# Function f(a)=a

1. Mar 27, 2007

### Doom of Doom

I find it interesting that some functions have specific values of a where f(a)=a.
Is there a way to find x that satisfies these following equations?:

e^-x=x or -log(x)=x
cos(x)=x

They seem like they should be interesting values, and I was trying to figure out if they could be represented using pi, e, or what not. Also, sin(x)=x has only the trivial value of 0 while tan(x)=x has infinitley many solutions.

2. Mar 27, 2007

### JasonRox

That's actually harder than it looks because you have a algebraic equation on one said and a exponential or trigonometric function on the other. Generally, they aren't solvable unless it is trivial, like you pointed out for sin(x).

3. Mar 27, 2007

### Integral

Staff Emeritus
I believe you are referring to what is called a fixed point of the function. If the derivative of the function is <1 around the fixed point then you may be able to find the fixed point by iteration. Make a guess, compute the functional value, repeat with the generated value as your new x.

4. Mar 28, 2007

### christianjb

for x=e^-x

Defn. y=e^-x - x

Obj. find where y(x)=0
Use Newton Raphson's method

x(n+1)=x(n)+(e^-x - x)/(e^-x +1)

x(0)=1, it then goes:
1 0.5378828427399903
2 0.5669869914054132
3 0.5671432859891229
4 0.5671432904097839
5 0.5671432904097839

Which can be confirmed by substitution e^-x=x

5. Mar 28, 2007

### christianjb

For cos(x)
x(n+1)=x+(cos(x)-x)/(sin(x)+1)

x(0)=1
1 0.7503638678402439
2 0.7391128909113617
3 0.7390851333852839
4 0.7390851332151607
5 0.7390851332151607
6 0.7390851332151607

6. Mar 28, 2007

### Integral

Staff Emeritus
Or as a simple fixed point iteration, its slower but a better demonstration of the idea of a fixed point

let $$x_{n+1} = e^{-x_{n}}$$

let $$x_0 = 1$$

1.000000000
0.367879441
0.692200628
0.500473501
0.606243535
0.545395786
0.579612336
0.560115461
0.571143115
0.564879347
0.568428725
0.566414733
0.567556637
0.566908912
0.567276232
0.567067898
0.567186050
0.567119040
0.567157044
0.567135490
0.567147714

now for the cos

let $$x_{n+1} = cos(x_n)$$

$$x_0 = 1$$

1
0.540302306
0.857553216
0.65428979
0.793480359
0.701368774
0.763959683
0.722102425
0.750417762
0.731404042
0.744237355
0.73560474
0.741425087
0.737506891
0.740147336
0.738369204

7. Mar 28, 2007

### Sourabh N

there is another way to do that. If u can sketch the graphs of two functions (using a graph software,etc.) u can find the point where the 2 curves meet.

8. Mar 28, 2007

### Doom of Doom

Thank you for telling me the name for this kind of math. I have now gone to my college library and checked out a book on Fixed Point Theory.

9. Mar 28, 2007

### ZioX

The Newton-Raphson method is a special kind of fixed point iteration as well.

Fixed points in practice are used to find the roots of a function/equation. Something like 2x+cos(x)sin(x)exp(-x/2)=0. It's easy to see that if the function g(x)=-cos(x)sin(x)exp(-x/2)/2 has a fixed point then this fixed point is the root of the original equation. Using theorems on fixed points we might prove the existence of such a point.