Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$?
$begingroup$
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
asked Jan 3 at 15:01
sejysejy
1589
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 15:01
sejysejy
1589
$endgroup$
2
$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10
1
$begingroup$
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.
$endgroup$
– GEdgar
Jan 3 at 15:34
add a comment |
$begingroup$
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
asked Jan 3 at 15:01
sejysejy
1589
$endgroup$
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
asked Jan 3 at 15:01
sejysejy
1589
asked Jan 3 at 15:01
sejysejy
1589
edited Jan 4 at 3:23
sejy
asked Jan 3 at 15:01
sejysejy
1589
asked Jan 3 at 15:01
sejysejy
1589
asked Jan 3 at 15:01
sejysejy
1589
1589
2
$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10
1
$begingroup$
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.
$endgroup$
– GEdgar
Jan 3 at 15:34
add a comment |
2
$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10
1
$begingroup$
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.
$endgroup$
– GEdgar
Jan 3 at 15:34
2
2
$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10
$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10
1
1
$begingroup$
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.
$endgroup$
– GEdgar
Jan 3 at 15:34
$begingroup$
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.
$endgroup$
– GEdgar
Jan 3 at 15:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $x,yin mathbb{C}$.
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $x,yin mathbb{C}$.
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
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 $x,yin mathbb{C}$.
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
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 $x,yin mathbb{C}$.
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
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 $x,yin mathbb{C}$.
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
edited Jan 3 at 15:19
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 3 at 15:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
10.6k41741
add a comment |
add a comment |
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$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10
1
$begingroup$
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.
$endgroup$
– GEdgar
Jan 3 at 15:34