Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$?
Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.
It was a multiple choice questions with
a) $4sinh(3) sinh(1)$
b) $4sinh(1)sinh(3)$
I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .
Am I correct both option is correct ?
If wrong please explain why ?
complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral
add a comment |
Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.
It was a multiple choice questions with
a) $4sinh(3) sinh(1)$
b) $4sinh(1)sinh(3)$
I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .
Am I correct both option is correct ?
If wrong please explain why ?
complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral
2
You're correct. Multiplication is commutative. The question was badly designed.
– Michael Lugo
yesterday
1
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
– GEdgar
yesterday
add a comment |
Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.
It was a multiple choice questions with
a) $4sinh(3) sinh(1)$
b) $4sinh(1)sinh(3)$
I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .
Am I correct both option is correct ?
If wrong please explain why ?
complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral
Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.
It was a multiple choice questions with
a) $4sinh(3) sinh(1)$
b) $4sinh(1)sinh(3)$
I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .
Am I correct both option is correct ?
If wrong please explain why ?
complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral
complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral
edited yesterday
asked yesterday
sejy
1149
1149
2
You're correct. Multiplication is commutative. The question was badly designed.
– Michael Lugo
yesterday
1
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
– GEdgar
yesterday
add a comment |
2
You're correct. Multiplication is commutative. The question was badly designed.
– Michael Lugo
yesterday
1
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
– GEdgar
yesterday
2
2
You're correct. Multiplication is commutative. The question was badly designed.
– Michael Lugo
yesterday
You're correct. Multiplication is commutative. The question was badly designed.
– Michael Lugo
yesterday
1
1
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
– GEdgar
yesterday
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
– GEdgar
yesterday
add a comment |
1 Answer
1
active
oldest
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Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.
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1 Answer
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1 Answer
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active
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Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.
add a comment |
Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.
add a comment |
Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.
Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.
edited yesterday
answered yesterday
Antonios-Alexandros Robotis
9,40741640
9,40741640
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2
You're correct. Multiplication is commutative. The question was badly designed.
– Michael Lugo
yesterday
1
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
– GEdgar
yesterday