jpat1023 Posted September 1, 2004 Share Posted September 1, 2004 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. Link to comment Share on other sites More sharing options...
Primarygun Posted September 1, 2004 Share Posted September 1, 2004 Can I use the log sign>?kidding Link to comment Share on other sites More sharing options...
Gauss Posted September 1, 2004 Share Posted September 1, 2004 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 Link to comment Share on other sites More sharing options...
bloodhound Posted September 1, 2004 Share Posted September 1, 2004 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] Link to comment Share on other sites More sharing options...
jpat1023 Posted September 2, 2004 Author Share Posted September 2, 2004 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. Link to comment Share on other sites More sharing options...
NSX Posted September 2, 2004 Share Posted September 2, 2004 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? Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 2, 2004 Share Posted September 2, 2004 There is a unique solution : approx : .318 - 1.337 I Mandrake Link to comment Share on other sites More sharing options...
Gauss Posted September 2, 2004 Share Posted September 2, 2004 You are right Bloodhound I did not consider the case when [math]x \in \mathbb {C} [/math] Link to comment Share on other sites More sharing options...
Gauss Posted September 2, 2004 Share Posted September 2, 2004 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. Link to comment Share on other sites More sharing options...
Gauss Posted September 2, 2004 Share Posted September 2, 2004 Mandrake, how did you arrive at your solution? Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 2, 2004 Share Posted September 2, 2004 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 Link to comment Share on other sites More sharing options...
bloodhound Posted September 2, 2004 Share Posted September 2, 2004 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) Link to comment Share on other sites More sharing options...
bloodhound Posted September 2, 2004 Share Posted September 2, 2004 i cant seem to get the software to get me the approx value of LamW(-1) Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 3, 2004 Share Posted September 3, 2004 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 Link to comment Share on other sites More sharing options...
Primarygun Posted September 3, 2004 Share Posted September 3, 2004 What's the answer? Link to comment Share on other sites More sharing options...
bloodhound Posted September 3, 2004 Share Posted September 3, 2004 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 Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 3, 2004 Share Posted September 3, 2004 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 Link to comment Share on other sites More sharing options...
Teotihuacan Posted September 3, 2004 Share Posted September 3, 2004 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! Link to comment Share on other sites More sharing options...
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