# Partial zeta curlicues

I asked a question at MSE and Raymond Manzoni provided an excellent answer that included these visuals:

I made a very basic attempt at recreation with

y = 10000; ListLinePlot[Table[{Im[E^(-I y Log[k])/Sqrt[k]], Re[E^(-I y
Log[k])/Sqrt[k]]}, {k, 1, Round[Sqrt[10000]]}]]


but a very messy plot resulted. I tried various parametric plots, but was equally disappointed with the results. Could someone please point me in the right direction to achieve the curlicues (that were apparently applet-generated) in the answer listed?

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You need to define partial sums (I will use Simon's definition in the comments which is faster than my older one):

Manzoni[n_, y_] := Transpose@{Re@#, Im@#} &@Accumulate[Range[n]^-(.5 + y I)]


The thing worth attention here is 1/k^(.5 + I y) - meaning 0.5 which makes it automatically numeric before the sum is taken. If you would keep 1/2 and wrap N[...] outside it would be much slower.

ListLinePlot[Manzoni[3000, 10000], Frame -> True, GridLines -> Automatic,
PlotStyle -> Directive[Red, Opacity[.7]]]


ListLinePlot[Manzoni[4000, 10000], Frame -> True, GridLines -> Automatic,
PlotStyle -> Directive[Red, Opacity[.2]],
PlotRange -> {{-.3398, -.339}, {-.0378, -.0365}}, AspectRatio -> 1]


From pure artistic point of view I think a spline can make it real pretty, perhaps even tattoo-ready ;-)

Graphics[BSplineCurve[Manzoni[3000, 10000]]]


Other parameters:

Graphics[BSplineCurve[Manzoni[5000, 20000]]]


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This is incredible! Thank you so much for writing this - I hope this answer gets the attention it deserves!! – martin Jun 2 '14 at 7:24
You could use something like data = Transpose@{Re@#, Im@#} &@ Accumulate[Range[3000]^-(0.5 + 10000 I)]; to calculate the data – Simon Woods Jun 2 '14 at 16:04
+1 @SimonWoods - much better of course - thank you - I updated the post with your definition. – Vitaliy Kaurov Jun 2 '14 at 16:15
+1. BTW I think there is an L-system generating a similar pattern, but I failed to find the exact name. – Silvia Jun 11 '14 at 17:59