I was wondering how fast the Taylor Series for $\exp(\pi i)$ would converge. According to Taylor,

$$\exp(\pi i) = 1 + (\pi i) + (\pi i)^2/2! + (\pi i)^3/3! + (\pi i)^4/4! + \ldots = \sum_{j=0}^\infty \frac{(\pi i)^j}{j!}.$$

According to Euler, $\exp(\pi i)= -1$. I wanted to estimate the convergence rate, so I just fired up Mathematica and plotted the values of $$error(k) = \left| 1+ \sum_{j=0}^\infty \frac{(\pi i)^j}{j!}\right|.$$ The Mathematica plot is shown below.

It occurred to me that Mathematica knows thousands of digits of pi. I could force it to only use the first 16 digits by using double precision floating point numbers. If I do that, then the log(Error) plot is “wrong” because the approximation for pi is not accurate enough. The erroneous plot is shown below.

If you want a good estimate of the truncated Taylor series error using $k$ terms, you could use the absolute value of term $k+1$ which is

$$error(k)\approx f(k)= \frac{\pi^{k+1}}{(k+1)!}$$.

Here is a plot of $r(k)= error(k)/f(k)$. If the estimator is good, then the $r(k)$ values (y-coordinates) should be near 1.

Creating these plots requires around 30 digits of pi! If I wanted to look at more terms, I would need more digits of pi.

Here is the code I used to generate the plots.

SetOptions[ListPlot, GridLines -> Automatic, Frame -> True, ImageSize -> 250]; error[k_] := Abs[ 1 + Sum[ (I Pi)^j/j!, {j, 0, k}]]; plt0 = ListPlot[ Array[error, 10], Frame -> True, FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", "Error"})]; plt1 = ListPlot[Log[ Array[error, 40]], FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", "log(Error)"})]; error[k_] := Abs[ 1 + Sum[ (I N[Pi])^j/j!, {j, 0, k}]]; plt3 = ListPlot[Log[ Array[error, 40]], Frame -> True, FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", "log(Error)", "Bad Error plot Caused by using only 16 digits of pi", " "}), PlotLabel -> Style["Bad Error plot Caused by using\n only 16 digits of pi", 16]]; errorRat[k_] := Abs[ 1 + Sum[ (I Pi)^j/j!, {j, 0, k}]]/ (Pi^(k + 1)/(k + 1)! ); plt4 = ListPlot[Array[errorRat, 40], Frame -> True, FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", "error(k)/f(k)"}), PlotLabel -> Style["Plot error(k) divided by estimated\n error(k) for \ k=1,2,..., 40", 16]];