# Proof by induction and reduction formulae

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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

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?

**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?

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

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

**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

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

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....

**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....

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

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

**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_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

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

**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

(2n!)/(2^n)(n!)

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(Original post by

So what is the typo in c)?

**mqb2766**)So what is the typo in c)?

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#10

(Original post by

Still not sure!

**jc768**)Still not sure!

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(Original post by

So what is the typo in c)?

**mqb2766**)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

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!

**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!

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(Original post by

That's correct . When n=2, you were out by a factor of 4.

**mqb2766**)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

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

**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

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#15

(Original post by

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

**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

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