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Figuring out zeros and divergence of a double series
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I’m trying to figure out an interesting double series, but can’t figure out how to represent it as a single series, and know little about double series. $$ sum_n=1^∞ sum_m=1^∞ fraccos (yln(n/m))(n^x)(m^x) $$ is the double series. I can see that the central diagonal, if one were to plot this out on a plane, is equal to the Riemann zeta function of 2x, as any number divided by itself is one, the natural log of one is zero, zero times any value of y is zero, and cosine of zero is one. I am trying to figure out for what values of y and x does this series diverge, where it converges, and where it equals zero.
sequences-and-series summation
asked Jul 28 at 18:22
MathPonderer
113
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up vote
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I’m trying to figure out an interesting double series, but can’t figure out how to represent it as a single series, and know little about double series. $$ sum_n=1^∞ sum_m=1^∞ fraccos (yln(n/m))(n^x)(m^x) $$ is the double series. I can see that the central diagonal, if one were to plot this out on a plane, is equal to the Riemann zeta function of 2x, as any number divided by itself is one, the natural log of one is zero, zero times any value of y is zero, and cosine of zero is one. I am trying to figure out for what values of y and x does this series diverge, where it converges, and where it equals zero.
sequences-and-series summation
asked Jul 28 at 18:22
MathPonderer
113
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I’m trying to figure out an interesting double series, but can’t figure out how to represent it as a single series, and know little about double series. $$ sum_n=1^∞ sum_m=1^∞ fraccos (yln(n/m))(n^x)(m^x) $$ is the double series. I can see that the central diagonal, if one were to plot this out on a plane, is equal to the Riemann zeta function of 2x, as any number divided by itself is one, the natural log of one is zero, zero times any value of y is zero, and cosine of zero is one. I am trying to figure out for what values of y and x does this series diverge, where it converges, and where it equals zero.
sequences-and-series summation
asked Jul 28 at 18:22
MathPonderer
113
I’m trying to figure out an interesting double series, but can’t figure out how to represent it as a single series, and know little about double series. $$ sum_n=1^∞ sum_m=1^∞ fraccos (yln(n/m))(n^x)(m^x) $$ is the double series. I can see that the central diagonal, if one were to plot this out on a plane, is equal to the Riemann zeta function of 2x, as any number divided by itself is one, the natural log of one is zero, zero times any value of y is zero, and cosine of zero is one. I am trying to figure out for what values of y and x does this series diverge, where it converges, and where it equals zero.
sequences-and-series summation
asked Jul 28 at 18:22
MathPonderer
113
edited Jul 28 at 20:28
asked Jul 28 at 18:22
MathPonderer
113
asked Jul 28 at 18:22
MathPonderer
113
asked Jul 28 at 18:22
MathPonderer
113
113
add a comment |Â
add a comment |Â
1 Answer
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For $x > 1$ this series is easily seen to be absolutely convergent to $C^2(x, y) + S^2(x, y)$, where $C(x, y) = displaystylesum_n = 1^infty fraccos(yln n)n^x$ and $S(x, y) = displaystylesum_n = 1^infty fracsin(yln n)n^x$. Essentially, the sum is $|zeta(x+iy)|^2$, and it is known that for $x > 1$ there are no zeros.
If there is some form of conditional (non-absolute) convergence to be analysed, the question is what exactly this form is (for multiple series, this is important - there's no widespread conventions on that).
answered Jul 29 at 1:38
metamorphy
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For $x > 1$ this series is easily seen to be absolutely convergent to $C^2(x, y) + S^2(x, y)$, where $C(x, y) = displaystylesum_n = 1^infty fraccos(yln n)n^x$ and $S(x, y) = displaystylesum_n = 1^infty fracsin(yln n)n^x$. Essentially, the sum is $|zeta(x+iy)|^2$, and it is known that for $x > 1$ there are no zeros.
If there is some form of conditional (non-absolute) convergence to be analysed, the question is what exactly this form is (for multiple series, this is important - there's no widespread conventions on that).
answered Jul 29 at 1:38
metamorphy
7358
add a comment |Â
up vote
0
down vote
For $x > 1$ this series is easily seen to be absolutely convergent to $C^2(x, y) + S^2(x, y)$, where $C(x, y) = displaystylesum_n = 1^infty fraccos(yln n)n^x$ and $S(x, y) = displaystylesum_n = 1^infty fracsin(yln n)n^x$. Essentially, the sum is $|zeta(x+iy)|^2$, and it is known that for $x > 1$ there are no zeros.
If there is some form of conditional (non-absolute) convergence to be analysed, the question is what exactly this form is (for multiple series, this is important - there's no widespread conventions on that).
answered Jul 29 at 1:38
metamorphy
7358
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For $x > 1$ this series is easily seen to be absolutely convergent to $C^2(x, y) + S^2(x, y)$, where $C(x, y) = displaystylesum_n = 1^infty fraccos(yln n)n^x$ and $S(x, y) = displaystylesum_n = 1^infty fracsin(yln n)n^x$. Essentially, the sum is $|zeta(x+iy)|^2$, and it is known that for $x > 1$ there are no zeros.
If there is some form of conditional (non-absolute) convergence to be analysed, the question is what exactly this form is (for multiple series, this is important - there's no widespread conventions on that).
answered Jul 29 at 1:38
metamorphy
7358
For $x > 1$ this series is easily seen to be absolutely convergent to $C^2(x, y) + S^2(x, y)$, where $C(x, y) = displaystylesum_n = 1^infty fraccos(yln n)n^x$ and $S(x, y) = displaystylesum_n = 1^infty fracsin(yln n)n^x$. Essentially, the sum is $|zeta(x+iy)|^2$, and it is known that for $x > 1$ there are no zeros.
If there is some form of conditional (non-absolute) convergence to be analysed, the question is what exactly this form is (for multiple series, this is important - there's no widespread conventions on that).
answered Jul 29 at 1:38
metamorphy
7358
edited Jul 29 at 15:42
answered Jul 29 at 1:38
metamorphy
7358
answered Jul 29 at 1:38
metamorphy
7358
answered Jul 29 at 1:38
metamorphy
7358
7358
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