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Math fundamental: can linearity generate reciprocity?

  1. Aug 3, 2003 #1
    Does a nontrivial function f(cx) exist such that

    (d/dx)f(cx)=(1/c)f(cx)

    and c is constant? "Linearity" here requires that c and x in the function argument preserve their product to the first power.
     
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  3. Aug 3, 2003 #2

    marcus

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    try f(cx) = exp(x/c)

    c can be a constant
     
  4. Aug 3, 2003 #3

    marcus

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    Let's test it out. For any number c not equal to zero,
    define f(cx) = exp(x/c)

    Now check your equation

    (d/dx)f(cx) = (d/dx)exp(x/c) = (1/c)exp(x/c) = (1/c)f(cx)

    Therefore (d/dx)f(cx)=(1/c)f(cx)

    so it works.

    So the answer is YES a nontrivial function exists satisfying the equation with c a constant.
    And any nonzero number c will do for the constant.
     
  5. Aug 3, 2003 #4
    Re: Re: Math fundamental: can linearity generate reciprocity?

    I think if f(x) = exp(x) then
    f(cx)= exp(cx) and df/dx = c * exp(cx)
     
  6. Aug 3, 2003 #5
    Re: Re: Math fundamental: can linearity generate reciprocity?

    Loren, may I ask why you are looking for this function?
     
  7. Aug 3, 2003 #6

    HallsofIvy

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    1. c here is a contant and so has nothing to do with the "linearity" of f.


    2. I think you are confusing things by always talking about "the function f(cx)". f(cx) does not represent a function, it represents a value of a function. f(x), f(y), f(cx) all refer to the same function, f.

    Let y= cx. Then df/dx= df/dy dy/dx= c df/dx= c((1/c)f(cx))
    = f(y) so you are requiring that c df/dy= f(y) or that
    df/dy= (1/c)f(y). df/f= (1/c)dy so ln(f)= y/c+ C or

    f(y)== C' ey/c.

    That is, f(x)= C' ex/c as Marcus said.
     
  8. Aug 3, 2003 #7

    HallsofIvy

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    I forgot to say:

    Of course, that function is not linear so the answer to the original question is "no".

    In general, f satisfying df/dx= (constant) f(x) is exponential, not linear.
     
  9. Aug 3, 2003 #8
    Don't you mean df/dx=c*df/dy?
    d[f(cx)]/dx=1/c[f(cx)]
    if y=cx, dy=cdx
    cd[f(cx)]/(cdx)=1/c[f(cx)]
    cd[f(y)]/dy=1/c[f(y)]
    df/dy=1/c2*f
    df/f=dy/c2
    lnf=y/c2+C1
    f(y)=C2ey/c2
    f(cx)=C2ex/c
     
    Last edited: Aug 3, 2003
  10. Aug 3, 2003 #9
    I am confused. Are you folks saying that (d/dx)f(c0x)=(1/c0)f(c0x) for all Aexp(c0x)=f(c0x)?

    What about the falsehood (d/dx)Aexp(4x)=(1/4)Aexp(4x), for instance, where c0=4, or in general, where c0 is other than 1?
     
  11. Aug 3, 2003 #10
    No, we are saying that d[f(cx)]/dx=1/c[f(cx)] only if f(cx) is of the form
    f(cx)=Cex/c

    d[Cex/c]/dx=(1/c)Cex/c=1/c[f(cx)]

    There is no way to obtain the reciprocal constant upon differentiation.
     
    Last edited: Aug 3, 2003
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