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Find the Error: Differentiation!

  1. May 17, 2009 #1
    This is just a pretty simple "riddle" that I have always liked a lot. I didn't come up with it, I actually got it off of a website a few years ago. I'm sure for some of you, it won't be new, but here goes..

    x = x
    x = 1 + 1 + ... + 1 (x times)
    x(x) = x(1 + 1 + ... + 1)
    x2 = x + x + ... + x
    D(x2) = D(x + x + ... + x)
    D(x2) = D(x) + D(x) + ... + D(x)
    2x = 1 + 1 + ... + 1 (x times)
    2x = x
    2 = 1

    Uh oh! What's going on with that real number line? :yuck:
     
  2. jcsd
  3. May 17, 2009 #2
    I'm gonna take a stab at it.

    I think the mistake came when you did the derivative. Since your differentiating with respect for x.

    2x=(1+1+1 xtimes)+(x+x+x+x)
    the ennd result after you take the derivative should be 2x=x+x not 2x=x
     
  4. May 17, 2009 #3
    The problem arises when you take the derivative of

    x+x+...+x

    and get

    1+1+...+1

    You see, the 1+1+...+1 can be represented as a summation over j with limits 1 and x. But since you differentiate with respect to x, and x is one of the limits, you can't just differentiate the summand.
     
  5. May 17, 2009 #4
    That is a cute one. :tongue2:

    But... (1 + 1 + ... + 1) "x times"?
    The right hand side is only defined for non-negative integer values of x.
    How can you write the symbol 1 "x times" when x is 1.5, or square root of 2, or pi? How can you add 1 to itself "x times" when x is pi? If you cannot define this for all real numbers, then the function is not from R to R and is therefore not differentiable (in the sense of real functions.)
    We see the problem more clearly if we differentiate earlier in the process:
    x = 1 + 1 + ... + 1
    D(x)=D(1 + 1 + ... + 1)
    1 = D(1) + D(1) + ... + D(1)
    1 = 0

    You must be able to effectively compute the action 1 + 1 + ... + 1 "x times".
    You could try to defined it for non integer values.
    But if you define it as meaning x, that is
    1 + 1 + ... + 1 "x times" = x,
    then that is the effective form that we would differentiate. To take the derivative of this new version we would first need to write it in its effective form. So, in this case we get D(1 + 1 + .... + 1 "x times") = D(x) = 1 by definition.
     
  6. May 17, 2009 #5

    Hurkyl

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    For fun abuse of notation,

    D(x + x + ... + x (x times)) = (1 + 1 + ... + 1) (x times) + (x + x + ... + x) D(x times)
     
  7. May 17, 2009 #6
    :approve: Which, of course, we have D(x times) = (1 times), so...

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

    (1 + 1 + ... + 1) (x times) + (x + x + ... + x)(1 times)= x + (x) = 2x
     
  8. May 17, 2009 #7
    Haha I should have known you all would have chopped this up so quickly! The reason I like this one so much is because it is very easy to remember, too.
     
  9. May 18, 2009 #8
    so i'm just wondering. was what i posted right? lol :p

    I haven't done this stuff in like 2 years nearly.
     
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