dr. nirav vyas legendres function.pdf
TRANSCRIPT
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Function
N. B. Vyas
Department of MathematicsAtmiya Institute of Technology and Science
Department of Mathematics
N. B. Vyas Legendre’s Function
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
The differential equation
N. B. Vyas Legendre’s Function
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
The differential equation
(1−x 2)y −2xy + n (n + 1) y = 0
N. B. Vyas Legendre’s Function
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
The differential equation
(1−x 2)y −2xy + n (n + 1) y = 0is called Legendre’s differential equation ,n is real constant
N. B. Vyas Legendre’s Function
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Polynomials:⇒
P 0(x ) = 1
N. B. Vyas Legendre’s Function
L d ’ P l i l
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Polynomials:⇒
P 0(x ) = 1
⇒ P 1(x ) = x
N. B. Vyas Legendre’s Function
L d ’ P l i l
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Polynomials:⇒
P 0(x ) = 1
⇒ P 1(x ) = x
⇒ P 2(x ) =
12(3x 2 −1)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
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Legendre s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Polynomials:⇒
P 0(x ) = 1
⇒ P 1(x ) = x
⇒ P 2(x ) =
12(3x 2 −1)
⇒ P 3(x ) =
12
(5x 3 −3x )
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
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Legendre s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Polynomials:⇒
P 0(x ) = 1
⇒ P 1(x ) = x
⇒ P 2(x ) =
12(3x 2 −1)
⇒ P 3(x ) =
12
(5x 3 −3x )
⇒ P 4(x ) =
18(35x 3 −30x 2 + 3)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
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Legendre s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Legendre’s Polynomials:⇒
P 0(x ) = 1
⇒ P 1(x ) = x
⇒ P 2(x ) =
12(3x 2 −1)
⇒ P 3(x ) =
12
(5x 3 −3x )
⇒ P 4(x ) =
18(35x 3 −30x 2 + 3)
⇒ P 5(x ) =
18
(63x 5 −70x 3 + 15 x )
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
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Legendre s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Ex.1 Express f (x ) in terms of Legendre’s polynomials wheref (x ) = x3 + 2x 2 −x −3.
N. B. Vyas Legendre’s Function
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Legendre’s Polynomials
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Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
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Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
⇒ P 1(x ) = x∴ x = P
1(x )
N. B. Vyas Legendre’s Function
Legendre’s Polynomialsl f d l l
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Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
⇒ P 1(x ) = x∴ x = P
1(x )
⇒ P 3(x ) =
12
(5x 3 −3x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsE l f L d ’ P l i l
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Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
⇒ P 1(x ) = x∴ x = P 1(x )
⇒ P 3(x ) =
12
(5x 3 −3x )
∴ 2P 3(x ) = (5 x 3 −3x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
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Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
⇒ P 1(x ) = x∴ x = P 1(x )
⇒ P 3(x ) =
12
(5x 3 −3x )
∴ 2P 3(x ) = (5 x 3 −3x )∴ 2P 3(x ) + 3 x = 5x 3
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
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Examples of Legendre s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
⇒ P 1(x ) = x∴ x = P 1(x )
⇒ P 3(x ) =
12
(5x 3 −3x )
∴ 2P 3(x ) = (5 x 3 −3x )∴ 2P 3(x ) + 3 x = 5x 3
∴ 2P 3(x ) + 3 P 1(x ) = 5 x 3 {∵ x = P 1(x )}
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
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Examples of Legendre s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:⇒
P 0(x ) = 1∴ 1 = P 0(x )
⇒ P 1(x ) = x∴ x = P 1(x )
⇒ P 3(x ) =
12
(5x 3 −3x )
∴ 2P 3(x ) = (5 x 3 −3x )∴ 2P 3(x ) + 3 x = 5x 3
∴ 2P 3(x ) + 3 P 1(x ) = 5 x 3 {∵ x = P 1(x )}∴ x3 =
25
P 3(x ) + 35
P 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
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Examples of Legendre s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
⇒ P 2(x ) = 1
2(3x 2 −1)
N. B. Vyas Legendre’s Function
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
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p g yGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
⇒ P 2(x ) = 1
2(3x 2 −1)
∴ 2P 2(x ) = (3 x 2 −1)∴ 2P 2(x ) + 1 = 3 x 2
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
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Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
⇒ P 2(x ) = 1
2(3x 2 −1)
∴ 2P 2(x ) = (3 x 2 −1)∴ 2P 2(x ) + 1 = 3 x 2
∴ 2P 2(x ) + P 0(x ) = 3 x 2 {∵ 1 = P 0(x )}
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
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Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
⇒ P 2(x ) = 1
2(3x 2 −1)
∴ 2P 2(x ) = (3 x 2 −1)∴ 2P 2(x ) + 1 = 3 x 2
∴ 2P 2(x ) + P 0(x ) = 3 x 2 {∵ 1 = P 0(x )}∴ x2 =
23
P 2(x ) + 13
P 0(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
G i F i f P ( )
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Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
⇒ P 2(x ) = 1
2(3x 2 −1)
∴ 2P 2(x ) = (3 x 2 −1)∴ 2P 2(x ) + 1 = 3 x 2
∴ 2P 2(x ) + P 0(x ) = 3 x 2 {∵ 1 = P 0(x )}∴ x2 =
23
P 2(x ) + 13
P 0(x )
Now, f (x ) = x3 + 2 x 2 −x −3
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
G ti g F ti f P ( x )
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Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
⇒ P 2(x ) = 1
2(3x 2 −1)
∴ 2P 2(x ) = (3 x 2 −1)∴ 2P 2(x ) + 1 = 3 x 2
∴ 2P 2(x ) + P 0(x ) = 3 x2
{∵ 1 = P 0(x )}∴ x2 =
23
P 2(x ) + 13
P 0(x )
Now, f (x ) = x3 + 2 x 2 −x −3f (x ) = x3 + 2 x 2
−x
−3
= 25
P 3(x ) + 35
P 1(x ) + 43
P 2(x ) + 23
P 0(x ) −P 1(x ) −3P 0(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P ( x )
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Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Ex.2 Express x3
−5x 2
+ 6x
+ 1 interms of Legendre’s polynomial.
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P ( x )
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Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Ex.3 Express 4x 3
−2x 2
−3x
+ 8 interms of Legendre’s polynomial.
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
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Generating Function for n ( )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Generating Function for P n (x )
∞
n =0P n (x )t n = 1√ 1 −2xt + t2
= (1 −2xt
+ t2
)−12
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
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Ge e at g u ct o o n ( )Rodrigue’s Formula
Recurrence Relations for P n ( x )
The function (1
−2xt + t2)−1
2 iscalled Generating function of Legendre’s polynomial P n (x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
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g ( )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Ex Show that
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
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( )Rodrigue’s Formula
Recurrence Relations for P n ( x )
Ex Show that
(i)P n (1) = 1
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
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Rodrigue’s FormulaRecurrence Relations for P n ( x )
Ex Show that
(i)P n (1) = 1(ii)P n (−1) = (−1)n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
d ’ l
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Rodrigue’s FormulaRecurrence Relations for P n ( x )
Ex Show that
(i)P n (1) = 1(ii)P n (−1) = (−1)n
(iii)P n (
−x) = (
−1)n P n (x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
R d i ’ F l
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Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:
(i) We have∞
n =0
P n (x )t n = (1 −2xt + t2)− 12
Putting x = 1 in eq(1), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
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Rodrigue’s FormulaRecurrence Relations for P n ( x )
Solution:
(i) We have∞
n =0
P n (x )t n = (1 −2xt + t2)− 12
Putting x = 1 in eq(1), we get∞
n =0
P n (1) t n = (1
−2t + t2)− 1
2 = (1
−t )− 1
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
![Page 37: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/37.jpg)
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Rodrigue s FormulaRecurrence Relations for P n ( x )
Solution:
(i) We have∞
n =0
P n (x )t n = (1 −2xt + t2)− 12
Putting x = 1 in eq(1), we get∞
n =0
P n (1) t n = (1
−2t + t2)− 1
2 = (1
−t )− 1
∴
∞
n =0
P n (1) t n = 11 −t
= 1 + t + t2 + t3 + ...
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
![Page 38: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/38.jpg)
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Rodrigue s FormulaRecurrence Relations for P n ( x )
Solution:
(i) We have∞
n =0
P n (x )t n = (1 −2xt + t2)− 12
Putting x = 1 in eq(1), we get∞
n =0
P n (1) t n = (1
−2t + t2)− 1
2 = (1
−t )− 1
∴
∞
n =0
P n (1) t n = 11 −t
= 1 + t + t2 + t3 + ...
∴
∞
n =0P n (1) t n =
∞
n =0t n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
![Page 39: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/39.jpg)
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Rodrigue s FormulaRecurrence Relations for P n ( x )
Solution:
(i) We have∞
n =0
P n (x )t n = (1 −2xt + t2)− 12
Putting x = 1 in eq(1), we get∞
n =0
P n (1) t n = (1
−2t + t2)− 1
2 = (1
−t )− 1
∴
∞
n =0
P n (1) t n = 11 −t
= 1 + t + t2 + t3 + ...
∴
∞
n =0P n (1) t n =
∞
n =0t n
Comparing the coefficient of tn both the sides, we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
![Page 40: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/40.jpg)
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gRecurrence Relations for P n ( x )
Solution:
(i) We have∞
n =0
P n (x )t n = (1 −2xt + t2)− 12
Putting x = 1 in eq(1), we get∞
n =0
P n (1) t n = (1
−2t + t2)− 1
2 = (1
−t )− 1
∴
∞
n =0
P n (1) t n = 11 −t
= 1 + t + t2 + t3 + ...
∴
∞
n =0P n (1) t
n
=
∞
n =0t
n
Comparing the coefficient of tn both the sides, we getP n (1) = 1
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
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Recurrence Relations for P n ( x )
(ii) Putting x = −1 in eq(1), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
![Page 42: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/42.jpg)
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Recurrence Relations for P n ( x )
(ii) Putting x = −1 in eq(1), we get∞
n =0
P n (−1)t n = (1 + 2 t + t2)− 12 = (1 + t)− 1
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s Formula
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Recurrence Relations for P n ( x )
(ii) Putting x = −1 in eq(1), we get∞
n =0
P n (−1)t n = (1 + 2 t + t2)− 12 = (1 + t)− 1
∴
∞
n =0P n (−1)t
n
= 11 + t = 1 −t + t2 −t 3 + ...
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaR R l i f P ( )
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Recurrence Relations for P n ( x )
(ii) Putting x = −1 in eq(1), we get∞
n =0
P n (−1)t n = (1 + 2 t + t2)− 12 = (1 + t)− 1
∴
∞
n =0P n (−1)t
n
= 11 + t = 1 −t + t2 −t 3 + ...
∴
∞
n =0
P n (−1)t n =∞
n =0
(−1)n t n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaR R l ti f P ( x )
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Recurrence Relations for P n ( x )
(ii) Putting x = −1 in eq(1), we get∞
n =0
P n (−1)t n = (1 + 2 t + t2)− 12 = (1 + t)− 1
∴
∞
n =0P n (−1)t
n
= 11 + t = 1 −t + t2 −t 3 + ...
∴
∞
n =0
P n (−1)t n =∞
n =0
(−1)n t n
Comparing coefficients of tn
, we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P ( x )
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Recurrence Relations for P n ( x )
(ii) Putting x = −1 in eq(1), we get∞
n =0
P n (−1)t n = (1 + 2 t + t2)− 12 = (1 + t)− 1
∴
∞
n =0P n (−1)t
n
= 11 + t = 1 −t + t2 −t 3 + ...
∴
∞
n =0
P n (−1)t n =∞
n =0
(−1)n t n
Comparing coefficients of tn
, we getP n (−1) = ( −1)n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P ( x )
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Recurrence Relations for P n ( x )
(iii) Now replacing x by
−x in eq(1), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
![Page 48: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/48.jpg)
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Recurrence Relations for n ( )
(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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n ( )
(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
Now, replacing t by −t in eq(1), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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( )
(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
Now, replacing t by −t in eq(1), we get∞
n =0P n (x )(−t )n = (1 + 2 xt + t2)−
12 —(b)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
Now, replacing t by −t in eq(1), we get∞
n =0P n (x )(−t )n = (1 + 2 xt + t2)−
12 —(b)
from equation (a) and (b)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
Now, replacing t by −t in eq(1), we get∞
n =0P n (x )(−t )n = (1 + 2 xt + t2)−
12 —(b)
from equation (a) and (b)∞
n =0
P n (
−x )( t )n =
∞
n =0
P n (x )(
−1)n (t )n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
Now, replacing t by −t in eq(1), we get∞
n =0P n (x )(−t )n = (1 + 2 xt + t2)−
12 —(b)
from equation (a) and (b)∞
n =0
P n (
−x )( t )n =
∞
n =0
P n (x )(
−1)n (t )n
Comparing the coefficients of tn , both sides, we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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(iii) Now replacing x by
−x in eq(1), we get
∞
n =0
P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)
Now, replacing t by −t in eq(1), we get∞
n =0P n (x )(−t )n = (1 + 2 xt + t2)−
12 —(b)
from equation (a) and (b)∞
n =0
P n (
−x )( t )n =
∞
n =0
P n (x )(
−1)n (t )n
Comparing the coefficients of tn , both sides, we getP n (−x ) = ( −1)n P n (x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Rodrigue’s Formula
P n (x ) = 12n n !
dn
dx n [(x 2 −1)n ]
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
Differentiating wit respect to x
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
Differentiating wit respect to x∴ y1 = n (x 2
−1)n − 1(2x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
Differentiating wit respect to x∴ y1 = n (x 2
−1)n − 1(2x )
∴ y1 = 2nx (x 2 −1)n
(x 2 −1) =
2nxyx 2 −1
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
Differentiating wit respect to x∴ y1 = n (x 2
−1)n − 1(2x )
∴ y1 = 2nx (x 2 −1)n
(x 2 −1) =
2nxyx 2 −1
∴ (x 2 −1)y1 = 2nxy
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
Differentiating wit respect to x∴ y1 = n (x 2
−1)n − 1(2x )
∴ y1 = 2nx (x 2 −1)n
(x 2 −1) =
2nxyx 2 −1
∴ (x 2 −1)y1 = 2nxy
Differentiating with respect to x,
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Proof:Let y = ( x 2 −1)n
Differentiating wit respect to x∴ y1 = n (x 2
−1)n − 1(2x )
∴ y1 = 2nx (x 2 −1)n
(x 2 −1) = 2nxy
x 2 −1∴ (x 2 −1)y1 = 2nxy
Differentiating with respect to x,(x 2 −1)y2 +2 xy 1 = 2nxy 1 + 2ny
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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dn
dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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dn
dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V
→ dn
dx n ((x 2−1)y2) = nC 0(x 2−1)yn +2 + nC 1(2x )yn +1 + nC 2(2)yn
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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dn
dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V
→ dn
dx n ((x 2−1)y2) = nC 0(x 2−1)yn +2 + nC 1(2x )yn +1 + nC 2(2)yn
→ dn
dx n(2xy
1) = nC
0(2x )yn
+1 + nC
1(2)yn
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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dn
dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V
→ dn
dx n ((x 2−1)y2) = nC 0(x 2−1)yn +2 + nC 1(2x )yn +1 + nC 2(2)yn
→ dn
dx n(2xy 1) = nC 0(2x )yn +1 + nC 1(2)yn
→ dn
dx n (2nxy 1) = nC 0(2nx )yn +1 + nC 1(2n )yn
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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dn
dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V
→ dn
dx n ((x 2−1)y2) = nC 0(x 2−1)yn +2 + nC 1(2x )yn +1 + nC 2(2)yn
→ dn
dx n(2xy 1) = nC 0(2x )yn +1 + nC 1(2)yn
→ dn
dx n (2nxy 1) = nC 0(2nx )yn +1 + nC 1(2n )yn
→ dn
dx n (2ny ) = 2 ny n
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
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dn
dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V
→ dn
dx n ((x 2−1)y2) = nC 0(x 2−1)yn +2 + nC 1(2x )yn +1 + nC 2(2)yn
→ dn
dx n(2xy 1) = nC 0(2x )yn +1 + nC 1(2)yn
→ dn
dx n (2nxy 1) = nC 0(2nx )yn +1 + nC 1(2n )yn
→ dn
dx n (2ny ) = 2 ny n
Also nC 0 = 1 , nC 1 = n, nC 2 = n (n −1)2!
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
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∴ (x 2
−1)yn +2 + 2nxy n +1 + n (n
−1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
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∴ (x 2
−1)yn +2 + 2nxy n +1 + n (n
−1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
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∴ (x 2
−1)yn +2 + 2nxy n +1 + n (n
−1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2
−1)yn +2 + 2nxy n +1 + n (n
−1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2
−1)yn +2 + 2nxy n +1 + n (n
−1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
Let v = yn = dn
ydx n
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
Let v = yn = dn
ydx n
∴ (1 −x 2)v2 −2xv 1 + n (n + 1) v = 0 ——(2)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
Let v = yn = dn
ydx n
∴ (1 −x 2)v2 −2xv 1 + n (n + 1) v = 0 ——(2)Equation (2) is a Legendre’s equation in variables v and x
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
Let v = yn = dn
ydx n
∴ (1 −x 2)v2 −2xv 1 + n (n + 1) v = 0 ——(2)Equation (2) is a Legendre’s equation in variables v and x
⇒ P n (x ) is a solution of equation (2)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
Let v = yn = dn
ydx n
∴ (1 −x 2)v2 −2xv 1 + n (n + 1) v = 0 ——(2)Equation (2) is a Legendre’s equation in variables v and x
⇒ P n (x ) is a solution of equation (2)Also, v = f (x ) is a solution of equation (2)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn
+2 xy n +1 + 2ny n = 2nxy n +1 + n (2n )yn + 2ny n
∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0∴ (x 2 −1)yn +2 + 2xy n +1 −n (n + 1) yn = 0∴ (1 −x 2)yn +2 −2xy n +1 + n (n + 1) yn = 0
Let v = yn = dn
ydx n
∴ (1 −x 2)v2 −2xv 1 + n (n + 1) v = 0 ——(2)Equation (2) is a Legendre’s equation in variables v and x
⇒ P n (x ) is a solution of equation (2)Also, v = f (x ) is a solution of equation (2)P n = cv where c is constant
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
Now y = ( x 2 −1)n
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
Now y = ( x 2 −1)n
= ( x + 1) n (x −1)n
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
Now y = ( x 2 −1)n
= ( x + 1) n (x −1)n
∴
dn ydx n = ( x + 1)
n dn
dx n ((x −1)n
)
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
Now y = ( x 2 −1)n
= ( x + 1) n (x −1)n
∴
dn ydx n = ( x + 1)
n dn
dx n ((x −1)n
)
+ n (x + 1) n − 1 dn − 1
dx n − 1 ((x −1)n )
N. B. Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
Now y = ( x 2 −1)n
= ( x + 1) n (x −1)n
∴
dn ydx n = ( x + 1)
n dn
dx n ((x −1)n
)
+ n (x + 1) n − 1 dn − 1
dx n − 1 ((x −1)n )
+ ...
N B Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ P n (x ) = c dn
ydx n = c dn
dx n (x 2 −1)n ——(3)
Now y = ( x 2 −1)n
= ( x + 1) n (x −1)n
∴
dn ydx n = ( x + 1)
n dn
dx n ((x −1)n
)
+ n (x + 1) n − 1 dn − 1
dx n − 1 ((x −1)n )
+ ...
+dn ((x + 1) n )
dx n (x −1)n
N B Vyas Legendre’s Function
Legendre’s Polynomials
Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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Recurrence Relations for P n (x ) :
−
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
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N B Vyas Legendre’s Function
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )P f W h
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )P f W h
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to t, weget
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )P f W h
![Page 91: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/91.jpg)
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12(1 −2xt + t2)− 32 (−2x + 2 t )
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )Proof: We have
![Page 92: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/92.jpg)
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12(1 −2xt + t2)− 32 (−2x + 2 t )
= (1 −2xt + t2)− 1(1 −2xt + t2)− 12 (x −t )
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )Proof: We have
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12(1 −2xt + t2)− 32 (−2x + 2 t )
= (1 −2xt + t2)− 1(1 −2xt + t2)− 12 (x −t )
= (1 −2xt + t2)− 1
2
(1 −2xt + t2)
(x
−t )
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )Proof: We have
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12(1 −2xt + t2)− 32 (−2x + 2 t )
= (1 −2xt + t2)− 1(1 −2xt + t2)− 12 (x −t )
= (1 −2xt + t2)− 1
2
(1 −2xt + t2)
(x
−t )
from (i)
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )Proof: We have
![Page 95: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/95.jpg)
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Proof: We have∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12(1 −2xt + t2)− 32 (−2x + 2 t )
= (1 −2xt + t2)− 1(1 −2xt + t2)− 12 (x −t )
= (1 −2xt + t2)− 1
2
(1 −2xt + t2)
(x
−t )
from (i)
(1 −2xt + t2)∞
n =1
nP n (x )t n − 1 = ( x −t )∞
n = 0
P n (x )t n
N B Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
∴
∞
nP (x )t n − 1 2x∞
nP (x )t n +∞
nP (x )t n +1 =
![Page 96: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/96.jpg)
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∴
n =1
nP n (x )t −2x
n =1
nP n (x )t +n =1
nP n (x )t =
x∞
n =0
P n (x )t n
−∞
n =0
P n (x )t n +1
N B V L g d ’ F ti
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
∴
∞
nP n (x )t n − 1 2x∞
nP n (x )t n +∞
nP n (x )t n +1 =
![Page 97: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/97.jpg)
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∴
n =1
nP n (x )t −2x
n =1
nP n (x )t +n =1
nP n (x )t =
x∞
n =0
P n (x )t n
−∞
n =0
P n (x )t n +1
replacing n by n+1 in 1 st term, n by n-1 in 3 rd term in
L.H.S.
N B V L d ’ F ti
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
∴
∞
nP n (x )t n − 1 2x∞
nP n (x )t n +∞
nP n (x )t n +1 =
![Page 98: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/98.jpg)
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∴
n =1
nP n (x )t −2x
n =1
nP n (x )t +n =1
nP n (x )t
x∞
n =0
P n (x )t n
−∞
n =0
P n (x )t n +1
replacing n by n+1 in 1 st term, n by n-1 in 3 rd term in
L.H.S.replacing n by n-1 in 2 nd term in R.H.S
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
∴
∞
nP n (x )t n − 1 2x∞
nP n (x )t n +∞
nP n (x )t n +1 =
![Page 99: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/99.jpg)
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n =1
nP n (x )t −2x
n =1
nP n (x )t
n =1
nP n (x )t
x∞
n =0
P n (x )t n
−∞
n =0
P n (x )t n +1
replacing n by n+1 in 1 st term, n by n-1 in 3 rd term in
L.H.S.replacing n by n-1 in 2 nd term in R.H.S∞
n =0
(n + 1) P n +1 (x )t n
−2x∞
n =1
nP n (x )t n +∞
n =2
(n −1)P n − 1(x )t n = x
∞
n =0
P n (x )t n −∞
n =1
P n − 1(x )t n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
∴
∞
nP n (x )t n − 1 2x∞
nP n (x )t n +∞
nP n (x )t n +1 =
![Page 100: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/100.jpg)
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n =1
nP n (x )t −2x
n =1
nP n (x )t
n =1
nP n (x )t
x∞
n =0
P n (x )t n
−∞
n =0
P n (x )t n +1
replacing n by n+1 in 1 st term, n by n-1 in 3 rd term in
L.H.S.replacing n by n-1 in 2 nd term in R.H.S∞
n =0
(n + 1) P n +1 (x )t n
−2x∞
n =1
nP n (x )t n +∞
n =2
(n −1)P n − 1(x )t n = x
∞
n =0
P n (x )t n −∞
n =1
P n − 1(x )t n
comparing the coefficients of tn on both the sides
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (n + 1) P n +1 (x ) −2xnP n (x ) + ( n −1)P n − 1(x ) =xP n (x ) −P n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (n + 1) P n +1 (x ) −2xnP n (x ) + ( n −1)P n − 1(x ) =xP n (x ) −P n − 1(x )
∴ (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −(n −1 + 1) P n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
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∴ (n + 1) P n +1 (x ) −2xnP n (x ) + ( n −1)P n − 1(x ) =xP n (x ) −P n − 1(x )
∴ (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −(n −1 + 1) P n − 1(x )∴ (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
1
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∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
1
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∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to x, weget
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
∞ 1
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∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to x, weget
∞
n =0P n (x )t
n
= −12(1 −2xt + t2)
− 32 (−2t )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
∞ 1
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∞
n =0
P n (x )t n = (1 −2xt + t2)−
12 ——–(i)
Differentiating equation (i) partially with respect to x, weget
∞
n =0P n (x )t
n
= −12(1 −2xt + t2)
− 32 (−2t )
∴
∞
n =0
P n (x )t n = t(1 −2xt + t2)− 1
2
(1 −2xt + t2) ————-(ii)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
∞ 1
![Page 110: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/110.jpg)
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n =0
P n (x )t n = (1 −2xt + t2)−
2 ——–(i)
Differentiating equation (i) partially with respect to x, weget
∞
n =0P n (x )t
n
= −12(1 −2xt + t
2)
− 3
2 (−2t )
∴
∞
n =0
P n (x )t n = t(1 −2xt + t2)− 1
2
(1 −2xt + t2) ————-(ii)
⇒
Differentiating equation (i) partially with respect to t, weget
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
∞ 1
![Page 111: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/111.jpg)
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n =0
P n (x )t n = (1 −2xt + t2)−
2 ——–(i)
Differentiating equation (i) partially with respect to x, weget
∞
n =0P n (x )t
n
= −12(1 −2xt + t
2)
− 3
2 (−2t )
∴
∞
n =0
P n (x )t n = t(1 −2xt + t2)− 1
2
(1 −2xt + t2) ————-(ii)
⇒
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12
(1 −2xt + t2)− 32 (−2x + 2 t )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have
∞ 1
![Page 112: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/112.jpg)
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n =0
P n (x )t n = (1 −2xt + t2)−
2 ——–(i)
Differentiating equation (i) partially with respect to x, weget
∞
n =0P n (x )t
n
= −12(1 −2xt + t
2)
− 3
2 (−2t )
∴
∞
n =0
P n (x )t n = t(1 −2xt + t2)− 1
2
(1 −2xt + t2) ————-(ii)
⇒
Differentiating equation (i) partially with respect to t, weget
∞
n =1
nP n (x )t n − 1 = −12
(1 −2xt + t2)− 32 (−2x + 2 t )
∞
n − 1 (x t )(1 2xt + t2)− 12
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
∞
nP n (x )t n − 1 = (x −t )
t
∞
P n (x )t n
{by eq. (ii)
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n =1t
n =0{
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
∞
nP n (x )t n − 1 = (x −t )
t
∞
P n (x )t n
{by eq. (ii)
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n =1t
n =0{
∴ t∞
n =1
nP n (x )t n − 1 = ( x −t )∞
n =0
P n (x )t n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
∞
nP n (x )t n − 1 = (x −t )
t
∞
P n (x )t n
{by eq. (ii)
![Page 115: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/115.jpg)
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n =1 n =0{
∴ t∞
n =1
nP n (x )t n − 1 = ( x −t )∞
n =0
P n (x )t n
∴
∞
n =1
nP n (x )t n = x∞
n =0
P n (x )t n
−
∞
n =0
P n (x )t n +1
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
∞
nP n (x )t n − 1 = (x −t )
t
∞
P n (x )t n
{by eq. (ii)
![Page 116: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/116.jpg)
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n =1 n =0{
∴ t∞
n =1
nP n (x )t n − 1 = ( x −t )∞
n =0
P n (x )t n
∴
∞
n =1
nP n (x )t n = x∞
n =0
P n (x )t n
−
∞
n =0
P n (x )t n +1
Replacing n by n-1 in 2 nd term in R.H.S.
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
∞
nP n (x )t n − 1 = (x −t )
t
∞
P n (x )t n
{by eq. (ii)
![Page 117: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/117.jpg)
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n =1 n =0{
∴ t∞
n =1
nP n (x )t n − 1 = ( x −t )∞
n =0
P n (x )t n
∴
∞
n =1
nP n (x )t n = x∞
n =0
P n (x )t n
−
∞
n =0
P n (x )t n +1
Replacing n by n-1 in 2 nd term in R.H.S.
∴
∞
n =1
nP n (x )t n = x∞
n =0
P n (x )t n
−∞
n =1
P n − 1(x )t n
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
∞
n
nP n (x )t n − 1 = (x −t )
t
∞
n
P n (x )t n
{by eq. (ii)
![Page 118: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/118.jpg)
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n=1
n=0
∴ t∞
n =1
nP n (x )t n − 1 = ( x −t )∞
n =0
P n (x )t n
∴
∞
n =1
nP n (x )t n = x∞
n =0
P n (x )t n
−
∞
n =0
P n (x )t n +1
Replacing n by n-1 in 2 nd term in R.H.S.
∴
∞
n =1
nP n (x )t n = x∞
n =0
P n (x )t n
−∞
n =1
P n − 1(x )t n
comparing the coefficients of tn on both sides, we get
N. B. Vyas Legendre’s Function
![Page 119: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/119.jpg)
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Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
![Page 120: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/120.jpg)
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
![Page 121: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/121.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
![Page 122: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/122.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
![Page 123: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/123.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
![Page 124: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/124.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
∴
(2n +1) P n
(x )+(2 n +1) xP n (x ) = ( n +1) P n +1 (x )+ nP n − 1(x )—(b)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
![Page 125: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/125.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
∴
(2n +1) P n
(x )+(2 n +1) xP n (x ) = ( n +1) P n +1 (x )+ nP n − 1(x )—(b)Also from relation (2)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P
n (x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
( ) 1 ( ) ( ) ( ) 1( )
![Page 126: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/126.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
∴
(2n
+1)P n
(x
)+(2n
+1)xP
n (x
) = (n
+1)P
n +1 (x
)+nP
n − 1(x
)—(b)Also from relation (2)nP n (x ) = xP n (x ) −P n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n
(x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
( 1) P n +1 ( ) (2 1) P n ( ) P n 1( )
![Page 127: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/127.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
∴
(2n
+1)P n
(x
)+(2n
+1)xP
n (x
) = (n
+1)P
n +1 (x
)+nP
n − 1(x
)—(b)Also from relation (2)nP n (x ) = xP n (x ) −P n − 1(x )
∴ xP n (x ) = nP n (x ) + P n − 1(x )—– (c)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n
(x
)Rodrigue’s FormulaRecurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
( 1) P n +1 ( ) (2 1) P n ( ) P n − 1( )
![Page 128: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/128.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
∴
(2n
+1)P n
(x
)+(2n
+1)xP
n (x
) = (n
+1)P
n +1 (x
)+nP
n − 1(x
)—(b)Also from relation (2)nP n (x ) = xP n (x ) −P n − 1(x )
∴ xP n (x ) = nP n (x ) + P n − 1(x )—– (c)Substituting the value of (c) in equation (b), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n
Proof: We have ( from relation (1) )
( + 1) P n +1 ( ) (2 + 1) P n ( ) P n − 1( )
![Page 129: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/129.jpg)
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(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n 1(x )∴ (2n + 1) xP n (x ) = ( n + 1) P n +1 (x ) + nP n − 1(x ) — (a)
differentiating equation (a) partially with respect to x, Weget
∴ (2n +1) P n (x )+(2 n +1) xP n (x ) = ( n +1) P n +1
(x )+ nP n − 1
(x )—(b)Also from relation (2)nP n (x ) = xP n (x ) −P n − 1(x )
∴ xP n (x ) = nP n (x ) + P n − 1(x )—– (c)Substituting the value of (c) in equation (b), we get
∴ (2n + 1) P n (x ) + (2 n + 1)[nP n (x ) + P n − 1(x )] =(n + 1) P n +1 (x ) + nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
![Page 130: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/130.jpg)
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∴ (2n + 1)( n + 1) P n (x ) + (2 n + 1) P n − 1(x ) =(n + 1) P n +1 (x ) + nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
![Page 131: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/131.jpg)
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∴ (2n + 1)( n + 1) P n (x ) + (2 n + 1) P n − 1(x ) =(n + 1) P n +1 (x ) + nP n − 1(x )
∴ (2n + 1)( n + 1) P n (x ) = ( n + 1) P n +1 (x ) −(n + 1) P n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
![Page 132: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/132.jpg)
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∴ (2n + 1)( n + 1) P n (x ) + (2 n + 1) P n − 1(x ) =(n + 1) P n +1 (x ) + nP n − 1(x )
∴ (2n + 1)( n + 1) P n (x ) = ( n + 1) P n +1 (x ) −(n + 1) P n − 1(x )∴ dividing by ( n + 1), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
![Page 133: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/133.jpg)
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∴ (2n + 1)( n + 1) P n (x ) + (2 n + 1) P n − 1(x ) =(n + 1) P n +1 (x ) + nP n − 1(x )
∴ (2n + 1)( n + 1) P n (x ) = ( n + 1) P n +1 (x ) −(n + 1) P n − 1(x )∴ dividing by ( n + 1), we get∴ (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
![Page 134: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/134.jpg)
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) )
![Page 135: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/135.jpg)
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Proof: We have (from relation (3) ),
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) )
![Page 136: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/136.jpg)
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Proof: We have (from relation (3) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) )
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Proof: We have (from relation (3) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) )
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Proof: We have (from relation (3) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
∴ nP n (x ) = xP n (x )
−P n − 1(x ) ——(b)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) ),
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Proof: We have (from relation (3) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
∴ nP n (x ) = xP n (x )
−P n − 1(x ) ——(b)
Taking (a) - (b), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) ),
![Page 140: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/140.jpg)
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Proof: We have (from relation (3) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
∴ nP n (x ) = xP n (x ) −P n − 1(x ) ——(b)
Taking (a) - (b), we get∴ (n + 1) P n (x ) = P n +1 (x ) −xP n (x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) ),
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( ( ) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
∴ nP n (x ) = xP n (x ) −P n − 1(x ) ——(b)
Taking (a) - (b), we get∴ (n + 1) P n (x ) = P n +1 (x ) −xP n (x )
replacing n by n −1, we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) ),
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( ( ) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
∴ nP n (x ) = xP n (x ) −P n − 1(x ) ——(b)
Taking (a) - (b), we get∴ (n + 1) P n (x ) = P n +1 (x ) −xP n (x )
replacing n by n −1, we get∴ nP n − 1(x ) = P n (x )
−xP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )
Proof: We have (from relation (3) ),
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( ( ) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)Also we have (from relation (2) ),
∴ nP n (x ) = xP n (x ) −P n − 1(x ) ——(b)
Taking (a) - (b), we get∴ (n + 1) P n (x ) = P n +1 (x ) −xP n (x )
replacing n by n −1, we get∴ nP n − 1(x ) = P n (x )
−xP n − 1(x )
∴ P n (x ) = xP n − 1(x ) + nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
![Page 144: Dr. Nirav Vyas legendres function.pdf](https://reader030.vdocuments.site/reader030/viewer/2022021101/577cb1031a28aba7118b6b7b/html5/thumbnails/144.jpg)
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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( ( ) )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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( ( ) )P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )nP n (x ) = xP n (x )
−P n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )nP n (x ) = xP n (x )
−P n − 1(x )
xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )nP n (x ) = xP n (x )
−P n − 1(x )
xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)taking (a) - x X (b), we get
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s Formula
Recurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )nP n (x ) = xP n (x )
−P n − 1(x )
xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)taking (a) - x X (b), we get(1−x 2)P n (x ) = xP n − 1(x )+ nP n − 1(x )−nxP n (x )−xP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )nP n (x ) = xP n (x )
−P n − 1(x )
xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)taking (a) - x X (b), we get(1−x 2)P n (x ) = xP n − 1(x )+ nP n − 1(x )−nxP n (x )−xP n − 1(x )(1
−x 2)P n (x ) = n [P n −
1(x )
−xP n (x )]
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(5) (1
−x 2)P n (x ) = n [P n − 1(x )
−xP n (x )]
Proof: We have (from relation (4) )
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P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )nP n (x ) = xP n (x )
−P n − 1(x )
xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)taking (a) - x X (b), we get(1−x 2)P n (x ) = xP n − 1(x )+ nP n − 1(x )−nxP n (x )−xP n − 1(x )(1
−x 2)P n (x ) = n [P n −
1(x )
−xP n (x )]
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )
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N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 x 2)P n (x ) = n [P n − 1(x ) xP n (x )] ——(a)
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(1 −x )P n (x ) n [P n 1(x ) −xP n (x )] (a)
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 x 2)P n (x ) = n [P n − 1(x ) xP n (x )] ——(a)
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(1 −x )P n (x ) n [P n 1(x ) −xP n (x )] (a)also we have (from relation (1) )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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( − ) n ( ) [ 1( ) − ( )] (a)also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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( − ) n ( ) [ 1( ) − ( )] ( )also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + nxP n (x ) −nP n − 1(x )
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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( ) ( ) [ ( ) ( )] ( )also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + nxP n (x ) −nP n − 1(x )(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + n [xP n (x ) −P n − 1(x )]
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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( ) ( ) ( ) ( ) ( )also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + nxP n (x ) −nP n − 1(x )(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + n [xP n (x ) −P n − 1(x )]n (P n − 1(x ) −xP n (x )) = ( n + 1)( xP n (x ) −P n +1 (x ))
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + nxP n (x ) −nP n − 1(x )(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + n [xP n (x ) −P n − 1(x )]n (P n − 1(x ) −xP n (x )) = ( n + 1)( xP n (x ) −P n +1 (x ))from equation (a),
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + nxP n (x ) −nP n − 1(x )(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + n [xP n (x ) −P n − 1(x )]n (P n − 1(x ) −xP n (x )) = ( n + 1)( xP n (x ) −P n +1 (x ))from equation (a),
(1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
N. B. Vyas Legendre’s Function
Legendre’s PolynomialsExamples of Legendre’s Polynomials
Generating Function for P n ( x )
Rodrigue’s FormulaRecurrence Relations for P n ( x )
(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
Proof: We have (from relation (5) )(1 −x 2)P n (x ) = n [P n − 1(x ) −xP n (x )] ——(a)
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8/20/2019 Dr. Nirav Vyas legendres function.pdf
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also we have (from relation (1) )(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )
(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + nxP n (x ) −nP n − 1(x )(n + 1) P n +1 (x ) = ( n + 1) xP n (x ) + n [xP n (x ) −P n − 1(x )]n (P n − 1(x ) −xP n (x )) = ( n + 1)( xP n (x ) −P n +1 (x ))from equation (a),
(1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]
N. B. Vyas Legendre’s Function