How to express a term of multiplicationJoin all square root expressions?Mathematica rule to express exponentials as multiplicationManipulating form of output reduxHow to make TransformationFunctions work with SimplifyExpress roots of quadratic equation in a specific formFactor out imaginary unit of root of obviously negative termProblem with simplification of $sqrt1+fracy^2x^2$Simplifying an equation into a certain formRemove a term which includes a defined variable from an equationExpress variable in terms other variables when lots of equations are involved
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How to express a term of multiplication
Join all square root expressions?Mathematica rule to express exponentials as multiplicationManipulating form of output reduxHow to make TransformationFunctions work with SimplifyExpress roots of quadratic equation in a specific formFactor out imaginary unit of root of obviously negative termProblem with simplification of $sqrt1+fracy^2x^2$Simplifying an equation into a certain formRemove a term which includes a defined variable from an equationExpress variable in terms other variables when lots of equations are involved
$begingroup$
I have the following term which I would like to express correctly in Mathematica:
$$
prod_j=1,jneq i^m(rho_i-rho_j)
$$
Can you please help?
simplifying-expressions expression-manipulation products
edited 7 hours ago
Roman
9,92511640
asked 8 hours ago
user65940user65940
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have the following term which I would like to express correctly in Mathematica:
$$
prod_j=1,jneq i^m(rho_i-rho_j)
$$
Can you please help?
simplifying-expressions expression-manipulation products
edited 7 hours ago
Roman
9,92511640
asked 8 hours ago
user65940user65940
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have the following term which I would like to express correctly in Mathematica:
$$
prod_j=1,jneq i^m(rho_i-rho_j)
$$
Can you please help?
simplifying-expressions expression-manipulation products
edited 7 hours ago
Roman
9,92511640
asked 8 hours ago
user65940user65940
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have the following term which I would like to express correctly in Mathematica:
$$
prod_j=1,jneq i^m(rho_i-rho_j)
$$
Can you please help?
simplifying-expressions expression-manipulation products
simplifying-expressions expression-manipulation products
edited 7 hours ago
Roman
9,92511640
asked 8 hours ago
user65940user65940
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Roman
9,92511640
asked 8 hours ago
user65940user65940
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Roman
9,92511640
edited 7 hours ago
Roman
9,92511640
edited 7 hours ago
Roman
9,92511640
9,92511640
asked 8 hours ago
user65940user65940
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
user65940user65940
111
asked 8 hours ago
user65940user65940
111
111
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user65940 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Long form:
Product[(ρ[i] - ρ[j]), i, 1, m, j, i + 1, m] Product[(ρ[i] - ρ[j]), i, 1, m, j, 1, i - 1]
By observing that each factor occurs twice (up to sign):
Product[ - (ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
By counting the signs:
(-1)^(m (m - 1)/2) Product[(ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
$endgroup$
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Long form:
Product[(ρ[i] - ρ[j]), i, 1, m, j, i + 1, m] Product[(ρ[i] - ρ[j]), i, 1, m, j, 1, i - 1]
By observing that each factor occurs twice (up to sign):
Product[ - (ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
By counting the signs:
(-1)^(m (m - 1)/2) Product[(ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
$endgroup$
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
$begingroup$
Long form:
Product[(ρ[i] - ρ[j]), i, 1, m, j, i + 1, m] Product[(ρ[i] - ρ[j]), i, 1, m, j, 1, i - 1]
By observing that each factor occurs twice (up to sign):
Product[ - (ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
By counting the signs:
(-1)^(m (m - 1)/2) Product[(ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
$endgroup$
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
$begingroup$
Long form:
Product[(ρ[i] - ρ[j]), i, 1, m, j, i + 1, m] Product[(ρ[i] - ρ[j]), i, 1, m, j, 1, i - 1]
By observing that each factor occurs twice (up to sign):
Product[ - (ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
By counting the signs:
(-1)^(m (m - 1)/2) Product[(ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
$endgroup$
Long form:
Product[(ρ[i] - ρ[j]), i, 1, m, j, i + 1, m] Product[(ρ[i] - ρ[j]), i, 1, m, j, 1, i - 1]
By observing that each factor occurs twice (up to sign):
Product[ - (ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
By counting the signs:
(-1)^(m (m - 1)/2) Product[(ρ[i] - ρ[j])^2, i, 1, m, j, i + 1, m]
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
edited 7 hours ago
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
answered 8 hours ago
Henrik SchumacherHenrik Schumacher
63.7k590178
63.7k590178
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Thank you so much. You mean without the ^2 cous its not in the question, but it helped.
$endgroup$
– user65940
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
Yes the square is in the question: every term occurs twice (with opposite signs). There should be a $(-1)^m(m-1)/2$ factor too, to account for these signs.
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
@Roman Huh, I completely forgot about the sign. Thanks for pointing this out!
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
user65940 is a new contributor. Be nice, and check out our Code of Conduct.
user65940 is a new contributor. Be nice, and check out our Code of Conduct.
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