# how to integrate this ??

• Aug 6th 2009, 09:39 AM
Amer
how to integrate this ??
if Bernstein polynomials defined by

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

how I can integrate it

this is my work

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

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

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

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

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

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

Thanks very much .
• Aug 6th 2009, 10:00 AM
Bruno J.
Why is there an $\displaystyle x$ in what you say is the answer? Certainly the answer is not independent of the bounds $\displaystyle a,b$. Perhaps that is the indefinite integral?
• Aug 6th 2009, 10:41 AM
Amer
Quote:

Originally Posted by Bruno J.
Why is there an $\displaystyle x$ in what you say is the answer? Certainly the answer is not independent of the bounds $\displaystyle a,b$. Perhaps that is the indefinite integral?

I see

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

from this

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

I need just to know how to integrate

$\displaystyle \int x^k(1-x)^{n-k} dx$
• Aug 6th 2009, 11:27 AM
Bruno J.
Have you tried induction on $\displaystyle n$? It might very well work with the right argument. For instance you have

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

Induction and integration by parts might work from there...
• Aug 6th 2009, 11:47 AM
Amer
Quote:

Originally Posted by Bruno J.
Have you tried induction on $\displaystyle n$? It might very well work with the right argument. For instance you have

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

Induction and integration by parts might work from there...

you mean to begin with

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

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

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

I walk in the right direction right
• Aug 7th 2009, 11:45 AM
Amer
any hints to integrate

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

Thanks
• Aug 9th 2009, 01:56 AM
simplependulum
$\displaystyle \int x^k (1-x)^{n-k}~dx$

integration by parts

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

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

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

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

Sub $\displaystyle j=0$

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

therefore , the integral =

$\displaystyle \frac{1}{(n+1)\binom{n}{k}} \sum_{j=0}^{n-k} B_{k+j+1,n+1}(x)$
• Aug 9th 2009, 08:27 AM
Amer
Quote:

Originally Posted by simplependulum
$\displaystyle \int x^k (1-x)^{n-k}~dx$

integration by parts

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

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

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

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

Sub $\displaystyle j=0$

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

therefore , the integral =

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

Thanks (very)x(10^infinity) much
• Aug 9th 2009, 08:14 PM
simplependulum
Quote:

Originally Posted by Amer
Thanks (very)x(10^infinity) much

What is the sign of infinity ? is it negative $\displaystyle - \infty$ (Rofl)