Dr. Nirav Vyas legendres function.pdf

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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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The differential equation


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(1−x 2)y −2xy + n (n + 1) y = 0


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(1−x 2)y −2xy + n (n + 1) y = 0is called Legendre’s differential equation ,n is real constant


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Legendre’s Polynomials:⇒

P 0(x ) = 1


L d ’ P l i l

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P 0(x ) = 1

⇒ P 1(x ) = x


L d ’ P l i l

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P 0(x ) = 1

⇒ P 1(x ) = x

⇒ P 2(x ) =

12(3x 2 −1)


Legendre’s Polynomials

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Legendre s PolynomialsExamples of 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 )



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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)



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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 )



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Ex.1 Express f (x ) in terms of Legendre’s polynomials wheref (x ) = x3 + 2x 2 −x −3.


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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 )



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Solution:⇒

P 0(x ) = 1∴ 1 = P 0(x )

⇒ P 1(x ) = x∴ x = P

1(x )


Legendre’s Polynomialsl f d l l

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Solution:⇒

P 0(x ) = 1∴ 1 = P 0(x )

⇒ P 1(x ) = x∴ x = P

1(x )

⇒ P 3(x ) =

12

(5x 3 −3x )


Legendre’s PolynomialsE l f L d ’ P l i l

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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 )



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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



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Examples of Legendre s PolynomialsGenerating Function 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 )}



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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 )



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⇒ P 2(x ) = 1

2(3x 2 −1)


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p g yGenerating Function 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



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⇒ 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 )}



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⇒ 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 )



G i F i f P ( )

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⇒ 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



G ti g F ti f P ( x )

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⇒ 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 )



Generating Function for P ( x )

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Ex.2 Express x3

−5x 2

+ 6x

+ 1 interms of Legendre’s polynomial.



Generating Function for P ( x )

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Ex.3 Express 4x 3

−2x 2

−3x

+ 8 interms of Legendre’s polynomial.



Generating Function for P n ( x )

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Generating Function for n ( )Rodrigue’s Formula


Generating Function for P n (x )

∞

n =0P n (x )t n = 1√ 1 −2xt + t2

= (1 −2xt

+ t2

)−12




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Ge e at g u ct o o n ( )Rodrigue’s Formula


The function (1

−2xt + t2)−1

2 iscalled Generating function of Legendre’s polynomial P n (x )


Legendre’s PolynomialsExamples of Legendre’s PolynomialsGenerating Function for P n ( x )

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g ( )Rodrigue’s Formula


Ex Show that



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( )Rodrigue’s Formula


Ex Show that

(i)P n (1) = 1



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Ex Show that

(i)P n (1) = 1(ii)P n (−1) = (−1)n



d ’ l

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Ex Show that

(i)P n (1) = 1(ii)P n (−1) = (−1)n

(iii)P n (

−x) = (

−1)n P n (x )



R d i ’ F l

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Solution:

(i) We have∞

n =0

P n (x )t n = (1 −2xt + t2)− 12

Putting x = 1 in eq(1), we get



Rodrigue’s Formula

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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




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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


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 + ...




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Solution:

(i) We have∞

n =0

P n (x )t n = (1 −2xt + t2)− 12


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




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Solution:

(i) We have∞

n =0

P n (x )t n = (1 −2xt + t2)− 12


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




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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


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




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(ii) Putting x = −1 in eq(1), we get




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(ii) Putting x = −1 in eq(1), we get∞

n =0

P n (−1)t n = (1 + 2 t + t2)− 12 = (1 + t)− 1




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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 + ...



Rodrigue’s FormulaR R l i f P ( )

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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



Rodrigue’s FormulaR R l ti f P ( x )

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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



Rodrigue’s FormulaRecurrence Relations for P ( x )

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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




Recurrence Relations for P ( x )

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(iii) Now replacing x by

−x in eq(1), we get





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Recurrence Relations for n ( )



∞

n =0

P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)





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n ( )



∞

n =0

P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)

Now, replacing t by −t in eq(1), we get





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( )



∞

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)





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∞

n =0

P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)


n =0P n (x )(−t )n = (1 + 2 xt + t2)−

12 —(b)

from equation (a) and (b)





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∞

n =0

P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)


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





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∞

n =0

P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)


n =0P n (x )(−t )n = (1 + 2 xt + t2)−

12 —(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





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∞

n =0

P n (−x )t n = (1 + 2 xt + t2)− 12 —(a)


n =0P n (x )(−t )n = (1 + 2 xt + t2)−

12 —(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 )





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P n (x ) = 12n n !

dn

dx n [(x 2 −1)n ]





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Proof:Let y = ( x 2 −1)n





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Differentiating wit respect to x





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Differentiating wit respect to x∴ y1 = n (x 2

−1)n − 1(2x )





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−1)n − 1(2x )

∴ y1 = 2nx (x 2 −1)n

(x 2 −1) =

2nxyx 2 −1





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−1)n − 1(2x )

∴ y1 = 2nx (x 2 −1)n

(x 2 −1) =

2nxyx 2 −1

∴ (x 2 −1)y1 = 2nxy





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−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,





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−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





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dn

dx n (UV ) = nC 0UV n + nC 1U 1V n − 1 + ... + nC n U n V





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dn


→ dn

dx n ((x 2−1)y2) = nC 0(x 2−1)yn +2 + nC 1(2x )yn +1 + nC 2(2)yn





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dn


→ dn


→ dn

dx n(2xy

1) = nC

0(2x )yn

+1 + nC

1(2)yn





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dn


→ dn


→ 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





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dn


→ dn


→ dn

dx n(2xy 1) = nC 0(2x )yn +1 + nC 1(2)yn

→ dn


→ dn

dx n (2ny ) = 2 ny n





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dn


→ dn


→ dn

dx n(2xy 1) = nC 0(2x )yn +1 + nC 1(2)yn

→ dn


→ dn

dx n (2ny ) = 2 ny n

Also nC 0 = 1 , nC 1 = n, nC 2 = n (n −1)2!





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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





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∴ (x 2

−1)yn +2 + 2nxy n +1 + n (n

−1)yn


∴ (x 2 −1)yn +2 + 2xy n +1 + ( n 2 −n + 2 n −2n 2 −2n )yn = 0





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∴ (x 2

−1)yn +2 + 2nxy n +1 + n (n

−1)yn


∴ (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



Examples of Legendre’s PolynomialsGenerating Function for P n ( x )Rodrigue’s Formula


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∴ (x 2

−1)yn +2 + 2nxy n +1 + n (n

−1)yn


∴ (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





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∴ (x 2

−1)yn +2 + 2nxy n +1 + n (n

−1)yn



Let v = yn = dn

ydx n





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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn



Let v = yn = dn

ydx n

∴ (1 −x 2)v2 −2xv 1 + n (n + 1) v = 0 ——(2)





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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn



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





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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn



Let v = yn = dn

ydx n


⇒ P n (x ) is a solution of equation (2)





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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn



Let v = yn = dn

ydx n


⇒ P n (x ) is a solution of equation (2)Also, v = f (x ) is a solution of equation (2)





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∴ (x 2 −1)yn +2 + 2nxy n +1 + n (n −1)yn



Let v = yn = dn

ydx n


⇒ 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





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∴ P n (x ) = c dn

ydx n = c dn

dx n (x 2 −1)n ——(3)





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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





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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





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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

)





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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 )





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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




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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





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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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(1) (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )P f W h



Proof: We have∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)






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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






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Proof: We have∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)


∞

n =1

nP n (x )t n − 1 = −12(1 −2xt + t2)− 32 (−2x + 2 t )





(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)


∞

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 )






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Proof: We have∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)


∞

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 )






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Proof: We have∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)


∞

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)






http://find/

http://goback/



Proof: We have∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)


∞

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





∴

∞

nP (x )t n − 1 2x∞

nP (x )t n +∞

nP (x )t n +1 =

http://find/



∴

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




∴

∞

nP n (x )t n − 1 2x∞

nP n (x )t n +∞

nP n (x )t n +1 =

http://find/



∴

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




∴

∞

nP n (x )t n − 1 2x∞

nP n (x )t n +∞

nP n (x )t n +1 =

http://find/



∴

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


L.H.S.replacing n by n-1 in 2 nd term in R.H.S





∴

∞

nP n (x )t n − 1 2x∞

nP n (x )t n +∞

nP n (x )t n +1 =

http://find/



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


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





∴

∞

nP n (x )t n − 1 2x∞

nP n (x )t n +∞

nP n (x )t n +1 =

http://find/



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


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





http://goforward/

http://find/

http://goback/



∴ (n + 1) P n +1 (x ) −2xnP n (x ) + ( n −1)P n − 1(x ) =xP n (x ) −P n − 1(x )





http://find/




∴ (n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −(n −1 + 1) P n − 1(x )





http://find/




∴ (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 )





(2) nP n (x ) = xP n (x ) −P n − 1(x )

http://find/







(2) nP n (x ) = xP n (x ) −P n − 1(x )Proof: We have

http://find/








1

http://find/



∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)






1

http://find/



∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)

Differentiating equation (i) partially with respect to x, weget






∞ 1

http://find/



∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)


∞

n =0P n (x )t

n

= −12(1 −2xt + t2)

− 32 (−2t )



Generating Function for P n ( x )Rodrigue’s FormulaRecurrence Relations for P n ( x )


∞ 1

http://find/



∞

n =0

P n (x )t n = (1 −2xt + t2)−

12 ——–(i)


∞

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)





∞ 1

http://find/



n =0

P n (x )t n = (1 −2xt + t2)−

2 ——–(i)


∞

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)

⇒






∞ 1

http://find/



n =0

P n (x )t n = (1 −2xt + t2)−

2 ——–(i)


∞

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)

⇒


∞

n =1

nP n (x )t n − 1 = −12

(1 −2xt + t2)− 32 (−2x + 2 t )





∞ 1

http://find/



n =0

P n (x )t n = (1 −2xt + t2)−

2 ——–(i)


∞

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)

⇒


∞

n =1

nP n (x )t n − 1 = −12

(1 −2xt + t2)− 32 (−2x + 2 t )

∞

n − 1 (x t )(1 2xt + t2)− 12




∞

nP n (x )t n − 1 = (x −t )

t

∞

P n (x )t n

{by eq. (ii)

http://find/



n =1t

n =0{




∞

nP n (x )t n − 1 = (x −t )

t

∞

P n (x )t n

{by eq. (ii)

http://find/



n =1t

n =0{

∴ t∞

n =1

nP n (x )t n − 1 = ( x −t )∞

n =0

P n (x )t n




∞

nP n (x )t n − 1 = (x −t )

t

∞

P n (x )t n

{by eq. (ii)

http://find/



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




∞

nP n (x )t n − 1 = (x −t )

t

∞

P n (x )t n

{by eq. (ii)

http://find/



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.




∞

nP n (x )t n − 1 = (x −t )

t

∞

P n (x )t n

{by eq. (ii)

http://find/



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 =1

nP n (x )t n = x∞

n =0

P n (x )t n

−∞

n =1

P n − 1(x )t n




∞

n

nP n (x )t n − 1 = (x −t )

t

∞

n

P n (x )t n

{by eq. (ii)

http://find/



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 =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


http://find/




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) )






n (x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


http://find/



(n + 1) P n +1 (x ) = (2 n + 1) xP n (x ) −nP n − 1(x )




n (x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


http://find/



(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 (x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


http://find/




differentiating equation (a) partially with respect to x, Weget




n (x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


http://find/





∴

(2n +1) P n

(x )+(2 n +1) xP n (x ) = ( n +1) P n +1 (x )+ nP n − 1(x )—(b)




n (x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


http://find/





∴

(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 (x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


( ) 1 ( ) ( ) ( ) 1( )

http://find/





∴

(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 )



Generating Function for P n

(x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


( 1) P n +1 ( ) (2 1) P n ( ) P n 1( )

http://find/

http://goback/





∴

(2n

+1)P n

(x

)+(2n

+1)xP

n (x

) = (n

+1)P

n +1 (x

)+nP

n − 1(x


∴ xP n (x ) = nP n (x ) + P n − 1(x )—– (c)



Generating Function for P n

(x


(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


( 1) P n +1 ( ) (2 1) P n ( ) P n − 1( )

http://find/



(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)


∴

(2n

+1)P n

(x

)+(2n

+1)xP

n (x

) = (n

+1)P

n +1 (x

)+nP

n − 1(x


∴ xP n (x ) = nP n (x ) + P n − 1(x )—– (c)Substituting the value of (c) in equation (b), we get





(3) (2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x )n


( + 1) P n +1 ( ) (2 + 1) P n ( ) P n − 1( )

http://find/



(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)


∴ (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 )





http://find/



∴ (2n + 1)( n + 1) P n (x ) + (2 n + 1) P n − 1(x ) =(n + 1) P n +1 (x ) + nP n − 1(x )





http://find/

http://goback/




∴ (2n + 1)( n + 1) P n (x ) = ( n + 1) P n +1 (x ) −(n + 1) P n − 1(x )





http://find/

http://goback/




∴ (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





http://find/




∴ (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 )





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )

http://find/







(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )

Proof: We have (from relation (3) )

http://find/



Proof: We have (from relation (3) ),





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/

http://goback/



Proof: We have (from relation (3) ),(2n + 1) P n (x ) = P n +1 (x ) −P n − 1(x ) —–(a)





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/



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) ),





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/




∴ nP n (x ) = xP n (x )

−P n − 1(x ) ——(b)





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/




∴ nP n (x ) = xP n (x )

−P n − 1(x ) ——(b)

Taking (a) - (b), we get





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/




∴ 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 )





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/



( ( ) ),(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)


replacing n by n −1, we get





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/




∴ nP n (x ) = xP n (x ) −P n − 1(x ) ——(b)


replacing n by n −1, we get∴ nP n − 1(x ) = P n (x )

−xP n − 1(x )





(4) P n (x ) = xP n − 1(x ) + nP n − 1(x )


http://find/

http://goback/




∴ nP n (x ) = xP n (x ) −P n − 1(x ) ——(b)


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 )





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]

http://find/







(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/



( ( ) )





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/



( ( ) )P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/



P n (x ) = xP n − 1(x ) + nP n − 1(x ) ——- (a)also we have (from relation (2) )





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/

http://goback/



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 )





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/




−P n − 1(x )

xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/

http://goback/




−P n − 1(x )

xP n (x ) = nP n (x ) + P n − 1(x ) ——– (b)taking (a) - x X (b), we get





(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/




−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 )




(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/




−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 )]




(5) (1

−x 2)P n (x ) = n [P n − 1(x )

−xP n (x )]


http://find/

http://goback/




−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 )]




(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]

http://find/







(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/







(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)

http://find/



(1 −x )P n (x ) n [P n 1(x ) −xP n (x )] (a)





(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)

http://find/



(1 −x )P n (x ) n [P n 1(x ) −xP n (x )] (a)also we have (from relation (1) )





(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)

http://find/



( − ) 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 )





(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/

http://goback/



( − ) 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 )





(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/



( ) ( ) [ ( ) ( )] ( )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 )]





(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/



( ) ( ) ( ) ( ) ( )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 ))





(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/



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),





(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/





(1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]





(6) (1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/

http://goback/





(1 −x 2)P n (x ) = ( n + 1)[xP n (x ) −P n +1 (x )]


http://find/

Dr. Nirav Vyas legendres function.pdf

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