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Math Help - Math Induction Help

  1. #1
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    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.
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  2. #2
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    I tried putting over common denominator and get:

    (n^3 + 3n^2 + 2n) / (2n) * (-n^2-2n+2)
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  3. #3
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    Quote Originally Posted by thekrown View Post
    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.
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  4. #4
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    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].
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    Quote Originally Posted by thekrown View Post
    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.
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    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
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  7. #7
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    Quote Originally Posted by thekrown View Post
    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.

    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.
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  8. #8
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    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.
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  9. #9
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    Quote Originally Posted by thekrown View Post
    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?
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    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.
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  11. #11
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    Quote Originally Posted by thekrown View Post
    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).
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  12. #12
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    I will keep working on this. Thank you for your help!
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  13. #13
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    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!
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