# An interesting pattern comes out of Riemann sum

Posted 11 months ago
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 An interesting pattern comes out of Riemann sum sq=Table[j,{j,1000}] n=Select[sq,PrimeQ,(100)] sq2=Table[k,{k,100}] n3=sq2*-1 r=Table[k1,{k1,100}] f=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n])) fsolution=((((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n])))*-1/12 ff=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r*1/n]))*1/2 fff=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r*1/2*-n3])) ffff=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/2*n3])) ffffsolution=((((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/-2*n3])))/-2 ffffa=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n3])) ffffab=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/-0.083])) h=f/ff hb=f/(ff*2) h2=ff/ffff hh=f-ff hh2=ffff-fff z=(h2-h)*h bb=Im[f] bb1=Im[ff] bb2=Im[fff] bb3=ReIm[ffff] bb3a=ReIm[ffffa] cc=Re[ffffa] cc2=Im[ffffa] cc2b=Im[ffffab] bb4=ReIm[fsolution] bb5=Im[ffffsolution] vva=Round[f] vv1=Round[ff] bba=Round[fff] vvc=Round[ffff] vvb=Round[ffffsolution] vv=Round[fsolution] g=2*Pi*(r+f) gg=2*Pi*r rr=(z-z*Power[z, (z)^-1])*2*n3 s1c=((1/2)+bb*r*Sqrt[-1]) s1cc=(((1)+bb*r*Sqrt[-1])+((0)+bb*r*Sqrt[-1]))/2 s1b=((1/2)+fsolution*bb4) s1=(1/2)+fsolution*r ss1=(1/2)+fsolution*r*Sqrt[-1] ss2=(1/2)+bb1*r*Sqrt[-1] ssss1=(1/2)+ff*r ssss1b=((1/2)+bb2*r*Sqrt[-1])/-2 sss2=Re[ssss1] ss3=(1/2)+fff*r*Sqrt[-1] ssss=(1/2)+fff*r sssa=(1/2)+cc2*Sqrt[-1*-n3] sssab=(1/2)+cc2b*Sqrt[-1*-n3] sssb=((1/2)+ffffsolution*r)/-2 sssb1=((1/2)+bb5*Sqrt[r*-1])/-2 sssc=(1/2)+ffff*r*Sqrt[-1] sss3=Re[ssss] s=ReIm[s1] s2=ReIm[ss1] sa=ReIm[ssss1] sb=ReIm[ssss] sc=ReIm[sssa] sd=ReIm[sssb] se=ReIm[sssc] s2=(1/2)+vvc*r*Sqrt[-1] zz=-n3 zx=n x1c1c=\!$$\*UnderoverscriptBox[\(\[Sum]$$, $$zx = 100000$$, $$100100$$]$$1/zx*zx^s1cc$$\) x1cc=ReIm[x1c1c] x1cc1=Re[x1c1c] x1cc2=Im[x1c1c] ListLinePlot[x1cc] 
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Posted 11 months ago
 The graph at the begining is like this
Posted 11 months ago
 Real part and imaginary part are complementary:
 As the numbers increase new patterns emerge: x1c1c=\!$$\*UnderoverscriptBox[\(\[Sum]$$, $$zx = 1000000000000$$, $$1000000000100$$]$$1/zx*zx^s1cc$$\)! 
 I don't know how to show the manipulate so I am placing the whole program lines, sorry!! It is about squaring a circle... sq=Table[j,{j,1000}] n=Select[sq,PrimeQ,(100)] sq2=Table[k,{k,100}] n3=sq2*-1 r=Table[k1,{k1,100}] f=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n])) fsolution=((((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n])))*-1/12 ff=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r*1/n]))*1/2 fff=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r*1/2*-n3])) ffff=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/2*n3])) ffffsolution=((((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/-2*n3])))/-2 ffffa=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/n3])) ffffab=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r/-0.083])) h=f/ff hb=f/(ff*2) h2=ff/ffff hh=f-ff hh2=ffff-fff z=(h2-h)*h bb=Im[f] bb1=Im[ff] bb2=Im[fff] bb3=ReIm[ffff] bb3a=ReIm[ffffa] cc=Re[ffffa] cc2=Im[ffffa] cc2b=Im[ffffab] bb4=ReIm[fsolution] bb5=Im[ffffsolution] vva=Round[f] vv1=Round[ff] bba=Round[fff] vvc=Round[ffff] vvb=Round[ffffsolution] vv=Round[fsolution] g=2*Pi*(r+f) gg=2*Pi*r rr=(z-z*Power[z, (z)^-1])*2*n3 s1c=((1/2)+bb*r*Sqrt[-1]) s1cc=(((1)+bb*r*Sqrt[-1])+((0)+bb*r*Sqrt[-1]))/2 s1b=((1/2)+fsolution*bb4) s1=(1/2)+fsolution*r ss1=(1/2)+fsolution*r*Sqrt[-1] ss2=(1/2)+bb1*r*Sqrt[-1] ssss1=(1/2)+ff*r ssss1b=((1/2)+bb2*r*Sqrt[-1])/-2 sss2=Re[ssss1] ss3=(1/2)+fff*r*Sqrt[-1] ssss=(1/2)+fff*r sssa=(1/2)+cc2*Sqrt[-1*-n3] sssab=(1/2)+cc2b*Sqrt[-1*-n3] sssb=((1/2)+ffffsolution*r)/-2 sssb1=((1/2)+bb5*Sqrt[r*-1])/-2 sssc=(1/2)+ffff*r*Sqrt[-1] sss3=Re[ssss] s=ReIm[s1] s2=ReIm[ss1] sa=ReIm[ssss1] sb=ReIm[ssss] sc=ReIm[sssa] sd=ReIm[sssb] se=ReIm[sssc] s2=(1/2)+vvc*r*Sqrt[-1] zz=-n3 zx=n x1c1c=\!$$\*UnderoverscriptBox[\(\[Sum]$$, $$zx = 100000$$, $$100100$$]$$1/zx*zx^s1cc$$\) x=ReIm[x1c1c] x1cc1=Re[x1c1c] x1cc2=Im[x1c1c] ListLinePlot[x1cc] Manipulate[ ParametricPlot[{r Cos[x], r Sin[x]}, {x, 1, m}, {r, 0, k}, Mesh -> Full, PlotRange -> {{-2, 2}, {-2, 2}}], {m, 0, 5 Pi}, {k, 1, 3}]