Re: [math-fun] Fun with math: Dividing one by 998001 yields a surprising result
What am I doing wrong? (seriously): << In[3]:= N[1/99801,500] Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . .
--Dan ________________________________________________________________________________________ It goes without saying that .
You want 998001. Two zeros. On Fri, Jan 27, 2012 at 1:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
What am I doing wrong? (seriously):
<< In[3]:= N[1/99801,500]
Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . .
--Dan
________________________________________________________________________________________ It goes without saying that .
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Anyway, if y'all like this sort of thing, look at 1/998999. On Fri, Jan 27, 2012 at 1:15 PM, Allan Wechsler <acwacw@gmail.com> wrote:
You want 998001. Two zeros.
On Fri, Jan 27, 2012 at 1:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
What am I doing wrong? (seriously):
<< In[3]:= N[1/99801,500]
Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . .
--Dan
________________________________________________________________________________________ It goes without saying that .
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Anyway, if y'all like this sort of thing, look at 1/998999.
Between some trivia I found some nice fractions in http://www.geometer.org/mathcircles/fractions.pdf (by Tom Davis) e.g. 1/998 = .001002004008016032064128256... and a fraction for .000100040009001600250036... (which?) what I could not find there but it also quite well-known: The decimal expansions of the multiples of 1/7 (and of inverses of so-called full reptend primes) are cyclic shifted versions of each other. There exists a simple card trick (which I found in a book of Martin Gardner) based on this fact. Christoph
SetAccuracy[1/998001, 4000] gives the same problem (no 998) for me. Bill C. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Friday, January 27, 2012 11:16 AM To: Dan Asimov; math-fun Subject: [EXTERNAL] Re: [math-fun] Fun with math: Dividing one by 998001 yields a surprising result You want 998001. Two zeros. On Fri, Jan 27, 2012 at 1:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
What am I doing wrong? (seriously):
<< In[3]:= N[1/99801,500]
Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . .
--Dan
________________________________________________________________________________________ It goes without saying that .
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I find 998 at location 2399, using only 3000 digits! WFL On 1/27/12, Cordwell, William R <wrcordw@sandia.gov> wrote:
SetAccuracy[1/998001, 4000] gives the same problem (no 998) for me.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Friday, January 27, 2012 11:16 AM To: Dan Asimov; math-fun Subject: [EXTERNAL] Re: [math-fun] Fun with math: Dividing one by 998001 yields a surprising result
You want 998001. Two zeros.
On Fri, Jan 27, 2012 at 1:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
What am I doing wrong? (seriously):
<< In[3]:= N[1/99801,500]
Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . .
--Dan
________________________________________________________________________________________ It goes without saying that .
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Let's see, 1000 occurs almost right away, and 1001 occurs in the context 010011. 1002 occurs in 001002, but I am suspecting that 1003 never occurs. Have I mis-thunk? On Fri, Jan 27, 2012 at 3:25 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
I find 998 at location 2399, using only 3000 digits! WFL
On 1/27/12, Cordwell, William R <wrcordw@sandia.gov> wrote:
SetAccuracy[1/998001, 4000] gives the same problem (no 998) for me.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Friday, January 27, 2012 11:16 AM To: Dan Asimov; math-fun Subject: [EXTERNAL] Re: [math-fun] Fun with math: Dividing one by 998001 yields a surprising result
You want 998001. Two zeros.
On Fri, Jan 27, 2012 at 1:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
What am I doing wrong? (seriously):
<< In[3]:= N[1/99801,500]
Out[3]= 0.0000100199396799631266219777356940311219326459654712878628470656606 . . .
--Dan
It goes without saying that .
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hello, these expansions are the reflect that for example 1/49 = 0.02040816326530612244897959... we clearly see powers of 2, but in general we have this Maple procedure : dec:=proc(p) local n, p1, p2, p3, p4, ex; ex := trunc(log10(p) + 1); p2 := trunc(10^ex/p); p3 := trunc(10^ex) - p*trunc(10^ex/p); p4 := ex*n; print(1/p, "can be expanded with this serie"); return 1/p = Sum(p2*p3^(n - 1)/10^p4, n = 1 .. infinity) end; I did that a long time ago ! another example with 1/997. 0.00100300902708124373119358074222668004012036108324974924774322968906720160\ 481444332998996990972918756268806419257773319959879638916750250752256770
dec(997); 1/997, "can be expanded with this serie"
infinity ----- (n - 1) \ 3 1/997 = ) -------- / (3 n) ----- 10 n = 1 in that case where we have 1/999^2 then it is the fact that 1/(1-x)^2 is 1 -------- 2 (1 - x)
series(%,x,23); 2 3 4 5 6 7 8 9 10 1 + 2 x + 3 x + 4 x + 5 x + 6 x + 7 x + 8 x + 9 x + 10 x + 11 x + 12
11 12 13 14 15 16 17 18 19 x + 13 x + 14 x + 15 x + 16 x + 17 x + 18 x + 19 x + 20 x 20 21 22 23 + 21 x + 22 x + 23 x + O(x ) The expansion in base 1000 gives the expansion. Best regards, Simon Plouffe
participants (6)
-
Allan Wechsler -
Cordwell, William R -
Dan Asimov -
Fred lunnon -
Pacher Christoph -
Simon Plouffe