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    I'm going over the proof by induction.
    How is;

     \displaystyle \sum_{r=0}^{m} \binom{m}{r}u^{(r+1)}v^{(m-r)} = \sum_{r=1}^{m+1} \binom{m}{r-1}u^{(r)}v^{(m-r+1)} ?

    Where u, v are functions of x and u^(r) is the rth derivative of u.
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    (Original post by NotNotBatman)
    I'm going over the proof by induction.
    How is;

     \displaystyle \sum_{r=0}^{m} \binom{m}{r}u^{(r+1)}v^{(m-r)} = \sum_{r=1}^{m+1} \binom{m}{r-1}u^{(r)}v^{(m-r+1)} ?

    Where u, v are functions of x and u^(r) is the rth derivative of u.
    r \mapsto (r-1) then the m+1 on top is needed in order to keep the same amount of terms.
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    (Original post by NotNotBatman)
    I'm going over the proof by induction.
    How is;

     \displaystyle \sum_{r=0}^{m} \binom{m}{r}u^{(r+1)}v^{(m-r)} = \sum_{r=1}^{m+1} \binom{m}{r-1}u^{(r)}v^{(m-r+1)} ?

    Where u, v are functions of x and u^(r) is the rth derivative of u.
    I'm not sure if this is right, but you can see it if you expand the sum

     \text{LHS} = \displaystyle \binom{m}{0} uv^m + \binom{m}{1} u^2v^{m-1}+...

    \displaystyle + \binom{m}{m-1} u^{n}v+ \binom{m}{m} u^{m+1}

    Accordingly, the RHS expanded should give an equivalent expression. This might be an easier approach than using a substitution
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    To expand upon what rdk said, if you let r \mapsto (r-1), then you have,

    \displaystyle \sum_{r-1=0}^{r-1=m} \ldots = \sum_{r=1}^{m+1} \ldots
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    (Original post by RDKGames)
    r \mapsto (r-1) then the m+1 on top is needed in order to keep the same amount of terms.
    How comes you can have r \mapsto (r-1) ?
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    (Original post by Desmos)
    I'm not sure if this is right, but you can see it if you expand the sum

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \text{LHS} = \displaystyle \binom(m 0)uv^m + \binom(m1)u^2v^{m-1)+\binom(m (m-1))u^{n}v+\binom(m m)u^{m+1}
    you could expand but making a quick substitution is a lot simpler.

    Btw use \binom{n}{k} in latex which produces

    \binom{n}{k}.
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    (Original post by NotNotBatman)
    How comes you can have r \mapsto (r-1) ?
    for the index,
    (Original post by _gcx)
    To expand upon what rdk said, if you let r \mapsto (r-1), then you have,

    \displaystyle \sum_{r-1=0}^{r-1=m} \ldots = \sum_{r=1}^{m+1} \ldots
    once you've changed the index you can just replace the rs with (r-1)s and you'll see that they are equal. Provided we do this we are not changing the value of the series.
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    (Original post by _gcx)
    you could expand but making a quick substitution is a lot simpler.

    Btw use \binom{n}{k} in latex which produces

    \binom{n}{k}.
    Could you quote my post with the correct LaTeX? Cos I can't see what's wrong. And I thought of using a substitution but I didn't know if that was correct or not.
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    (Original post by Desmos)
    I'm not sure if this is right, but you can see it if you expand the sum

     \displaystyle \binom{m}{0} uv^m + \binom{m}{1} u^2v^{m-1}+\binom{m}{m-1} u^{n}v+ \binom{m}{m} u^{m+1}
    needed } not ) around m-1
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    Thanks all, I understand it now.

    (Original post by RDKGames)
    r \mapsto (r-1) then the m+1 on top is needed in order to keep the same amount of terms.
    PRSOM.
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    (Original post by NotNotBatman)
    How comes you can have r \mapsto (r-1) ?
    One of the properties of finite sums, you can 'shift along' a sequence of terms you're adding freely by these off-sets without altering the actual sum.

    Like \displaystyle \sum_{n=0}^{k}ar^n = a+ar+ar^2+...+ar^k

    and \displaystyle \sum_{n=1}^{k+1} ar^{n-1} = a+ar+ar^2+...+ar^k

    Exactly the same result.
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    (Original post by RDKGames)
    One of the properties of sums, you can 'shift along' a sequence of terms you're adding freely by these off-sets without altering the actual sum.

    Like \displaystyle \sum_{n=0}^{k}ar^n = a+ar+ar^2+...+ar^k

    and \displaystyle \sum_{n=1}^{k+1} ar^{n-1} = a+ar+ar^2+...+ar^k

    Exactly the same result.
    Ah, yeah, that makes sense.
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    (Original post by RDKGames)
    One of the properties of finite sums, you can 'shift along' a sequence of terms you're adding freely by these off-sets without altering the actual sum.

    Like \displaystyle \sum_{n=0}^{k}ar^n = a+ar+ar^2+...+ar^k

    and \displaystyle \sum_{n=1}^{k+1} ar^{n-1} = a+ar+ar^2+...+ar^k

    Exactly the same result.
    We just learnt this today in uni. (Although I can't remember if this result comes up in the FP1 chapter on sums.)

    Thanks RDK and _gcx and PRSOM.
 
 
 
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