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Math Help - how to integrate this ??

  1. #1
    MHF Contributor Amer's Avatar
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    how to integrate this ??

    if Bernstein polynomials defined by

    B_{k,n}(x)=\left(\begin{array}{c}n\\k\end{array}\r  ight)x^k (1-x)^{n-k}

    how I can integrate it

    this is my work

    \int_{a}^{b} \left(\begin{array}{c}n\\k\end{array}\right) x^k (1-x)^{n-k}dx

    \int_{a}^{b} \left(\begin{array}{c}n\\k\end{array}\right) x^k \sum_{i=0}^{n-k} \left(\begin{array}{c}n-k\\i\end{array}\right) (-1)^i x^i dx

    since the summation is the Taylor series of (1-x)^{n-k}

      \sum_{i=0}^{n-k}\left(\begin{array}{c}n\\k\end{array}\right)  \left(\begin{array}{c}n-k\\i\end{array}\right) (-1)^i \int_{a}^{b} x^{k+i} dx

     \sum_{i=0}^{n-k}\left(\begin{array}{c}n\\k\end{array}\right) \left(\begin{array}{c}n-k\\i\end{array}\right) (-1)^i \left(\frac{x^{k+i+1}}{k+i+1}\mid_a^b\right)

    my work is correct but I think I can't reach the true answer if I continue
    the answer is \frac{1}{n+1}\sum_{i=k+1}^{n+1} B_{i,n+1}(x)


    Thanks very much .
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Why is there an x in what you say is the answer? Certainly the answer is not independent of the bounds a,b. Perhaps that is the indefinite integral?
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Why is there an x in what you say is the answer? Certainly the answer is not independent of the bounds a,b. Perhaps that is the indefinite integral?
    I see

    ok consider it was indefinite how I can get this \frac{1}{n+1}\sum_{i=k+1}^{n+1} B_{i,n+1}(x)

    from this


    \int \left(\begin{array}{c}n\\k\end{array}\right) x^k (1-x)^{n-k}dx

    I need just to know how to integrate

    \int x^k(1-x)^{n-k} dx
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Have you tried induction on n? It might very well work with the right argument. For instance you have

    B_{k,n}(x) = {n \choose k} x^k(1-x)^{n-k}
    = {n-1 \choose k} x^k(1-x)^{n-k}+{n-1 \choose k-1} x^k(1-x)^{n-k} = (1-x)B_{k,n-1}+xB_{k-1,n-1}

    Induction and integration by parts might work from there...
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Have you tried induction on n? It might very well work with the right argument. For instance you have

    B_{k,n}(x) = {n \choose k} x^k(1-x)^{n-k}
    = {n-1 \choose k} x^k(1-x)^{n-k}+{n-1 \choose k-1} x^k(1-x)^{n-k} = (1-x)B_{k,n-1}+xB_{k-1,n-1}

    Induction and integration by parts might work from there...
    you mean to begin with

    \int B_{1,2} = \int (1-x)B_{1,1} + \int x B_{0,1}

    \int B_{1,2} = \int \left(\begin{array}{c}1\\1\end{array}\right) (1-x)x^1 (1-x)^{0} + \int \left(\begin{array}{c}1\\0\end{array}\right)x(x)^0 (1-x)^1

    \int B_{1,2} = \int \left(\begin{array}{c}1\\1\end{array}\right) (1-x)x  + \int \left(\begin{array}{c}1\\0\end{array}\right)x (1-x)

    I walk in the right direction right
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  6. #6
    MHF Contributor Amer's Avatar
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    any hints to integrate

    \int  x^k (1-x)^{n-k}

    Thanks
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  7. #7
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     \int x^k (1-x)^{n-k}~dx

    integration by parts

     = \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \frac{(n-k)x^{k+2} (1-x)^{n-k-1}}{(k+1)(k+2)} + .... + \frac{ (n-k)(n-k-1)...(1) x^{k + n-k + 1}}{(k+1)(k+2)...(k+n-k+1)}

     = \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \frac{(n-k)x^{k+2} (1-x)^{n-k-1}}{(k+1)(k+2)} + .... + \frac{ (n-k)(n-k-1)...(1) x^{n + 1}}{(k+1)(k+2)...(n+1)}

     = \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \sum_{j=1}^{n-k}[\frac{(n-k)(n-k-1)...(n-k-j+1) x^{k+j+1} (1-x)^{n-k-j}}{(k+1)...(k+j+1)}]<br />

     \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \frac{1}{(n+1)\binom{n}{k} }\sum_{j=1}^{n-k}[ \binom{n+1}{k+j+1} x^{k+j+1} (1-x)^{n-k-j}]

    Sub j=0

     = \frac{x^{k+1} (1-x)^{n-k}}{k+1}

    therefore , the integral =

     \frac{1}{(n+1)\binom{n}{k}} \sum_{j=0}^{n-k} B_{k+j+1,n+1}(x)
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  8. #8
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by simplependulum View Post
     \int x^k (1-x)^{n-k}~dx

    integration by parts

     = \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \frac{(n-k)x^{k+2} (1-x)^{n-k-1}}{(k+1)(k+2)} + .... + \frac{ (n-k)(n-k-1)...(1) x^{k + n-k + 1}}{(k+1)(k+2)...(k+n-k+1)}

     = \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \frac{(n-k)x^{k+2} (1-x)^{n-k-1}}{(k+1)(k+2)} + .... + \frac{ (n-k)(n-k-1)...(1) x^{n + 1}}{(k+1)(k+2)...(n+1)}

     = \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \sum_{j=1}^{n-k}[\frac{(n-k)(n-k-1)...(n-k-j+1) x^{k+j+1} (1-x)^{n-k-j}}{(k+1)...(k+j+1)}]<br />

     \frac{x^{k+1}(1-x)^{n-k}}{k+1} + \frac{1}{(n+1)\binom{n}{k} }\sum_{j=1}^{n-k}[ \binom{n+1}{k+j+1} x^{k+j+1} (1-x)^{n-k-j}]

    Sub j=0

     = \frac{x^{k+1} (1-x)^{n-k}}{k+1}

    therefore , the integral =

     \frac{1}{(n+1)\binom{n}{k}} \sum_{j=0}^{n-k} B_{k+j+1,n+1}(x)

    Thanks (very)x(10^infinity) much
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  9. #9
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    Quote Originally Posted by Amer View Post
    Thanks (very)x(10^infinity) much
    What is the sign of infinity ? is it negative  - \infty
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