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Question-196793




Question Number 196793 by MrGHK last updated on 31/Aug/23
Answered by Mathspace last updated on 02/Sep/23
B(x,y)=((Γ(x)Γ(y))/(Γ(x+y)))  ⇒(∂B/∂x)(x,y)=Γ(y).((Γ^′ (x)Γ(x+y)−Γ(x)Γ^′ (x+y))/(Γ^2 (x+y)))  =Γ(y)((Ψ(x)Γ(x)Γ(x+y)−Γ(x)Ψ(x+y)Γ(x+y))/(Γ^2 (x+y)))  =Γ(y)((Ψ(x)Γ(x)−Γ(x)Ψ(x+y))/(Γ(x+y)))  =((Γ(y)Γ(x))/(Γ(x+y))){Ψ(x)−Ψ(x+y)}⇒  (∂B/∂x)(x,y)  =B(x,y){Ψ(x)−Ψ(x+y)} ⇒  and  (∂^2 B/(∂y∂x))(x,y)  =(∂B/∂y)(Ψ(x)−Ψ(x+y))  +B(x,y){−Ψ^′ (x+y)}  =B(x,y)(Ψ(y)−Ψ(x+y))(Ψ(x)−Ψ(x+y)  −B(x,y)Ψ^′ (x+y)  =B(x,y){(Ψ(x)−Ψ(x+y)(Ψ(y)−Ψ(x+y)−Ψ^′ (x+y)}
$${B}\left({x},{y}\right)=\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$\Rightarrow\frac{\partial{B}}{\partial{x}}\left({x},{y}\right)=\Gamma\left({y}\right).\frac{\Gamma^{'} \left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Gamma^{'} \left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)} \\ $$$$=\Gamma\left({y}\right)\frac{\Psi\left({x}\right)\Gamma\left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Psi\left({x}+{y}\right)\Gamma\left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)} \\ $$$$=\Gamma\left({y}\right)\frac{\Psi\left({x}\right)\Gamma\left({x}\right)−\Gamma\left({x}\right)\Psi\left({x}+{y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$=\frac{\Gamma\left({y}\right)\Gamma\left({x}\right)}{\Gamma\left({x}+{y}\right)}\left\{\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right\}\Rightarrow \\ $$$$\frac{\partial{B}}{\partial{x}}\left({x},{y}\right) \\ $$$$={B}\left({x},{y}\right)\left\{\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right\}\:\Rightarrow \\ $$$${and} \\ $$$$\frac{\partial^{\mathrm{2}} {B}}{\partial{y}\partial{x}}\left({x},{y}\right) \\ $$$$=\frac{\partial{B}}{\partial{y}}\left(\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right) \\ $$$$+{B}\left({x},{y}\right)\left\{−\Psi^{'} \left({x}+{y}\right)\right\} \\ $$$$={B}\left({x},{y}\right)\left(\Psi\left({y}\right)−\Psi\left({x}+{y}\right)\right)\left(\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\right. \\ $$$$−{B}\left({x},{y}\right)\Psi^{'} \left({x}+{y}\right) \\ $$$$={B}\left({x},{y}\right)\left\{\left(\Psi\left({x}\right)−\Psi\left({x}+{y}\right)\left(\Psi\left({y}\right)−\Psi\left({x}+{y}\right)−\Psi^{'} \left({x}+{y}\right)\right\}\right.\right. \\ $$
Commented by MrGHK last updated on 02/Sep/23
Great solution
$$\boldsymbol{{G}}{reat}\:{solution} \\ $$

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