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Math Help - Derivative at f1(0)

  1. #16
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    Let me state how I define exponentials.

    Define \ln x = \int_1^x \frac{1}{t} dt \mbox{ for }x>0

    It is not hard to show that,
    1) \ln x is an increasing function.
    2) \ln x is differenciable and therefore continous for (0,\infty).
    3) (\ln x)' = \frac{1}{x} \mbox{ for }x>0.
    4)There exists a number, called e, so that \ln e = 1.

    We see that \ln x is a one-to-one function. Define \exp x to be its inverse function on its range.

    It is not hard to show that \exp x is increasing, differenciable and so continous, and furthermore, (\exp x)' = \exp x.

    We have the following supprising property that if q is a rational number then \exp q = e^q. So we define e^r, to be \exp r. With that we generalize exponents as follows a^x = e^{\ln a x} = \exp (\ln a x) \mbox{ for }a>0.

    The difficutly besides for using the definition is to have a formal definition of what an exponent means.
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  2. #17
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let me state how I define exponentials.

    Define \ln x = \int_1^x \frac{1}{t} dt \mbox{ for }x>0

    It is not hard to show that,
    1) \ln x is an increasing function.
    2) \ln x is differenciable and therefore continous for (0,\infty).
    3) (\ln x)' = \frac{1}{x} \mbox{ for }x>0.
    4)There exists a number, called e, so that \ln e = 1.

    We see that \ln x is a one-to-one function. Define \exp x to be its inverse function on its range.

    It is not hard to show that \exp x is increasing, differenciable and so continous, and furthermore, (\exp x)' = \exp x.

    We have the following supprising property that if q is a rational number then \exp q = e^q. So we define e^r, to be \exp r. With that we generalize exponents as follows a^x = e^{\ln a x} = \exp (\ln a x) \mbox{ for }a>0.

    The difficutly besides for using the definition is to have a formal definition of what an exponent means.
    Your definitions of the natural logerithm are completely valid, but I'm confused by something. Why would you define exp q to be anything other than e^q? I thought exp q was simply another notation for e^q. Am I mistaken?
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  3. #18
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    Quote Originally Posted by ThePerfectHacker View Post
    Define \ln x = \int_1^x \frac{1}{t} dt \mbox{ for }x>0
    This is a standard definition for the logarithm function.
    Using u-substitution we prove that: ln(a+b)=ln(a)+ln(b).
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  4. #19
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Plato View Post
    This is a standard definition for the logarithm function.
    Using u-substitution we prove that: ln(a+b)=ln(a)+ln(b).
    Did you mean ln(ab) = ln(a) + ln(b)?
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  5. #20
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    Quote Originally Posted by ecMathGeek View Post
    Your definitions of the natural logerithm are completely valid, but I'm confused by something. Why would you define exp q to be anything other than e^q? I thought exp q was simply another notation for e^q. Am I mistaken?
    Okay,

    1)We define \ln x as the integral above.

    2)Using FTC and other things we develop important properties for \ln x such as: \ln ab = \ln a + \ln b \mbox{ for }a,b>0.

    3)I said that \ln x is continous and increasing as one of its properties so it has an inverse function which we define as \exp x.

    Note, #3 does not have anything to do with e from the way it is defined above.

    4)It can be shown that if q is a rational number then \exp q = e^q i.e. if q=a/b then \exp q = (\sqrt[b]{e} )^a.

    5)So we see that the \exp function evaluated at q is the same thing as raising the number e (which is define to be so that \ln e) to that power.

    6)Now is seems "natural" to write \exp (x) rather as e^x because for rational powers it makes sense, i.e. same as raising exponents. But for irrational powers (though they were still not define) this function still makes sense! Because \exp x is defined for all real x.

    I think this seems strange to you is that you never learned about what an "exponent" means. Yes a^n = e\cdot a\cdot ... \cdot a n-times. That is how we define them for positive integers. [math[a^0=1[/tex] that is a mere definition. But what about a^q where q is rational? You defined it as finding deminator roots. But how do you define a^r where r is any arbitrary number? That takes works. The approach above is the standard and simplest and nicest approach to this problem. Meaning, we define the exponent of e to be L provided that \ln L = e.


    ---
    My point is that the limit \lim_{x\to c} \frac{a^x - a^c}{x-c} is hard to find because we never appropriately defined what a^x means for a>0.
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