# Why does fixed point iteration have to be in radians?

This might seem like a dumb question but i can't find an explanation anywahere.

I was working on this problem:

Apply the fixed-point iteration method to this function $$f(x)=2(\sin x+\cos x)−x$$ with $x_1=2$

I rearranged the equation into an "fixed point equation":

$$f(x)=0\Rightarrow f(x)=2(\sin{x}+\cos{x})−x=0$$ $$x=2(\sin{⁡x}+\cos{⁡x})$$ $$\frac{x}{2}=\sin⁡{x}+\cos⁡{x}$$ $$x=\sin{x}+\cos{x}+\frac{x}{2}$$ $$g(x)=x⇒g(x)=\sin x+\cos x+\frac{x}{2}$$

I iterated a sufficient amount of times to find $x \approx 1.71$

I had done all of the above in radians for simplicity however i then tried it with my calculator working in degrees and $$r=2, x=2.034$$

I iterated a sufficient amount of times to find $x \approx 2.71$ which differs from the radians answer by 1. This seams suspicious as the real convergence does go to $x \approx 1.71$, why can (or cant if i'm wrong) fixed point only be in radians? and why was my degrees result exactly 1 greater?

What definition of trig functions (also called "circular functions") are you using?

The definitions in terms of right angles cannot be used in general as that requires the angle to be between 0 and 90 degrees but, in order to use them as functions we want them defined for all real numbers.

Of the "circle" definition is used: Construct a unit circle (center at (0, 0), radius 1) in an xy-coordinate system. Given non-negative number, t, measure, counter clockwise, around the circumference a distance t (for negative t, measure clockwise). The endpoint has coordinates $(\cos(t), \sin(t))$. That is, whatever the end point is, the x coordinate is, by definition, $\cos(t)$, the y coordinate is $\sin(t)$.

Notice that "$t$" is NOT in degrees because it is not an angle at all, it is a distance around the circumference. We can think of that as an angle by identifying the angle with the distance subtended on a unit circle which is simply the definition of "radian".

• thank you for the answer, i looked at your answer in edit mode but i cant find it, why is it like a scroll wheel?? – Sonny Da Silva-Peters Jun 18 '17 at 12:30

Yo can also find the solution in degrees if you make a change of variables: $$x\rightarrow \frac{\pi}{180}y$$

But you have to change all $x$ not only in the trigonometric functions. That is probably the reason you can not reach the same result seeking the root in radians or degrees.

equation in degrees: $$2\left[\sin{\left(\frac{\pi}{180}y\right)}+\cos{\left(\frac{\pi}{180}y\right)}\right]-\frac{\pi}{180}y=0$$

• okay, i just tried it out and i cant seem to make it work if i set $x\rightarrow \frac{\pi}{180}y$ in the first iteration can i then use the answer for the next iteration, or do have to use $\frac{\pi\mbox{answer}}{180}$ both aren't working for me at the moment – Sonny Da Silva-Peters Jun 18 '17 at 12:21
• Your new equation is a function of a new variable (in degrees) $y$. Please write the equation to solve as a fixed point: $$y=\frac{360}{\pi}\left[\sin{\left(\frac{\pi}{180}y\right)}+\cos{\left(\frac{\pi}{180}y\right)}\right]$$ and solve for $y$ with a seed for example $y^0=0$ – HBR Jun 18 '17 at 12:28