e^(-x) = 1-x/3 ?

[bansidhe]

Hi

I have been trying to solve for X in this equation. (I did plot and got two values for x...)

Anyway, I thought I would proceed by expanding e^(-x) and then solving for x..

I got:


1 - x + x^2/2 - x^3/6 +... ~ 1-x/3

I wind up after simplification of getting rid of one and dividing through by x ]w
ith a quadratic equation which gives an imaginary number..

(x^2) / 3 - x + 4/3=0

(sqrt of # < 0) CLearly this is wrong

what am I doing wrong here? This is driving me nuts!

thanks

bansidhe

 

[HallsofIvy]

Hi

I have been trying to solve for X in this equation. (I did plot and got two values for x...)

Anyway, I thought I would proceed by expanding e^(-x) and then solving for x..

I got:


1 - x + x^2/2 - x^3/6 +... ~ 1-x/3

I wind up after simplification of getting rid of one and dividing through by x ]w
ith a quadratic equation which gives an imaginary number..

(x^2) / 3 - x + 4/3=0

(sqrt of # < 0) CLearly this is wrong

what am I doing wrong here? This is driving me nuts!

Yes, you can write e^{-x} as 1- x+ x^2/2- x^3/6+...= 1- x/3, then subtracting 1 and adding x/3 to both sides
-2x/3+ x^2/2- x^3/6+ ...= 0. Apparently what you did was drop the rest of the sum: x^4/4!- x^5/5!+ x^6/6!+ ... which means that you are not solving the equation you are given.

thanks

bansidhe

This equation, like most equations that involve both a transcendental function, cannot be solved using elementary functions. It looks like you ought to be able to put it into a form in which you could use the "Lambert W function" which is defined as the inverse function to f(x)= xe^x.

 

[bansidhe]

e^(-x)=1-x/3

Thanks... It has been over 12 years since I have done any calculus at all. Anyway, I am glad it wasn't as straightforward as all that...
I did drop the rest of the series... making the assumption that later terms would not be significant. I suppose that was my first mistake!
Thanks for the Lambert function tip. I'll have to give it a try.