[Sasukekun]

Set , then



Hence, as f'(z) = 0, f(z) is constant, this constant is equal to;



Then set





Then




Intuitively it's obvious that , but I have an issue with that proof. It assumed, when differentiating w.r.t. z, that c was not a function of z. However, when we put then c is a function of , so differentiating it w.r.t would not produce a zero gradient. Any help?

[latentcorpse]

(Original post by Sasukekun)
Set , then



Hence, as f'(z) = 0, f(z) is constant, this constant is equal to;



Then set





Then




Intuitively it's obvious that , but I have an issue with that proof. It assumed, when differentiating w.r.t. z, that c was not a function of z. However, when we put then c is a function of , so differentiating it w.r.t would not produce a zero gradient. Any help?

The point is that c is not a function of z I think. In other words, you set c to be z1+z2 and it doesn't change from that as z varies.

z1 is not a variable. It is a particular value of z.

To be honest, I'm not sure I fully understood your question so if this post is useless let me know!

[Bobifier]

I am not sure why you have introduced your z1 and z2. Assuming c is a constant, the obvious thing to do would simply be to declare your function equal to e^c. This continues to be clear when you achieve f = e^(z1 + z2), which if you refer back to your definition of your new variables is clearly equal to e^c.

[Zii]

if you change c to c(z) then your original assumption of c being constant is no longer correct, instead you now have the function exp(c(z)) which differentiates to c'(z)exp(c(z)). So what? if c(z) = c = constant we recover the original solution: 0.