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My friend Boris (Boryan) gave me a task, and completely refuses to give the answer what's wrong here. $$x^2=\overbrace{x+\cdots+x} ^{x\text{ times}}$$ $$(x^2)'=(x+\cdots+x)'$$ $$2x=1+\cdots+1$$ $$2x=x$$ $$2=1$$ Yeah! I've succesfully copypasted latex formulas!


I think the problem is in non-formal symbols. It brings me to question, what is the result for $(\sum_{i=1}^{x}x)' = ?$ That's usually an obstacle for those who memorised many things without clear understanding of definitions. So, I'm interested in fundamental mistake of this equations, because I want to get out of this mess)

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    $\begingroup$ Calculus alliterates with continuous not discrete. $\endgroup$
    – evil999man
    Apr 18, 2014 at 6:46
  • $\begingroup$ There was a typo in previous one $\endgroup$
    – evil999man
    Apr 18, 2014 at 6:47
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    $\begingroup$ Intuitively, both the summands and the length of the sum change with respect to $x$. The differentiation step here only takes account of the change in the summands. $\endgroup$
    – mudri
    Apr 18, 2014 at 10:44
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    $\begingroup$ One cannot differentiate a function that is only defined at integer values at all. $\endgroup$ Apr 18, 2014 at 11:29
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    $\begingroup$ Note that if you also differentiate the "$x$ times" part and add it on, you get the correct answer. See this MO Post. $\endgroup$
    – ronno
    Apr 18, 2014 at 13:48

7 Answers 7

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the mistake is at the beginning $$x^2=\overbrace{x+\cdots+x} ^{x\text{ times}}$$ fails miserably when $x$ is not a natural number as the expression $$\overbrace{x+\cdots+x} ^{x\text{ times}}$$ is meaningless for example in case $x=1.37$ what does $$\overbrace{1.37+\cdots+1.37} ^{1.37\text{ times}}$$ mean?

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  • $\begingroup$ I didn't thinked about it 46 mins ago. It's really important to define $D(X)$ at first. But I can't upvote right now) $\endgroup$
    – Ralor
    Apr 18, 2014 at 5:55
  • $\begingroup$ It's precisely as meaningless as fractional-order differention or fractional order Fourier transformation. That is, none -- all of these operations are well defined. The one you propose is a complicated way of writing 1.8769. Just because you're used to something only for integer arguments doesn't mean it doesn't extend (unambiguously). $\endgroup$ Apr 19, 2014 at 0:00
  • $\begingroup$ @EricTowers thank you Eric you are correct, i should rephrase, $$\overbrace{1.37+\cdots+1.37} ^{1.37\text{ times}}$$ is utterly meaningless unless a proper extension is defined [unless you can prove there is ONLY one such extension possible..].. for real numbers.. as it stands i still see the first line as the problem as it is not well defined, i did not say it is not well definable by some extension.. i welcome your comments.. thx $\endgroup$
    – userX
    Apr 19, 2014 at 4:26
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The first misconception is the definition of $x$-times applied operations.
If one ignores that, there still remains the error in the derivation $$\frac d{dx}\sum_{i=1}^x x \neq \sum_{i=0}^x \frac d{dx} x = x$$ You have to use Leibniz integral rule, seeing the sum as a special type of integral, namely $$\frac d{dx} \sum_{i=1}^x x = \frac d{dx} \int_0^x x d\#(i) = x|_{i=x} \cdot 1 - x|_{i=0} \cdot 0 + \int_0^x 1 d\#(i) = x + x = 2x$$ Where $\#$ denotes the counting measure. If you understand this, I can elaborate on the RHS.

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  • $\begingroup$ You added "counting measure" because x is integer? I'm not familiar with this step, could you explain it a bit? $\endgroup$
    – Ralor
    Apr 18, 2014 at 5:27
  • $\begingroup$ @Ralor The counting measure is a special measure defined on $(\mathbb Z, \mathcal P(\mathbb Z))$ by "counting the elements", i.e. $$\#(A) = \text{number of elements in } A$$ This uses measure theory, though. $\endgroup$
    – AlexR
    Apr 18, 2014 at 5:30
  • $\begingroup$ I believe it is straghtforward answer, but I still don't understand it for 100%. But now I'm sure how "measure theory" translates into Russian, so I will try to do so. $\endgroup$
    – Ralor
    Apr 18, 2014 at 5:51
  • $\begingroup$ @Ralor this is the english wikipedia entry. At the left you can see a list of other languages (Ukrainian was among them, for example) - maybe this will help find the appropriate term. $\endgroup$
    – AlexR
    Apr 18, 2014 at 5:56
  • $\begingroup$ I think you know that translations Russian -> Russian is often possible, especially when you're talking about some non-trivial things having only partial knowledges (just some of Discrete analysis in my case) $\endgroup$
    – Ralor
    Apr 18, 2014 at 6:03
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If $x$ is any real positive number then the equality should be written in the form $x^2=\int_0^xxdy$ which is evidently true.

Then the derivatives are $(x^2)'=2x$ and $\left( x\int_0^x dy\right)'=\int_0^xdy+x=x+x=2x$

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The problem is that $(x + \cdots + x)' \not= 1 + \cdots +1$

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    $\begingroup$ Well it is, but not if the amount of x's is variable too. $\endgroup$
    – Jori
    Apr 18, 2014 at 8:30
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$$(x^2)'=\bigg(\sum_1^xx\bigg)'=\sum_1^xx'+\sum_1^{x'}x=\sum_1^x1+\sum_1^1x=x+x=2x.$$

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Visualize the RHS graph. You will see that it is discontinuous. So can't differentiate. If you take $x^3$ instead of $x^2$, yon can also prove 1=2=3! Surely try that too!

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You have a genuine result that $(f+g)' = f' + g'$.

It follows from this that you can do the same for any finite sum of functions.

But now you seek to do the same for sums with a variable number of terms according to the value of $x$. That's not an application of the rule you have, because you're no longer looking at a function that's a sum of finitely many functions. Rather you're looking at a function whose value at a given point has been written as a sum of finitely many values. But so what if it is? Every real number is the sum of finitely many real numbers, and for that matter every integer is the sum of finitely many 1s! This tells us nothing about the function and so we shouldn't expect it to tell us anything about the derivative.

It's also a problem of course trying to differentiate a function of integers in the first place, and if $x$ isn't an integer then the RHS makes no sense. But you could overcome that objection by choosing instead to differentiate $x$ floor($x$) as the same sum with floor($x$) terms. The answer would still be wrong, indeed the function isn't differentiable or even continuous at integer values of $x$. For any of the piecewise continuous and differentiable regions between the integers, within which the value of floor($x$) is a constant, it would now give you the correct derivative, floor($x$).

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