Legendary Polynomial & Bessel Functions MCQ Quiz - Objective Question with Answer for Legendary Polynomial & Bessel Functions - Download Free PDF

Explanation:

From the identity of Bessel functions

\(\rm \int_0^x J_0(t)J_1(x-t)dt\) = J0(x) - cos x

Option (3) is true.

Explanation:

We know that

Pn(x) Qn-2(x) - Qn(x) Pn-2(x) = \((2n-1)x\over n(n-1)\), where Pn(x) and Qn(x) are Legendre’s polynomials of first and second kind respectively.

Putting n = 5 we get

P5(x) Q3(x) - Q5(x) P3(x) = \(9x\over 5\times 4\) = \(\rm\frac{9}{20}x\)

Option (2) is true.

Explanation:

Putting x = 1 in the differential equation

(1 - x²)P_n''(x) - 2xP_n'(x) + n(n + 1)P_n(x) = 0

we get

0 × Pn''(1) - 2Pn'(1) + n(n + 1)Pn(1) = 0

⇒ 2Pn'(1) = n(n + 1)Pn(1) 

⇒ 2Pn'(1) = n(n + 1) (as Pn(1) = 1)

⇒ Pn'(1) = \(\rm \frac{n(n+1)}{2}\)

Option (3) is true.

Formula used:

\(\frac{d}{dx}[x^nJ_n(x)]\ =\ x^nJ_{n-1}(x)\)

This can be also written as

∫ xⁿJ_n-1(x)dx = xⁿJ_n(x)

Calculation:

We know that, from recurrence relations 

∫ xnJn-1(x)dx = xnJn(x)

Put n = 2

∫ x2J1(x)dx = x2J2(x) + C

Additional Information

Bessel’s Equation:

The differential equation, recurrence relations 

\(x^2\frac{d^2y}{dx^2}\ +\ x\frac{dy}{dx}\ +\ (x^2\ -\ n^2)y\ =\ 0\)

known as Bessel’s equation of order n and its particular solutions are called Bessel’s functions.

Formula used:

\(\frac{d}{dx}[x^nJ_n(x)]\ =\ x^nJ_{n-1}(x)\)

This can be also written as

∫ xⁿJ_n-1(x)dx = xⁿJ_n(x)

Calculation:

We know that, from recurrence relations 

∫ xnJn-1(x)dx = xnJn(x)

Put n = 2

∫ x2J1(x)dx = x2J2(x) + C

Additional Information

Bessel’s Equation:

The differential equation, recurrence relations 

\(x^2\frac{d^2y}{dx^2}\ +\ x\frac{dy}{dx}\ +\ (x^2\ -\ n^2)y\ =\ 0\)

known as Bessel’s equation of order n and its particular solutions are called Bessel’s functions.

Explanation:

From the identity of Bessel functions

\(\rm \int_0^x J_0(t)J_1(x-t)dt\) = J0(x) - cos x

Option (3) is true.

Explanation:

We know that

Pn(x) Qn-2(x) - Qn(x) Pn-2(x) = \((2n-1)x\over n(n-1)\), where Pn(x) and Qn(x) are Legendre’s polynomials of first and second kind respectively.

Putting n = 5 we get

P5(x) Q3(x) - Q5(x) P3(x) = \(9x\over 5\times 4\) = \(\rm\frac{9}{20}x\)

Option (2) is true.

Explanation:

Putting x = 1 in the differential equation

(1 - x²)P_n''(x) - 2xP_n'(x) + n(n + 1)P_n(x) = 0

we get

0 × Pn''(1) - 2Pn'(1) + n(n + 1)Pn(1) = 0

⇒ 2Pn'(1) = n(n + 1)Pn(1) 

⇒ 2Pn'(1) = n(n + 1) (as Pn(1) = 1)

⇒ Pn'(1) = \(\rm \frac{n(n+1)}{2}\)

Option (3) is true.

Concept Used:-

Beta function, or the Euler integral of first type, is a different type of function which is closely related to the gamma function and to binomial coefficients. The beta function for binomial coefficients m and n is given as,

\(\begin{aligned} & \beta(m, n)=\int_0^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x \ \ \ \text{or}\\ & \beta(m, n)=\int_0^{\infty} \frac{x^{n-1}}{(1+x)^{m+n}} d x \\ \end{aligned}\)

Explanation:-

We have to find the value of integral \(\int_{\rm{0}}^\infty {\frac{{{{\rm{x}}^{{\rm{m}}\,{\rm{ - }}\,{\rm{1}}}}\,{\rm{ + }}\,{{\rm{x}}^{{\rm{n - 1}}}}}}{{{{\left( {{\rm{1 + x}}} \right)}^{{\rm{m + n}}}}}}} {\rm{dx}}\).

We know that,

\(\begin{aligned} & \beta(m, n)=\int_0^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x \ \ \ \ .....(1)\\ & \beta(m, n)=\int_0^{\infty} \frac{x^{n-1}}{(1+x)^{m+n}} d x \ \ \ \ .....(2)\\ \end{aligned}\)

On adding both the equations, we get 

\(\begin{aligned} &2 \beta(m, n)=\int_0^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x +\int_0^{\infty} \frac{x^{n-1}}{(1+x)^{m+n}} d x \\ &2 \beta(m, n)=\int_0^{\infty} \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} d x \\ \end{aligned}\)

\(\begin{aligned} & \Rightarrow \int_1^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x=\int_1^0 \frac{\left(\frac{1}{t}\right)^{m-1}}{\left(1+\frac{1}{t}\right)^{m+n}}\left(\frac{-d t}{t^2}\right) \\ & \Rightarrow \int_1^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x=-\int_1^0 \frac{1}{t^{m-1}} \cdot \frac{t^{m+n}}{(1+t)^{m+n}} \cdot \frac{d t}{t^2} \\ & \Rightarrow \int_1^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x=\int_0^1 \frac{t^{n-1}}{(1+t)^{m+n}} d t \\ & \Rightarrow \int_1^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} d x=\int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}} d x\ \ \ \ \ \ .....(1) \end{aligned} \)

Now a beta function can be given as,

\(\beta(m, n)=\frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}\)

Put this value in equation (1),

\(\begin{aligned} & \Rightarrow 2\beta(m, n)=\int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} d x \\ & \Rightarrow \frac{2\Gamma(m) \Gamma(n)}{\Gamma(m+n)}=\int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} d x \end{aligned}\)

So, the correct option is 2.