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How Euler calculate the value of gamma number
Clash Royale CLAN TAG#URR8PPP
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Gamma number is the limit of (sum(f, i = k .. n)1/n - Ln(n) ) when n approaches to ∞ and need a large number of calculation. What's the way that Euler found
algebra-precalculus
asked Jul 24 at 6:14
user359791
331
add a comment |Â
up vote
0
down vote
favorite
Gamma number is the limit of (sum(f, i = k .. n)1/n - Ln(n) ) when n approaches to ∞ and need a large number of calculation. What's the way that Euler found
algebra-precalculus
asked Jul 24 at 6:14
user359791
331
Interesting fact: Euler calculated the first 5 decimal places in 1734. but needed additional 47 years to add 11 more.
– Oldboy
Jul 24 at 6:57
@Oldboy Wikipedia says that he published 15 digits already in 1735.
– A.Γ.
Jul 24 at 8:11
@A.Γ. You're right, somehow I skipped that row.
– Oldboy
Jul 24 at 8:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Gamma number is the limit of (sum(f, i = k .. n)1/n - Ln(n) ) when n approaches to ∞ and need a large number of calculation. What's the way that Euler found
algebra-precalculus
asked Jul 24 at 6:14
user359791
331
Gamma number is the limit of (sum(f, i = k .. n)1/n - Ln(n) ) when n approaches to ∞ and need a large number of calculation. What's the way that Euler found
algebra-precalculus
asked Jul 24 at 6:14
user359791
331
asked Jul 24 at 6:14
user359791
331
asked Jul 24 at 6:14
user359791
331
asked Jul 24 at 6:14
user359791
331
331
Interesting fact: Euler calculated the first 5 decimal places in 1734. but needed additional 47 years to add 11 more.
– Oldboy
Jul 24 at 6:57
@Oldboy Wikipedia says that he published 15 digits already in 1735.
– A.Γ.
Jul 24 at 8:11
@A.Γ. You're right, somehow I skipped that row.
– Oldboy
Jul 24 at 8:20
add a comment |Â
Interesting fact: Euler calculated the first 5 decimal places in 1734. but needed additional 47 years to add 11 more.
– Oldboy
Jul 24 at 6:57
@Oldboy Wikipedia says that he published 15 digits already in 1735.
– A.Γ.
Jul 24 at 8:11
@A.Γ. You're right, somehow I skipped that row.
– Oldboy
Jul 24 at 8:20
Interesting fact: Euler calculated the first 5 decimal places in 1734. but needed additional 47 years to add 11 more.
– Oldboy
Jul 24 at 6:57
Interesting fact: Euler calculated the first 5 decimal places in 1734. but needed additional 47 years to add 11 more.
– Oldboy
Jul 24 at 6:57
@Oldboy Wikipedia says that he published 15 digits already in 1735.
– A.Γ.
Jul 24 at 8:11
@Oldboy Wikipedia says that he published 15 digits already in 1735.
– A.Γ.
Jul 24 at 8:11
@A.Γ. You're right, somehow I skipped that row.
– Oldboy
Jul 24 at 8:20
@A.Γ. You're right, somehow I skipped that row.
– Oldboy
Jul 24 at 8:20
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
In $1736$, Euler used the Euler-Maclaurin summation formula, applied to the harmonic numbers.
This allowed him to establish
$$H_n=gamma+log n+frac12n-sum_k=1^infty fracB_2k2k,n^2k$$ where the $B_2k$ are Bernouilli numbers.
He applied the formula with $n=10$ up to $k=7$, to obtain the first $16$ decimals.
$$gammaapprox 1+frac12+frac13+frac14+frac15+frac16+frac17+frac18+frac19+frac110−log(10)
\−frac120
+frac11200
−frac11200000
+frac1252000000
−frac124000000000
+frac11320000000000
−frac69132760000000000000
+frac11200000000000000
\=frac103776500111525069933603600000000000000-log(10)
\=colorgreen0.577215664901532902118cdots$$
answered Jul 24 at 7:06
Yves Daoust
111k665203
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
1
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
add a comment |Â
up vote
1
down vote
To compute
$$ sum_k=1^inftyleft(frac1k-lnfrack+1kright), $$
note that from $ln(1+x)=x-frac12x^2+frac13x^3mpldots$ we can use (for $k$ big enough, if necessary)
$$frac12k^2-frac13k^3< frac1k-lnfrack+1k<frac12k^2.$$
Hence the error from stopping after $N$ summands is bounded from above by
$$ tag1sum_k=N+1^inftyfrac12k^2<frac12sum_k=N+1^inftyleft(frac1k-1-frac1kright)=frac12N$$
and from below by
$$tag3beginalignsum_k=N+1^inftyleft(frac12k^2-frac13k^3right)&>left(frac12-frac13(N+1)right)sum_k=N+1^inftyleft(frac1k-frac1k+1right)\&=frac12(N+1)-frac13(N+1)^2endalign $$
While at first glance, $(1)$ seems to imply that we need to compute up to $N=500000$ in order to make the error $<10^-6$, it is really the difference between $(1)$ and $(2)$ that we need to control. And that is already $<10^-6$ for $N=1000$, a great improvement!
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
0
down vote
Here is the original Eueler's paper that mentiones $gamma$ (denoted with $C$) for the first time:
Observations on harmonic Progressions, Leonhard Euler (english adaptation)
In that paper Euler computed the first 6 decimals of $gamma$ constant, and the number premiers on page 8:
Therefore, we detected the value of this constant C, which is C = 0.577218.
The same part from the original Euler's paper:
This paper does not mention Bernoulli numbers. The original paper can be found here: http://eulerarchive.maa.org/docs/originals/E043.pdf
answered Jul 24 at 7:15
Oldboy
2,6101316
2
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
In $1736$, Euler used the Euler-Maclaurin summation formula, applied to the harmonic numbers.
This allowed him to establish
$$H_n=gamma+log n+frac12n-sum_k=1^infty fracB_2k2k,n^2k$$ where the $B_2k$ are Bernouilli numbers.
He applied the formula with $n=10$ up to $k=7$, to obtain the first $16$ decimals.
$$gammaapprox 1+frac12+frac13+frac14+frac15+frac16+frac17+frac18+frac19+frac110−log(10)
\−frac120
+frac11200
−frac11200000
+frac1252000000
−frac124000000000
+frac11320000000000
−frac69132760000000000000
+frac11200000000000000
\=frac103776500111525069933603600000000000000-log(10)
\=colorgreen0.577215664901532902118cdots$$
answered Jul 24 at 7:06
Yves Daoust
111k665203
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
1
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
add a comment |Â
up vote
2
down vote
In $1736$, Euler used the Euler-Maclaurin summation formula, applied to the harmonic numbers.
This allowed him to establish
$$H_n=gamma+log n+frac12n-sum_k=1^infty fracB_2k2k,n^2k$$ where the $B_2k$ are Bernouilli numbers.
He applied the formula with $n=10$ up to $k=7$, to obtain the first $16$ decimals.
$$gammaapprox 1+frac12+frac13+frac14+frac15+frac16+frac17+frac18+frac19+frac110−log(10)
\−frac120
+frac11200
−frac11200000
+frac1252000000
−frac124000000000
+frac11320000000000
−frac69132760000000000000
+frac11200000000000000
\=frac103776500111525069933603600000000000000-log(10)
\=colorgreen0.577215664901532902118cdots$$
answered Jul 24 at 7:06
Yves Daoust
111k665203
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
1
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In $1736$, Euler used the Euler-Maclaurin summation formula, applied to the harmonic numbers.
This allowed him to establish
$$H_n=gamma+log n+frac12n-sum_k=1^infty fracB_2k2k,n^2k$$ where the $B_2k$ are Bernouilli numbers.
He applied the formula with $n=10$ up to $k=7$, to obtain the first $16$ decimals.
$$gammaapprox 1+frac12+frac13+frac14+frac15+frac16+frac17+frac18+frac19+frac110−log(10)
\−frac120
+frac11200
−frac11200000
+frac1252000000
−frac124000000000
+frac11320000000000
−frac69132760000000000000
+frac11200000000000000
\=frac103776500111525069933603600000000000000-log(10)
\=colorgreen0.577215664901532902118cdots$$
answered Jul 24 at 7:06
Yves Daoust
111k665203
In $1736$, Euler used the Euler-Maclaurin summation formula, applied to the harmonic numbers.
This allowed him to establish
$$H_n=gamma+log n+frac12n-sum_k=1^infty fracB_2k2k,n^2k$$ where the $B_2k$ are Bernouilli numbers.
He applied the formula with $n=10$ up to $k=7$, to obtain the first $16$ decimals.
$$gammaapprox 1+frac12+frac13+frac14+frac15+frac16+frac17+frac18+frac19+frac110−log(10)
\−frac120
+frac11200
−frac11200000
+frac1252000000
−frac124000000000
+frac11320000000000
−frac69132760000000000000
+frac11200000000000000
\=frac103776500111525069933603600000000000000-log(10)
\=colorgreen0.577215664901532902118cdots$$
answered Jul 24 at 7:06
Yves Daoust
111k665203
edited Jul 24 at 7:22
answered Jul 24 at 7:06
Yves Daoust
111k665203
answered Jul 24 at 7:06
Yves Daoust
111k665203
answered Jul 24 at 7:06
Yves Daoust
111k665203
111k665203
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
1
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
add a comment |Â
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
1
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
It seems to be true according to Euler's paper E47 from 1735 Inventio summae cuiusque seriei ex dato termino generali (English translation here, page 9).
– A.Γ.
Jul 24 at 10:32
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
@A.Γ.: er, do you think I invented this ?
– Yves Daoust
Jul 24 at 10:35
1
1
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
You have not provided any reference, so how do I know?
– A.Γ.
Jul 24 at 11:41
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
I have got my answer,thanks a lot
– user359791
Jul 25 at 6:48
add a comment |Â
up vote
1
down vote
To compute
$$ sum_k=1^inftyleft(frac1k-lnfrack+1kright), $$
note that from $ln(1+x)=x-frac12x^2+frac13x^3mpldots$ we can use (for $k$ big enough, if necessary)
$$frac12k^2-frac13k^3< frac1k-lnfrack+1k<frac12k^2.$$
Hence the error from stopping after $N$ summands is bounded from above by
$$ tag1sum_k=N+1^inftyfrac12k^2<frac12sum_k=N+1^inftyleft(frac1k-1-frac1kright)=frac12N$$
and from below by
$$tag3beginalignsum_k=N+1^inftyleft(frac12k^2-frac13k^3right)&>left(frac12-frac13(N+1)right)sum_k=N+1^inftyleft(frac1k-frac1k+1right)\&=frac12(N+1)-frac13(N+1)^2endalign $$
While at first glance, $(1)$ seems to imply that we need to compute up to $N=500000$ in order to make the error $<10^-6$, it is really the difference between $(1)$ and $(2)$ that we need to control. And that is already $<10^-6$ for $N=1000$, a great improvement!
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
To compute
$$ sum_k=1^inftyleft(frac1k-lnfrack+1kright), $$
note that from $ln(1+x)=x-frac12x^2+frac13x^3mpldots$ we can use (for $k$ big enough, if necessary)
$$frac12k^2-frac13k^3< frac1k-lnfrack+1k<frac12k^2.$$
Hence the error from stopping after $N$ summands is bounded from above by
$$ tag1sum_k=N+1^inftyfrac12k^2<frac12sum_k=N+1^inftyleft(frac1k-1-frac1kright)=frac12N$$
and from below by
$$tag3beginalignsum_k=N+1^inftyleft(frac12k^2-frac13k^3right)&>left(frac12-frac13(N+1)right)sum_k=N+1^inftyleft(frac1k-frac1k+1right)\&=frac12(N+1)-frac13(N+1)^2endalign $$
While at first glance, $(1)$ seems to imply that we need to compute up to $N=500000$ in order to make the error $<10^-6$, it is really the difference between $(1)$ and $(2)$ that we need to control. And that is already $<10^-6$ for $N=1000$, a great improvement!
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
up vote
1
down vote
To compute
$$ sum_k=1^inftyleft(frac1k-lnfrack+1kright), $$
note that from $ln(1+x)=x-frac12x^2+frac13x^3mpldots$ we can use (for $k$ big enough, if necessary)
$$frac12k^2-frac13k^3< frac1k-lnfrack+1k<frac12k^2.$$
Hence the error from stopping after $N$ summands is bounded from above by
$$ tag1sum_k=N+1^inftyfrac12k^2<frac12sum_k=N+1^inftyleft(frac1k-1-frac1kright)=frac12N$$
and from below by
$$tag3beginalignsum_k=N+1^inftyleft(frac12k^2-frac13k^3right)&>left(frac12-frac13(N+1)right)sum_k=N+1^inftyleft(frac1k-frac1k+1right)\&=frac12(N+1)-frac13(N+1)^2endalign $$
While at first glance, $(1)$ seems to imply that we need to compute up to $N=500000$ in order to make the error $<10^-6$, it is really the difference between $(1)$ and $(2)$ that we need to control. And that is already $<10^-6$ for $N=1000$, a great improvement!
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
To compute
$$ sum_k=1^inftyleft(frac1k-lnfrack+1kright), $$
note that from $ln(1+x)=x-frac12x^2+frac13x^3mpldots$ we can use (for $k$ big enough, if necessary)
$$frac12k^2-frac13k^3< frac1k-lnfrack+1k<frac12k^2.$$
Hence the error from stopping after $N$ summands is bounded from above by
$$ tag1sum_k=N+1^inftyfrac12k^2<frac12sum_k=N+1^inftyleft(frac1k-1-frac1kright)=frac12N$$
and from below by
$$tag3beginalignsum_k=N+1^inftyleft(frac12k^2-frac13k^3right)&>left(frac12-frac13(N+1)right)sum_k=N+1^inftyleft(frac1k-frac1k+1right)\&=frac12(N+1)-frac13(N+1)^2endalign $$
While at first glance, $(1)$ seems to imply that we need to compute up to $N=500000$ in order to make the error $<10^-6$, it is really the difference between $(1)$ and $(2)$ that we need to control. And that is already $<10^-6$ for $N=1000$, a great improvement!
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
answered Jul 24 at 6:38
Hagen von Eitzen
265k20258477
265k20258477
add a comment |Â
add a comment |Â
up vote
0
down vote
Here is the original Eueler's paper that mentiones $gamma$ (denoted with $C$) for the first time:
Observations on harmonic Progressions, Leonhard Euler (english adaptation)
In that paper Euler computed the first 6 decimals of $gamma$ constant, and the number premiers on page 8:
Therefore, we detected the value of this constant C, which is C = 0.577218.
The same part from the original Euler's paper:
This paper does not mention Bernoulli numbers. The original paper can be found here: http://eulerarchive.maa.org/docs/originals/E043.pdf
answered Jul 24 at 7:15
Oldboy
2,6101316
2
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
add a comment |Â
up vote
0
down vote
Here is the original Eueler's paper that mentiones $gamma$ (denoted with $C$) for the first time:
Observations on harmonic Progressions, Leonhard Euler (english adaptation)
In that paper Euler computed the first 6 decimals of $gamma$ constant, and the number premiers on page 8:
Therefore, we detected the value of this constant C, which is C = 0.577218.
The same part from the original Euler's paper:
This paper does not mention Bernoulli numbers. The original paper can be found here: http://eulerarchive.maa.org/docs/originals/E043.pdf
answered Jul 24 at 7:15
Oldboy
2,6101316
2
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is the original Eueler's paper that mentiones $gamma$ (denoted with $C$) for the first time:
Observations on harmonic Progressions, Leonhard Euler (english adaptation)
In that paper Euler computed the first 6 decimals of $gamma$ constant, and the number premiers on page 8:
Therefore, we detected the value of this constant C, which is C = 0.577218.
The same part from the original Euler's paper:
This paper does not mention Bernoulli numbers. The original paper can be found here: http://eulerarchive.maa.org/docs/originals/E043.pdf
answered Jul 24 at 7:15
Oldboy
2,6101316
Here is the original Eueler's paper that mentiones $gamma$ (denoted with $C$) for the first time:
Observations on harmonic Progressions, Leonhard Euler (english adaptation)
In that paper Euler computed the first 6 decimals of $gamma$ constant, and the number premiers on page 8:
Therefore, we detected the value of this constant C, which is C = 0.577218.
The same part from the original Euler's paper:
This paper does not mention Bernoulli numbers. The original paper can be found here: http://eulerarchive.maa.org/docs/originals/E043.pdf
answered Jul 24 at 7:15
Oldboy
2,6101316
edited Jul 24 at 7:24
answered Jul 24 at 7:15
Oldboy
2,6101316
answered Jul 24 at 7:15
Oldboy
2,6101316
answered Jul 24 at 7:15
Oldboy
2,6101316
2,6101316
2
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
add a comment |Â
2
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
2
2
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
According to this source, page 3, Euler failed to recognize Bernoulli's numbers first in his formula for summation.
– A.Γ.
Jul 24 at 11:47
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
@A.Γ. Wonderful piece of history, +1
– Oldboy
Jul 24 at 11:58
add a comment |Â
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Interesting fact: Euler calculated the first 5 decimal places in 1734. but needed additional 47 years to add 11 more.
– Oldboy
Jul 24 at 6:57
@Oldboy Wikipedia says that he published 15 digits already in 1735.
– A.Γ.
Jul 24 at 8:11
@A.Γ. You're right, somehow I skipped that row.
– Oldboy
Jul 24 at 8:20