e^x=x

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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.

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Can I use the log sign>?kidding

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

Let $y = e^x$ and $y = x$

Hence $e^x = x$

$e^x - x = 0$

Since the two function do not intersect there is no solution

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

Let $y = e^x$ and $y = x$

Hence $e^x = x$

$e^x - x = 0$

Since the two function do not intersect there is no solution

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

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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.

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

What does the C mean?

n/m

Does it mean complex?

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

Mandrake

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You are right Bloodhound I did not consider the case when $x \in \mathbb {C}$

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For the case when $x \in \mathbb{C}$

Using two terms of the taylor series expansion we have

$e^x - x = 0$

$1 + x + \frac {x^2}{2} - x = 0$

$1 + \frac {x^2}{2} = 0$

$x_1 = (0 + i\sqrt {2})$

$x_2 = (0 - i\sqrt {2})$

Using three terms of the taylor series expansion we have

$1 + x + \frac {x^2}{2} + \frac {x^3}{6} - x = 0$

$x_1 = (0.246017 + i1.28751)$

$x_2 = (0.246017 - i1.28751)$

$x_3 = -3.49203$

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

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Mandrake, how did you arrive at your solution?

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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

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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.

$\frac{-LambertW(-ln(e))}{ln(e)}$

dont know why it doesnt automatically simplify

so its just $LambertW$(-1)

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

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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

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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

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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

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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!

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