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Thread: question about induction

  1. #1
    Member Jones's Avatar
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    question about induction

    Hi,

    I had this induction problem which i couldn't figure out how to solve.

    for all n $\displaystyle \geq 1$ show that: $\displaystyle (1+x)^n \geq 1+nx$

    The solution looked like this:
    $\displaystyle \begin{array}{lcr}(1+x)^{k+1} = (1+x)(1+x)^k \\
    \geq ~(1+x){\color{red}(1+kx)} \\
    =~1+kx+x+kx^2\\
    \geq~1+kx+x\\
    =~1+(k+1)x \end{array}$

    $\displaystyle \therefore (1+x)^{k+1} \geq 1+(k+1)x$
    How can the stuff marked in red be valid?
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  2. #2
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    That's precisely the induction step -- you assumed that $\displaystyle (1+x)^n \geq 1 + nx$, so you get $\displaystyle (1+x)(1+x)^k \geq (1+x)(1 + kx)$

    And the rest follows easily..
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  3. #3
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    Quote Originally Posted by Jones View Post
    Hi,

    I had this induction problem which i couldn't figure out how to solve.

    for all n $\displaystyle \geq 1$ show that: $\displaystyle (1+x)^n \geq 1+nx$

    The solution looked like this:
    $\displaystyle \begin{array}{lcr}(1+x)^{k+1} = (1+x)(1+x)^k \\
    \geq ~(1+x){\color{red}(1+kx)} \\
    =~1+kx+x+kx^2\\
    \geq~1+kx+x\\
    =~1+(k+1)x \end{array}$

    $\displaystyle \therefore (1+x)^{k+1} \geq 1+(k+1)x$
    How can the stuff marked in red be valid?
    This is a rather well known thing called Bernoulli's Inequality, and it MUST be that $\displaystyle x>-1$

    Tonio
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  4. #4
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    Quote Originally Posted by Jones View Post
    Hi,

    I had this induction problem which i couldn't figure out how to solve.

    for all n $\displaystyle \geq 1$ show that: $\displaystyle (1+x)^n \geq 1+nx$

    The solution looked like this:
    $\displaystyle \begin{array}{lcr}(1+x)^{k+1} = (1+x)(1+x)^k \\
    \geq ~(1+x){\color{red}(1+kx)} \\
    =~1+kx+x+kx^2\\
    \geq~1+kx+x\\
    =~1+(k+1)x \end{array}$

    $\displaystyle \therefore (1+x)^{k+1} \geq 1+(k+1)x$
    How can the stuff marked in red be valid?

    at that part you bolded they are replacing the (1+x)^k with what we know.

    we know that $\displaystyle (1+x)^n \geq 1+nx$.


    at the part you bolded, $\displaystyle (1+x)(1+x)^k$ at this part we replace$\displaystyle (1+x)^k$ with $\displaystyle (1+kx) $ .

    leaving us with $\displaystyle (1+x)(1+kx)$.
    Last edited by p00ndawg; Oct 19th 2009 at 02:52 PM.
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