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Thread: Mathematical induction

  1. #1
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    Mathematical induction

    Hi. I need help with this:
    Show that if f is concave up on [a,b] and if there exists numbers
    $\displaystyle a\leq x_{j}\leq b$
    $\displaystyle 1\leq j\leq p$
    and
    $\displaystyle 0\leq \lambda _{j}\leq 1$
    $\displaystyle 1\leq j\leq p$
    such that
    $\displaystyle \sum \lambda _{j}=1$
    then
    $\displaystyle f(\sum ^{p}_{j=1}\lambda _{j}x_{j})\leq \sum ^{p}_{j=1}\lambda _{j}f(x_{j})$

    True for j=2, prove for j=p+1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ralf View Post
    Hi. I need help with this:
    Show that if f is concave up on [a,b] and if there exists numbers
    $\displaystyle a\leq x_{j}\leq b$
    $\displaystyle 1\leq j\leq p$
    and
    $\displaystyle 0\leq \lambda _{j}\leq 1$
    $\displaystyle 1\leq j\leq p$
    such that
    $\displaystyle \sum \lambda _{j}=1$
    then
    $\displaystyle f(\sum ^{p}_{j=1}\lambda _{j}x_{j})\leq \sum ^{p}_{j=1}\lambda _{j}f(x_{j})$

    True for j=2, prove for j=p+1

    $\displaystyle j$ is not the induction variable $\displaystyle p$ is.

    You will have proven a base case where $\displaystyle p=2$, now you need to show
    that if this is true for $\displaystyle p=k$ for some $\displaystyle k$, then it is true for $\displaystyle p=k+1$.

    RonL
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  3. #3
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    I'm not sure how to show this. Can I maybe substitute
    $\displaystyle \sum ^{k}_{j=1}\lambda _{j}x_{j}=\lambda _{u}x_{u}$
    and then use the known case p=2 like this
    $\displaystyle f(\sum ^{k+1}_{j=1}\lambda _{j}x_{j})=f(\sum ^{k}_{j=1}\lambda _{j}x_{j}+\lambda _{k+1}x_{k+1})=f(\lambda _{u}x_{u}+\lambda _{k+1}x_{k+1})\leq$$\displaystyle \lambda _{u}f(x_{u})+\lambda _{k+1}f(x_{k+1})=\sum ^{k+1}_{j=1}\lambda _{j}f(x_{j})$

    I would really appreciate it if someone could give me a hint. If it's correct to do it the way I did it now (which i doubt), then how do I know if I can do the substitution I just did?
    Last edited by Ralf; Mar 20th 2008 at 12:41 PM.
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