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![Page 1: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/1.jpg)
Lecture 1 Sujin Khomrutai – 1 / 28
Method of Applied MathLecture 2: Legendre’s Equation and Gamma
function
Sujin Khomrutai, Ph.D.
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Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 2 / 28
Plan
✔ Legendre’s equation.✔ Gamma function✔ Applications
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Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 3 / 28
Definition For a number n, the equation of the form
(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0
is called the Legendre equation of order n.
• n = 0: (1− x2)y′′ − 2xy′ = 0
• n = 1: (1− x2)y′′ − 2xy′ + 2y = 0
• n = 2: (1− x2)y′′ − 2xy′ + 6y = 0
• n = 3: (1− x2)y′′ − 2xy′ + 12y = 0
In most phenomena, n is a non-negative integer. So we restrict tothat n ∈ {0, 1, 2, . . . , }.
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Example 1: (Electrostatics)
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 4 / 28
Example. A two metallic spherical caps are placed so that theupper part stayed at a constant potential 110 V and the lowerpart is grounded. The electrostatic potential at any point in thespace can be calculated by solving the Legendre’s equation.
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Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 5 / 28
Power series method. a = 0 is an ordinary point. Set solutiony to the Legendre’s equation
(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0,
as
y =∞∑
k=0
bkxk
y′ =∞∑
k=0
bk+1(k + 1)xk
y′′ =∞∑
k=0
bk+2(k + 2)(k + 1)xk.
![Page 6: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/6.jpg)
Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 6 / 28
The first term
(1− x2)y′′ = y′′ − x2y′′
=∞∑
k=0
bk+2(k + 2)(k + 1)xk −∞∑
k=0
bk+2(k + 2)(k + 1)xk+2
=∞∑
k=0
bk+2(k + 2)(k + 1)xk −∞∑
j=2
bjj(j − 1)xj
= b2 · 2 · 1 + b3 · 3 · 2x
+∞∑
k=2
[bk+2(k + 2)(k + 1)− bkk(k − 1)]xk
where we have use shifting and splitting.
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Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 7 / 28
The second term
2xy′ = 2x∞∑
k=0
bk+1(k + 1)xk
=∞∑
k=0
2bk+1(k + 1)xk+1
=∞∑
k=1
2bkkxk
Third term
n(n+ 1)y = n(n+ 1)∞∑
k=0
bkxk =
∞∑
k=0
n(n+ 1)bkxk
![Page 8: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/8.jpg)
Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 8 / 28
Thus
b2 · 2 · 1 + b3 · 3 · 2x
+∞∑
k=2
[bk+2(k + 2)(k + 1)− bkk(k − 1)]xk
−∞∑
k=1
2bkkxk +
∞∑
k=0
n(n + 1)bkxk = 0
∴ [2b2 + n(n+ 1)b0] + [6b3 − 2b1 + n(n+ 1)b1]x
+∞∑
k=2
[bk+2(k + 2)(k + 1)− bkk(k − 1)
− 2bkk + n(n+ 1)bk]xk = 0
![Page 9: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/9.jpg)
Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 9 / 28
Finally, we get
[2b2 + n(n+ 1)b0] + [6b3 + (n− 1)(n+ 2)b1]x
+∞∑
k=2
[(k + 2)(k + 1)bk+2 − (n− k)(n+ k + 1)bk]xk
Apply the fact:∑
∞
k=0 ck(x− a)k = 0 ⇔ ck = 0 for all k:
b2 = −n(n+ 1)
2b0
b3 = −(n− 1)(n+ 2)
6b1
bk+2 = −(n− k)(n+ k + 1)
(k + 2)(k + 1)bk ∀ k = 2, 3, . . . .
![Page 10: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/10.jpg)
Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 10 / 28
Let us consider the case n = 3. Recurrence equations are
b2 = −3 · 42
b0, b3 = −2 · 53!
b1
bk+2 = −(3− k)(3 + k + 1)
(k + 2)(k + 1)bk ∀ k ≥ 2
k = 2 : b4 = −1 · 64 · 3b2 =
1 · 6 · 3 · 44!
b0
k = 3 : b5 = −0 · 75 · 4b3 = 0
k = 4 : b6 = −(−1) · 86 · 5 b4 = −(−1) · 8 · 1 · 6 · 3 · 4
6!b0
k = 5 : b7 = −(−2) · 97 · 6 b5 = 0
![Page 11: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/11.jpg)
Series solutions of Legendre’s equation
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 11 / 28
So in the case n = 3, we obtain the solution
y = b0
[
1− 3 · 42!
x2 +1 · 6 · 3 · 4
4!x4 + · · ·
]
+ b1
[
x− 2 · 53!
x3
]
This is the general solution with fundamental solutions
y1 = 1− 3 · 42!
x2 +1 · 6 · 3 · 4
4!x4 + · · · an infinite series
y2 = x− 2 · 53!
x3 = −5
3x3 + x a polynomial degree 3
For a general n ∈ {0, 1, 2, . . .}, the Legendre equation has onepolynomial solution of degree n and an infinite series solution.
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Legendre functions
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 12 / 28
Theorem. For n ∈ {0, 1, 2, . . .}, the general solutions to theLegendre’s equation
(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0
is
y = C1Pn(x) + C2Qn(x)
where Pn is a polynomial of degree n, Qn an infinite series.
• Pn = the Legendre polynomial.
• Qn = the Legendre function of the second kind.
![Page 13: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/13.jpg)
Legendre functions
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 13 / 28
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Legendre functions
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 14 / 28
![Page 15: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/15.jpg)
Rodrigues’ formula
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 15 / 28
The Legendre’s equation with n = 0 is
(1− x2)y′′ − 2xy′ = 0.
A polynomial of degree n = 0 is constant. It can be easily checkthat y = 1 is a solution. So we put
P0(x) = 1.
Rodrigues’ formula If n ∈ {1, 2, 3, . . .} then
Pn(x) =1
2nn!
dn
dxn(x2 − 1)n
![Page 16: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/16.jpg)
Example 2
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 16 / 28
EX. Verify Rodigues’ solution formula for n = 1, 2, 3.
n = 1 : P1(x) =1
2
d
dx(x2 − 1) = x
(1− x2)x′′ − 2x · x′ + 2x = 0
(1− x2) · 0− 2x · 1 + 2x = 0 True
n = 2 : P2(x) =1
22 · 2!d2
dx2(x2 − 1)2
=1
8
d2
dx2(x4 − 2x2 + 1) =
3
2x2 − 1
2
(1− x2)(3
2x2 − 1
2)′′ − 2x(
3
2x2 − 1
2)′ + 6(
3
2x2 − 1
2) = 0
(1− x2) · 3− 2x · 3x+ (9x2 − 3) = 0
(3− 3x2)− 6x2 + (9x2 − 3) = 0 True
![Page 17: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/17.jpg)
Example 2
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 17 / 28
n = 3 : P3(x) =1
23 · 3!d3
dx3(x2 − 1)3
=1
48
d3
dx3(x6 − 3x4 + 3x2 − 1)
=5
2x3 − 3
2x
(1− x2)(5
2x3 − 3
2x)′′ − 2x(
5
2x3 − 3
2x)′ + 12(
5
2x3 − 3
2x) = 0
(1− x2)(15x)− 2x(15
2x2 − 3
2) + (30x3 − 18x) = 0
(15x− 15x3)− (15x3 − 3x) + (30x3 − 18x) = 0 True
![Page 18: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/18.jpg)
Example 3
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 18 / 28
EX. Solve the Legendre’s equation
(1− x2)y′′ − 2xy′ + 2y = 0.
Sol. We get a solution P1(x) = x. Use the reduction of order
Q1(x) = P1(x)
∫
1
(P1(x))2e−
∫−2x
1−x2dxdx
= x
∫
1
x2e− ln(1−x2)dx
= x
∫
1
x2(1− x2)dx =
x
2ln
(
1 + x
1− x
)
− 1
∴ y = C1x+ C2
[
x
2ln
(
1 + x
1− x
)
− 1
]
![Page 19: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/19.jpg)
The Gamma function
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 19 / 28
The factorials n! are
0! = 1! = 1
2! = 2 · 1 = 2
3! = 3 · 2 · 1 = 6,
...
n! = n(n− 1)(n− 2) · · · 3 · 2 · 1
Observe that∫
∞
0
e−tdt =
∫
∞
0
e−ttdt = 1,
∫
∞
0
e−tt2dt = 2 = 2!,
∫
∞
0
e−tt3dt = 6 = 3!.
![Page 20: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/20.jpg)
The Gamma function
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 20 / 28
Definition For x > 0, the integral
∫
∞
0
e−ttx−1 dt is called the Gamma function denoted Γ(x).
If x < 0 and x 6∈ {−1,−2, . . .} we put
Γ(x) =Γ(x+ n)
(x+ n− 1)(x+ n− 2) · · · (x+ 1)x
where n ∈ {1, 2, . . .} satisfies x+ n > 0.
Consecutive product formula
Γ(x+ n) = (x+ n− 1)(x+ n− 2) · · · (x+ 1)xΓ(x).
![Page 21: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/21.jpg)
The Gamma function
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 21 / 28
The graph of Gamma function is as shown below
![Page 22: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/22.jpg)
Properties of Gamma function
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 22 / 28
(1) Γ(1) = 1. For x > 0,
Γ(x+ 1) = xΓ(x)
(2) Generally, for n ∈ {1, 2, . . .} we get
Γ(x+ n) = (x+ n− 1) · · · (x+ 1)xΓ(x).
(3) In particular, if n ∈ {0, 1, 2, . . .} then
Γ(n+ 1) = n!.
![Page 23: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/23.jpg)
Properties of Gamma function
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 23 / 28
Proof. (1) For Γ(1), consider
Γ(1) =
∫
∞
0
e−tt1−1dt =
∫
∞
0
e−tdt
= (−e−t)∣
∣
∣
∞
t=0= 0− (−e0) = 1.
Next, for x > 0 consider
Γ(x+ 1) =
∫
∞
0
e−tt(x+1)−1dt =
∫
∞
0
e−ttxdt
= (−e−ttx)∣
∣
∣
∞
t=0−∫
∞
0
(−e−txtx−1)dt (by parts)
= x
∫
∞
0
e−ttx−1dt = xΓ(x).
![Page 24: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/24.jpg)
Properties of Gamma function
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 24 / 28
(2) For n = 2, we use (1) (x → x+ 1):
Γ(x+ 2) = Γ((x+ 1) + 1) = (x+ 1)Γ(x+ 1),
and then use (1) one more time:
Γ(x+ 1) = xΓ(x).
Thus
Γ(x+ 2) = (x+ 1)xΓ(x)
The argument can be extended to n = 3, n = 4, . . ..
(3) Use x = 1 in (2) and that Γ(1) = 1.
![Page 25: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/25.jpg)
Example 4
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 25 / 28
EX. Evaluate
Γ(1) + Γ(4),Γ(2.8)
Γ(0.8).
![Page 26: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/26.jpg)
Example 5
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 26 / 28
EX. Given that Γ(0.5) =√π. Evaluate
Γ(−2.5).
![Page 27: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/27.jpg)
Example 6
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 27 / 28
EX. Express
1
(ν + 1)(ν + 2)(ν + 3)
as a ratio of the Gamma function.
![Page 28: Method of Applied Mathpioneer.netserv.chula.ac.th/~ksujin/Slide02(ISE,317).pdfLecture1 SujinKhomrutai–1/28 Method of Applied Math Lecture 2: Legendre’s Equation and Gamma function](https://reader034.vdocuments.site/reader034/viewer/2022042403/5f17901b6f849a710f56d2e3/html5/thumbnails/28.jpg)
Example 7
Legendre’s eqn
Legendre’s eqn
EX 1.
Series Sol
Legendre func
Rodrigues’ form
EX 2.
EX 3.
Gamma function
Properties
EX 4.
EX 5.
EX 6.
EX 7.
Lecture 1 Sujin Khomrutai – 28 / 28
EX. Evaluate the integral
∫
∞
0
t1.35e−tdt.
Express the value in terms of the Gamma function.