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Hi,
I am working my way through Q12 shown:
https://ibb.co/4pNhy5z
I have been able to show part a)
and using my answer to part a) I have correctly found the answer to part b) to be 3pi/16
However I am sure part c) must be an error, because the formula they ask us to prove doesn't work, for example we know that when n=2 we should get 3pi/16 from part b) but we don't we get 3pi/4
this is surely an error right? and if it is, does anyone know what the correct question would be? eg is it just a typo somewhere?
as always, any answers are much appreciated!
I am working my way through Q12 shown:
https://ibb.co/4pNhy5z
I have been able to show part a)
and using my answer to part a) I have correctly found the answer to part b) to be 3pi/16
However I am sure part c) must be an error, because the formula they ask us to prove doesn't work, for example we know that when n=2 we should get 3pi/16 from part b) but we don't we get 3pi/4
this is surely an error right? and if it is, does anyone know what the correct question would be? eg is it just a typo somewhere?
as always, any answers are much appreciated!
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#2
Not worked it out, but for n even/odd you could iterate out the formula in a) and write out in closed form. The starting values n=0,1 would be easy?
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(Original post by mqb2766)
Not worked it out, but for n even/odd you could iterate out the formula in a) and write out in closed form. The starting values n=0,1 would be easy?
Not worked it out, but for n even/odd you could iterate out the formula in a) and write out in closed form. The starting values n=0,1 would be easy?
I_0 =pi/2
I_2=pi/4
I_4=3pi/16
I_6=5pi/32
these are the terms in the sequence
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#4
(Original post by jc768)
sorry I don't know what "iterate out the formula and write in closed form" means
I_0 =pi/2
I_2=pi/4
I_4=3pi/16
I_6=5pi/32
these are the terms in the sequence
sorry I don't know what "iterate out the formula and write in closed form" means
I_0 =pi/2
I_2=pi/4
I_4=3pi/16
I_6=5pi/32
these are the terms in the sequence
Don't cancel terms, easier to spot the pattern. Write them out as factorials close to the given solution.
I think I know where the typo in c) is, but....
Last edited by mqb2766; 1 month ago
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(Original post by mqb2766)
Just plug some numbers in n=1,2,3,4, like you've done and spot the pattern.
Don't cancel terms, easier to spot the pattern. Write them out as factorials close to the given solution.
I think I know where the typo in c) is, but....
Just plug some numbers in n=1,2,3,4, like you've done and spot the pattern.
Don't cancel terms, easier to spot the pattern. Write them out as factorials close to the given solution.
I think I know where the typo in c) is, but....
1/2 x pi/2 , 3/4 x 1/2 x pi/2, 5/6 x 3/4 x 1/2 x pi/2
but not good at turning this into factorial notation
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#6
(Original post by jc768)
I can see we get something like:
1/2 x pi/2 , 3/4 x 1/2 x pi/2, 5/6 x 3/4 x 1/2 x pi/2
but not good at turning this into factorial notation
I can see we get something like:
1/2 x pi/2 , 3/4 x 1/2 x pi/2, 5/6 x 3/4 x 1/2 x pi/2
but not good at turning this into factorial notation
I_8 = (7*5*3*1)/(8*6*4*2) * pi/2
The denominator is
(2^4)*4! = (2^n)*n!
Which is looking good?
The numerator is
8*7*6*5*4*3*2*1 / (8*6*4*2) = ...
Can you finish it off
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(Original post by mqb2766)
Ok. For n=4 and I_8, you have
I_8 = (7*5*3*1)/(8*6*4*2) * pi/2
The denominator is
(2^4)*4! = (2^n)*n!
Which is looking good?
The numerator is
8*7*6*5*4*3*2*1 / (8*6*4*2) = ...
Can you finish it off
Ok. For n=4 and I_8, you have
I_8 = (7*5*3*1)/(8*6*4*2) * pi/2
The denominator is
(2^4)*4! = (2^n)*n!
Which is looking good?
The numerator is
8*7*6*5*4*3*2*1 / (8*6*4*2) = ...
Can you finish it off
(2n!)/(2^n)(n!)
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(Original post by mqb2766)
So what is the typo in c)?
So what is the typo in c)?
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#10
(Original post by jc768)
Still not sure!
Still not sure!
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(Original post by mqb2766)
So what is the typo in c)?
So what is the typo in c)?
(2n)!/(2^2n)(n!)^2
we then need to multiply the denominator by 2 which leads to
(2n)!/(2^2n+1)(n!)^2
thanks for the pointers!
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#12
(Original post by jc768)
when putting the numerator and denominator together we get:
(2n)!/(2^2n)(n!)^2
we then need to multiply the denominator by 2 which leads to
(2n)!/(2^2n+1)(n!)^2
thanks for the pointers!
when putting the numerator and denominator together we get:
(2n)!/(2^2n)(n!)^2
we then need to multiply the denominator by 2 which leads to
(2n)!/(2^2n+1)(n!)^2
thanks for the pointers!
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(Original post by mqb2766)
That's correct . When n=2, you were out by a factor of 4.
That's correct . When n=2, you were out by a factor of 4.
When n=k+1
We need to find I_(2k+2)
And from the earlier definition this is
(2k+2)I_(2k+2)=(2k+1)I_2k
Which leads to:
I_(2k+2)=(2k+1)(I_2k)/(2k+2)
and then
I_(2k+2)=(2k+1)(2k!)(pi)/(2k+2)(2^2k+1)(k!)^2
but not sure where to go next
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(Original post by jc768)
However I'm now struggling with the inductive step!
When n=k+1
We need to find I_(2k+2)
And from the earlier definition this is
(2k+2)I_(2k+2)=(2k+1)I_2k
Which leads to:
I_(2k+2)=(2k+1)(I_2k)/(2k+2)
and then
I_(2k+2)=(2k+1)(2k!)(pi)/(2k+2)(2^2k+1)(k!)^2
but not sure where to go next
However I'm now struggling with the inductive step!
When n=k+1
We need to find I_(2k+2)
And from the earlier definition this is
(2k+2)I_(2k+2)=(2k+1)I_2k
Which leads to:
I_(2k+2)=(2k+1)(I_2k)/(2k+2)
and then
I_(2k+2)=(2k+1)(2k!)(pi)/(2k+2)(2^2k+1)(k!)^2
but not sure where to go next
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#15
(Original post by jc768)
Well I know I need to show that this is equal to (2k+2)!(pi)/(2^2k+3)(k+1)!^2 but this is what I am struggling with
Well I know I need to show that this is equal to (2k+2)!(pi)/(2^2k+3)(k+1)!^2 but this is what I am struggling with
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