e^x=x

 

Anybody got a solution for this?

 

I'll post the solution later, unless someone comes up with a better one and some sort of proof for it.

Can I use the log sign>?kidding

The problem is to find the real zeros of a nonlinear function f(x)=0.

 

Let [math]y = e^x[/math] and [math]y = x[/math]

 

Hence [math]e^x = x[/math]

 

[math]e^x - x = 0[/math]

 

Since the two function do not intersect there is no solution

The problem is to find the real zeros of a nonlinear function f(x)=0.

 

Let [math]y = e^x[/math] and [math]y = x[/math]

 

Hence [math]e^x = x[/math]

 

[math]e^x - x = 0[/math]

 

Since the two function do not intersect there is no solution

 

buts thats only for [math]x \in \mathbb{R}[/math] . there might still be a solution where [math]x \in \mathbb{C}[/math]

Guass,

thats the only thing i could come up with too. The only reason i posted it was to see if someone could come up with something else because i have notice that there are some pretty good math people here.

buts thats only for [math]x \in \mathbb{R}[/math] . there might still be a solution where [math']x \in \mathbb{C}[/math]

 

What does the C mean?

 

[edit]

n/m

 

Does it mean complex?

There is a unique solution : approx : .318 - 1.337 I

 

Mandrake

You are right Bloodhound I did not consider the case when [math]x \in \mathbb {C} [/math]

For the case when [math] x \in \mathbb{C} [/math]

 

Using two terms of the taylor series expansion we have

 

[math] e^x - x = 0 [/math]

 

[math] 1 + x + \frac {x^2}{2} - x = 0 [/math]

 

[math] 1 + \frac {x^2}{2} = 0 [/math]

 

[math]x_1 = (0 + i\sqrt {2})[/math]

 

[math]x_2 = (0 - i\sqrt {2})[/math]

 

Using three terms of the taylor series expansion we have

 

[math] 1 + x + \frac {x^2}{2} + \frac {x^3}{6} - x = 0 [/math]

 

[math]x_1 = (0.246017 + i1.28751)[/math]

 

[math]x_2 = (0.246017 - i1.28751)[/math]

 

[math]x_3 = -3.49203[/math]

 

The number of solutions you require depends on how many terms you use in the taylor series expansion.

Mandrake, how did you arrive at your solution?

In fact i used the so called Lambert W functions, the property of this function W(z) being

W(z)exp(W(z)) = z;

 

Taking z = -1 here will gives a solution to the original equation, being x= -W(-1) =(approx) .318 - 1.337 I

 

I think this is the only solution.

 

Mandrake

i tried to use MAPLE to solve it. and it gave me this

 

the command was

 

solve(e^x=x,x); which basically says solve e^x=x for x

 

and the output was.

 

[math]\frac{-LambertW(-ln(e))}{ln(e)}[/math]

 

dont know why it doesnt automatically simplify

 

so its just [math]$LambertW$[/math](-1)

i cant seem to get the software to get me the approx value of LamW(-1)

How about using evalf or something like that ?

That is a long time ago i used maple but i think that is a valid command right ?

 

Mandrake

What's the answer?

evalf think uses floating point arithmetic . tried that. seems to give error

 

tried evalc which seemed to be the more correct command. and got some long giberrish with some functions i havent even seen before

evalf think uses floating point arithmetic . tried that. seems to give error

 

tried evalc which seemed to be the more correct command. and got some long giberrish with some functions i havent even seen before

 

Isnt the whole point of what you are trying to do : approximating the lambertW function with some complex number ?

Analytical expressions would simply give x = -LambertW(-1) as a solution.

 

have you tried using command line maple also ? evalf(-LambertW(-1)); ?

 

Mandrake

undefinedundefinedundefinedFunny, in a second year Stats course, we were taught that the Sumation of e^x = the function of u^n.

 

The "proof", was of course, empirical. And, I think, every student concurred!