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Math Help - PLEASE HELP (Again) Induction!!

  1. #1
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    Exclamation PLEASE HELP (Again) Induction!!

    Hi, i am clear on how to answer many questions on proof by induction, however i dont understand how to do this question! My lecturer has not explained it at all and i am so panicking!!!!

    Calculate several first values of the products
    n
    Π
    j=2
    (1 4/(2j 1) ) = (1 4/3)(1-4/5 )(1 4/7)・ ・ ・(1 4/(2n 1) )
    and guess the general expression for it. Prove your guess by mathematical induction.
    Nobody has to tell me the answer, but could please someone explain how to do it??

    PLEASE!!!!!!!!
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  2. #2
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    Hello, Bexii!

    Calculate several first values of the product:

    \prod ^n_{j=2}\left[1 - \tfrac{4}{(2j-1)^2}\right] \;=\;\left(1 - \tfrac{4}{3^2}\right)\left(1 - \tfrac{4}{5^2}\right)\left(1 - \tfrac{4}{7^2}\right)\hdots\left(1 - \tfrac{4}{(2n-1)^2}\right)

    and guess the general expression for it.

    Prove your guess by mathematical induction.

    \begin{array}{ccccccc}P_2 &=& 1 - \frac{4}{3^2} &=& \frac{5}{9} &=& \frac{1}{3}\cdot\frac{5}{3} \\ \\[-4mm]<br />
P_3 &=&\frac{5}{9}(1 - \frac{4}{5^2}) &=& \frac{7}{15} &=& \frac{1}{3}\cdot\frac{7}{5} \\ \\[-4mm]<br />
P_4 &=&\frac{7}{15}(1 - \frac{4}{7^2}) &=& \frac{3}{7} &=& \frac{1}{3}\cdot\frac{9}{7} \\ \\[-4mm]<br />
P_5 &=& \frac{3}{7}(1 - \frac{4}{9^2}) &=& \frac{11}{27} &=& \frac{1}{3}\cdot \frac{11}{9} \\ <br />
& & & \vdots\end{array}

    I would conjecture that: . P_n \;=\;\frac{1}{3}\cdot\frac{2n+1}{2n-1}

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  3. #3
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    Hiya how do you prove it by induction? Or have you already done that sorry not very good at the subject.....
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  4. #4
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    Hello, Bexii!


    We want to prove that:

    S(n) \;=\;\left(1-\frac{4}{3^2}\right)\left(1 - \frac{4}{5^2}\right)\left(1-\frac{4}{7^2}\right)\cdots\left(1 - \frac{4}{(2n-1)^2}\right) \;=\;\frac{1}{3}\cdot\frac{2n+1}{2n-1} . .for n \geq 2



    Verify the base case: . S(2)

    . . 1 - \frac{4}{3^2} \;=\;\frac{1}{3}\cdot\frac{2(2)+1}{2(2)-1} \quad\Rightarrow\quad \frac{5}{9} \:=\:\frac{5}{9} . . . True



    Assume S(k)

    . . \left(1 - \frac{4}{3^2}\right)\left(1 - \frac{4}{5^2}\right)\left(1-\frac{4}{7^2}\right) \cdots\left(1 - \frac{4}{(2k-1)^2}\right) \;=\;\frac{1}{3}\cdot\frac{2k+1}{2k-1}



    Multiply both sides by \left(1 - \frac{4}{(2k+1)^2}\right)

    . . \underbrace{\left(1-\frac{4}{3^2}\right)\left(1 - \frac{4}{5^2}\right) \cdots\left(1 - \frac{4}{(2k+1)^2}\right)}_{\text{The left side of }S(k+1)}    \;=\;\frac{1}{3}\cdot\frac{2k+1}{2k-1}\cdot\left(1 - \frac{4}{(2k+1)^2}\right)



    We must show that the right side is the right side of S(k+1).

    We have: . \frac{1}{3}\cdot\frac{2k+1}{2k-1}\cdot\frac{(2k+1)^2-4}{(2k+1)^2} \;=\;\frac{1}{3}\cdot\frac{2k+1}{2k-1}\cdot\frac{4k^2 + 4k + 1 - 4}{(2k+1)^2}

    . . = \;\frac{1}{3}\cdot\frac{2k+1}{2k-1}\cdot\frac{4k^2 + 4k-3}{(2k+1)^2} \;=\;\frac{1}{3}\cdot\frac{2k+1}{2k-1}\cdot\frac{(2k-1)(2k+3)}{(2k+1)^2}

    . . = \;\frac{1}{3}\cdot\frac{2k+3}{2k+1} \;=\;\underbrace{\frac{1}{3}\cdot\frac{2(k+1)+1}{2  (k+1)-1}}_{\text{Right side of }S(k+1)}


    The inductive proof is complete.

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  5. #5
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    Quote Originally Posted by Bexii View Post
    Hiya how do you prove it by induction? Or have you already done that sorry not very good at the subject.....
    Try to follow the induction proof that I showed you before. Again you need to do 3 steps. The first step is to check that the conjecture Soroban gave you is true at j=2, which he already did for you. Then assume that it's is true for j=n. Final step is to prove that the formula is true for j=n+1.
    Hint: for j=n+1, we have
    \left(1 - \tfrac{4}{3^2}\right)\left(1 - \tfrac{4}{5^2}\right)\left(1 - \tfrac{4}{7^2}\right)\hdots\left(1 - \tfrac{4}{(2(n+1)-1)^2}\right) =\left(1 - \tfrac{4}{3^2}\right)\left(1 - \tfrac{4}{5^2}\right)\left(1 - \tfrac{4}{7^2}\right)\hdots\left(1 - \tfrac{4}{(2n+1)^2}\right)
    =\left(1 - \tfrac{4}{3^2}\right)\left(1 - \tfrac{4}{5^2}\right)\left(1 - \tfrac{4}{7^2}\right)\hdots+\left(1-\tfrac{4}{(2n-1)^2}\right)\left(1 - \tfrac{4}{(2n+1)^2}\right). Use the hypothesis that the formula is true for j=n on the last expression.

    Sorry: I didn't see you posted the solution a bit before me, Soroban.
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  6. #6
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    THANKYOU SO MUCH YOU GUYS!!! I understand it!!!
    Im getting so frustrated with my classes, because my assessment has questions on induction i dont even remember being mentioned!
    Hypothetically, if your doing proof by induction for inequalities which involves a sequence, you would just need to focus on the part of the inequality which isnt the sequence?????
    Dont know what id do without this site at the moment!!!!!And you guys!!!
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