I don't know what proof technique Danish was going after. I'd be willing to say that being irrational the number has "somewhat random" digits that well into the decimal representation. Or maybe it has something to do with the grouping-by-threes method of taking cube roots.
But the computer method is doubly unsatisfying: 1) it is just using the computer as a calculator; 2) I don't see a way of getting MATLAB actually to show the number past about 3200 or 3300 digits.
But it does let you sample substrings past the millionth digit:
>> digits 1500000
>> n=vpa(3)
n =
3.0
>> a=n^(1/3)
a =
1.4422495703074083823216383107801095883918692534993505775464161945416875968299973398547554797056452566
... over 3000 lines '''
176731971522554815525185125921221782431712421585426965243491795671262792916278224032726104873355805200
14437742082220231915... Output truncated. Text exceeds maximum line length for Command Window display.
>> c=char(a)
c =
'1.44224957030740838232163831078010958839186925349935057754641619454168759682999733985475547970564525668683
... ditto ...
96940472384729779188970878085259038417131674516532260879974343736270427029820131645151470885563288321316
62930723917673197152255481552518512592122178243171242158542696524349179567126279291627822403272610487335
43774208222023191... Output truncated. Text exceeds maximum line length for Command Window display.
>> c(1:10)
ans =
'1.44224957'
...shows how to get substrings in MATLAB.
>> c(1000000:10000010)
Index exceeds the number of array elements (1500000).
...and that MATLAB will report length asked for exceeded.
>> c(1000000:1000010)
ans =
'55494463209'
>>
but at least the part beginning 4944... is in the specified range after the decimal point (the 55 at the beginning is 1000001 and 100002 starting counting with the "1." at the beginning of the whole cube root). Most of the shown digits are non-2.
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Posted by Charlie
on 2020-11-14 20:40:24 |
unsatisfying computer solution
