# Points of Intersection

1. Jan 21, 2009

### ACE_99

1. The problem statement, all variables and given/known data
Determine the points of intersection for the following equations

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

2. Relevant equations

None

3. 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.

2. Jan 22, 2009

### Staff: Mentor

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.

3. Jan 22, 2009

### ACE_99

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

4. Jan 22, 2009

### Mentallic

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

5. Jan 22, 2009

### gabbagabbahey

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.

6. Jan 22, 2009

### Mentallic

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?

7. Jan 22, 2009

### gabbagabbahey

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

8. Jan 22, 2009

### 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?

9. Jan 22, 2009

### Staff: Mentor

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.

10. Jan 22, 2009

### Mentallic

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.

11. Jan 22, 2009

### gabbagabbahey

How's this?

12. Jan 22, 2009

### 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?

13. Jan 22, 2009

### gabbagabbahey

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.

Last edited: Jan 22, 2009
14. Jan 23, 2009

### Mentallic

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.