Math Induction Help

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December 5th 2009, 04:29 PM
thekrown

Math Induction Help

2. b) Show by mathematical induction that:

(1 - 1/4) (1 - 1/9) (1 - 1/16) ... (1 - 1/n^2) = n+1 / 2n, for n >= 2

I did p(1) and it is true

I assumed p(k) is true and moved onto p(k+1) to see if it is true. I'm stuck here.

My work:

Since p(1) true the left side is replaced with the right, I get:

n+1 / 2n = (n+1/2n) (1-1/(n+1)^2)

A lil' kick in the right direction and I should be able to solve this.
December 5th 2009, 04:52 PM
thekrown

I tried putting over common denominator and get:

(n^3 + 3n^2 + 2n) / (2n) * (-n^2-2n+2)
December 5th 2009, 06:14 PM
mr fantastic

Quote:

Originally Posted by thekrown

2. b) Show by mathematical induction that:

(1 - 1/4) (1 - 1/9) (1 - 1/16) ... (1 - 1/n^2) = n+1 / 2n, for n >= 2

I did p(1) and it is true

I assumed p(k) is true and moved onto p(k+1) to see if it is true. I'm stuck here.

My work:

Since p(1) true the left side is replaced with the right, I get:

n+1 / 2n = (n+1/2n) (1-1/(n+1)^2)

A lil' kick in the right direction and I should be able to solve this.

$\left( \frac{k + 1}{2k} \right) \cdot \left( 1 - \frac{1}{(k+1)^2}\right) = \left( \frac{k + 1}{2k} \right) \cdot \left( \frac{k^2 + 2k}{(k+1)^2} \right)$

and this readily simplifies to the required result.
December 5th 2009, 07:07 PM
thekrown

It seems I forgot to add the [1-1/(k+1)^2] to the left side. Also, would you mind explaining what happened to the right? I don't recognize [k^2+2k / (k+1)^2].
December 6th 2009, 01:07 AM
mr fantastic

Quote:

Originally Posted by thekrown

It seems I forgot to add the [1-1/(k+1)^2] to the left side. Also, would you mind explaining what happened to the right? I don't recognize [k^2+2k / (k+1)^2].

The right hand side follows by getting a common denominator in the second bracket of the left hand side and then simplifying. Your (simple) job is to simplify the right hand side a bit more to demonstrate the required result.
December 6th 2009, 08:10 AM
thekrown

I got it thank you! The last answer you gave is the final answer?

We would finish with something like...

=> p(k+1) is true
p(1) true
Assume p(k) true => p(k+1) true

Thus p is true for all n by math induction
December 6th 2009, 06:16 PM
HallsofIvy

Quote:

Originally Posted by thekrown

2. b) Show by mathematical induction that:

(1 - 1/4) (1 - 1/9) (1 - 1/16) ... (1 - 1/n^2) = n+1 / 2n, for n >= 2

I did p(1) and it is true

I very much hope you mean you did p(2)! You are told that this is only true for n>= 2.

Quote:

I assumed p(k) is true and moved onto p(k+1) to see if it is true. I'm stuck here.

My work:

Since p(1) true the left side is replaced with the right, I get:

n+1 / 2n = (n+1/2n) (1-1/(n+1)^2)

A lil' kick in the right direction and I should be able to solve this.
December 6th 2009, 06:20 PM
thekrown

Yes I meant p(2), typo on my part. =)

But is the answer your gave with two terms on each side, but are not identical, the answer? I tried multiplying it out and am not able to factor it so that I get the two groups that are on the left side of the equation.
December 6th 2009, 06:39 PM
mr fantastic

Quote:

Originally Posted by thekrown

Yes I meant p(2), typo on my part. =)

But is the answer your gave with two terms on each side, but are not identical, the answer? I tried multiplying it out and am not able to factor it so that I get the two groups that are on the left side of the equation.

Do you realise that step 2 requires you to show that P(k) true => P(k+1) true, that is, you need to get on the right hand side $\frac{k+2}{2(k+1)}$ ? And that $\left( \frac{k + 1}{2k} \right) \cdot \left( \frac{k^2 + 2k}{(k+1)^2} \right)$ reduces to this?
December 6th 2009, 06:40 PM
thekrown

Hmmm... the way I was taught was do perform the series with p(2) true, then assume* p(k) is true and go right ahead to p(k+1) and prove that.

I am capable of applying math induction on a simpler type of series, except this particular one I cannot seem to simplify the right hand side and get the left hand side.
December 6th 2009, 06:46 PM
mr fantastic

Quote:

Originally Posted by thekrown

Hmmm... the way I was taught was do perform the series with p(2) true, then assume* p(k) is true and go right ahead to p(k+1) and prove that.

I am capable of applying math induction on a simpler type of series, except this particular one I cannot seem to simplify the right hand side and get the left hand side.

Go back and read my reply (post #2). That is what I have done! I have used the result assumed by P(k) and given the result that follows from this for P(k+1).
December 6th 2009, 06:48 PM
thekrown

I will keep working on this. Thank you for your help!
December 6th 2009, 07:19 PM
thekrown

Holy makarel! This was a lot simpler than I had thought. Somehow my brain has been jammed and I was unable to compute. What can you expect studying 6+ hours of math a day?

Oh boy wish me luck on my exams and thank you enormously for your help and patience!