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Math Help - Induction proof?

  1. #1
    Super Member fardeen_gen's Avatar
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    Induction proof?

    If \prod_{n = 1}^{n} x_{n} denotes x_{1}\cdot x_{2}\cdot x_{3}\cdot ...\cdot x_{n} then prove, by induction, that:
    \left\{\prod_{n = 1}^{n} f_{n}(x)\right\}' = \sum_{i = 1}^{n} \{f_{1}(x)\cdot f_{2}(x)\cdot ...\cdot f_{i}'(x)\cdot ...\cdot f_{n}(x)\}

    where ' denotes derivative w.r.t x
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  2. #2
    Junior Member Infophile's Avatar
    Joined
    May 2009
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    Have you tried ?

    It's not difficult, you know that (f_1f_2)'=f_1'f_2+f_1f_2' (initialisation).

    So, suppose this egality true for n functions, and show it's true for n+1 functions too :

    \left(\prod_{i=1}^{n+1}f_i\right)'=\left(f_{n+1}\p  rod_{i=1}^{n}f_i\right)'=f_{n+1}'\prod_{i=1}^{n}f_  i+f_{n+1}{\color{red}\left(\prod_{i=1}^{n}f_i\righ  t)'}

    Now substitute term in red (hypothesis of induction) :

    \left(\prod_{i=1}^{n+1}f_i\right)'=\left(f_{n+1}\p  rod_{i=1}^{n}f_i\right)'=f_{n+1}'\prod_{i=1}^{n}f_  i+f_{n+1}\sum_{i=1}^{n}\left\{f_1...f_i'...f_n\rig  ht\}

    i.e : \left(\prod_{i=1}^{n+1}f_i\right)'=f_{n+1}'\prod_{  i=1}^{n}f_i+\sum_{i=1}^{n}\left\{f_1...f_i'...f_nf  _{n+1}\right\}

    And finally : \color{blue}\boxed{\left(\prod_{i=1}^{n+1}f_i\righ  t)'=\sum_{i=1}^{{\color{red}n+1}}\left\{f_1...f_i'  ...f_{n+1}\right\}}

    What ends the prove by induction.

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