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PostPosted: Sat, 20 Dec 2008 00:05:43 UTC 
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Let $ B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1} \ dx, where a>0, \ b>0. Suppose a>0, \ b> 0, \ c>0 and n \in \mathbb{N} are given such that $ B(a,b)=c. Directly * evaluate B(a,b+n).


* You may only use the definition of B(a,b) and integration techniques!


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PostPosted: Sat, 20 Dec 2008 19:21:54 UTC 
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Does that include cplx. analysis integration techniques?

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PostPosted: Fri, 26 Dec 2008 04:31:21 UTC 
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This problem can be done by using not so simple integration techniques
c.f. Advanced Calculus, Buck, 1965, pp214-219:

wherein: B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} and \Gamma(x+1)=x\Gamma(x) are derived.
Now induction on n would lead to the solution.

My solution uses integration by parts once:

Spoiler:
$ \mbox{ B(a,b) } = \int_0^1 x^{a-1}(1-x)^{b-1}\ dx= c

1=x + (1-x) \ \Rightarrow\ c=B(a,b)= B(a+1,b) + B(a,b+1)

$ So $ \exists k: \ B(a+1,b)=(1-k)c $ and $  B(a,b+1)=kc

Now integration by parts yields:

$ B(a,b+1)=  \int_0^1 x^{a-1}(1-x)^{b}\ dx=\left \frac{x^a(1-x)^b}{a}\right]_0^1 + \frac{b}{a} \int_0^1 x^{a}(1-x)^{b-1}\ dx

B(a,b+1)=\frac{b}{a} B(a+1,b) \iff \ kc=\frac{b}{a} (1-k)c \iff k=\frac{b}{a+b}

Thus B(a,b+1)=\frac{b}{a+b}\cdot c; $ (Generally: $B(p,q+1)=\frac{q}{p+q}B(p,q)\ )

B(a,b+2)=\frac{b+1}{a+b+1}B(a,b+1);\  B(a,b+3)=\frac{b+2}{a+b+2}B(a,b+2);

$ and $  \cdots\ B(a,b+n)=\frac{b+n}{a+b+n} B(a,b+n-1)

$ B(a,b+n) =\frac{(b+n)(b+n-1)\ \cdots\ (b+1)b}{(a+b+n)(a+b+n-1)\ \cdots\ (a+b+1)(a+b)}\cdot c

Funny what the brain conjures while one bikes 1 1/2 hours on a cold day!
Correction:

per commutative's observation the last two lines should have been:

$ and $  \cdots\ B(a,b+n)=\frac{b+n-1}{a+b+n-1} B(a,b+n-1)

$ B(a,b+n) =\frac{(b+n-1)(b+n-2)\ \cdots\ (b+1)b}{(a+b+n-1)(a+b+n-2)\ \cdots\ (a+b+1)(a+b)}\cdot c

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Last edited by robbwrr on Fri, 26 Dec 2008 22:25:54 UTC, edited 1 time in total.

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PostPosted: Fri, 26 Dec 2008 09:56:14 UTC 
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robbwrr wrote:

My solution uses integration by parts once:

Spoiler:
$ \mbox{ B(a,b) } = \int_0^1 x^{a-1}(1-x)^{b-1}\ dx= c

1=x + (1-x) \ \Rightarrow\ c=B(a,b)= B(a+1,b) + B(a,b+1)

$ So $ \exists k: \ B(a+1,b)=(1-k)c $ and $  B(a,b+1)=kc

Now integration by parts yields:

$ B(a,b+1)=  \int_0^1 x^{a-1}(1-x)^{b}\ dx=\left \frac{x^a(1-x)^b}{a}\right]_0^1 + \frac{b}{a} \int_0^1 x^{a}(1-x)^{b-1}\ dx

B(a,b+1)=\frac{b}{a} B(a+1,b) \iff \ kc=\frac{b}{a} (1-k)c \iff k=\frac{b}{a+b}

Thus B(a,b+1)=\frac{b}{a+b}\cdot c; $ (Generally: $B(p,q+1)=\frac{q}{p+q}B(p,q)\ )

B(a,b+2)=\frac{b+2}{a+b+2}B(a,b+1);\  B(a,b+3)=\frac{b+3}{a+b+3}B(a,b+2);
$ and $  \cdots\ B(a,b+n)=\frac{b+n}{a+b+n} B(a,b+n-1)

$ B(a,b+n) =\frac{(b+n)(b+n-1)\ \cdots\ (b+1)b}{(a+b+n)(a+b+n-1)\ \cdots\ (a+b+1)(a+b)}\cdot c

Funny what the brain conjures while one bikes 1 1/2 hours on a cold day!


nice work! except that your final answer is actually B(a,b+n+1) instead of B(a,b+n). that, of course, is not important.


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