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
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
$endgroup$
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
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
$endgroup$
2
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
Apr 3 at 15:26
1
$begingroup$
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
Apr 3 at 15:59
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
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
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
$endgroup$
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
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
asked Apr 3 at 14:54
Tyler22AlexTyler22Alex
321212
321212
2
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
Apr 3 at 15:26
1
$begingroup$
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
Apr 3 at 15:59
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
add a comment |
2
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
Apr 3 at 15:26
1
$begingroup$
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
Apr 3 at 15:59
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
2
2
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
Apr 3 at 15:26
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
Apr 3 at 15:26
1
1
$begingroup$
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
Apr 3 at 15:59
$begingroup$
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
Apr 3 at 15:59
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
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 $gamma$.
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
$endgroup$
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
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 $gamma$.
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
$endgroup$
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
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 $gamma$.
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
$endgroup$
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
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 $gamma$.
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
$endgroup$
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 Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
answered Apr 3 at 15:22
Gareth McCaughan♦Gareth McCaughan
67.2k3170261
67.2k3170261
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
add a comment |
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
1
1
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
That was fast! Nice job!
$endgroup$
– Tyler22Alex
Apr 3 at 16:00
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
The second hint made it pretty hard not to get :-).
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
$begingroup$
(Though I think I'd probably have got it quickly even without.)
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44
add a comment |
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2
$begingroup$
Really March? Really 2018?
$endgroup$
– Gareth McCaughan♦
Apr 3 at 15:26
1
$begingroup$
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
Apr 3 at 15:59
$begingroup$
Ah, OK. Fair enough.
$endgroup$
– Gareth McCaughan♦
Apr 3 at 16:44