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Question-127508

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Question Number 127508 by mohammad17 last updated on 30/Dec/20
Commented by mohammad17 last updated on 30/Dec/20
such that β beta function
$${such}\:{that}\:\beta\:{beta}\:{function} \\ $$
Answered by Ar Brandon last updated on 30/Dec/20
β(a,b)β(a+b,c) =((Γ(a)Γ(b))/(Γ(a+b)))∙((Γ(a+b)Γ(c))/(Γ(a+b+c))) =((Γ(a)Γ(b)Γ(c))/(Γ(a+b+c))) =((Γ(a)Γ(b+c))/(Γ(a+b+c)))∙((Γ(b)Γ(c))/(Γ(b+c))) =β(a,b+c)β(b,c)
$$\beta\left(\mathrm{a},\mathrm{b}\right)\beta\left(\mathrm{a}+\mathrm{b},\mathrm{c}\right) \\ $$$$=\frac{\Gamma\left(\mathrm{a}\right)\Gamma\left(\mathrm{b}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}\right)}\centerdot\frac{\Gamma\left(\mathrm{a}+\mathrm{b}\right)\Gamma\left(\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$=\frac{\Gamma\left(\mathrm{a}\right)\Gamma\left(\mathrm{b}\right)\Gamma\left(\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$=\frac{\Gamma\left(\mathrm{a}\right)\Gamma\left(\mathrm{b}+\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}\centerdot\frac{\Gamma\left(\mathrm{b}\right)\Gamma\left(\mathrm{c}\right)}{\Gamma\left(\mathrm{b}+\mathrm{c}\right)} \\ $$$$=\beta\left(\mathrm{a},\mathrm{b}+\mathrm{c}\right)\beta\left(\mathrm{b},\mathrm{c}\right) \\ $$
Commented by mohammad17 last updated on 30/Dec/20
thank you sir can you help me in all question
$${thank}\:{you}\:{sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{all}\:{question} \\ $$
Answered by Ar Brandon last updated on 30/Dec/20
β(a,b)β(a+b,c) =((Γ(a)Γ(b))/(Γ(a+b)))∙((Γ(a+b)Γ(c))/(Γ(a+b+c))) =((Γ(a)Γ(b)Γ(c))/(Γ(a+b+c))) =((Γ(b)Γ(a+c))/(Γ(a+b+c)))∙((Γ(a)Γ(c))/(Γ(a+c))) =β(a,c)β(b,c+a)
$$\beta\left(\mathrm{a},\mathrm{b}\right)\beta\left(\mathrm{a}+\mathrm{b},\mathrm{c}\right) \\ $$$$=\frac{\Gamma\left(\mathrm{a}\right)\Gamma\left(\mathrm{b}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}\right)}\centerdot\frac{\Gamma\left(\mathrm{a}+\mathrm{b}\right)\Gamma\left(\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$=\frac{\Gamma\left(\mathrm{a}\right)\Gamma\left(\mathrm{b}\right)\Gamma\left(\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$=\frac{\Gamma\left(\mathrm{b}\right)\Gamma\left(\mathrm{a}+\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}\centerdot\frac{\Gamma\left(\mathrm{a}\right)\Gamma\left(\mathrm{c}\right)}{\Gamma\left(\mathrm{a}+\mathrm{c}\right)} \\ $$$$=\beta\left(\mathrm{a},\mathrm{c}\right)\beta\left(\mathrm{b},\mathrm{c}+\mathrm{a}\right) \\ $$

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