[math-fun] Borweins' high precision fraud
From http://www.math.grinnell.edu/~chamberl/courses/444/worksheets/high-precision... In[1]:= ContinuedFraction[Sum[Floor[E n]/2^n,{n,250000}],199]/.t_:>N@t /;t>9^99 Out[1]= {4,1,6,2,8,33818640,128,4294967296,4.034765434510795*10^118,2361183241434822606848,9.526820527087379*10^139,4.669359298953362*10^2270,2.143017214372535*10^301,1.000651735774753*10^2572,2.068391310070504*10^51881,2,1,12,1,6,3,2,5,7,1,230,3,2,1,1,1,3,1,1,9,4,2,4,2,15,1,66,1,3,2,5,1,124,2,1,5,1,1,1,6,14,9,5,5,19,2,6,1,2,2,3,1,6,4,2,1,15,8,1,5,1,2,1,18,1,3,1,2,1,2,2,1,8,1,3,5,6,2,1,3,1,2,1,3,1,3,1,1,2,6,5,2,1,1,10,6,5,1,1,22,1,7,4,5,1,2,1,3,6,1,1,14,1,1,4,1,10,1,10,5,8,7,1,2,3,1,1,7,1,1,18,1,1,21,1,5,8,3,6,21,3,4,1,58,2,2,1,2,47,235,1,1,1,1,1,3,66,6,2,8,1,2,1,2,2,3,2,1,1,34,7,8,1,10,5,2,1,3,2,1,2,3,1,1} In[476]:= ContinuedFraction[Sum[Floor[n π]/2^n,{n,500000}],199]/.n_:>N@n /;n>9^99//tim During evaluation of In[476]:= 37.877844,199 Out[476]= {6,127,638816050508714029100700827906,128,4.770094238300255*10^9930,10384593717069655257060992658440192,4.953549065688299*10^9964,5.144059450474297*10^9998,3.340033906683899*10^49925,1,126,638816050508714029100700827906,128,4.770094238300255*10^9930,10384593717069655257060992658440192,4.953549065688299*10^9964,1,6,33,1,7,4,1,1,950,3,1,1,1,5,1,1,36,7,1,4,4,1,4,1,1,2,2,1,6,8,2,1042,3,1,5,1,1,15,1,67,2,1,34,1,1,2,8,1,3,4,3,3,47,2,4,1,8,14,3,2,3,84,3,12,4,1,1,1,1,2,2,1,2,2,1,8,3,1,1,1,1,3,1,2,1,1,3,2,1,15,1,1,1,2,2,4,1,2,3,2,1,1,3,4,1,7,2,5,4,1,3,1,8,1,31,2,2,3,6,1,8,2,9,10,2,1,11,2,128,7,2,15,2,3,2,2,1,1,1,26,7,1,2,1,5,8,5,2,6,1,1,3,2,1,5,6,1,1,3,1,1,16,24,1,12,2,1,56,9,2,1,1,1,2,1,1,1,2,1,1,8,4,11} In[477]:= ContinuedFraction[Sum[Floor[n GoldenRatio]/2^n,{n,500000}],199]/.t_:>N@t/;t>9^99//tim During evaluation of In[477]:= 38.432713,199 Out[477]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968,618970019642690137449562112,22300745198530623141535718272648361505980416,13803492693581127574869511724554050904902217944340773110325048447598592,3.078281734093319*10^113,4.249103942534137*10^183,1.307993905256674*10^297,5.557802059636756*10^480,7.269571220627866*10^777,4.040283790268164*10^1258,2.937113076490272*10^2036,1.186677035312830*10^3295,3.485404637988021*10^5331,4.136049642673213*10^8626,1.441580660752191*10^13958,5.962449176788715*10^22584,1,4.682698892358224*10^32253,2,8,1,1,6,1,2,16,1,6,1,10,1,2,1,1,6,1,14,1,1,16,2,1,2,1,2,1,3,1,3,1,6,2,1,2,3,2,3,20,2,2,2,1,4,12,1,2,2,1,24,1,6,2,2,3,8,1,2,2,45,2,255,1,1,2,1,7,1,1,1,1,1,2,1,3,2,3,2,6,1,11,2,1,6,1,1,1,1,1,1,3,2,1,4,2,15,1,8,2,1,1,7,66,1,2,1,3,1,1,16,4,1,51,2,1,3,3,1,1,2,2,1,1,4,2,2,1,1,2,1,7,6,1,5,3,4,4,1,5,1,5,1,1,1,12,1,1,1,13,20,2,1,14,34,1,5,1,3,1,2,3,1,6,5,219,1,3,1,41} In[479]:= 2^Fibonacci@Range[-1,15] Out[479]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968,618970019642690137449562112,22300745198530623141535718272648361505980416,13803492693581127574869511724554050904902217944340773110325048447598592,307828173409331868845930000782371982852185463050511302093346042220669701339821957901673955116288403443801781174272,4249103942534136789516705652419749018636744941816255385595553105603228478886817941913300018121834285351114635889972008122772634701221657915276159830132698815550650166683145752253825024} (Shallit's?) --rwg
%479, below, is Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 0 + 1/2 + 3/4 + 1/ 2 + 3/8 + 1/4 + 9/64 + 11/128 + 3/64 + . . . = 2.709803442861291314629 G,Knuth,&P, , Exercise 6.49, gives Sum[2^-Floor[n GoldenRatio],{n,0,∞}] = 1+1/2+1/8+1/16+1/64+1/256+1/512+1/2048+1/4096+...= 1.709803442861291314642 ! --rwg On Sat, Sep 30, 2017 at 12:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
From http://www.math.grinnell.edu/~chamberl/courses/444/ worksheets/high-precision-fraud.pdf
In[1]:= ContinuedFraction[Sum[Floor[E n]/2^n,{n,250000}],199]/.t_:>N@t /;t>9^99 Out[1]= {4,1,6,2,8,33818640,128,4294967296,4.034765434510795* 10^118,2361183241434822606848,9.526820527087379*10^139,4. 669359298953362*10^2270,2.143017214372535*10^301,1. 000651735774753*10^2572,2.068391310070504*10^51881,2,1, 12,1,6,3,2,5,7,1,230,3,2,1,1,1,3,1,1,9,4,2,4,2,15,1,66,1,3, 2,5,1,124,2,1,5,1,1,1,6,14,9,5,5,19,2,6,1,2,2,3,1,6,4,2,1, 15,8,1,5,1,2,1,18,1,3,1,2,1,2,2,1,8,1,3,5,6,2,1,3,1,2,1,3,1, 3,1,1,2,6,5,2,1,1,10,6,5,1,1,22,1,7,4,5,1,2,1,3,6,1,1,14,1, 1,4,1,10,1,10,5,8,7,1,2,3,1,1,7,1,1,18,1,1,21,1,5,8,3,6,21, 3,4,1,58,2,2,1,2,47,235,1,1,1,1,1,3,66,6,2,8,1,2,1,2,2,3,2, 1,1,34,7,8,1,10,5,2,1,3,2,1,2,3,1,1}
In[476]:= ContinuedFraction[Sum[Floor[n π]/2^n,{n,500000}],199]/.n_:>N@n /;n>9^99//tim During evaluation of In[476]:= 37.877844,199 Out[476]= {6,127,638816050508714029100700827906,128,4.770094238300255*10^ 9930,10384593717069655257060992658440192,4.953549065688299*10^ 9964,5.144059450474297*10^9998,3.340033906683899*10^49925,1,126, 638816050508714029100700827906,128,4.770094238300255*10^9930, 10384593717069655257060992658440192,4.953549065688299*10^ 9964,1,6,33,1,7,4,1,1,950,3,1,1,1,5,1,1,36,7,1,4,4,1,4,1,1, 2,2,1,6,8,2,1042,3,1,5,1,1,15,1,67,2,1,34,1,1,2,8,1,3,4,3,3, 47,2,4,1,8,14,3,2,3,84,3,12,4,1,1,1,1,2,2,1,2,2,1,8,3,1,1,1, 1,3,1,2,1,1,3,2,1,15,1,1,1,2,2,4,1,2,3,2,1,1,3,4,1,7,2,5,4, 1,3,1,8,1,31,2,2,3,6,1,8,2,9,10,2,1,11,2,128,7,2,15,2,3,2, 2,1,1,1,26,7,1,2,1,5,8,5,2,6,1,1,3,2,1,5,6,1,1,3,1,1,16,24, 1,12,2,1,56,9,2,1,1,1,2,1,1,1,2,1,1,8,4,11}
In[477]:= ContinuedFraction[Sum[Floor[n GoldenRatio]/2^n,{n,500000}], 199]/.t_:>N@t/;t>9^99//tim During evaluation of In[477]:= 38.432713,199 Out[477]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968, 618970019642690137449562112,22300745198530623141535718272648361505980416, 138034926935811275748695117245540509049022179443407731103250 48447598592,3.078281734093319*10^113,4.249103942534137*10^ 183,1.307993905256674*10^297,5.557802059636756*10^480,7. 269571220627866*10^777,4.040283790268164*10^1258,2. 937113076490272*10^2036,1.186677035312830*10^3295,3. 485404637988021*10^5331,4.136049642673213*10^8626,1. 441580660752191*10^13958,5.962449176788715*10^22584,1,4. 682698892358224*10^32253,2,8,1,1,6,1,2,16,1,6,1,10,1,2,1,1, 6,1,14,1,1,16,2,1,2,1,2,1,3,1,3,1,6,2,1,2,3,2,3,20,2,2,2,1, 4,12,1,2,2,1,24,1,6,2,2,3,8,1,2,2,45,2,255,1,1,2,1,7,1,1,1, 1,1,2,1,3,2,3,2,6,1,11,2,1,6,1,1,1,1,1,1,3,2,1,4,2,15,1,8, 2,1,1,7,66,1,2,1,3,1,1,16,4,1,51,2,1,3,3,1,1,2,2,1,1,4,2,2, 1,1,2,1,7,6,1,5,3,4,4,1,5,1,5,1,1,1,12,1,1,1,13,20,2,1,14, 34,1,5,1,3,1,2,3,1,6,5,219,1,3,1,41}
In[479]:= 2^Fibonacci@Range[-1,15] Out[479]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968, 618970019642690137449562112,22300745198530623141535718272648361505980416, 13803492693581127574869511724554050904902217944340773110325048447598592, 307828173409331868845930000782371982852185463050511302093346 042220669701339821957901673955116288403443801781174272, 424910394253413678951670565241974901863674494181625538559555 310560322847888681794191330001812183428535111463588997200812 2772634701221657915276159830132698815550650166683145752253825024} (Shallit's?) --rwg
Bill, Define the function f(x) as follows: f(x) := Sum[2^-k Sinc[x] Product[ If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {k, 0, Infinity}] Then the integral between 0 and infinity of f(x) agrees to pi for slightly more than 10^2000 digits. [Regarding the discussion of pyramids, you really want every face to be an equilateral triangle. Then, from a distance, the reflection in the sand will give a mirage of a floating octahedron.] Best wishes, Adam P. Goucher
Sent: Friday, October 06, 2017 at 1:05 PM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Borweins' high precision fraud
%479, below, is Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 0 + 1/2 + 3/4 + 1/ 2 + 3/8 + 1/4 + 9/64 + 11/128 + 3/64 + . . . = 2.709803442861291314629 G,Knuth,&P, , Exercise 6.49, gives Sum[2^-Floor[n GoldenRatio],{n,0,∞}] = 1+1/2+1/8+1/16+1/64+1/256+1/512+1/2048+1/4096+...= 1.709803442861291314642 ! --rwg
On Sat, Sep 30, 2017 at 12:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
From http://www.math.grinnell.edu/~chamberl/courses/444/ worksheets/high-precision-fraud.pdf
In[1]:= ContinuedFraction[Sum[Floor[E n]/2^n,{n,250000}],199]/.t_:>N@t /;t>9^99 Out[1]= {4,1,6,2,8,33818640,128,4294967296,4.034765434510795* 10^118,2361183241434822606848,9.526820527087379*10^139,4. 669359298953362*10^2270,2.143017214372535*10^301,1. 000651735774753*10^2572,2.068391310070504*10^51881,2,1, 12,1,6,3,2,5,7,1,230,3,2,1,1,1,3,1,1,9,4,2,4,2,15,1,66,1,3, 2,5,1,124,2,1,5,1,1,1,6,14,9,5,5,19,2,6,1,2,2,3,1,6,4,2,1, 15,8,1,5,1,2,1,18,1,3,1,2,1,2,2,1,8,1,3,5,6,2,1,3,1,2,1,3,1, 3,1,1,2,6,5,2,1,1,10,6,5,1,1,22,1,7,4,5,1,2,1,3,6,1,1,14,1, 1,4,1,10,1,10,5,8,7,1,2,3,1,1,7,1,1,18,1,1,21,1,5,8,3,6,21, 3,4,1,58,2,2,1,2,47,235,1,1,1,1,1,3,66,6,2,8,1,2,1,2,2,3,2, 1,1,34,7,8,1,10,5,2,1,3,2,1,2,3,1,1}
In[476]:= ContinuedFraction[Sum[Floor[n π]/2^n,{n,500000}],199]/.n_:>N@n /;n>9^99//tim During evaluation of In[476]:= 37.877844,199 Out[476]= {6,127,638816050508714029100700827906,128,4.770094238300255*10^ 9930,10384593717069655257060992658440192,4.953549065688299*10^ 9964,5.144059450474297*10^9998,3.340033906683899*10^49925,1,126, 638816050508714029100700827906,128,4.770094238300255*10^9930, 10384593717069655257060992658440192,4.953549065688299*10^ 9964,1,6,33,1,7,4,1,1,950,3,1,1,1,5,1,1,36,7,1,4,4,1,4,1,1, 2,2,1,6,8,2,1042,3,1,5,1,1,15,1,67,2,1,34,1,1,2,8,1,3,4,3,3, 47,2,4,1,8,14,3,2,3,84,3,12,4,1,1,1,1,2,2,1,2,2,1,8,3,1,1,1, 1,3,1,2,1,1,3,2,1,15,1,1,1,2,2,4,1,2,3,2,1,1,3,4,1,7,2,5,4, 1,3,1,8,1,31,2,2,3,6,1,8,2,9,10,2,1,11,2,128,7,2,15,2,3,2, 2,1,1,1,26,7,1,2,1,5,8,5,2,6,1,1,3,2,1,5,6,1,1,3,1,1,16,24, 1,12,2,1,56,9,2,1,1,1,2,1,1,1,2,1,1,8,4,11}
In[477]:= ContinuedFraction[Sum[Floor[n GoldenRatio]/2^n,{n,500000}], 199]/.t_:>N@t/;t>9^99//tim During evaluation of In[477]:= 38.432713,199 Out[477]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968, 618970019642690137449562112,22300745198530623141535718272648361505980416, 138034926935811275748695117245540509049022179443407731103250 48447598592,3.078281734093319*10^113,4.249103942534137*10^ 183,1.307993905256674*10^297,5.557802059636756*10^480,7. 269571220627866*10^777,4.040283790268164*10^1258,2. 937113076490272*10^2036,1.186677035312830*10^3295,3. 485404637988021*10^5331,4.136049642673213*10^8626,1. 441580660752191*10^13958,5.962449176788715*10^22584,1,4. 682698892358224*10^32253,2,8,1,1,6,1,2,16,1,6,1,10,1,2,1,1, 6,1,14,1,1,16,2,1,2,1,2,1,3,1,3,1,6,2,1,2,3,2,3,20,2,2,2,1, 4,12,1,2,2,1,24,1,6,2,2,3,8,1,2,2,45,2,255,1,1,2,1,7,1,1,1, 1,1,2,1,3,2,3,2,6,1,11,2,1,6,1,1,1,1,1,1,3,2,1,4,2,15,1,8, 2,1,1,7,66,1,2,1,3,1,1,16,4,1,51,2,1,3,3,1,1,2,2,1,1,4,2,2, 1,1,2,1,7,6,1,5,3,4,4,1,5,1,5,1,1,1,12,1,1,1,13,20,2,1,14, 34,1,5,1,3,1,2,3,1,6,5,219,1,3,1,41}
In[479]:= 2^Fibonacci@Range[-1,15] Out[479]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968, 618970019642690137449562112,22300745198530623141535718272648361505980416, 13803492693581127574869511724554050904902217944340773110325048447598592, 307828173409331868845930000782371982852185463050511302093346 042220669701339821957901673955116288403443801781174272, 424910394253413678951670565241974901863674494181625538559555 310560322847888681794191330001812183428535111463588997200812 2772634701221657915276159830132698815550650166683145752253825024} (Shallit's?) --rwg
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Gack. So you're saying someone actually needs to _prove_ Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 1 + Sum[2^-Floor[n GoldenRatio],{n,0,∞}] . Imponderable: How can a mirage have solid angles? In any case, I worked out that the solid angle at the apex of a regular n-gon pyramid with vertex angles a: pyramidSolidAngle[n_, a_] := n 2 ArcSin[Sec[a/2] Sin[2π/n] (1 - √(1 - Csc[π/n]^2 Sin[a/2]^2))/2] along with three sadistically dissimilar alternatives, to frustrate FullSimplify and trigonometry students. --rwg Another imponderable: If radishes grow underground, why do they take the trouble to be red? On Fri, Oct 6, 2017 at 8:24 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Bill,
Define the function f(x) as follows:
f(x) := Sum[2^-k Sinc[x] Product[ If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {k, 0, Infinity}]
Then the integral between 0 and infinity of f(x) agrees to pi for slightly more than 10^2000 digits.
[Regarding the discussion of pyramids, you really want every face to be an equilateral triangle. Then, from a distance, the reflection in the sand will give a mirage of a floating octahedron.]
Best wishes,
Adam P. Goucher
Sent: Friday, October 06, 2017 at 1:05 PM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Borweins' high precision fraud
%479, below, is Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 0 + 1/2 + 3/4 + 1/ 2 + 3/8 + 1/4 + 9/64 + 11/128 + 3/64 + . . . = 2.709803442861291314629 G,Knuth,&P, , Exercise 6.49, gives Sum[2^-Floor[n GoldenRatio],{n,0,∞}] = 1+1/2+1/8+1/16+1/64+1/256+1/512+1/2048+1/4096+...= 1.709803442861291314642 ! --rwg
On Sat, Sep 30, 2017 at 12:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
From http://www.math.grinnell.edu/~chamberl/courses/444/ worksheets/high-precision-fraud.pdf
[CHOP]
Hello, here is one that I like a lot : 1304*sum(1/(4*n^2-4*n+4*163^2+1),n=1..infinity); is equal to Pi at 443 digits. if 163 is replaced by 1000 : the precision is 2727 Digits. The trick is the expression : 8*sum(1/(4*n^2-4*n+4*k^2+1),n=1..infinity) = = 1/k*Pi*tanh(Pi*k) and since tanh(X) is very close to 1 when X >> 1 then the approximation is very good. The underlying explanation is that the sum is complex but the imaginary parts cancel each other and gives birth to the tanh( ) argument. (classical result). Simon Plouffe Le 2017-10-06 à 19:32, Bill Gosper a écrit :
Gack. So you're saying someone actually needs to _prove_ Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 1 + Sum[2^-Floor[n GoldenRatio],{n,0,∞}] .
Imponderable: How can a mirage have solid angles? In any case, I worked out that the solid angle at the apex of a regular n-gon pyramid with vertex angles a: pyramidSolidAngle[n_, a_] := n 2 ArcSin[Sec[a/2] Sin[2π/n] (1 - √(1 - Csc[π/n]^2 Sin[a/2]^2))/2]
along with three sadistically dissimilar alternatives, to frustrate FullSimplify and trigonometry students. --rwg Another imponderable: If radishes grow underground, why do they take the trouble to be red?
On Fri, Oct 6, 2017 at 8:24 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Bill,
Define the function f(x) as follows:
f(x) := Sum[2^-k Sinc[x] Product[ If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {k, 0, Infinity}]
Then the integral between 0 and infinity of f(x) agrees to pi for slightly more than 10^2000 digits.
[Regarding the discussion of pyramids, you really want every face to be an equilateral triangle. Then, from a distance, the reflection in the sand will give a mirage of a floating octahedron.]
Best wishes,
Adam P. Goucher
Sent: Friday, October 06, 2017 at 1:05 PM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Borweins' high precision fraud
%479, below, is Sum[Floor[n GoldenRatio]/2^n,{n,0,∞}] = 0 + 1/2 + 3/4 + 1/ 2 + 3/8 + 1/4 + 9/64 + 11/128 + 3/64 + . . . = 2.709803442861291314629 G,Knuth,&P, , Exercise 6.49, gives Sum[2^-Floor[n GoldenRatio],{n,0,∞}] = 1+1/2+1/8+1/16+1/64+1/256+1/512+1/2048+1/4096+...= 1.709803442861291314642 ! --rwg
On Sat, Sep 30, 2017 at 12:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
From http://www.math.grinnell.edu/~chamberl/courses/444/ worksheets/high-precision-fraud.pdf
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