Gamma Funk'tion

Algebra Level 1

Evaluate

Γ ( 16 3 ) Γ ( 10 3 ) , \dfrac{\Gamma\left(\frac {16}{3}\right)}{\Gamma\left(\frac{10}{3}\right)},

where Γ ( z ) \Gamma (z) denotes the gamma function.

If the value of the above expression can be expressed in the form of a b \frac{a}{b} , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 139.

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3 solutions

Rohit Udaiwal
Dec 5, 2015

We use here the recursive formula, Γ ( z ) = ( z 1 ) Γ ( z 1 ) . \Gamma(z)=(z-1)\Gamma (z-1). Now we have, Γ 16 3 Γ 10 3 = ( 16 3 1 ) Γ ( 16 3 1 ) Γ 10 3 = 13 3 Γ 13 3 Γ 10 3 = 13 3 ( 13 3 1 ) Γ ( 13 3 1 ) Γ 10 3 = 13 3 10 3 Γ 10 3 Γ 10 3 = 130 9 . \dfrac{\Gamma \frac{16}{3}}{\Gamma\frac{10}{3}} \\ =\dfrac{(\frac {16}{3}-1)\Gamma(\frac {16}{3}-1)}{\Gamma\frac{10}{3}} \\ =\dfrac{13}{3}\dfrac{\Gamma\frac{13}{3}}{\Gamma\frac{10}{3}} \\ =\dfrac{13}{3}\dfrac{(\frac {13}{3}-1)\Gamma (\frac{13}{3}-1)}{\Gamma\frac{10}{3}} \\ =\dfrac {13}{3} \cdot \dfrac{10}{3} \cdot \dfrac{\Gamma\frac{10}{3}}{\Gamma\frac{10}{3}} \\ =\dfrac{130}{9}.

Please do change the subject of the problem to Calculus.

Swapnil Das - 5 years, 6 months ago
Otto Bretscher
Dec 5, 2015

Γ ( z ) = ( z 1 ) Γ ( z 1 ) = ( z 1 ) ( z 2 ) Γ ( z 2 ) \Gamma(z)=(z-1)\Gamma(z-1)=(z-1)(z-2)\Gamma(z-2) so Γ ( z ) Γ ( z 2 ) = ( z 1 ) ( z 2 ) \frac{\Gamma(z)}{\Gamma(z-2)}=(z-1)(z-2)

For z = 16 3 z=\frac{16}{3} this ratio is ( 16 3 1 ) ( 16 3 2 ) = 130 9 (\frac{16}{3}-1)(\frac{16}{3}-2)=\frac{130}{9} so that the sum we seek is 139 \boxed{139}

Moderator note:

What can we do if the difference of the variables was not an integer? In such a case, must we evaluate the Gamma function?

Gia Hoàng Phạm
Nov 23, 2018

Γ ( 16 3 ) Γ ( 10 3 ) = 13 3 × Γ ( 13 3 ) Γ ( 10 3 ) = 10 9 × 13 9 × Γ ( 10 3 ) Γ ( 10 3 ) = 130 9 \frac{\Gamma \left(\frac{16}{3} \right)}{\Gamma \left(\frac{10}{3} \right)}=\frac{\frac{13}{3} \times \Gamma \left(\frac{13}{3} \right)}{\Gamma \left(\frac{10}{3} \right)}=\frac{\frac{10}{9} \times \frac{13}{9} \times \Gamma \left(\frac{10}{3} \right)}{\Gamma \left(\frac{10}{3} \right)}=\frac{130}{9}

130 + 9 = 139 130+9=\boxed{\large{139}}

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