Unlock my phone! March 2018












4












$begingroup$


New month, new password, but I've already forgotten! Help!



What is my phone's password this month?



enter image description here



Hint #1:




As per usual, my password is pretty long (maybe around 40 digits?)




Hint #2:




In March I took up an interest in analytical number theory











share|improve this question







New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Really March? Really 2018?
    $endgroup$
    – Gareth McCaughan
    14 hours ago










  • $begingroup$
    If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    Ah, OK. Fair enough.
    $endgroup$
    – Gareth McCaughan
    12 hours ago
















4












$begingroup$


New month, new password, but I've already forgotten! Help!



What is my phone's password this month?



enter image description here



Hint #1:




As per usual, my password is pretty long (maybe around 40 digits?)




Hint #2:




In March I took up an interest in analytical number theory











share|improve this question







New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Really March? Really 2018?
    $endgroup$
    – Gareth McCaughan
    14 hours ago










  • $begingroup$
    If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    Ah, OK. Fair enough.
    $endgroup$
    – Gareth McCaughan
    12 hours ago














4












4








4





$begingroup$


New month, new password, but I've already forgotten! Help!



What is my phone's password this month?



enter image description here



Hint #1:




As per usual, my password is pretty long (maybe around 40 digits?)




Hint #2:




In March I took up an interest in analytical number theory











share|improve this question







New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




New month, new password, but I've already forgotten! Help!



What is my phone's password this month?



enter image description here



Hint #1:




As per usual, my password is pretty long (maybe around 40 digits?)




Hint #2:




In March I took up an interest in analytical number theory








mathematics knowledge visual






share|improve this question







New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 14 hours ago









Tyler22AlexTyler22Alex

182111




182111




New contributor




Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Tyler22Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Really March? Really 2018?
    $endgroup$
    – Gareth McCaughan
    14 hours ago










  • $begingroup$
    If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    Ah, OK. Fair enough.
    $endgroup$
    – Gareth McCaughan
    12 hours ago














  • 1




    $begingroup$
    Really March? Really 2018?
    $endgroup$
    – Gareth McCaughan
    14 hours ago










  • $begingroup$
    If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    Ah, OK. Fair enough.
    $endgroup$
    – Gareth McCaughan
    12 hours ago








1




1




$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan
14 hours ago




$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan
14 hours ago












$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
13 hours ago




$begingroup$
If you look at my other puzzles you’ll see I’m starting January of 2018 and going month by month to the present day. Sorry for confusion!
$endgroup$
– Tyler22Alex
13 hours ago












$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan
12 hours ago




$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan
12 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Your password is




the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.




So it will begin




either 5772 or 05772 or 0.5772.




Since




to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.




Explanation in case the above doesn't make it clear:




the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.







share|improve this answer









$endgroup$









  • 1




    $begingroup$
    That was fast! Nice job!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    The second hint made it pretty hard not to get :-).
    $endgroup$
    – Gareth McCaughan
    12 hours ago










  • $begingroup$
    (Though I think I'd probably have got it quickly even without.)
    $endgroup$
    – Gareth McCaughan
    12 hours ago












Your Answer





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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Your password is




the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.




So it will begin




either 5772 or 05772 or 0.5772.




Since




to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.




Explanation in case the above doesn't make it clear:




the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.







share|improve this answer









$endgroup$









  • 1




    $begingroup$
    That was fast! Nice job!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    The second hint made it pretty hard not to get :-).
    $endgroup$
    – Gareth McCaughan
    12 hours ago










  • $begingroup$
    (Though I think I'd probably have got it quickly even without.)
    $endgroup$
    – Gareth McCaughan
    12 hours ago
















4












$begingroup$

Your password is




the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.




So it will begin




either 5772 or 05772 or 0.5772.




Since




to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.




Explanation in case the above doesn't make it clear:




the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.







share|improve this answer









$endgroup$









  • 1




    $begingroup$
    That was fast! Nice job!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    The second hint made it pretty hard not to get :-).
    $endgroup$
    – Gareth McCaughan
    12 hours ago










  • $begingroup$
    (Though I think I'd probably have got it quickly even without.)
    $endgroup$
    – Gareth McCaughan
    12 hours ago














4












4








4





$begingroup$

Your password is




the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.




So it will begin




either 5772 or 05772 or 0.5772.




Since




to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.




Explanation in case the above doesn't make it clear:




the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.







share|improve this answer









$endgroup$



Your password is




the first 40 digits, or perhaps decimal places, of the Euler-Mascheroni constant, sometimes denoted $gamma$.




So it will begin




either 5772 or 05772 or 0.5772.




Since




to 41 decimal places the constant's value is 0.57721566490153286060651209008240243104215, the last digit will (if you've used 40 decimal places rather than 40 digits including the leading zero) depend on how you've chosen to round it.




Explanation in case the above doesn't make it clear:




the harmonic series is $1+frac12+frac13+frac14+cdots$ and the sum of its first $n$ terms is $log n+gamma+varepsilon(n)$ where $varepsilon(n)rightarrow0$ as $nrightarrowinfty$. So the limiting difference between the harmonic series and the natural logarithm is $gamma$.








share|improve this answer












share|improve this answer



share|improve this answer










answered 14 hours ago









Gareth McCaughanGareth McCaughan

66.2k3167259




66.2k3167259








  • 1




    $begingroup$
    That was fast! Nice job!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    The second hint made it pretty hard not to get :-).
    $endgroup$
    – Gareth McCaughan
    12 hours ago










  • $begingroup$
    (Though I think I'd probably have got it quickly even without.)
    $endgroup$
    – Gareth McCaughan
    12 hours ago














  • 1




    $begingroup$
    That was fast! Nice job!
    $endgroup$
    – Tyler22Alex
    13 hours ago










  • $begingroup$
    The second hint made it pretty hard not to get :-).
    $endgroup$
    – Gareth McCaughan
    12 hours ago










  • $begingroup$
    (Though I think I'd probably have got it quickly even without.)
    $endgroup$
    – Gareth McCaughan
    12 hours ago








1




1




$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
13 hours ago




$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
13 hours ago












$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan
12 hours ago




$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan
12 hours ago












$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan
12 hours ago




$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan
12 hours ago










Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.










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Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.












Tyler22Alex is a new contributor. Be nice, and check out our Code of Conduct.
















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