Need help getting exp equation to lambert form

[nalo]

I have an equation that I have numerically solved in MATLAB - but can't figure out how MatLab has done it! It has used a Lambert W equation to get an answer for my formula which is in the style:

z[1-e^(-1/z)]=1

which MuPad simplifies down to:

z = 1/(W_k(-1/e¹) + 1)

But I've been trying to figure out how to even get z into the correct form for Lambert W [Xe^X = Y -> X = W(Y)] and can't figure it out.

Hopefully someone can open my eyes as I can't help but think it should be a simple enough solution.

Thanks for any help!

Nalo

 

[LCKurtz]

I have an equation that I have numerically solved in MATLAB - but can't figure out how MatLab has done it! It has used a Lambert W equation to get an answer for my formula which is in the style:

z[1-e^(-1/z)]=1

which MuPad simplifies down to:

z = 1/(W_k(-1/e¹) + 1)

But I've been trying to figure out how to even get z into the correct form for Lambert W [Xe^X = Y -> X = W(Y)] and can't figure it out.

Hopefully someone can open my eyes as I can't help but think it should be a simple enough solution.

Thanks for any help!

Nalo

Rewrite your equation as:

$$e^{-\frac 1 z} = 1 - \frac 1 z$$

Let u = 1 - 1/z giving e^u-1 = u. This gives:

1 = ue^1-u) = ue¹e^-u. Now let v = -u, giving ve^v= -1/e. So v = W(-1/e). Now you can just back substitute for z.

 

[Dick]

Back substitute for z if you dare. (-1)e^(-1)=(-1/e) so W(-1/e)=(-1). Your original equation doesn't have any real roots.

 

[nalo]

Thank you very much LCKurtz! My maths brain is mucho rusty.

Back substitute for z if you dare. (-1)e^(-1)=(-1/e) so W(-1/e)=(-1). Your original equation doesn't have any real roots.

Ah I noticed that just after I posted. It's just a style example of a more long winded question I've got.

Just worked through my actual question there and got the same result as MATLAB (give or take a very little) so thank you all for your help!