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Question Number 40891 by abdo.msup.com last updated on 28/Jul/18
let x>0 and y>0 and B(x,y) =∫_0 ^1 t^(x−1) (1−t)^(y−1) dt 1)prove that B(x,y)=B(y,x) 2)B(x+1,y)=(x/y) B(x,y+1) 3)B(x+1,y)=(x/(x+y))B(x,y) 4)B(x,n+1)=((n!)/(x(x+1)....(x+n))) 5)B(n,p) = (1/((n+p−1)C_(n+p−2) ^(p−1) ))
letx>0andy>0andB(x,y)=∫01tx−1(1−t)y−1dt1)provethatB(x,y)=B(y,x)2)B(x+1,y)=xyB(x,y+1)3)B(x+1,y)=xx+yB(x,y)4)B(x,n+1)=n!x(x+1)….(x+n)5)B(n,p)=1(n+p−1)Cn+p−2p−1
Commented by maxmathsup by imad last updated on 03/Aug/18
1) B(x,y) = ∫_0 ^1 t^(x−1) (1−t)^(y−1) dt =_(1−t=u) ∫_0 ^1 (1−u)^(x−1) u^(y−1) du=B(y,x) 2)B(x+1,y) = ∫_0 ^1 t^x (1−t)^(y−1) dt by parts u=t^x and v^′ =(1−t)^(y−1) B(x+1,y) =[−(1/y)(1−t)^y t^x ]_0 ^1 +∫_0 ^1 xt^(x−1) (1/y)(1−t)^y dt =(x/y) ∫_0 ^1 t^(x−1) (1−t)^y dt = (x/y) B(x,y+1)
1)B(x,y)=∫01tx−1(1−t)y−1dt=1−t=u∫01(1−u)x−1uy−1du=B(y,x)2)B(x+1,y)=∫01tx(1−t)y−1dtbypartsu=txandv′=(1−t)y−1B(x+1,y)=[−1y(1−t)ytx]01+∫01xtx−11y(1−t)ydt=xy∫01tx−1(1−t)ydt=xyB(x,y+1)

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