Unlock my phone! March 2018
$begingroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
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
New contributor
$endgroup$
add a comment |
$begingroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
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
New contributor
$endgroup$
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
add a comment |
$begingroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
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
New contributor
$endgroup$
New month, new password, but I've already forgotten! Help!
What is my phone's password this month?
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
mathematics knowledge visual
New contributor
New contributor
New contributor
asked 14 hours ago
Tyler22AlexTyler22Alex
182111
182111
New contributor
New contributor
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $gamma$.
$endgroup$
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
add a comment |
Your Answer
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1 Answer
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$begingroup$
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 $gamma$.
$endgroup$
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
add a comment |
$begingroup$
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 $gamma$.
$endgroup$
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
add a comment |
$begingroup$
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 $gamma$.
$endgroup$
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 $gamma$.
answered 14 hours ago
Gareth McCaughan♦Gareth 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
add a comment |
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
add a comment |
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.
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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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