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- Thread starter jacks
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- Jan 26, 2012

- 66

This does not make sense because on the one hand I assume x is meant to be continuous over the interval [0, 1] and yet on the other hand the combinatorial as you have defined it is only valid for integer values of x.for a non nagative integer $n$, If $\displaystyle I_{n}=\int_{0}^{1}\binom{x}{n}dx$, then $I_{n}=$

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$

- Jan 26, 2012

- 890

May be for real \(x\) and \(r\) a natural number he intends:This does not make sense because on the one hand I assume x is meant to be continuous over the interval [0, 1] and yet on the other hand the combinatorial as you have defined it is only valid for integer values of x.

\[{x \choose r}=\frac{\Gamma(x+1)}{r!\;\Gamma(x-r+1)}\]

Though my money is on asking the wrong question.

CB

- Jan 26, 2012

- 890

Can you please post the original, or full question? As MrF points out as asked this makes no sense so we suspect this is not the full or actual question.for a non nagative integer $n$, If $\displaystyle I_{n}=\int_{0}^{1}\binom{x}{n}dx$, then $I_{n}=$

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$

CB

- Feb 13, 2012

- 1,704

http://mathworld.wolfram.com/BinomialCoefficient.html

... the definition of the factorial function as...

$\displaystyle z!=\int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... allows the definition of binomial coefficient as...

$\displaystyle \binom {x}{y}= \frac{x!}{y!\ (x-y)!}$

... where x and y are, in most general case, complex numbers...

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

Which is the same thing as replacing them by gamma functions.

http://mathworld.wolfram.com/BinomialCoefficient.html

... the definition of the factorial function as...

$\displaystyle z!=\int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... allows the definition of binomial coefficient as...

$\displaystyle \binom {x}{y}= \frac{x!}{y!\ (x-y)!}$

... where x and y are, in most general case, complex numbers...

Kind regards

$\chi$ $\sigma$

CB

\( \binom{x}{n} = \frac{x (x-1) (x-2) \cdots (x-n+1)} {n!} \)

See "Binomial coefficients as polynomials" in

http://en.wikipedia.org/wiki/Binomial_coefficient

- Feb 13, 2012

- 1,704

$\displaystyle (1+t)^{x}= \sum_{n=0}^{\infty} \binom{x}{n}\ t^{n}$ (1)

... and then we use (1) to arrive to the identity...

$\displaystyle \int_{0}^{1} (1+t)^{x}\ dx= \frac{t}{\ln (1+t)} = \sum_{n=0}^{\infty} t^{n}\ \int_{0}^{1} \binom{x}{n}\ dx$ (2)

But the (2) is the McLaurin expansion of $\displaystyle \frac{t}{\ln (1+t)}$ so that is...

$\displaystyle I_{n}=\int_{0}^{1} \binom{x}{n}\ dx= \frac{1}{n!}\ \lim_{t \rightarrow 0} \frac{d^{n}}{d x^{n}}\ \frac{t}{\ln (1+t)}$ (3)

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

$\displaystyle \frac{\ln (1+t)}{t} = 1 -\frac{t}{2}+\frac{t^{2}}{3}-...$ (1)

... a comfortable way to compute them is to impose the identity...

$\displaystyle (I_{0}+I_{1}\ t +I_{2}\ t^{2}+...)\ (1 -\frac{t}{2}+\frac{t^{2}}{3}-...)=1$ (2)

... and from (2) we derive recursively...

$\displaystyle I_{0}=1$

$\displaystyle I_{1}=\frac{I_{0}}{2}=\frac{1}{2}$

$\displaystyle I_{2}=\frac{I_{1}}{2}-\frac{I_{0}}{3}=-\frac{1}{12}$

$\displaystyle I_{3}=\frac{I_{2}}{2}-\frac{I_{1}}{3}+\frac{I_{0}}{4}=\frac{1}{24}$

... and so one...

Kind regards

$\chi$ $\sigma$