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If $f(x)$ definition be $$\large {f(x)=\underbrace{x+x+x+...+x}_{x-times}}$$ what is the domain of this function ?

Obviously for natural numbers $f(x)$ equal to $x^2$ but is it definable for other numbers ? Is there a way to proof that does not have derivative ?

Any clue ?

Thanks in advanced .

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    If $x=\frac{5}{4}$, what does it mean *$\frac{5}{4}$ times* ? This is not a definition, and the domain of a function should be given in advance. – 2017-02-09

4 Answers 4

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When you write $x$-times you are implicitly saing that $x\in \Bbb N$ and on that case the function $f$ is a discrete function (domain in $\Bbb N$) and then there is no derivative for it.

An interesting thing (which is not correct, it is just an dirty trick) is suppose that you can derivate that. On one hand we have:

$$f(x)=x+x+x+...+x\to f'(x)=1+1+...+1=x\quad (1)$$

but on the other hand we have

$$f(x)=x^2\to f'(x)=2x\quad (2)$$

and we have a contradiction looking to $(1)$ and $(2)$.

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    Is that not because the first "method" of differentiating there forgets the chain rule? – 2017-02-09
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The formula $x^2$ is indeed defined for other numbers.

However, that's not the formula you were given. The phrase "$x$-times" implicitly restricts your formula to the natural numbers, and so that's the domain.

It's true that your function is the restriction to the natural numbers $\mathbb{N}$ of a function $F(x)=x^2$ whose domain is all real numbers, and certainly that function with domain equal to the real numbers is differentiable. However, the domain of a differentiable function is usually not allowed to be a discrete set like $\mathbb{N}$.

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The domain of a function is part of its definition; it is meaningless to ask what the domain of a function is. Asking this question is like asking "I'm thinking of a number that's prime. What is it?"

For the related question of "What is the largest domain that this partially defined function might have?", there are a variety of answers.

If we consider domains as subsets of $\mathbb{R}$ or $\mathbb{C}$, then Arnaldo's answer is correct; it is meaningless to talk about adding things $1.7$-times. In order to take a derivative, the function needs to be defined on an open interval of the reals; since this one isn't, it doesn't have a derivative.

However the domain doesn't have to be $\mathbb{N}$; it could be defined on infinite cardinals as well.

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It makes no sense to tell of derivability of a function of the form $f : \mathbb{N} \to \mathbb{R}$.