Proof by induction and reduction formulae

Watch
jc768
Badges: 7
Rep:
?
#1
Report Thread starter 1 month ago
#1
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!
0
reply
mqb2766
Badges: 19
Rep:
?
#2
Report 1 month ago
#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?
0
reply
jc768
Badges: 7
Rep:
?
#3
Report Thread starter 1 month ago
#3
(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?
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
0
reply
mqb2766
Badges: 19
Rep:
?
#4
Report 1 month ago
#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
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....
Last edited by mqb2766; 1 month ago
0
reply
jc768
Badges: 7
Rep:
?
#5
Report Thread starter 1 month ago
#5
(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....
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
0
reply
mqb2766
Badges: 19
Rep:
?
#6
Report 1 month ago
#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
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
0
reply
jc768
Badges: 7
Rep:
?
#7
Report Thread starter 1 month ago
#7
(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
the numerator is

(2n!)/(2^n)(n!)
0
reply
mqb2766
Badges: 19
Rep:
?
#8
Report 1 month ago
#8
(Original post by jc768)
the numerator is

(2n!)/(2^n)(n!)
So what is the typo in c)?
0
reply
jc768
Badges: 7
Rep:
?
#9
Report Thread starter 1 month ago
#9
(Original post by mqb2766)
So what is the typo in c)?
Still not sure!
0
reply
mqb2766
Badges: 19
Rep:
?
#10
Report 1 month ago
#10
(Original post by jc768)
Still not sure!
Really? What do you have on the numerator and denominator and how does it compare to c)? Everything is written out in the same way.
0
reply
jc768
Badges: 7
Rep:
?
#11
Report Thread starter 1 month ago
#11
(Original post by mqb2766)
So what is the typo in c)?
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!
0
reply
mqb2766
Badges: 19
Rep:
?
#12
Report 1 month ago
#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!
That's correct . When n=2, you were out by a factor of 4.
0
reply
jc768
Badges: 7
Rep:
?
#13
Report Thread starter 1 month ago
#13
(Original post by mqb2766)
That's correct . When n=2, you were out by a factor of 4.
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
0
reply
jc768
Badges: 7
Rep:
?
#14
Report Thread starter 1 month ago
#14
(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
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
0
reply
DFranklin
Badges: 18
Rep:
?
#15
Report 1 month ago
#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
Multiply numerator and denominator by k+1. Note also 2(k+1)= 2k+2.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Following the government's announcement, do you think you will be awarded a fair grade this year?

Yes (462)
51.45%
No (436)
48.55%

Watched Threads

View All
Latest
My Feed