I found this browsing the net. Any comments?
The derivative of x², with respect to x, is 2x. However, suppose we write x² as the sum of x x's, and then take the derivative:
Let f(x) = x + x + ... + x (x times)
Then f'(x) = d/dx[x + x + ... + x] (x times)
= d/dx[x] + d/dx[x] + ... + d/dx[x] (x times)
= 1 + 1 + ... + 1 (x times)
This argument appears to show that the derivative of x², with respect to x, is actually x. Where is the fallacy?
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
I don't know...the only thing I can think is that
x+x+x+.....+x (x times) is only defined for x in Z+
So that is isn't continuous anywhere or differentiable, really...
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
Also the equation cannot be written down without ellipsis (...) in it. Or you could write a summation.
But the number of terms changes with x, so the equation keeps changing, plus for negative x or fractional x's,
how many terms are there?
igloo myrtilles fourmis
The fallacy is that you're treating the (x times) bit as a constant.
What you're actually doing there is differentiating kx to get k.
Why did the vector cross the road?
It wanted to be normal.