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Limit involved in derivative of exponential function

  1. Feb 8, 2009 #1

    symbolipoint

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    Can a convenient value for a be found without resorting to substituting numerical values for h in this expression?

    EDIT: I am trying to indicate, "as h approaches zero".
    EDIT: neither of the formattings worked; hopefully someone understands what I am asking?

    Lim[tex]_{h\rightarrow\0}[/tex][tex]\frac{ax-1}{h}[/tex]

    In case that formatting failed, an attempt at rewriting it is:
    Limh[tex]\rightarrow[/tex]0[tex]\frac{ax-1}{h}[/tex]

    The most desired value for this limit is 1, and the suitalbe value for a would need to be a = e. I have seen this accomplished using numerical value substitutions , but can the same be accomplished using purely symbolic steps, without any numerical value subsitutions?
     
    Last edited: Feb 8, 2009
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  3. Feb 8, 2009 #2

    mathman

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    It looks like you meant x and h to be the same thing. ah=ehln(a).

    Expand in a power series to get 1+hln(a)+O(h2). Therefore the limit for h->0 will be ln(a).
     
  4. Feb 10, 2009 #3

    symbolipoint

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    There is another way to achieve the derivation for derivative of the exponential function, relying on a bit of clever algebra with logarithms and implicit differentiation of y=a^x.

    I still wish I could find a clear way to understand the limit of (a^h - 1)/h as h approaches zero; without using numeric value substitutions.
    [tex]lim_{h to 0}\frac{a^h-1}{h} [/tex]

    edit: that typesetting is better than what I accomplished earlier, but I'd sure like to put in that right-pointing arrow instead of "to"
     
    Last edited: Feb 10, 2009
  5. Feb 11, 2009 #4
    Use \to for the arrow.
     
  6. Feb 11, 2009 #5

    HallsofIvy

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    How you do that depends on exactly how you define the exponential and, in particular, how you define e. If you define e as "limit of (1+ 1/n)n as n goes to infinity" then you can say that e is approximately equal to (1+ 1/n)n for large n so that e1/n is approximately 1+ 1/n.

    Setting h= 1/n, h goes to 0 as n goes to infinity and that says that eh is approximately equal to 1+ h so that eh-1 is approximately equal to h and (eh-1)/h goes to 1 as h goes to 0.

    Of course, for the general case, use the fact that ah= eh ln(a).
     
  7. Feb 11, 2009 #6

    lurflurf

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    so we define exp(x) to be a function such that
    exp(x+y)=exp(x)*exp(y)
    this property does not define a unique function
    there are an infinite number of both nice and non nice functions having this property
    there are several ways of picking out one
    in particular the classical exponential (the "nice" one for which exp(1)=e) has
    exp'(0)=1
    exp'(0)=lim [exp(h+0)-exp(0)]/h=lim [exp(h)-1]/h
    the general nice exponential is
    exp(c*x)
    {[exp(a*x)]'|x=0}=c
    in the exponential notation we may write
    exp(c*x)=exp(c)^x
    let a=exp(1)
    exp(c*x)=a^x
    we may ask the relation between
    exp(1)=a and c
    clearly
    lim [a^h-1]/h=c
    we may (some justification required) invert the relation into
    a=lim (1+h*c)^(1/h)
    this requires a definition for x^y such as
    x^y:=exp(y*log(x))
    an adjustment is needed to avoid circular reasoning
    we may define integer exponents in the obvious inductive way (x^(n+1+=x*x^n)
    then consider the restricted form of the limit
    a=lim (1+h*c)^(1/h)
    that is let h=1,1/2,1/3,1/4,...
    a=lim{n=1,2,...} (1+c/n)^n=exp(c)
    lim{n=1,2,...} (1+1/n)^n=exp(1)=e
    this gives as desired a symbolic form for e, how useful this form is depends on the application
     
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