• Oct 18th 2012, 12:52 AM
vamosromil
Let's say f(x) = x^2.
This would give f'(x) = 2x.
Now we change the definition of f(x) = x^2 to f(x) = x + x + x... up to x times.
If we now differentiate it with respect to x, we will get f'(x) = d(x)/dx + d(x)/dx + d(x)/dx ....up to x times. And this sums up to x. Thus according to new definition of f(x), the derivative f'(x) would be x and not 2x.
Though we all know that there is some gap in calculating the derivative in second method, what is the missing thing here? How can it be captured to cover every aspect?
• Oct 18th 2012, 02:19 AM
chiro
Hey vamosromil.

The subtle difference is that 1) this only applies if x is an integer and 2) such a summation assumes that we sum up a known amount of times (i.e. n*x where n is fixed).

If you add up d(x)/dx n times you get n which is what we expect.

The derivative of nx is n and does not depend on x at all.

You can't do what you did because what you are doing (and the way you are doing it) is ill-defined. If you are summing up a fixed amount of times is just (x + x + ... + x) n times and not "x" times (what if x is 1.123123798123978123 or pi)?

You have to be careful with how you specify things.
• Oct 18th 2012, 02:21 AM
FernandoRevilla
Quote:

Originally Posted by vamosromil
Now we change the definition of f(x) = x^2 to f(x) = x + x + x... up to x times.

So, $\displaystyle f(\sqrt{2})=\sqrt{2}+\sqrt{2}+\ldots +\sqrt{2}$ up to $\displaystyle \sqrt{2}$ times? :)

Edited: Sorry, I didn't see chiro's post.
• Oct 18th 2012, 02:40 AM
MathoMan
I wonder how can you write down 2.453+2.435+?... exactly 2.435 times?
• Oct 18th 2012, 02:44 AM
vamosromil