Proof e x = exp ⁡ ( x )?Define ln ⁡ ( x ) = ∫ 1 x 1 t Assume I have proven that ln ⁡ x is one-to-one and therefore has an inverse exp ⁡ ( x )Define e as: ln ⁡ e = 1Now, if you have no other notion of exponentials, or logarithms, how could define what e x means and show that its the inverse of ln ⁡ x?You are allowed to assume the logarithmic product and quotient property.Thanks for the help.

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Proof       e    x    =  exp  ⁡  (  x  )?Define  ln  ⁡  (  x  )  =      ∫          1              x            1    t  Assume I have proven that   ln  ⁡  x is one-to-one and therefore has an inverse   exp  ⁡  (  x  )Define   e as:  ln  ⁡  e  =  1Now, if you have no other notion of exponentials, or logarithms, how could define what       e    x   means and show that its the inverse of   ln  ⁡  x?You are allowed to assume the logarithmic product and quotient property.Thanks for the help.

Hayley Mccarthy · Accepted Answer

Writing   f for   ln and   g for its inverse, you can show easily that   g is infinitely differentiable and that       g          (      n      )        (  0  )  =  1. This gives you the Taylor series   g  (  x  )  =      ∑          n      =      0              ∞            x    n        /    n  !. After that, everything follows from the classical analysis of   g that is performed in every elementary real variables text (see Rudin&#039;s Real &amp;amp; Complex Analysis, for example).As I recall, the introductory &quot;Chapter 0&quot; of that text is a marvel of succinct mathematics that fully constructs the exponential function from scratch. It&#039;s really a pleasure to read and I&#039;m always awed at his insight every time I read it.

logiski9s · Accepted Answer

Full disclosure: this is essentially a rewrite of MPW&#039;s answer.Defining       e    x   as       ∑          k      =      0              ∞                          x        k                    k        !             makes sense, in a way it&#039;s the most fundamental/general definition because it can be applied to any system for which addition, multiplication and scaling are defined. Reals, complex numbers, quaternions, matrices, etc.Now, from calculus we have this result:      d          d      x            [          f              −        1              (    x    )    ]    =            1                        f          ′                (                  f                      −            1                          (        x        )        )            By fundamental theorem of calculus we obtain       ln    ′    ⁡  (  x  ) as       1          x      , hence:      d          d      x            [          ln              −        1              ⁡    (    x    )    ]    =      ln          −      1        ⁡  (  x  )It follows (formally by induction) that the   nth derivative of       ln          −      1        ⁡  (  x  ) is       ln          −      1        ⁡  (  x  ) and hence the   nth derivative at   x  =  0 is       ln          −      1        ⁡  (  0  )  =  1Thus we get the Taylor series:      ln          −      1        ⁡  (  x  )  =      ∑          k      =      0              ∞                          x        k                    k        !              =      e    x

Proof e x </msup> = exp ⁡ ( x ) ? Define

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