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Thread: What's wrong with this?

  1. #1 What's wrong with this? 
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    Well here's a silly thing, but my high school mathematics doesn't provide me with a convincing answer can you give me one?

    For any positive integer x:

    x + x + x + ........ + x (x times) = x^2

    Now, differentiating the above equation with respect to x.

    1 + 1 + 1 + ........ + 1 (x times) = 2x

    or

    x = 2x

    or 1=2
    Thanks in advance.
    Nentut.


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  3. #2  
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    How does x + x + x + ........ + x (x times) = x^2 for all real x? Now as you need a complete space to do most calculus your problem lies there.


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3  
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    How does x + x + x + ........ + x (x times) = x^2 for all real x? Now as you need a complete space to do most calculus your problem lies there.
    Not sure I understand your reasoning... but isn't is satisfied with well... positive integers?

    Nentut.
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  5. #4  
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    Yes it is, but can you do calculus on the positive integers?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  6. #5  
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    That is not the problem.

    The problem is that x*x =/= kx since x is not k. You derivate a function, not the numbers in a growing serie, since the numbers are growing, that's the definition of x^q in this case. It is the function you derivate and not the value it equals at any given time.
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  7. #6  
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    LeavingQuietly - what is the definition of multiplication on the naturals? a*b means b+b+...+b (a times) so on the naturals x^2 = x+x+...+x (x times). So i have no idea what you are trying to get at here. The problem is two fold, you cant do calculus on the naturals (as the space is not complete) and if you extend multiplication to the reals (nothing fancy here) the required identity x^2 = x+x+...+x does not hold (i.e. what is pi<sup>2</sup>? It is definitly not pi+pi+...+pi pi times!)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  8. #7  
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    Aaah, to derivate a constant, is the most satisfying brainfart I can accomplish. It always becomes zero.

    x = 1 (x times).
    Derivate both sides.
    1 = 0+0+0+0....

    You know now, where the problem i know have, is, don't you? X is not a serie of constants, it is a variable. You derivate the variable as is. The variable cannot be described as constants, since a varying constant is a variable and not a constant. How the variable varies depends on the variable. You cannot say "how the constants varies depends on the constants" derivating a reall serie is defined (I think) as advanced. Perhaps I can find something in the library that I can publish here. The maclaurin series are derivable, but that's because the series is actually defined by dn(f(x)) that is, the n:th derivata's f(0) value. The definition of your terms, is that they form a single factor. That factor is x^2. Derived it is 2x
    If 2 or more unproportional gradual functions exist in the same "macro" function, then these can be used to describe the macro function in a more fundamental scale.

    1~1, n~k, hence they are proportional and belong to the same fundamental function.
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  9. #8  
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    huh?

    How is f(x) = 1+1+...+1 (floor(x) times) a constant?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  10. #9  
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    Quote Originally Posted by river_rat
    huh?

    How is f(x) = 1+1+...+1 (floor(x) times) a constant?
    It isn't, that's the point. You cannot put x in one side and a konstant in the other and claim that they are equal unless the konstant is a value that the function can have and thus you do not in such a case derive the constant, you derive the function and for the value a the function f(x) = a has the derivata f'(x) = p where p depends on the a value for f(x) and the relation between f(x) and f'(x)
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  11. #10  
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    So how is x^2 a constant?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  12. #11  
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    Quote Originally Posted by river_rat
    So how is x^2 a constant?
    For any given x, x^2 is a constant. for instance, 5^5 = 25.
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  13. #12  
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    for any given x, f(x) is constant for any function f. Now how is g(x) = x^2 constant but f(x) = x+x+...+x (floor(x) times) not?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  14. #13  
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    Quote Originally Posted by river_rat
    for any given x, f(x) is constant for any function f. Now how is g(x) = x^2 constant but f(x) = x+x+...+x (floor(x) times) not?
    It is, but you still derivate the function and not the constant it equals for a given x-value. The both upper sides are equal, we derive the function as it is defined. f(x) = x+x+...+x (floor(x) times) is a variable sum over x. it cannot be derived unless you divide the sum sign (n-->x) with x and multiply the elements with x, hence get x^2, terms = 1. Now you've solved it. The variable sum in itself has to be treated for what it is.
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  15. #14  
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    huh again? what constants are you talking about? Okay so the sum i gave is piece-wise differentiable but so what (and i don't think that is what you are referring to but anyway).
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  16. #15  
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    How would you differentiate it?
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  17. #16  
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    Let f(x) = x+x+...+x (floor(x) times)

    Let 0<n<x<n+1 then f(x) = x+x+...+x (n times) = nx

    so f'(x) = n
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  18. #17  
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    Quote Originally Posted by river_rat
    Let f(x) = x+x+...+x (floor(x) times)

    Let 0<n<x<n+1 then f(x) = x+x+...+x (n times) = nx

    so f'(x) = n
    n = int(x)

    D(int(x)*x)= int(x) + x*int(x)'

    x*int(x)' varies, since int(x)' is a jumpy function. It's not that simple.

    as you can see, this closes 2x.

    when x = int(x), the f'(x) is neverending.
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  19. #18  
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    by int(x) i assume you mean the integer part of x - that is piecewise constant so is piecewise differentiable, in fact its derivative is zero everywhere it is defined.

    If n<x<n+1 then int(x) = n, so int'(x) = 0 so what is the problem? I have no idea what you are on about here

    as you can see, this closes 2x.

    when x = int(x), the f'(x) is neverending.
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  20. #19  
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    Yes, in the interval you choose the differential can be expressed in those term. However, why should 0<n<x<n+1? Why not leave it as int(x) + x*int(x)'? In that way, the equation is defined in all points, even if sometimes defined as undefined.
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  21. #20  
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    I did not choose an interval, for any real x (not an integer) you can find a natural number n such that n<x<n+1. Using that fact allows you to simplify the sum.

    You cant define something as undefined, either it is defined everywhere or it is not.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  22. #21  
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    Look, you cannot ignore the undefined points of a function just because they are undefined, that's what I meant. They still exist. Seriously, if that was acceleration in a system that you existed in, then you would not feel so well. Natural numbers or not.
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  23. #22  
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    The function f(x) = x+x+...+x (floor(x) times) is defined everywhere - its derivative defined everywhere except on the integers and you cant extend the derivative of the function continuously so what exactly do you want to do there?

    Very few functions are differentiable LeavingQuietly, its just something you have to live with.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  24. #23  
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    Yeah. Still, with the definition f'(x) = (f(x+h)-f(x-h))/2h, h-->0, f'(x) sometimes has a value =/= n. Purely mathematical. That's what you would write if you only had a graph. I can see your arguments are plenty and strong. But when in doubt, return to source of it. If you cannot go that last step, then don't. Saying d(x*int(x)) = int(x), IMHO won't do.
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  25. #24  
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    Yes, but that definition is a local one, given a point x you work out the derivative quotient and then take the limit. So lets take a point x which is not an integer (as the function f(x) = x+x+...+x (floor(x) times) is not continuous there and thus cannot have a derivative at any integer point).

    Okay, so f(x) = floor(x) x and so (f(x+h)-f(x))/h = floor(x) for suffiently small h so that floor(x+h) = floor(x). Then as h -> 0 we have that f'(x) trivially is just floor(x).

    Notice i am not saying that f'(x) = floor(x) for all x, just for all noninteger x.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  26. #25  
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    Now when we have gotten that out of our sholders I suggest we spend some quality time with our loved ones. Merry christmas in advance.

    It was very interesting to have this discussion.
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  27. #26 Simplify? 
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    Key to solution is perhaps to simplify so as to avoid complications.

    x + x + x ... + x = x^2

    Thus

    0 = 0

    Differentiating both sides yield same results.

    The problem here is that we are not differentiating a depending variable wrt independent variable, thus problem arises.
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  28. #27  
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    Quote Originally Posted by enoughsaid05
    Key to solution is perhaps to simplify so as to avoid complications.

    x + x + x ... + x = x^2

    Thus

    0 = 0

    Differentiating both sides yield same results.

    The problem here is that we are not differentiating a depending variable wrt independent variable, thus problem arises.
    How do you get 0=0?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  29. #28  
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    Well since the function is only valid for integers it is not continuous and for a function to be differentiable it needs have a defined limit and be continous. Therefore the function does not have a derivitave anywhere it is defined and that is the problem i think
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