Ordinary Differential Equations Multiple Choice Questions on “Special Functions -1 (Gamma)”.
1. Which of the following is true?
a) Γ(n+1) = nΓ(n) for any real number
b) Γ(n) = nΓ(n+1) for any real number
c) Γ(n+1) = nΓ(n) for n>1
d) Γ(n) = nΓ(n+1) for n>1
Answer: c
Explanation: Γ(n+1) = n! = n. (n-1)! = n.Γ(n). Hence Γ(n+1) = nΓ(n) for n>1.
2. Γ(n+1) = n! can be used when ____________
a) n is any integer
b) n is a positive integer
c) n is a negative integer
d) n is any real number
Answer: b
Explanation: ( int_{0}^{infty} x^{n} e^{-x}dx )
= ( mid e^{-x} x^{n} mid_{0}^{infty} + n int_{0}^{infty} x^{n-1} e^{-x}dx )
= ( n Gamma(n) ).
3. Which of the following is not a definition of Gamma function?
a) (Gamma(n) = n!)
b) (Gamma(n) = int_{0}^{infty} x^{n-1} e^{-x}dx)
c) (Gamma(n+1) = nGamma(n))
d) (Gamma(n) = int_{0}^{1} log left({1 atop y}right)^{n-1})
Answer: a
Explanation: Each and every option represents the definition of Gamma function except Γ(n) = n! as Γ(n+1) = n! if n is a positive number.
4. Gamma function is said to be as Euler’s integral of second kind.
a) True
b) False
Answer: a
Explanation: Euler’s integral of first kind is nothing but the Beta function and Euler’s integral of second kind is nothing but Gamma function. These integrals were considered by L.Euler.
5. What is the value of (Gammaleft(frac{1}{2}right))?
a) (sqrt{pi})
b) (left(frac{sqrt{pi}}{sqrt{2}}right))
c) (left(frac{sqrt{pi}}{2}right))
d) (frac{pi}{2})
Answer: a
Explanation: (Gammaleft(frac{1}{2}right) = int_{0}^{infty} x^{frac{-1}{2}} e^{-x}dx)
= (int_{0}^{infty} e^{-y^2} dy)
= (int_{0}^{infty} e^{-x^2} dx)
= (Gammaleft(frac{1}{2}right)^2 = int_{0}^{frac{pi}{2}} int_{0}^{infty} e^{-r^2} rdrdtheta)
= (4 * frac{pi}{2} * frac{1}{2})
= (pi)
= (Gammaleft(frac{1}{2}right) = sqrt{pi}).
6. Is the given statement true or false?
(displaystylebeta(m, n) = frac{Gamma(m).Gamma(n)}{Gamma(m+n)})
a) True
b) False
Answer: a
Explanation: We know, (Gamma(n) int_{0}^{infty} x^{n-1} e^{-x}dx.) So, the product of two factorials is:
(Gamma(m).Gamma(n) = int_{0}^{infty} x^{m-1} e^{-x}dx int_{0}^{infty} y^{n-1} e^{-y}dy).
= ( int_{0}^{infty} int_{0}^{infty} x^{m-1} y^{n-1} e^{-x} e^{-y} dxdy )
Now, we do a change of variables where x= uv and y= u(1-v) which implies u varies from 0 to ∞ and v varies from 0 to 1. Jacobian of this gives –u.
( Gamma(m).Gamma(n) = int_{0}^{1} int_{0}^{infty} e^{-u} u^{m-1} v^{m-1} u^{n-1} (1-v)^{n-1} ududv).
= ( Gamma(m + n).beta(m, n) )
Therefore, (displaystylebeta(m, n) = frac{Gamma(m).Gamma(n)}{Gamma(m+n)}).
7. What is the value of (Gamma(5.5))?
a) (displaystylefrac{11*9*7*5*3*1*sqrt{pi}}{32} )
b) (displaystylefrac{9*7*5*3*1*sqrt{pi}}{32} )
c) (displaystylefrac{9*7*5*3*1*sqrt{pi}}{64} )
d) (displaystylefrac{11*9*7*5*3*1*sqrt{pi}}{64})
Answer: b
Explanation: (Gammaleft(frac{11}{2}right) = frac{9}{2} * Gammaleft(frac{9}{2}right) = frac{9}{2} * frac{7}{2} * Gammaleft(frac{7}{2}right) = frac{9}{2} * frac{7}{2} * frac{5}{2} * Gammaleft(frac{5}{2}right))
= (frac{9}{2} * frac{7}{2} * frac{5}{2} * frac{3}{2} * Gammaleft(frac{3}{2}right))
= (frac{9}{2} * frac{7}{2} * frac{5}{2} * frac{3}{2} * frac{1}{2} * Gammaleft(frac{1}{2}right))
= (displaystylefrac{9*7*5*3*1*sqrt{pi}}{32}).
8. What is the value of (int_0^∞ e^{-x^2} dx)?
a) ( sqrt{pi} )
b) (frac{sqrt{pi}}{sqrt{2}} )
c) (frac{sqrt{pi}}{2} )
d) (frac{pi}{2} )
Answer: c
Explanation: Substitute (x^2 = y)
(2xdx = dy)
= (int_0^∞ x^{frac{-1}{2}} e^{-x} dx )
This is of the form of Gamma function. Here, (n-1 = frac{-1}{2} ). Therefore (n = frac{1}{2}.)
Therefore, (Gamma(frac{frac{1}{2}}{2}) = frac{sqrtpi}{2}.)
9. What is the value of the integral (int_0^{π⁄2}sqrt{tan(θ) } , dθ)?
a) (frac{Gamma(frac{3}{4})^2}{sqrt{pi}} )
b) (frac{Gamma(frac{1}{4})^2}{sqrt{pi}} )
c) (frac{Gamma(frac{3}{4})^2}{pi} )
d) (frac{Gamma(frac{1}{4})^2}{pi} )
Answer: a
Explanation: (int_0^{π⁄2}sqrt{tan(θ)} , dθ )
= (int_0^{π⁄2}sqrt{(sin(θ)cos(θ)} , dθ )
= (frac{1}{2} * beta(frac{3}{4}, frac{3}{4}) )
= (frac{frac{1}{2} * Gamma(frac{3}{4}) * Gamma(frac{3}{4}) }{Gamma(frac{3}{2}) })
= (frac{Gamma(frac{3}{4})^2}{sqrtpi} ).
10. What is the value of (int_0^1 frac{x^2}{sqrt{(1-x^4 )}} )?
a) (frac{2sqrt{pi} Gamma(frac{5}{4})}{Γ(frac{1}{4})} )
b) (frac{2piGamma(frac{3}{4})}{Gamma(frac{1}{4})} )
c) (frac{2sqrt{pi} Gamma(frac{3}{4})}{Gamma(frac{1}{4})} )
d) (frac{2sqrt{pi} Gamma(frac{3}{4})}{Gamma(frac{5}{4})} )
Answer: c
Explanation: Substitute (x^2 = sin(θ))
(2xdx = cos(θ)dθ)
= (int_0^{π⁄2}frac{sin(θ)}{cos(θ)} frac{cos(θ)}{2sqrt{sin(θ)}} dθ )
= (frac{1}{2} * beta( frac{3}{4} , frac{1}{2}) )
= (frac{frac{1}{2} * Gamma(frac{3}{4}) * Gamma(frac{1}{2}) }{Gamma(frac{5}{4})} )
= (frac{2sqrt{pi} Gamma(frac{3}{4})}{Gamma(frac{1}{4})} ).
11. What is the value of (int_0^1 log(y)^8 dy)?
a) 5!
b) 6!
c) 7!
d) 8!
Answer: d
Explanation: ( Gamma(n) = int_0^1 log left(1 atop yright)^{n-1} )
Here, the integral is ( int_0^1 log (y)^8 dy) which can also be written as ( int_0^1- log(y)^8 dy ) which is actually (int_0^1 log left(1 atop yright)^{9-1} = Gamma(9) = 8!.)
12. What is the value of ( Gamma(frac{9}{4}))?
a) (frac{5}{4} * frac{1}{4} * Gamma(frac{1}{4}) )
b) (frac{9}{4} * frac{5}{4} * frac{1}{4} * Gamma(frac{1}{4}) )
c) (frac{5}{4} * frac{1}{4} * Gamma(frac{5}{4}) )
d) (frac{1}{4} * Gamma(frac{1}{4}) )
Answer: a
Explanation: (Gamma(frac{9}{4}) = Gamma(1+frac{5}{4}) = frac{5}{4} * Gamma(frac{5}{4}) = frac{5}{4} * Gamma(1+ frac{1}{4}) = frac{5}{4} * frac{1}{4} * Gamma(frac{1}{4}). )
13. (Gamma(m) * Gamma(1-m) = frac{pi}{sin(mpi)}). Check if the statement is True or False?
a) True
b) False
Answer: a
Explanation: From the relation between Beta and Gamma function, we have,
(beta(m, n) = frac{Gamma(m).Gamma(n)}{Gamma(m+n)} )
Let (n = 1 – m )
(frac{Gamma(m).Gamma(1-m)}{Gamma(1)} )
= (beta(m, 1-m))
= (int_0^∞ frac{x^{m-1}}{(1+x)} dx )
= ( frac{pi}{sin(mπ)} ) ( by method of residues).
14. What is the value of (int_0^∞ frac{1}{(1+x^4 )} dx)?
a) (frac{sqrt{2} pi}{4} )
b) (frac{sqrt{3} pi}{6} )
c) (frac{sqrt{2} pi}{6} )
d) (frac{sqrt{3} pi}{4} )
Answer: a
Explanation: Substitute (x^2= tan(θ) )
(2xdx = (sec(θ))^2dθ )
= (frac{1}{2} * int_0^∞ frac{1}{sqrt{sin(θ)cos(θ)}} dθ )
= (frac{1}{4} * beta(frac{1}{4}, frac{3}{4}) )
= (frac{1}{4} * frac{Gamma(1⁄4).Gamma(3⁄4)}{Gamma(1)} )
= (frac{1}{4} * frac{pi}{sin(π/4)} )
= (frac{sqrt{2} pi}{4}.)
15. What is the value of the integral (int_0^∞ frac{1}{c^x} dx)?
a) (frac{1}{logc} )
b) (frac{2}{logc} )
c) (frac{pi}{logc} )
d) (frac{1}{2logc} )
Answer: a
Explanation: (int_0^∞ frac{1}{c^x} dx )
= (int_0^∞ e^{-xlogc} dx )
Substitute (xlogc = t )
(logc ,dx = dt )
(= int_0^∞ e^{-t} frac{dt}{logc} )
(= frac{1}{logc} * int_0^∞ e^{-t} dt )
(=frac{1}{logc}. )
Global Education & Learning Series – Ordinary Differential Equations.
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