Skip to main content

numerics - Does Mathematica get Pi wrong?


I happened to watch a Youtube video on Pi. According to the video, the 1 millionth digit of Pi is 1. And here is another page of the first 1 million digits of Pi.


You can get the same answer from WolframAlpha.


However, if you let Mathematica to calculate the digits you will get:


N[Pi, 1000005]
....65200102821**3**0222`1000005.


You will notice the 1 millionth digit of Pi in Mathematica is 3, not 1. Is anything wrong here?


Actually, if you move back 32 digits, you can see the exact digits as in the video.


**5779458151**309275628320845315846520010282130222`1000005.

Updates:


According to the answers, you can obtain the correct digits by following commands:




  1. Command from answer of @m_goldberg


    RealDigits[N[Pi, 1000001]][[1, -10 ;; -1]]





  2. Similarly, you can convert the number to string, which is the answer of @Daniel_Lichtblau


    str = ToString[N[Pi, 1000001], InputForm];
    Characters[str][[1000002 - 9 ;; 1000002]]




  3. RealDigits can extract specific digits as in the answer of @Mr.Wizard


    RealDigits[Pi, 10, 10, 9 - 1*^6]





The last one is much faster than others:


In[334]:= Timing[RealDigits[Pi, 10, 10, 9 - 1*^6]]

Out[334]= {0.036622, {{5, 7, 7, 9, 4, 5, 8, 1, 5, 1}, -999990}}

In[335]:= Timing[RealDigits[N[Pi, 1000001]][[1, -10 ;; -1]]]

Out[335]= {0.229211, {5, 7, 7, 9, 4, 5, 8, 1, 5, 1}}


If this problem is generalized to obtain 1 million digits after decimal mark, the first two commands may provide wrong results. As is mentioned in the answer of @Mr.Wizard, the result provided by RealDigits[Pi, 10, 10, 9 - 1*^6] is the 999,991 to 1,000,000 digits behind decimal mark, nevertheless how many digits before the decimal mark. But for the first two method these digits should be counted and subtracted from the result. For the second method, the decimal mark takes one character, which should be included in the calculation.


The first method can be modified to following codes to consider the digits before decimal mark, but enough digits must be obtained in the first command:


num = RealDigits @ N[Pi, 1000010];   
num[[1,999991+num[[2]];;1000000+num[[2]]]]

Conclusions




  1. Command N does not guarantee to output exactly number of digits as the command requiring. Moreover, different results may be obtained in different version of Mathematica.





  2. RealDigits with four arguments is the most efficient way to extract specific digits in a number.




  3. Converting number to InputForm is another possible way to obtain the digits without RealDigits command.





Answer



You have selected the wrong digit. Mathematica gets the digit in the million-th decimal place right if the calculation is performed correctly.


 q = N[Pi, 1000010];

RealDigits[q][[1, 1000001]]


1

I take the 1000001-th digit because RealDigits includes the integer part, 3.


Update


It is really important to use RealDigits to decide this question. Looking at the displayed full form number is not reliable because it shows extra digits added in the working precision needed to get the specified real precision. Consider


N[Pi, 20] // FullForm



  3.1415926535897932384626433832795028841971693993751058209749`20.

That's a lot more than the 20 digits asked for. However,


RealDigits @ N[Pi, 20]


{{3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4}, 1}

gives the actual set of correct digits.



Comments

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...