Finding Points of Intersection for f(x) = 2x/pi and f(x) = sin x

[ACE_99]

Homework Statement

Determine the points of intersection for the following equations

f(x) = 2x/pi
f(x) = sin x

Homework Equations

None

The Attempt at a Solution

In order to find the points of intersection you set the two equal to each other then find where x is equal to 0. I'm not sure how to isolate for x with a sin in the equation.

 

[Mark44]

Homework Statement

Determine the points of intersection for the following equations

f(x) = 2x/pi
g(x) = sin x

The Attempt at a Solution

In order to find the points of intersection you set the two equal to each other then find where x is equal to 0. I'm not sure how to isolate for x with a sin in the equation.

x is equal to 0 at x = 0. I'm being sarcastic, but my point is that you set f(x) = g(x) (you used the same letter for both functions, so I changed one of them) and solve that equation. Doing this gets you 2x/pi = sin x, which can't be solved exactly.

You should graph the two functions as carefully as you can to see how many points of intersection there are (I think there are two), and then use a calculator (in radian mode) to estimate these values.

It would be helpful to work with the equation x = pi/2 * sin(x), which is equivalent to the one above. Put in a value of x, save it to memory, take the sine of it, and then multiply by pi/2.

Is your answer close to the original (stored) value of x. See if you can make it closer by taking an x value a little larger or smaller. If you do this several times, your x value and your value of pi/2 * sin(x) should get pretty close.

Do the same for the other value of x.

 

[ACE_99]

Thanks a lot that really helped me out, I got the answer.

 

[Mentallic]

Doing this gets you 2x/pi = sin x, which can't be solved exactly.

I'm curious as to why this is unsolvable without a crude method of approximation.

 

[gabbagabbahey]

Doing this gets you 2x/pi = sin x, which can't be solved exactly.

Oh?

The key here is to realize a few things:

(1)sin x oscillates between -1 and 1, so any solutions to the above equation must be on the interval -1\leq \frac{2x}{\pi} \leq 1 or equivalently, \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}

(2)y=sin x is concave up on the interval \frac{-\pi}{2} \leq x \leq 0, so the it can only intersect a straight line (like y=2x/pi) at a maximum of two points on that interval and y=sin x is concave down on the interval 0 \leq x \leq \frac{\pi}{2}, so the it can only intersect a straight line (like y=2x/pi) at a maximum of two points on that interval.

(3)x=0,x=-pi/2 and x=pi/2 are all obvious points of intersection.

The combination of (1),(2) and (3) guarantee that the only solutions are x=0,x=-pi/2 and x=pi/2.

 

[Mentallic]

(3)x=0,x=-pi/2 and x=pi/2 are all obvious points of intersection.

ok but what if our line was something that created a less obvious intersection.
y=x/2 ?
It's not so obvious anymore. Can this one be solved?

 

[gabbagabbahey]

ok but what if our line was something that created a less obvious intersection.
y=x/2 ?
It's not so obvious anymore. Can this one be solved?

An exact solution is not possible in that case, only a numerical approximation.

 

[Mentallic]

Which brings me back to my original question
Is there a reason why there is no method to find exact values for these intersections?

 

[Mark44]

Which brings me back to my original question
Is there a reason why there is no method to find exact values for these intersections?

To solve an equation such as 2x/pi = sin(x), you have to be able to get an equation with x on one side and anything else on the other side. The usual operations (add/subtract/multiply/divide both sides), square/cube/etc., extract roots, trig functions, logs, etc. just won't do it. You'll still be left with x on each side, so you haven't solved for x.

 

[Mentallic]

To solve an equation such as 2x/pi = sin(x), you have to be able to get an equation with x on one side and anything else on the other side. The usual operations (add/subtract/multiply/divide both sides), square/cube/etc., extract roots, trig functions, logs, etc. just won't do it. You'll still be left with x on each side, so you haven't solved for x.

Are there any more advanced methods of operation for isolating x?

Also, if someone could please throw me a link that directs me to a technique that can find an approximate solution to this such question.

 

[gabbagabbahey]

Also, if someone could please throw me a link that directs me to a technique that can find an approximate solution to this such question.

How's this?

 

[Mentallic]

Ahh yes Newton's method

It didn't strike me at first because I thought it was only possible to find the zeros of a function, not the intersection.
Foolishly enough, I missed that \frac{x}{2}=sin(x) \Rightarrow f(x)=\frac{x}{2}-sin(x)
Thanks. But are there still any methods of finding exact values?

 

[gabbagabbahey]

Thanks. But are there still any methods of finding exact values?

In general, \sin x=m x has no exact solutions (edit: Other than x=0!) as is typical for a transcendental equation It is only for certain special values of m that exact solutions exist; and still the most applicable method is graphing the functions.

 

[Mentallic]

has no exact solutions.

I just can't accept this. There must be some (probably) complicated expression for its exact solution. What I can accept though, is that a method in finding this complicated solution has yet to be invented/discovered.