What is the flaw in this proof?

Correction. Its obviously only the discrete case, but you get the picture.

math.stackexchange.com/questions/248203/what-is-the-derivative-of-a-summation-with-respect-to-its-upper-limit

The first proper answer ends with what is probably a solution to this.

That's irrelevant. The derivative of f: Z+ --> R where f(x) = x^3 exists on the integers, even though there's no interpretation of "2.5 cubed" for this function.

>The derivative of f: Z+ --> R where f(x) = x^3 exists on the integers
Prove it.

Whoa, I'm not at this level of math yet

imagine being this much of a brainlet

Are you high?

[math]\lim_{h\rightarrow 0}\frac{f(a+h)-f(h)}{h}[/math] can be evaluated as long as you restrict a+h to Z

Addendum: this is generalizable to the usual domains. The same could be done if f was f : Q -> R.

>limh→0f(a+h)−f(h)h can be evaluated as long as you restrict a+h to Z
Are you just going to keep posting claims without any proofs?

Give a proof that
>The derivative of f: Z+ --> R where f(x) = x^3 exists on the integers

>limh→0f(a+h)−f(h)h can be evaluated as long as you restrict a+h to Z

Quit samefagging. Here's a proof:

Limits don't require infinitesimals, you idiot.

More rigorously:

A function ƒ is said to be continuous at c if it is both defined at c and its value at c equals the limit of f as x approaches c.

Suppose we can determine the limit of x^3 as x approaches -2 on the integers. Clearly if we can do so for -2, we can do so for all other integers for which f(x) = x^3 is defined, since we will be using epsilon-delta for limit determination.

Let e = epsilon, d = delta.

Given e > 0, choose d = min{1, e/19}. Note that d>0.

For every x with 0 < abs( x - -2) < d, we have abs(x + 2) < 1 so -1 < x + 2 < 1 and then -3 < x < -1.

By necessity, then, 1 < x^2 < 9 and 2 < -2x < 6, and we get 7 < x^2 - 2x + 4 < 19.

So we have abs(x^2 - 2x + 4) < 19.

Now abs(x^3 - -8) = abs(x^3 + 8) = abs(x+2) * abs(x^2 - 2x + 4). And if abs(x + 2) < d, then abs(x+2) < e/19 and abs(x^2 - 2x + 4) < 19.

Finally we get that if abs(x + 2) < d, then abs(x^3 - -8) = abs(x+2) * abs(x^2 - 2x + 4) < (e/19) * 19 = e.

Therefore we have shown that for any positive e there is a positive d such that for any x with 0 < abs(x - -2) < d, we have abs(x^3 - -8) < e.

By definition of the limit, we are finished.

It is trivial to show that the same thing can be done with g(x) = x with g: Z -> R.

Clearly the limit exists for all x in Z, and clearly f(x) is both defined and equal to said limit.

Which means we can compute the derivative of f by finding the limit of the difference ratio (f(a+g(x)) -f(g(x)))/g(x), which is defined.

Because "(x times)" is a function and you need to chain rule it.

d/dx[x+x+x+...(x times)]=1+1+1....(x times) + x+...(d/dx[x] times) = x + x (1 time) = 2x

>By definition of the limit, we are finished.
You don't even seem to know the definition of the limit, brainlet. Didn't you learn this in middle school?

You need to show that for any positive e there is a positive d such that for any x with 0 < abs(x - -2) < d, we have abs(f(x) - -8) < e.

Going through your proof, if e=1, then d=1/19.
And so taking x = -2 - (1/20), we have 0 < abs( -2 - (1/20) - -2) = 1/20 < 1/19 = d.
But now we don't have abs(f(x)- -8) < e, since f(-2 - (1/20)) isn't defined.
By definition of the limit, we're not finished.

Why did you add? Isn't chain rule multiplicative?

[math]\frac{d}{dx}x^2=\frac{d}{dx}[x+x+\ldots+x][/math] is the false step.

The core flaw is this:

Think in term of differences. df=f(x+h)-f(x)

d( x+x+x...+x, x times)

Is different from:
( dx+dx+dx...+dx, x times)

Because you totally ignored the "x times". That x changes as well so you cannot simply ignore it.

you can't define the derivative for g:Z->R. the limit doesn't exist in the usual topology, because Z is discrete: no sequence (this is enough for metric spaces) approaches a point in Z

the best you can do is some kind of discrete slope, dg(x) = g(x+1) - g(x). which is useful, mind you. sum{x in [1..n]} dg(x) = g(n+1) - g(1), it's like an integral.

bump

It can not

Taking the derivative requires a continuous limit. Your definition that x^2=x+x+x...(x-times) only works for discretized integers. Thus, the limit is impossible to carry out, and differentiation is invalid.

So is wrong? He chain rules the RHS derivative and gets the correct result.

He means product rule.

Indeed, it's quite easy to see this works for any polynomial.

dude you're so fucking retarded I don't even know where to start. Here's a challenge for you. The limit [math]\lim_{x \to -2}x^3[/math] is equal to every real number [math]K \in \mathbb{R}[/math] when considered over [math]\mathbb{Z}[/math]. Proof: let [math]\epsilon > 0[/math] be arbitrary and choose [math]\delta = \frac{1}{2}[/math]. Then for any [math]x \in \mathbb{Z}[/math] such that [math]0 < |-2 - x| < \delta[/math] it's certainly true that [math]|x^3 - K| < \epsilon[/math]. Prove me wrong brainlet.

>That's irrelevant.
prove it, faggot
(spoiler: it's not irrelevant)

>What is the flaw in this proof?
Everything.

(Because [x times] is literally multiplication by x).

I don't see the contradiction. Solve for x, x = 0, the whole sum of 1's can be ignored since x = 0

Botched manipulation of free/bound variables

[math]The same fake reasoning could be performed for the discrete derivative operator.
If $(a_n)_{n\in \N} \in \R^{\N}$ is any sequence, we define for every $n\in \N$, $(\Delta a)_n:=a_{n+1}-a_n$.
It is obvious that $\Delta$ is linear, i.e for any positive integer $k$ and any sequences $u_{(1)};...; u_{(k)}$, we have $\Delta \left( \sum_{i=1}^k u_{(i)}\right)= sum_{i=1}^k \Delta u_{(i)}$. Now consider the sequence of squares $b_n:=n^2$ for every $n$.
Then $\Delta b_n=2n+1$ by direct calculation. But we have also $\Delta b_n= Delta (n+n+n+...+n)=1+1+1+...+1=n$ so $2n+1=n$ so $n+1=0$ for every $n$.
[/math]
Actually it is a
>Botched manipulation of free/bound variables

Function is defined only for [math] x \in \mathbb{N}[/math] , so differentiotion is impossible.

ITT: completely wrong and/or contradictory answers

IT'S BECAUSE YOU ARE TAKING THE DERIVATIVE RELATIVE TO X BUT IGNORING THE FACT THAT YOU HAVE X OF THEM.

In other words when you measure the rate of change you are effectively ignoring the change in addends.

"x times" works for rationals (if not irrationals as well), and you can integrate over rationals.

[math]Please do not LaTeX your writing only use it on the equations Thank you.[/math]