Locating roots by iteration

[burgerandcheese]

how did they get "kx + e^-x has a gradient of about k - 0.5" ?

 

[ksdhart2]

The function \(f(x) = x - e^{-x}\) has a real root which can be expressed in terms of the Lambert-W/Product Log Function:

\(\displaystyle f(x) = 0 \implies x = W(1) \approx 0.5671 \dots\)

And since this root is a solution to the equation \(x = e^{-x}\), we can conclude that, at the point we're interested in, \(e^{-x} = W(1) \approx 0.5\). The gradiant (i.e. derivative) of \(e^{-x}\) at this point is...?

 

[HallsofIvy]

The gradient of f+ g is the gradient of f plus the gradient of g: (f+ g)'= f'+ g' where I am using ' to indicate the gradient of a function.

 

[burgerandcheese]

The function \(f(x) = x - e^{-x}\) has a real root which can be expressed in terms of the Lambert-W/Product Log Function:

\(\displaystyle f(x) = 0 \implies x = W(1) \approx 0.5671 \dots\)

And since this root is a solution to the equation \(x = e^{-x}\), we can conclude that, at the point we're interested in, \(e^{-x} = W(1) \approx 0.5\). The gradiant (i.e. derivative) of \(e^{-x}\) at this point is...?

I don't think I've learnt about that yet, but thank you

The gradient of f+ g is the gradient of f plus the gradient of g: (f+ g)'= f'+ g' where I am using ' to indicate the gradient of a function.

This is so embarrassing I can't believe I didn't realise that