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An interesting pattern comes out of Riemann sum

Posted 1 year 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]

PERFECT PATTERN OF RIEMANN SUM

4 Replies

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}]

As the numbers increase new patterns emerge:

x1c1c=\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(zx = 1000000000000\), \(1000000000100\)]\(1/zx*zx^s1cc\)\)!

At high numbers:

Real part and imaginary part are complementary:

real part

imaginary part

The graph at the begining is like thisenter image description here

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