tag:blogger.com,1999:blog-4912713243046142041.post5707916407171227302..comments2024-03-27T18:20:38.176-05:00Comments on TYWKIWDBI ("Tai-Wiki-Widbee"): 1/998001 generates a fascinating seriesMinnesotastanhttp://www.blogger.com/profile/01382888179579245181noreply@blogger.comBlogger20125tag:blogger.com,1999:blog-4912713243046142041.post-49197919194826867152013-02-07T00:43:02.450-06:002013-02-07T00:43:02.450-06:00there is nothing strange, nor glitch, it's mat...there is nothing strange, nor glitch, it's mathematical. People! Stop! Stop giving emotional content over something which might look as a coincidence but which is not. There is no special god-alike hidden SMS in fraction. No number-encrypted magic serial number and no free-pass to matrix heaven. <br /><br />Please, look at it rationally, with your logical brain. If I would tell you: <br /> <br /> 1-1 = 0<br /><br />Would you ask me where is the 1? Same thing here.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-69371846790713190362012-03-23T14:01:17.602-05:002012-03-23T14:01:17.602-05:00you have realized that this fraction and its decim...you have realized that this fraction and its decimal result briefly describe the initial terms of the Fibonacci sequence??????<br />What do you think??????????lenhttp://page3.comnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-20223110362833108942012-02-06T03:42:09.897-06:002012-02-06T03:42:09.897-06:00sorry, the second last one is really always missin...sorry, the second last one is really always missing, e.g. for 1/81: 0.01234567901234...Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-69037474013272052072012-02-06T03:39:53.007-06:002012-02-06T03:39:53.007-06:00yep, that's it. You have to write a square mat...yep, that's it. You have to write a square matrix of 001's (999 rows and columns). The nth digit block of 1/998001 is given by the sum over the nth diagonal, i.e. n x 001. Try it with the easiest one, 1/9 = 0.11111, 1/81 = (1/9)^2 = 0.012345678901234...Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-28874499993105966092012-02-02T04:51:56.570-06:002012-02-02T04:51:56.570-06:00Some people won't read. Some others can't ...Some people won't read. Some others can't think.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-58231897514681942482012-01-26T16:00:06.031-06:002012-01-26T16:00:06.031-06:00998 isn't missing. The next number 999 'ov...998 isn't missing. The next number 999 'overflowed' into 998. Likewise the next number 1000, the 1 was added to 999.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-45812043095175177082012-01-24T10:08:21.160-06:002012-01-24T10:08:21.160-06:00Well, that's simple, 'cause e to the power...Well, that's simple, 'cause e to the power of pi is equal to 23.140... and e to the power of iPi (is Apple behind all this?... holy mother of god... IT'S A CONSPIRACY!) in the Gaussian plain (Gaußsche Zahlenebene, sorry, I'm German) is just cos(pi), which is -1. It's not that strange. Only complex. :DAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-64857836431100859642012-01-24T09:55:35.940-06:002012-01-24T09:55:35.940-06:00It's not a roundoff thing: 998 really is missi...It's not a roundoff thing: 998 really is missing, according to the calculation at the link in my comment above. After 999, the whole thing starts over with 000 and continues repeating forever.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-84265078297598723682012-01-24T08:15:38.375-06:002012-01-24T08:15:38.375-06:00I don't think that this is due to numerical er...I don't think that this is due to numerical errors, I rather think it's a systematic pattern.<br /><br />As far as I can see, the second last number is always missing.<br />I have got no mathematical proof for it yet, but it is quite obvious:<br /><br />For 1/998001 the 998 is missing.<br />For 1/9801 the 98 is missing.<br />For 1/81 the 8 is missing.<br /><br />For 1/7601 in octal, the 76 is missing,<br />For 1/FE01 in hex, its FE.<br /><br />I just wrote a small program which calculated the series using long devision,<br />if anyone wants to check it out: http://pastebin.com/i5ShcfdaTillnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-90681620754545272042012-01-23T22:14:56.531-06:002012-01-23T22:14:56.531-06:00I am recurrently gobsmacked by the breadth and dep...I am recurrently gobsmacked by the breadth and depth of knowledge of readers here. Thanks, James.Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-19373178669716624882012-01-23T19:02:45.182-06:002012-01-23T19:02:45.182-06:00Oops! I should have said you can check the identit...Oops! I should have said you can check the identity by showing the two sides become equal after you divide by x^2 and integrate.Jamesnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-46171440095883967342012-01-23T19:00:20.215-06:002012-01-23T19:00:20.215-06:00Mark above is exactly right. If you use 99^2=9801 ...Mark above is exactly right. If you use 99^2=9801 instead, you get 0.0001020304..., and if you use 9^2=81, you get 0.01234567... If you use 9999^2=99980001, you get 0.00000001000200030004...<br /><br />The reason is that the sum of nx^(n+1) from n=0 to infinity equals (1/x - 1)^(-2). (You can show this by checking they have the same derivative.) If you plug in x=1/1000 (say), the infinite sum becomes the decimal expansion you had, and it's value is 999^(-2).<br /><br />A welcome chance for me to give back to TYWKIWDBI!Jamesnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-66140163171889939852012-01-23T18:24:28.315-06:002012-01-23T18:24:28.315-06:00Stan, thanks for this post. Heumpje, thanks for th...Stan, thanks for this post. Heumpje, thanks for the clarification about the roundoff. I was wondering why it skipped a digit!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-90834471413329759942012-01-23T16:39:07.836-06:002012-01-23T16:39:07.836-06:00Well, that makes sense. Thanks.Well, that makes sense. Thanks.Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-9833638232461841472012-01-23T15:44:30.776-06:002012-01-23T15:44:30.776-06:00Having some experience with numerical roundoff, I ...Having some experience with numerical roundoff, I think the 'glitch' in the end is indeed that. That last 9 stands for 8999, which rounds of to '9'.Heumpjehttps://www.blogger.com/profile/01993462463863993238noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-17410300387636915712012-01-23T13:31:29.148-06:002012-01-23T13:31:29.148-06:00Yeah, you just have to remember what we were all t...Yeah, you just have to remember what we were all taught in high school, (maybe earlier. (Really, I swear.)) that any repeating non terminating decimal is a rational number. If you start with the repeating number you want, you can recover its rational value. With a big enough calculator, in this case. Assign digit values to 'Minnesotastan' (or whatever), and you can find 'your' fraction.jkhttp://www.homeschoolmath.net/teaching/rational_numbers.phpnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-7080780915792556512012-01-23T12:48:58.605-06:002012-01-23T12:48:58.605-06:00I just realized this would make a nice bar bet. W...I just realized this would make a nice bar bet. Write down the fraction, and then bet someone that you can calculate it to a hundred decimal places before they can do it to ten.Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-56022187695396642472012-01-23T12:46:35.370-06:002012-01-23T12:46:35.370-06:00Including this apparent glitch at the end?
99299...Including this apparent glitch at the end? <br /><br />992993994995996997999 <br /><br />I haven't looked for other irregularities.Minnesotastanhttps://www.blogger.com/profile/01382888179579245181noreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-3963091012072747972012-01-23T12:37:18.354-06:002012-01-23T12:37:18.354-06:00I forwarded this item to a friend who replied, Is ...I forwarded this item to a friend who replied, Is the result correct? So I found an online calculator which is up to the task, and the result agrees. http://www.math.jerrywickey.net/?prol=y&first=n&r=0.0e0&a=1&b=998001&cal=d&bp=3200+dec+places&do=ComputeAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4912713243046142041.post-37243518043168830632012-01-23T11:59:54.989-06:002012-01-23T11:59:54.989-06:001/998001 = (1/999)^2
1/999 = 0.001001001....
Sor...1/998001 = (1/999)^2<br /><br />1/999 = 0.001001001....<br /><br />Sort of comes from there.<br /><br />Now what is really strange is <br /><br />23.140692632779269005729086367949... to the power of i (imaginary, i = sqrt(-1)) is -1.Markhttps://www.blogger.com/profile/05337533198347182909noreply@blogger.com