Quote:
Originally Posted by BrianTheMick2
I thought that the entire problem with the proof involved expanding x^2 to x*x and then expanding one of the x's* to be just a bunch of integers (of unknown quantity*) that add up to x instead of, I don't know, just remembering that x is a variable.
It turns out, I believe that x=x. He could have used the same "proof" to show that 2=0 by using the same technique on both of the x's*.
*I don't know the plural of x.
**we commonly call "unknown quantity" a "variable" and this "variable" is amenable to calculus.
The expression
(1) (X+X+...+X) [X times]
on the face of it, is only defined for positive integers. So Mason and others itt have a valid point. You can't take the derivative of such a function because you can't take the limit as h --> 0 in the expression
(2) [f(x+h) - f(x)] / h
However, as I point out in my post above (does anyone read my posts?), you can understand expression (1) to be defined for real numbers as I describe above. For example, if X = 3.5 then (1) becomes
(3.5 + 3.5 + 3.5) + 3.5*0.5 = 12.25
Under this understanding you can take the derivative of expression (1) and in that case the problem with OP's proof is that it fails to treat the [X times] part of expression (1) as a variable, as Brian and others itt point out. I show various ways itt how the derivative can be properly calculated if you do treat "[X times]" as depending on X.
Here's another way. Notice that under the understanding of being defined for real numbers (1) is a special case of
(3) (X+X+...+X) [Y times]
where both X and Y are real numbers. From multivariate calculus, if we have a differentiable function of two variables,
(4) f(x1,x2)
we can form the differentiable function of one variable
(5) G(X) = f(X,X)
Then the derivative of G is the sum of the partial derivatives of f.
(6) G'(X) = f'_x1(X,X) + f'_x2(X,X)
Doing this for the function G(X) = (X+X+...+X) [X times] we get
(7) G'(X) = (1+1+...+1) [X times] + (X+X+...+X)[1 time] = X+X = 2X
You can see from this that the OP proof leaves out the second summand in (7) which is where [X times] is treated as a variable dependent on X when taking that partial derivative.
PairTheBoard