Solving for a fixed point on an interval

[ChickysPusss]

Homework Statement

OK, I need to solve for the fixed point of the equation

2sinπx + x = 0

on the interval [1, 2]. I know the answer to be ~1.21... but I need to prove it.

The Attempt at a Solution

I really just need help solving for a proper equation of x. I tried x = -2sinπx, but that doesn't work on the interval for the method I am using to solve for the fixed point (Banach).

So I tried this

2sinπx = -x
sinπx = (-x/2)
πx = arcsin(-x/2)
x = (1/π)*arcsin(-x/2)

But this equation seems totally wrong, as anything past 0 is negative. Can anyone tell me what I'm doing wrong, or perhaps that there is another way to solve for x? Thanks.

 

[Simon Bridge]

$x+2\sin\pi x=0$

... you could use any numerical method.
... you could approximate the trig function.
... or is it more that you want to prove that the given x of about 1,21... is a fixed point of some function?

 

[ChickysPusss]

I need to use Banach fixed point algorithm to solve the equation I listed. I was just wondering if there was some other way to create an "x = f(x)" function with the equation.

For example if I use x = -2sinπx, and attempt to solve for the fixed point by starting at 1, we see that it solves to x = 0, which IS a fixed a point, but not the one on the interval I need. The reason for this is deals with the fact that on the interval with that equation, it is not converging.

Perhaps my real question is:

Why is x = -2sinπx not equal to x = (1/π)*arcsin(-x/2)?

 

[tiny-tim]

Hi ChickysPusss!

2sinπx + x = 0

Sorry, but there's no exact solution to this …

you'll need to find an approximation method. ​

 

[Simon Bridge]

Did you try plotting $y(x)=x+2\sin(\pi x)$ to help select a decent starting point.
What happens if you try different starting points?

 

[HallsofIvy]

I need to use Banach fixed point algorithm to solve the equation I listed. I was just wondering if there was some other way to create an "x = f(x)" function with the equation.

For example if I use x = -2sinπx, and attempt to solve for the fixed point by starting at 1, we see that it solves to x = 0, which IS a fixed a point, but not the one on the interval I need. The reason for this is deals with the fact that on the interval with that equation, it is not converging.

Perhaps my real question is:

Why is x = -2sinπx not equal to x = (1/π)*arcsin(-x/2)?

It is but I would not recommend that as arcsin is a more complicated function. The problem is that you tried using $x= -2sin(\pi x)$ once then gave up. Since $\pi 1= \pi$ and $sin(\pi)= 0$ obviously it is going to give 0.

You know that the is a solution around 1.2 so try that as a starting point.

 

[ChickysPusss]

I didn't give up, I tried, 1.1, 1.2, and a couple other starting points, but the only one that wielded a solution was when I typed in 1.1216... which was close enough to the solution that it didn't even have to start the algorithm. The reason it won't work for the function on that interval is because if we call g(x) = -2sinπx .

g'(x) = is not converging on the interval [1,2] (ie the maximum value is not less than one on that interval.)

Also, you say

x = -2sinπx = x = (1/π)*arcsin(-x/2)

so how come when I graph it, they don't look the same?

 

[haruspex]

if we call g(x) = -2sinπx, g'(x) = is not converging on the interval [1,2]

g'(x) = -2πcosπx, right? The usual expansion converges everywhere, so I don't understand what you're saying there.

x = -2sinπx = x = (1/π)*arcsin(-x/2)
so how come when I graph it, they don't look the same?

They are equivalent statements regarding a specific value x. They will produce the same value for that x. Graphing them may produce rather different pairs of curves, but the intersection in each case will be the same.

 

[ChickysPusss]

They are equivalent statements regarding a specific value x. They will produce the same value for that x.

OK, I mean absolutely no disrespect if this is a stupid question but I have to know what I am not understanding about this statement. You say they will produce the same value for that x.

Are you saying that "-2*sin(π * 1.216) = (1/π)*arcsin(-1.216/2)"? This is when x = 1.216, and they do not equal each other. It equates to "1.255 = -.2080"

So what I assume is that I am not understanding your statement, and I'm sorry for that, I just want to know what you mean by it, and perhaps an example of it? Thank you for your patience everyone.

 

[haruspex]

Are you saying that "-2*sin(π * 1.216) = (1/π)*arcsin(-1.216/2)"? This is when x = 1.216, and they do not equal each other. It equates to "1.255 = -.2080"

The arcsin function is defined to return a value between -pi/2 and +pi/2. But in the equation you have to consider other possibilities, e.g. pi-arcsin(x).
Btw, I don't get 1.216 - more like 1.206.

 

[Dick]

OK, I mean absolutely no disrespect if this is a stupid question but I have to know what I am not understanding about this statement. You say they will produce the same value for that x.

Are you saying that "-2*sin(π * 1.216) = (1/π)*arcsin(-1.216/2)"? This is when x = 1.216, and they do not equal each other. It equates to "1.255 = -.2080"

So what I assume is that I am not understanding your statement, and I'm sorry for that, I just want to know what you mean by it, and perhaps an example of it? Thank you for your patience everyone.

You are trying to solve x=(-2)sin(pi*x) numerically by fixed point iteration. That won't work because the derivative of 2sin(pi*x) is too large near the fixed point. As you discovered, it won't converge. Then you tried to change the form to x=(1/pi)*arcsin(-x/2). That won't work either, because if you take a value near your fixed point, like 1.2 then arcsin(sin(1.2*pi)) isn't equal to 1.2*pi. The two equations aren't equivalent. arcsin will only correctly invert sin on the range [-pi/2,pi/2]. The trick is to shift the argument of the sin so that it will. pi*x-pi is in the range [-pi/2,pi/2] And since sin(a-pi)=(-sin(a)), sin(pi*x)=(-sin(pi*x-pi)). Find the arcsin form using that shift and fixed point iterate again. It will work this time.

 

[ChickysPusss]

Wow, I had no idea about that rule, looks like I shouldn't have passed trig. :/ Haha, thanks everyone who helped me figure this out, very much appreciated. Going to post my final solution because I hate when I find a topic and there's no final sum up of everything to help it all make sense.

pi = the mathematical constant pi in the following equations.

So we are looking for the solution to: 2*sin(pi * x) + x = 0 on the interval [1,2] using Banach's fixed point theorem.

Solving for x, and concluding x = -2*sin(pi * x) will not wield a solution using the algorithm because the derivative is too large near the fixed point ( >= 1.0).

So we need to find another way to solve for x. Dick and haruspex pointed out that we can't use arcsin on the interval we are working with currently, because it will only work on the interval from [-pi/2, pi/2].

So we shift the argument of pi first:
2*sin(pi*x) + x = 2 * -sin(pi*x - pi) + x //uses trigonometric identities

NOW we solve for x:
2*-sin(pi*x - pi) + x = 0 //Initial equation
2*-sin(pi*x - pi) = -x //Bring the x to the right
-sin(pi*x - pi) = -x/2 //Bring the 2 to the right
sin(pi*x - pi) = x/2 //Drop the negative signs
pi*x - pi = arcsin(x/2) //Get rid of sin on the left using arcsin
pi(x - 1) = arcsin(x/2) //Factor out pi from the left
x - 1 = (1/pi) * arcsin(x/2) //Bring the pi to the right
x = (1/pi) * arcsin(x/2) + 1 //Bring the one to the right

so g(x) = (1/pi) * arcsin(x/2) + 1
g'(x) = 1/(pi * (sqrt(4-x^2))) //Used my calculator to solve this

1.) Prove the function g(x) maps from [1, 2] -> [1, 2]

We only need to solve g(1) and g(2) because these will be the min and max values, since there is no solution to the equation g'(x) = 0.
g(1) = 1.1666
g(2) = 1.5
OK, so that's proof.

2.) Prove g(x) is a contraction on [1, 2]
We have to show that maximum value from [1,2] of g'(x) is less than 1.
Oh, what the helllllll? No, I don't care I'm rewriting the range, the book says on that interval but its undefined at 2 because of the divide by zero. The range is now [1, 1.5] because sometimes you have to take math into your own hands.

We have to show that maximum value from [1,1.5] of g'(x) is less than 1.
Well, the MAXIMUM value would have to be at 1.5, because it gives the smalls possible number at the denominator, and that solution is g'(1.5) = .2406, so yes, it works on this range. Above, if you change the range it works too. As practice, try doing it yourself!

OK, so this will work with Banach's Fixed Point algorithm, and the solution is about 1.206035, so haruspex was correct. Woo math, thank you one and all who helped, you're truly gods among men.

 

[Dick]

Very welcome! Nice summary. It conveys well the narrative thrills of finding a solution. It's great when somebody picks up on hints so well. But "gods among men" is a little strong.

 

[Simon Bridge]

But "gods among men" is a little strong.

besides - it's the priests who get the tithes ;)