it\'s a long question, just need help for question a As we have seen, it is usua
ID: 3123286 • Letter: I
Question
it's a long question, just need help for question a
As we have seen, it is usually quite tricky to figure out how many terms of a convergent infinite series need to be added to be able to guarantee that this partial sum coincides with the true sum in the first k digits of its decimal expansion. However, for certain series this is a very easy task. Here is an example. We know that pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ... As you can easily check this series^3 has the following two properties: 1. The terms alternate between being positive and negative numbers; 2. the absolute values of the terms are decreasing; 3. the terms have the limit 0.^3 This series is of course much prettier than Ramanujan's. However, it converges VERY slowly and is absolutely useless when it comes to really approximating pi. Now it is quite easy to show (you don't have to do this) that any series that has these three properties converges and that if, as usual, S, S_n and a_n de notes its sum, its nth partial sum and its nth term, respectively, then |S + S_n|Explanation / Answer
The first 2000 digits of 1/ are:
0.318309886183789986450799312456667680370035994643
55914711875563296632077006466821840248895081249479
12560548543265194753809537933816633014208589648181
79240423070178347044029966705371947416879710093186
40235152047890057031577366743598527601543123554220
33950361445658778353194236518125075746925561174739
38592736411972378031983731990455477701942970278328
45397536071900322598823314007544570546317168348617
89896985258934764288258690431430998251555306661596
23213601605729442025449241172958026253758671530587
As given in the seemingly crazy formula,
when n = 0
1/ = [ 163096908 ] / [ 6403203/2 ] = 0.318309886183796652607706183787889598854733200434
99577265757664322059320118513690824594658037452742
74357549537695363633119805972526807289099551241308
77252483979893869649447762513739406139535567591951
55686631766657723015464351811052470555686647966896
28669035019215076922821272234291218839006997918801
62284088985031496456663306077929783955642489376304
17464192869417114193389937870149268498636032313600
91587952191173118660358005470092559889002595984985
52714534109109731875613123229702419387117209318194
So we find disparity in the 14th decimal place
when n=1
The second term
= [ (-1) * 12 * 6! (13591409 + 545140134) ] / [ 3! 1! 6403209/2 ]
= [ -8640 * 558731543 ] / [ 6 * 6403209/2 ]
= [ -804573421920] / [ 18095625621654356959022098935941777779064832000000
0001/2]
= 0.000000000000005981069938657074052378697026217828
50742959152122730842427987207054294832553747219318
90717928578674270606434349458713621488446223624574
40528818275371648675220825130478323870558659714087
48138138847477724341845004098931530923445389393452
22951674555808769756242456630031993466817523119264
61782827631873497944280668264444772652215585757540
55926359193248611112269394397009176702208611352896
86587373202491960257050854568009267865941170477128
97341428024404598776729865126294373037317662396656
Observe that the first fourteen decimal places are zeros.
The third term
= [ 12 * 12! (13591409 + 545140134*2) ] / [ 6! (2!)3 64032015/2 ]
= [ 12! * 13246460124] / [ 5760 * 64032015/2 ]
= [ 479001600 * 13246460124 ] / [ 5760 * 64032015/2 ]
= [ 1101575623911840 ] / [ 12472571560196196319592288504749153247964447949210
68388944443347008749568000000000000000 ]1/2
= 0.000000000000000000000000000031191503911212483199
57373094131992369267016532488344124638393247533229
67736367708837018105977911533451938568707235591951
37229844717808411098953461874324576088370744056940
78477263796212997460403696263765074046011728081417
79123591924923242358332126664878434359701766185551
24982082638270307736006621706873208853198025712591
75733639493606946127687455271082381067587997423867
15653940443639694034521344243095203685223322146761
69562360110634851932232557279843133390594117513478
This number has 28 leading zeros.
So each iteration corrects values to 14 decimal places.
[1000/14] = 71
So for n = 72, we can get a value for 1/ which is insanely accurate to 1000 decimal places.
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