fkampas
(Frank Kampas)
February 8, 2018, 6:27pm
1
A previous post caused me to think about how to visualize Euler’s Formula Exp[ i Pi ] = -1.
I use a (slowly) converging series for Pi.
t[n_] = (4 (-1)^n)/(2 n + 1);
Sum[t[n], {n, 0, \[Infinity]}] - \[Pi]
0
Calculating the exponent of i times the sum of the series for an increasing number of terms gives me a plot which converges toward -1
se = {Re[#], Im[#]} & /@
Table[Exp[I Sum[t[n], {n, 0, m}]], {m, 0, 1000}];
ListPlot[se, Joined -> True]
Great!!!
My try:
t[n_] := (4 (-1)^n)/(2 n + 1);
f[m_] := ReIm[Exp[I Sum[t[n], {n, 0, m}]]];
ParametricPlot[Evaluate[f[m]], {m, 0, 50}, PlotRange -> Full]
1 Like
fkampas
(Frank Kampas)
February 9, 2018, 12:46am
4
I don’t understand how ParametricPlot can work for a function which only evaluates for integer arguments. When I look at the entire plot using PlotRange → All I get
Hello Frank
I checked and it works fine on MMA 10.2(11.2) version.
I don’t know how ParametricPlot can work, maybe one of the WRI/SE jedi can join the party to provide even more insights on this.
Regards Mariusz.
fkampas
(Frank Kampas)
February 9, 2018, 1:09pm
6
Evaluating f generates a continuous function, which is why ParametricPlot can do a plot.
In[12]:= Evaluate[f[x]]
Out[12]= {Re[E^(I (\[Pi] + 2 (-1)^x LerchPhi[-1, 1, 3/2 + x]))],
Im[E^(I (\[Pi] + 2 (-1)^x LerchPhi[-1, 1, 3/2 + x]))]}
Our plots agree for integer arguments of f
fkampas
(Frank Kampas)
February 9, 2018, 9:04pm
7
One can also do the series of expansion of e^(i pi)
ListPlot[Table[ReIm[Sum[(I \[Pi])^n/n!, {n, 0, m}]], {m, 0, 10}],
Joined -> True]
or a ParametricPlot of the evaluated form
ParametricPlot[
Evaluate[ReIm[Sum[(I \[Pi])^n/n!, {n, 0, m}]]], {m, 0, 20},
PlotRange -> All]