# Inequalities

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#1

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1 month ago
#2
(Original post by Nimantha)

Any thoughts? You need to find n, so can you write
n = ...
Then replace the variables to make an inequality? What are the min/max values.

Last edited by mqb2766; 1 month ago
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#3

I tried but i have stuck here
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1 month ago
#4
(Original post by Nimantha)

I tried but i have stuck here
Id isolate n in the inequality, so
... <= n <= ...
Then argue about the min/max value of the expressions if you don't simplify them first.
Last edited by mqb2766; 1 month ago
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#5
Is it wrong if I take - 4pi and 12pi as min and max values for the above equation for all alpha?
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1 month ago
#6
(Original post by Nimantha)
Is it wrong if I take - 4pi and 12pi as min and max values for the above equation for all alpha?
It would help to see your working, but that's the general idea.The
Note you also have x to bound.
Last edited by mqb2766; 1 month ago
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#7

Please tell me where have I gone wrong...
1
1 month ago
#8
(Original post by Nimantha)

Please tell me where have I gone wrong...
I would have written
n = ...
To find the actual expression in x, alpha you want to find the min/max value of which satisfies the constraints. Maybe try that for starters?

In doing so, you may have to consider x=0 division, but it should give insight.
Last edited by mqb2766; 1 month ago
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#9
(Original post by mqb2766)
I would have written
n = ...
To find the actual expression in x, alpha you want to find the min/max value of which satisfies the constraints. Maybe try that for starters?
Ok thanks...
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1 month ago
#10
(Original post by Nimantha)
Ok thanks...
It's also worth noting that typically x, alpha will be at extreme values (their constraint boundariies) to make n as large/small as possible.
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1 month ago
#11
(Original post by 3672N)

Please tell me where have I gone wrong...
When I was looking at this last night, I thought one of the pis was an x, must have been tired. That working looks fine.
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#12
But it doesn't work for all alpha in the range like pi/6
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#13

The original question was this, and I got stuck in the second part. And I think there is a problem in this question as it doesn't give the same number of solutions for all alpha in the given range
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1 month ago
#14
(Original post by 3672N)
But it doesn't work for all alpha in the range like pi/6
The way you asked it was that alpha was a slack variable that just had to satisfy the inequality constant. So as long as there exists one alpha in that range, the solution is fine.

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct
2pi - alpha <= x <= 2pi+alpha
So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?
Last edited by mqb2766; 1 month ago
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#15
(Original post by mqb2766)
The way you asked it was that alpha was a slack variable that just had to satisfy the inequality constant. So as long as there exists one alpha in that range, the solution is fine.

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct
2pi - alpha <= x <= 2pi+alpha
So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?
h and b are not the same, and the final ineqality is correct. B=1/2sqrt((a-b)^2+4h^2), which we can take as positive, so the maximum must be at pi/2
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1 month ago
#16
(Original post by 3672N)
h and b are not the same, and the final ineqality is correct. B=1/2sqrt((a-b)^2+4h^2), which we can take as positive, so the maximum must be at pi/2
what does the final inequality represent?
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#17
(Original post by mqb2766)
The way you asked it was that alpha was a slack variable that just had to satisfy the inequality constant. So as long as there exists one alpha in that range, the solution is fine.

Edit, when looking at the original question, alpha is fixed (determined by a and b (h?)), not a free variable, which obviously changes the problem. Is the final inequality constraint correct
2pi - alpha <= x <= 2pi+alpha
So for instance if alpha=0, x=2pi and there cannot be a stationary point as the range of x is a point? Or do you want to take the range extreme value as the max value?

Note there appears to be some ambiguity in the question, depending on the sign of B the location of the maximum could be 3pi/2 rather than pi/2. I take it you've accounted for that?
I don't undestand what did you mean by the solution is fine as long as there exist one one alpha, cause there can't be a general solution for n as, the number of solutions for n changes as alpha increases or decreses by pi
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1 month ago
#18
(Original post by 3672N)
I don't undestand what did you mean by the solution is fine as long as there exist one one alpha, cause there can't be a general solution for n as, the number of solutions for n changes as alpha increases or decreses by pi
Thats how I was interpreting the problem last night. Seeing the full question, its obviously incorrect as alpha is fixed by the initial transformation.
Last edited by mqb2766; 1 month ago
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#19
(Original post by mqb2766)
Thats how I was interpreting the problem last night. Seeing the full question, its obviously incorrect as alpha is fixed by the initial transformation.
Ahh sorry i should have post the original question earlier
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#20
(Original post by mqb2766)
what does the final inequality represent?
2pi-alpha<=x<=2pi+alpha
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