This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer
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
View Answer
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) \(\Gamma(n+1) = n\Gamma(n)\)
d) \(\Gamma(n) = \int_{0}^{1} log \left({1 \atop y}\right)^{n-1}\)
View Answer
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
View Answer
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 \(\Gamma\left(\frac{1}{2}\right)\)?
a) \(\sqrt{\pi}\)
b) \(\left(\frac{\sqrt{\pi}}{\sqrt{2}}\right)\)
c) \(\left(\frac{\sqrt{\pi}}{2}\right)\)
d) \(\frac{\pi}{2}\)
View Answer
Explanation: \(\Gamma\left(\frac{1}{2}\right) = \int_{0}^{\infty} x^{\frac{-1}{2}} e^{-x}dx\)
= \(\int_{0}^{\infty} e^{-y^2} dy\)
= \(\int_{0}^{\infty} e^{-x^2} dx\)
= \(\Gamma\left(\frac{1}{2}\right)^2 = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} e^{-r^2} rdrd\theta\)
= \(4 * \frac{\pi}{2} * \frac{1}{2}\)
= \(\pi\)
= \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\).
6. Is the given statement true or false?
\(\displaystyle\beta(m, n) = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)}\)
a) True
b) False
View Answer
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, \(\displaystyle\beta(m, n) = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)}\).
7. What is the value of \(\Gamma(5.5)\)?
a) \(\displaystyle\frac{11*9*7*5*3*1*\sqrt{\pi}}{32} \)
b) \(\displaystyle\frac{9*7*5*3*1*\sqrt{\pi}}{32} \)
c) \(\displaystyle\frac{9*7*5*3*1*\sqrt{\pi}}{64} \)
d) \(\displaystyle\frac{11*9*7*5*3*1*\sqrt{\pi}}{64}\)
View Answer
Explanation: \(\Gamma\left(\frac{11}{2}\right) = \frac{9}{2} * \Gamma\left(\frac{9}{2}\right) = \frac{9}{2} * \frac{7}{2} * \Gamma\left(\frac{7}{2}\right) = \frac{9}{2} * \frac{7}{2} * \frac{5}{2} * \Gamma\left(\frac{5}{2}\right)\)
= \(\frac{9}{2} * \frac{7}{2} * \frac{5}{2} * \frac{3}{2} * \Gamma\left(\frac{3}{2}\right)\)
= \(\frac{9}{2} * \frac{7}{2} * \frac{5}{2} * \frac{3}{2} * \frac{1}{2} * \Gamma\left(\frac{1}{2}\right)\)
= \(\displaystyle\frac{9*7*5*3*1*\sqrt{\pi}}{32}\).
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) \(\frac{\pi}{2} \)
View Answer
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, \(\Gamma(\frac{\frac{1}{2}}{2}) = \frac{\sqrt\pi}{2}.\)
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) \(\frac{\Gamma(\frac{1}{4})^2}{\pi} \)
View Answer
Explanation: \(\int_0^{π⁄2}\sqrt{tan(θ)} \, dθ \)
= \(\int_0^{π⁄2}\sqrt{(sin(θ)cos(θ)} \, dθ \)
= \(\frac{1}{2} * \beta(\frac{3}{4}, \frac{3}{4}) \)
= \(\frac{\frac{1}{2} * \Gamma(\frac{3}{4}) * \Gamma(\frac{3}{4}) }{\Gamma(\frac{3}{2}) }\)
= \(\frac{\Gamma(\frac{3}{4})^2}{\sqrt\pi} \).
10. What is the value of \(\int_0^1 \frac{x^2}{\sqrt{(1-x^4 )}} \)?
a) \(\frac{2\sqrt{\pi} \Gamma(\frac{5}{4})}{Γ(\frac{1}{4})} \)
b) \(\frac{2\pi\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \)
c) \(\frac{2\sqrt{\pi} \Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \)
d) \(\frac{2\sqrt{\pi} \Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})} \)
View Answer
Explanation: Substitute \(x^2 = sin(θ)\)
\(2xdx = cos(θ)dθ\)
= \(\int_0^{π⁄2}\frac{sin(θ)}{cos(θ)} \frac{cos(θ)}{2\sqrt{sin(θ)}} dθ \)
= \(\frac{1}{2} * \beta( \frac{3}{4} , \frac{1}{2}) \)
= \(\frac{\frac{1}{2} * \Gamma(\frac{3}{4}) * \Gamma(\frac{1}{2}) }{\Gamma(\frac{5}{4})} \)
= \(\frac{2\sqrt{\pi} \Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \).
11. What is the value of \(\int_0^1 log(y)^8 dy\)?
a) 5!
b) 6!
c) 7!
d) 8!
View Answer
Explanation: \( \Gamma(n) = \int_0^1 log \left(1 \atop y\right)^{n-1} \)
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 \(\int_0^1 log \left(1 \atop y\right)^{9-1} = \Gamma(9) = 8!.\)
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) \(\frac{1}{4} * \Gamma(\frac{1}{4}) \)
View Answer
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. \(\Gamma(m) * \Gamma(1-m) = \frac{\pi}{sin(m\pi)}\). Check if the statement is True or False?
a) True
b) False
View Answer
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) \(\frac{\sqrt{3} \pi}{4} \)
View Answer
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) \(\frac{1}{2logc} \)
View Answer
Explanation: \(\int_0^∞ \frac{1}{c^x} dx \)
= \(\int_0^∞ e^{-xlogc} dx \)
Substitute \(xlogc = t \)
\(logc \,dx = dt \)
\(= \int_0^∞ e^{-t} \frac{dt}{logc} \)
\(= \frac{1}{logc} * \int_0^∞ e^{-t} dt \)
\(=\frac{1}{logc}. \)
Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
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