Mason, why did you ignore all the posts correctly pointing out the fault and only reply to the ones incorrectly doing it?

Of course Mason knew full well where the flaw in the proof was before he posted it. He was just giving us something to have a little fun with. He may have been out-leveling me but the only thing he didn't seem to understand was my comment above. I was making the joke that NOT dividing by zero was where the proof went wrong.

btw, It makes perfectly good sense to understand something like "add 2 to itself 0.5 times" to mean half of what you get if you add 2 to itself 1 time. So adding 2 to itself 1.5 times is understood as 3. If we understand "add X to itself X times" in this way you can then form the difference quotient as I did in the post where I made that comment. Letting the denominator go to 0 in the limit then gives the definition of the derivative. So my joke even had a bit of truth to it because the denominator going to 0 in the limit, which gives the correct result, is actually a little tiny bit like dividing by 0.


PairTheBoard

Hi Everyone:

X+X+X...+X is not a continuous function. That's the flaw, and to take a derivative we need a continuous function.

However, I always thought this was the most nifty proof of 2=1 that I ever saw and apparently very few other people have seen it.

Best wishes,
Mason

Yeah....thats what everybody told you. (and tbf its not only not continuous, its defined almost nowhere....so without some type of highbrow analytic continuation a derivative is doa) You're also using "rules" for a derivative rather than the definition. HW level problems in a freshman level calc book could probably convince you the proof isnt so "nifty" if you bother to put in the work.

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

Suppose we do restrict the OP expression

(1) (X+X+...+X)[X times]

to only integer values for X. Look at what happens if we form the difference quotient

[f(x+h) - f(x)]/h

with X=10 and h=1

The proper evaluation treating [X times] as variable is

( (11+11+...+11)[11 times] - (10+10+...+10[10 times]) / 1 =

= (121 - 100)/1 = 21 or about 2*10

While the OP's evaluation, treating [X times] as nonvarable would be,

( (11+11+...+11)[10 times] - (10+10+...+10)[10 times]) / 1 =

= (110 - 100)/1 = 10 or 1*10

So the problem with the OP proof has little to do with expression (1) being defined only for integers and everything to do with the expression [X times] being treated as nonvariable in the proof.

PairTheBoard

This might be what I was trying to say in Post #9. I don't think the problem has anything to do with continuity of functions.

That is what I said, right?

The OP defined a sum of Sum[n, for i=1 to n] where x=n.

You cannot define fractional sums (at least with how we historically use them because you can just define about anything of course) . You can define non-integer times something of course. But not non-integer number of terms.

Of course x^2 is x times x but not as in summation

Like imagine someone talking about Sum[1/n^2 but n=1 to 10.56] lol. What that would mean?

Let the domain of the function f(X) be the positive integers with

f(X) = X multiplied by itself.

Then, since the derivative of X is 1 the derivative of f(X) is

f'(X) = 1 multiplied by itself = 1

But X multiplied by itself is X^2 and the derivative of X^2 is 2X. Therefore

1 = 2X for every positive integer X.
------------------------

Now, technically you could say the problem with this proof is that you shouldn't ought've been talking about the derivative of f(X) to begin with because f(X) is not continuous, having domain only the positive integers. But I think that misses the point of what's really wrong with the proof.


PairTheBoard

In other words, f(X) = g(X)g(X), where g(X) = X.

Then, since the derivative of g is g'(X) = 1...

... the derivative of f(X) is, by the product rule, g'(X)g(X) + g(X)g'(X) = X + X. So your third line breaks the product rule.

If that was what the OP wrote then it would have been clear the function was only meant to be defined for Integers. But that's not what the OP wrote. The OP expression was

(1) (X+X+...+X) [X times]

which he immediately took the derivative of. While Sum[X, for i = 1 to X] makes no sense for X noninteger it's not much of a reach to understand e.g. adding X an additional half a time to be half of what you'd get if you added an additional X one time. Thus providing an understanding of (1) where it is defined for nonintegers and the OP's derivative of it makes sense. In which case there is a need to explain the real problem with the proof.

Interestingly, there is another way to understand (1) to be defined for nonintegers and make sense as an indexed sum. Define the function

Integer(X) = The largest integer less than or equal to X = The integer part of X.

Then we can look at

(2) Sum[X, for i = 1 to Integer(X) ] = (X+X+...+X) [Integer(X) times]

Then (2) is defined for positive reals and is differentiable at noninteger values of X. Furthermore, the derivative of (2) at noninteger values of X is exactly as the OP would calculate it,

(1+1+...+1) [Integer(X) times] = Integer(X)

and the reason this differs from the derivative of X^2 is because (2) is only equal to X^2 when X is an integer, and those are the places where (2) is not continuous (as Mason points out) and not differentiable.


PairTheBoard

Right, it could be any number x times. Half of an x times. Pi times. One gazillionth of x. Square root of i, or whatever.

What he failed to realize that ones added up to a number that exactly the number of times you need to add those number of ones up to equal x is ****ing x. That is actually the definition of x. It doesn't really break down here to something complicated.

We could go into what 2x means, but it seems a bit silly to do so.

And that with the understanding in bold, X added X amount of times is a legitimate expression for X^2 even when X is noninteger. The problem then in the OP's derivative is that it just gives the partial derivative of

(1) (X+X+...+X)[X amount of times]

with respect to the expression (X+X+...+X).

The OP omits the partial derivative of (1) with respect to the second expression [X amount of times]. The valid way to take the derivative of (1) is to add the two partial derivatives, which gives

(1+1+...+1)[X times] + (X+X+...+X)[1 time] = X+X = 2X


PairTheBoard

And exactly why is it it not easy to see that in a false proof one can find different points to make the wrong argument. If i allow him to create a sum which he did in OP by suggesting integer values what he called

"2 squared = 2+2

3 squared = 3+3+3

4 squared is equal to 4+4+4+4

so X squared is equal to X+X+X+...+X (X amount of times)

If we take the derivative "

Stop here no need for anything past that point.


This is instantly restricting to integers and derivative requires continuity. Why do we need the rest even?

The problem is not the definition of x^2 as x times x with fractional parts included and endless sequence of them in irrational cases that is ok for anything. The problem is in the writing as x+x+x....+x (notice it closes) that is not at all what x*x means if x is not an integer. He simply created a sum that leaves no room for anything other than non negative integers.

Sum (n, for k=1 to n) with n=x

You cannot define fractional sums, rational, irrational etc non natural number sums. That is the first point a mistake is made that after doing the not bad yet thing to define x^2 in integers you move to derivatives that require limits hence connected intervals.

I do not care for mistakes done after that in taking the derivative of x*x. I am already out of the discussion at the moment the concept of derivative of an index is introduced.