Is it possible to define $x+x+x+x...x $ times? I need to compute its derivative. It differs from the derivative of $x^2$. It evaluates to $x$ via sum of derivatives.
Asked
Active
Viewed 265 times
1

5If $x$ is an integer, we can: it's just $x^2$. For all other values of $x$, you'll need a suitable extension of the function. Might I recommend $x^2$? – Theo Bendit Jul 25 '15 at 10:10

1What you are computing is $x\cdot \dfrac{dx}{dx}$. Remember the product rule: $$\dfrac{d(x\cdot x)}{dx} = x\cdot\dfrac{dx}{dx} + \dfrac{dx}{dx}\cdot x = 2x$$ – Darth Geek Jul 25 '15 at 10:10

http://math.stackexchange.com/questions/1096/whereistheflawinthisargumentofaproofthat12derivativeofrepeated – Hans Lundmark Jul 25 '15 at 10:12

We can define it this way : $$\underbrace{x+x+\cdots+x}_{ y \text{times}} =xy$$ (I would suspect that , in history, somewhere mathematicians thought about defining multiplication of non integer numbers this way as a generalization of the integer property. Now the important thing this is **just a notation** , and you can define any notation you want in mathematics, just **don't apply some rules you did not prove**. For example: $$(\underbrace{f(x)+\cdots+f(x)}_{ g(x) \text{times}})'= \underbrace{f'(x)+\cdots+f'(x)}_{ g(x)\text{times}}+\underbrace{f(x)+\cdots+f(x)}_{ g'(x)\text{times}} $$ – Elaqqad Jul 25 '15 at 10:32

(continuation comment : in the previous comment property $\to$ propriety), the example is the correct rule of differentiation you have to apply if you defined the notation like I did, and you can prove it from the chain rule. There is no contradiction in this. Intuitively speaking this is what we do in mathematics , we defined $a^x$ not as the product of $a$ $x$ times (the intuitive definition) but from the exponential $x\to e^x$ (extension of integer properties to real numbers and even complex numbers) – Elaqqad Jul 25 '15 at 10:47
1 Answers
1
There are two problems with this expression.
 First: How would $\underbrace{x+x+...+x+x}_{x}$ make sense if $x$ is not an integer? How would you add 3.5 to it self 3.5 times?
 Second: This one is more related to calculus. Assume that $\underbrace{x+x+...+x+x}_{x}$ is well defined, then you don't only have to differentiate all the individual $x$es , but also the $x$ under the brace. However you can't really do that using this notation.
Just notate it with $x^2$ and find the correct derivative $2x$.
wythagoras
 24,314
 6
 55
 112

3In a way we could apply the chain rule to $\underbrace{u+u+\ldots +u}_{v}$ and need to consider $\partial \over \partial u$ as well as $\partial \over \partial v$. Even though the notation still doesn't make sense completely, for noninteger $v$ we obtain $u+v$, i.e., $2x$ as derivative – Hagen von Eitzen Jul 25 '15 at 10:15

couldn't we define it the same way we define multiplication? Multiplication is repeated addition right? – martianwars Jul 25 '15 at 10:20

@KalpeshKrishna Multiplication is repeated addition, but the difference here is that we repeat adding $x$ namely $x$ times. Because they both are $x$, we can't differentiate them – wythagoras Jul 25 '15 at 10:25