of intersection is quite straightforward, x=0.

You could use Newton's method to find the other two from $0$ to $2{\pi}$ , by solving

$cosx-\frac{1}{2^x}=0$

However, that will give "very close approximations" !

**Pupil** · November 18th 2010, 04:43 PM

Originally Posted by Archie Meade

The first point of intersection is quite straightforward, x=0.

You could use Newton's method to find the other two from $0$ to $2{\pi}$ , by solving

$cosx-\frac{1}{2^x}=0$

However, that will give "very close approximations" !

Okay, it seems similar to solving trig equations. However, since we can't use the unit circle, so how would I go about approximating the points? Oh and its $0.5^x$

Thanks in advance.

**Archie Meade** · November 19th 2010, 02:34 AM

Originally Posted by Pupil

Okay, it seems similar to solving trig equations. However, since we can't use the unit circle, so how would I go about approximating the points? Oh and its $0.5^x$

Thanks in advance.

The Newton-Raphson formula is

$\displaystyle\ x_{n+1}=x_n-\frac{f\left(x_n\right)}{f'\left(x_n\right)}$

$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$

$x_2=x_1-\frac{f\left(x_1\right)}{f'\left(x_1\right)}$

$x_3=x_2-\frac{f\left(x_2\right)}{f'\left(x_2\right)}$

It's an iterative process. $x_0$ is a first guess at a solution.

Then $x_1$ is the calculated first approximation. $x_2$ will be much closer, as will $x_3$

$\frac{1}{2^x}=2^{-x}=0.5^x$

$f(x)=cosx-0.5^x$

$f'(x)=-sinx+\left(0.5^x\right)ln(0.5)$

Take an initial guess at the next root of $cosx-0.5^x=0,$ say $x=\frac{\pi}{3}$ .

Then

$\displaystyle\ x_1=\frac{\pi}{3}-\frac{cos\left(\frac{\pi}{3}\right)-0.5^{\frac{\pi}{3}}}{-sin\left(\frac{\pi}{3}\right)+\left(0.5^{\frac{\pi }{3}}\right)ln(0.5)}$

Use the result as $x_1$ in the 2nd iteration, which should be enough.

Use $\frac{3{\pi}}{2}$ as an initial guess for the 3rd root of $cosx-0.5^x=0$

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