Mixing
First order approximation of $zeta$(s) at s=1
$begingroup$
I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.
sequences-and-series approximation riemann-zeta
asked Jan 18 at 8:29
KarthikKarthik
13212
$endgroup$
add a comment |
$begingroup$
I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.
sequences-and-series approximation riemann-zeta
asked Jan 18 at 8:29
KarthikKarthik
13212
$endgroup$
$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43
$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53
add a comment |
$begingroup$
I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.
sequences-and-series approximation riemann-zeta
asked Jan 18 at 8:29
KarthikKarthik
13212
$endgroup$
I was playing around with Wolfram Alpha.
I found one interesting thing when I asked it to evaluate this particular summation.$$Sigma_{n=1}^inftyfrac{1}{n^{1+10^{-
10}}}$$
It returned this$$ approx10^{10}+gamma$$
That's right the number itself plus the Euler–Mascheroni constant.
I tried it for various other values and it turned out to follow the same pattern. As the value got smaller the closer it followed that pattern.
Image of Wolfram Alpha result click here
So formally stating $$ lim _{s rightarrow0 } zeta(1+s) = 1/s+ gamma$$
Is this true?
And why is this true?
Could it be proven by taking the first order approximation of the zeta function.
sequences-and-series approximation riemann-zeta
sequences-and-series approximation riemann-zeta
asked Jan 18 at 8:29
KarthikKarthik
13212
asked Jan 18 at 8:29
KarthikKarthik
13212
edited Jan 18 at 8:42
Karthik
asked Jan 18 at 8:29
KarthikKarthik
13212
asked Jan 18 at 8:29
KarthikKarthik
13212
asked Jan 18 at 8:29
KarthikKarthik
13212
13212
$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43
$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53
add a comment |
$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43
$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53
$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43
$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43
$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53
$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
1
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
1
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
add a comment |
$begingroup$
$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
1
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
add a comment |
$begingroup$
$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$$sum_{n=1}^inftyfrac{1}{n^{1+epsilon}}=zeta (1+epsilon)=frac{1}{epsilon }+gamma -gamma _1 epsilon +frac{gamma _2 }{2}epsilon
^2+Oleft(epsilon ^3right)$$ where appear the generalized Stieltjes constants.
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
edited Jan 18 at 8:44
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 18 at 8:39
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
1
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
add a comment |
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
1
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
$begingroup$
Refer to the comment by Claude for more details.
$endgroup$
– Karthik
Jan 18 at 9:08
1
1
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
@JackD'Aurizio. Please, don't call me Sir. I shall work on this tomorrow morning. Cheers & thank for the problem.
$endgroup$
– Claude Leibovici
Jan 18 at 17:19
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
$begingroup$
$zeta(s) = sum_{n=1}^infty frac1n n^{1-s} = (s-1) int_1^infty ( sum_{n le x} frac1n) x^{-s}dx$ may help to show $lim_{s to 1} zeta(s) - frac1{s-1} $ $=lim_{s to 1} (s-1) int_1^infty ( sum_{n le x} frac1n- log(x)) x^{-s}dx=lim_{s to 1} (s-1) int_1^infty (gamma+o(1)) x^{-s}dx = lim_{s to 1} gamma+o(1) = gamma$
$endgroup$
– reuns
Jan 18 at 19:14
add a comment |
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$begingroup$
Better now after your edit ! I shall fix my answer. Cheers :-)
$endgroup$
– Claude Leibovici
Jan 18 at 8:43
$begingroup$
arxiv.org/pdf/1201.2633.pdf could be of interest.
$endgroup$
– Claude Leibovici
Jan 18 at 8:53