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Posers and Puzzles

Posers and Puzzles

  1. 18 Jun '07 20:43
    We usually denote derivative like this: f'(x) where f is a function and x the variable, right.
    Let me from here on use the following notation for simplicity: D(f). If f(x) = x^2 then I write directly D(x^2) for its derivative, may I do this?

    Okay, everyone knows that D(x^2) = 2*x, right. There is no argue about this.

    So if I 'prove' that D(x^2) = x it should obviously be some error in the 'proof'.

    But look at this:
    D(x^2) =
    = D(x + x + x + ...) {with x number of terms} =
    = D(x) + D(x) + D(x) + ... {with x number of terms} =
    = 1 + 1 + 1 + ... {with x number of terms} =
    = x
    Hence D(x^2) = x ???

    This 'proof' is obviously wrong somewhere, but where is the error to be found?
  2. 19 Jun '07 07:19
    Interesting. There seems to something funny with the assumption that X^2 = X + X + X .... (for X terms). I see that this works, sort of, but what if x was a measurement that had units attached, like meters. Then (5m)^2 is not equal to 5m + 5m + 5m + ... The units are inconsistent. And how would you have 5m terms?
  3. Standard member Palynka
    Upward Spiral
    19 Jun '07 08:42
    Originally posted by FabianFnas
    We usually denote derivative like this: f'(x) where f is a function and x the variable, right.
    Let me from here on use the following notation for simplicity: D(f). If f(x) = x^2 then I write directly D(x^2) for its derivative, may I do this?

    Okay, everyone knows that D(x^2) = 2*x, right. There is no argue about this.

    So if I 'prove' that D(x^2) = x ...[text shortened]... = x ???

    This 'proof' is obviously wrong somewhere, but where is the error to be found?
    That's because you don't derive the x number of terms.
  4. 19 Jun '07 08:52 / 1 edit
    Originally posted by Palynka
    That's because you don't derive the x number of terms.
    No? I did, didn't I?

    D(3*x) = D(x+x+x) = D(x) + D(x) + D(x) = 1 + 1 + 1 = 3 Right?
    This could be written
    D(three times x) = D(three terms of x) = D(x) three times = 1 three times = 3 Right?

    So of course we can derive x number of times.
  5. Standard member Palynka
    Upward Spiral
    19 Jun '07 09:17
    Originally posted by FabianFnas
    No? I did, didn't I?

    D(3*x) = D(x+x+x) = D(x) + D(x) + D(x) = 1 + 1 + 1 = 3 Right?
    This could be written
    D(three times x) = D(three terms of x) = D(x) three times = 1 three times = 3 Right?

    So of course we can derive x number of times.
    Sorry if I wasn't clear, I meant that the x number of times is also variable (unlike 3 in your example).
  6. 19 Jun '07 09:57
    Originally posted by Palynka
    Sorry if I wasn't clear, I meant that the x number of times is also variable (unlike 3 in your example).
    I'm not sure myself where the error is to be found in my 'proof', though I understand that there must be an error somwhere.

    So, this is the error, that x is a variable in itself, and not a constant? Then I thank you, this has boggled my mind for a while...
  7. Standard member Palynka
    Upward Spiral
    19 Jun '07 10:26 / 1 edit
    Originally posted by FabianFnas
    I'm not sure myself where the error is to be found in my 'proof', though I understand that there must be an error somwhere.

    So, this is the error, that x is a variable in itself, and not a constant? Then I thank you, this has boggled my mind for a while...
    Yes, D(x.x)=D(x+x.+.. (x times)) is different than x.D(x)=D(x)+D(x)+...(x times).

    If you look at it as a derivative of a product, you'll have D(x).x+x.D(x)=2x

    Edit - Intuitively, you can see that a change in x, has a double effect on x+x..(x times): first, each element of the sum changes and secondly the number of elements also changes.
  8. 19 Jun '07 11:22
    Palynka is correct, so I won't repeat him.
  9. 19 Jun '07 13:19 / 1 edit
    x^2=x+x+x+...+x (x times) only if x is a non-negative integer, but to define the derivative x must be continuous.
  10. 19 Jun '07 14:20
    Tahnk you Palynka for your explanation.
    When I present the problem ot som fellow math student, they just shake their heads.
  11. 19 Jun '07 22:33
    You assume that the D() has a distributive property with addition when such is not the case, nor does it make any sense.

    D() takes a function as it's argument, and as such, operations which require specific values cannot be used, rendering distribution meaningless.

    Now what is true is this:

    Suppose h(x) = f(x) + g(x), h'(x) = f'(x) + g'(x).