You are Here: Home

# Is this proof of the exponential function valid? Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Enter our travel-writing competition for the chance to win a Nikon 1 J3 camera 20-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. Is this proof of the exponential function valid?
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?
2. Re: Is this proof of the exponential function valid?
(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!
3. Re: Is this proof of the exponential function valid?
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.
4. Re: Is this proof of the exponential function valid?
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.