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If f(x+a) = f(x-a) where a is infinitesimally small then...

  1. May 23, 2015 #1
    if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result f'(X) = 0 follow?

    X is a point along the x axis.. Sorry about the fault in the question
     
    Last edited: May 23, 2015
  2. jcsd
  3. May 23, 2015 #2
    Yes it does. You're differentiating a constant with respect to a variable, which will always be 0 no matter how small or large the constant is.
     
  4. May 23, 2015 #3
    to be sure df(a)/dx means find df(x)/dx then input x=a
     
  5. May 23, 2015 #4

    Orodruin

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    No. Why would you insert the infinitesimal a in the derivative? It is part of the limit in taking the derivative. It is the derivative at x which is zero.
     
  6. May 23, 2015 #5

    micromass

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    Does ##f(x+a) = f(x-a)## hold for all ##x##? For all infinitesimals ##a##?
     
  7. May 23, 2015 #6

    micromass

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    Not necessarily. Unless perhaps if the equation in the OP holds for all ##x##.
     
  8. May 23, 2015 #7

    HallsofIvy

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    Is that really the question you intended to ask? PWiz's point is that f(a) for any specific number a is a constant so its derivative is 0. I suspect that you intended to ask "if f(x_0- a)= f(x_0+ a), for any infinitesimal a, is df/dx, at x= x_0, equal to 0?"

    Though I will point out that "df(a)/dx" is a common short form for "df/dx evaluated at x= a" so I think he is stretching the point a little! However, since it is "a" that you are taking to be infinitesimal, you certainly don't what to say "df/dx evaluated at x= a" either.
     
  9. May 23, 2015 #8
    i just realised that my effort to ditch writing the question in latex resulted in a problem... i will edit the question for it to make sense.. Sorry about that.. I meant with x as just some point not the variable.
     
  10. May 23, 2015 #9
    if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result df(X)/dx = 0 follow?

    X is a point along the x axis.. Sorry about the fault in the question
     
  11. May 23, 2015 #10

    Orodruin

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    Well, to be perfectly frank, the OP is mixing a lot of things.
     
  12. May 23, 2015 #11

    micromass

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    Assuming that ##f(x+a) = f(x-a)## holds for every infinitesimal, then the answer is no. Take ##f(x) = |x|## at ##X=0##.
     
  13. May 23, 2015 #12
    Yeah really sorry about that :/
     
  14. May 23, 2015 #13

    Orodruin

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    I strongly suggest that you do not use the notation df(X)/dx to mean the derivative of f wrt x evaluated at X. The more common interpretation would be the derivative of f(X) wrt x, which is zero regardless of x. What was wrong with f'(X)?
     
  15. May 23, 2015 #14
    Not necessarily. For example, let the function ##f(x)## equal ##\frac {sin x}{x}## . Then for "X" = 0 , you'll approach the same function output as ##a## tends to 0 satisfying the first condition, but the function at "X" will be undefined.
     
  16. May 23, 2015 #15

    micromass

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    If you assume ##f## differentiable at ##X##, then the answer is true. Indeed, the Newton quotient for an arbitrary infinitesimal ##a## is

    [tex]\frac{f(X+a) - f(X)}{a} = \frac{f(X+(-a)) - f(X)}{a}[/tex]

    Taking standard part of left and right hand sides yields (since the function is differentiable at ##X##):

    [tex]f^\prime(X) = - f^\prime(X)[/tex]

    which gives us the result.
     
    Last edited: May 23, 2015
  17. May 23, 2015 #16

    Orodruin

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    This function has a well defined limit and can easily be extended to a continuous and differentiable function at x = 0.

    Sign error in the last step, but the later conclusion is correct. As it stands, the right hand side would evaluate to f'(X).
     
  18. May 23, 2015 #17
    The OP didn't specify the function domain, so I took the liberty of using natural domains.
     
  19. May 23, 2015 #18
    No. It only means that the function is continuous in x.
     
  20. May 23, 2015 #19

    micromass

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    No it doesn't.
     
  21. May 23, 2015 #20

    Orodruin

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    No it does not. Consider the function f(x) = 0 if x ≠ 0 and f(0) = 1.
     
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