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Derivating a sum weird result. wtf

  1. Jul 24, 2008 #1
    Hey I was doing a standard derivative problem but I ran into something somewhat strange. Check it out:

    [tex]\frac{d}{dx}\sum_{i=1}^xx=\frac{d}{dx}x^2=2x[/tex]

    Okay, simple enough..but what if I write it like this?

    [tex]\frac{d}{dx}\sum_{i=1}^xx=\frac{d}{dx}(x+x+...+x)=1+1+...+1=x[/tex].

    Why am I getting two different answers???
     
  2. jcsd
  3. Jul 24, 2008 #2

    Defennder

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    [tex]x^2 \neq \sum^x x [/tex]. This definition of x^2 does not make sense. d/dx does not just apply to whole numbers for x. It applies to all real numbers of x.
     
  4. Jul 24, 2008 #3

    rock.freak667

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    In that defintion for x^2 which you have as a sum. x is not a variable, it's the last term of your series.
     
  5. Jul 24, 2008 #4
    I don't get it. Are you saying that because x is the upper bound of the summation that means it's limited to nonnegative integers and because of that, you can't take the derivative?

    How's it not a variable? I looked up the definition on Mathworld.com:
    and x can take on different values (0, 1, 2, etc.) and the function [tex]f(x)=\sum_{i=1}^x x[/tex] depends on it.
     
  6. Jul 24, 2008 #5

    Redbelly98

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    Derivatives are defined for functions of a continuous variable. Since x is a discrete, integer variable in your summation, it doesn't make any sense to differentiate with respect to x here.

    If you don't understand my answer, I'll ask you this question: what is that sum for, say, x=3.5?
     
  7. Jul 24, 2008 #6

    HallsofIvy

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    Another way of looking at it: in [itex]x+ x+ x+ x+\cdot\cdot\cdot+ x[/itex], "x times", the "x times" is itself a function of x- this can't be treated like you would x added to itself a fixed number of times. Just as the derivative of [itex]x^x[/itex] is nothing like the derivative of [itex]x^n[/itex] for fixed n.
     
  8. Jul 24, 2008 #7
    Just to add something to Halls's post, you can get round the fact that the x above the summation sign 'ought' to be an integer, by replacing it with, say, floor(x). But the analysis you use in your argument doesn't allow for the fact that the upper summation limit varies with x - you differentiated within the summation sign, but ignored the x (or floor(x)) appearing above it.
     
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