The function, f ( x ) = { ( e x − 1 ) / x x ≠ 0 1 x = 0 is continuous and differentiable at x = 0. By composition, the x ≠ 0 case is analytic everywhere it is defined, and integrating f around 0 in the complex plane yields 0 and suggests that there is no pole there. Furthermore, working from the Taylor series of e x , the x ≠ 0 case can be seen to have a series that is defined at 0 and equal to 1: e x − 1 x = − 1 + ∑ k = 0 ∞ x k / k ! x = ∑ k = 1 ∞ x k / k ! x = ∑ k = 0 ∞ 1 k + 1 x k k ! From this we can also see that f is analytic and infinitely differentiable.All taken together, it seems strange that such a nicely behaved function would have an irremovable "patch" necessary at x = 0. Nonetheless, I have not been able to come up with another way to write the function that allows it to be defined at zero without a special case there. Is it impossible to write this analytic function without a case block?

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The function,  f  (  x  )  =      {                            (                      e            x                    −          1          )                      /                    x                          x          ≠          0                                      1                          x          =          0                        is continuous and differentiable at   x  =  0. By composition, the   x  ≠  0 case is analytic everywhere it is defined, and integrating   f around 0 in the complex plane yields 0 and suggests that there is no pole there. Furthermore, working from the Taylor series of       e    x  , the   x  ≠  0 case can be seen to have a series that is defined at 0 and equal to 1:                    e        x            −      1        x    =            −      1      +              ∑                  k          =          0                ∞                    x        k                    /            k      !        x    =                    ∑                  k          =          1                ∞                    x        k                    /            k      !        x    =      ∑          k      =      0        ∞        1          k      +      1                  x              k                    k      !      From this we can also see that   f is analytic and infinitely differentiable.All taken together, it seems strange that such a nicely behaved function would have an irremovable &quot;patch&quot; necessary at   x  =  0. Nonetheless, I have not been able to come up with another way to write the function that allows it to be defined at zero without a special case there. Is it impossible to write this analytic function without a case block?

odmeravan5c · Accepted Answer

The answer to your question lies within your derivation of the Maclaurin series expansion of   f. You made a mistake in simplification. You wrote that      1    x        ∑          k      =      1              ∞                  x      k              k      !        =      ∑          k      =      1              ∞                  x              k        −        1                    k      !        =      ∑          k      =      0              ∞                  x      k              (      k      +      1      )      !        ,but this is not entirely correct. The simplification      1    x        ∑          k      =      1              ∞                  x      k              k      !        =      ∑          k      =      1              ∞                  x              k        −        1                    k      !      is only valid for   x  ≠  0. Nonetheless, because the function is piecewise defined, it just so happens that the Maclaurin series expansion also works at   x  =  0  , but that is not due to your derivation; rather, that is due to the patching itself done piecewise.Now, as you indicate, this does indeed demonstrate that   f is continuous and differentiable at 0. In fact, it is real-analytic at 0, and furthermore, it extends to an entire function. So 0 is merely a removable singularity of the function   g defined by  g  (  x  )  =                    e        x            −      1        x    .What this means is that      lim          t      →      0        g  (  t  )  =  1  =  f  (  0  )  .Since   g is continuous everywhere else, it follows that  g  (  x  )  =      lim          t      →              x              g  (  t  )  =  f  (  x  )  .So we can simply define  f  (  x  )  =      lim          t      →              x              g  (  t  )  =      lim          t      →              x                                e        t            −      1        t    .This is a form of representing   f without using cases.

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