Undergrad Leading digits of incomputable large numbers

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SUMMARY

The discussion centers on the leading digits of incomputable large numbers, specifically referencing Graham's number and its discoverer, Ronald Graham. While algorithms exist to determine the last digits of such numbers, the first digit remains elusive for many, particularly those not expressed as powers of ten. Techniques such as Benford's Law and Stirling's approximation are highlighted as methods to estimate leading digits for certain large numbers, including e^1,000,000 and (10^10)!. The conversation emphasizes the mathematical curiosity behind these calculations and their implications for future problem-solving.

PREREQUISITES
  • Understanding of Graham's number and its properties
  • Familiarity with Benford's Law and its application to leading digits
  • Knowledge of Stirling's approximation for factorials
  • Basic concepts of computability in mathematics
NEXT STEPS
  • Explore the algorithms for calculating last digits of large numbers
  • Research the application of Benford's Law in various mathematical contexts
  • Study Stirling's approximation in detail for estimating factorials
  • Investigate the implications of computability theory in mathematics
USEFUL FOR

Mathematicians, computer scientists, and enthusiasts interested in number theory, particularly those exploring the properties of large numbers and their leading digits.

BWV
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Was watching a youtube on Grahams number last night and its discoverer Ronald Graham. He talked about how many algorithms can calculate the ending digits of the number (its a power of 3) but the first digit is unknown. Guessing this is generally true of all incomputably large numbers that are not powers of 10? Anyway to know the first digit of say e^1,000,000 or 10^10! or 3^^^3?
 
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Why does anybody want to know? This is of similar interest as the 117th digit of ##\pi.##
 
fresh_42 said:
Why does anybody want to know? This is of similar interest as the 117th digit of ##\pi.##
Ronald Graham, for one, wanted to know, it may not be of interest to know the trillionth digit of pi, but an algorithm to calculate it that was not just brute force would be interesting. But if you arent interested in the topic, don't waste your time replying
 
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In attempting to do this problem, one might learn new math techniques that could be applied to future problems.
 
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For the curious with OCD tendencies, the 117th digit is: ...0

from this list of digits and excluding the decimal point:

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230...

https://www.piday.org/million/

(I pasted the string into a text editor, deleted the decimal point and went to column 117)
 
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jedishrfu said:
In attempting to do this problem, one might learn new math techniques that could be applied to future problems.
This is obviously the case for the last digits:
Wikipedia said:
Despite its unimaginable size, the last digits of Graham's number ##G_{64}## can be determined using elementary number theory. The last 500 digits of Graham's number are:
02425950695064738395657479136519351798334535362521
43003540126026771622672160419810652263169355188780
38814483140652526168785095552646051071172000997092
91249544378887496062882911725063001303622934916080
25459461494578871427832350829242102091825896753560
43086993801689249889268099510169055919951195027887
17830837018340236474548882222161573228010132974509
27344594504343300901096928025352751833289884461508
94042482650181938515625357963996189939679054966380
03222348723967018485186439059104575627262464195387

It can be shown that in the concatenated arrow notation Graham's number falls between ##3\rightarrow 3\rightarrow 64\rightarrow 2## and ##3\rightarrow 3\rightarrow 65\rightarrow 2##.

It is apparently not possible to cut down such numbers from behind.

To me, they are all ##O(1)##.
 
Leading digits of numbers not a power of ten do follow Benfords Law

where the leading digit d (d ∈ {1, ..., 9}) occurs with prob
1656533102958.png


so:
1​
0.301​
2​
0.176​
3​
0.125​
4​
0.097​
5​
0.079​
6​
0.067​
7​
0.058​
8​
0.051​
9​
0.046​
 
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##e^{1000000} = 10^{1000000/\ln(10)} \approx 3.033215396802087545086402141418114327... × 10^{434294}##
For the first digit it's sufficient to calculate 1000000/ln(10) with a precision of around 0.1, which is trivial for a computer, you need about 7 digits. WolframAlpha directly delivered a few extra digits of the result.
If you ask the same question about ##{{{e^{10}}^{10}}^{10}}^{10}## then we don't know because we don't have 10^10^10 digits of ln(10) (we do have 10^10 digits however).
 
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  • #10
Computability has a specific meaning in mathematics: Graham's number is not incomputable.
 
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  • #11
BWV said:
Was watching a youtube on Grahams number last night and its discoverer Ronald Graham. He talked about how many algorithms can calculate the ending digits of the number (its a power of 3) but the first digit is unknown. Guessing this is generally true of all incomputably large numbers that are not powers of 10? Anyway to know the first digit of say e^1,000,000 or 10^10! or 3^^^3?
e^10000000 is easy. (10^10)! can still be computed completely, and you can easily find the first digits with Stirlings approximation. 3^^^3 is hopeless. you'll need to iterate fn+1 = 3fn about 7.6*10^12 times. To compute the leading digit of fn+1 you need all the digits of fn.
 
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  • #12
pbuk said:
Computability has a specific meaning in mathematics: Graham's number is not incomputable.
No, it is not, but I wouldn't wait for the TM to stop either. :wink:
 

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