Leibniz Rule for Differentiating Products

Formula to find a high order derivative of product.Example:

d9

dx9(x sin x) = x

d9

dx9(sin x) + 9

d

dx(x)

d8

dx8(sin x)

+9 ⋅ 82!

d2

dx2(x)

d7

dx7(sin x) + ⋅ ⋅ ⋅

= xd9

dx9(sin x) + 9

d8

dx8(sin x)

+9 ⋅ 82!⋅ 0 ⋅ d7

dx7(sin x)

= x cos x + 9 sin x

Leibniz Rule for Differentiating Products

Binomial expansion

(a + b)9 = a0b9 + 9ab8 +9 ⋅ 82!

a2b7 + ⋅ ⋅ ⋅

Exercise: Try to differentiate this

d10

dx10(xex)

Rodrigues’ Formula

Another way of finding Legendre Polynomials

Pl(x) =1

2l l!

dl

dx l(x2 − 1)l

Rodrigues’ Formula: Cont’dExample: when l = 0

P0(x) =1

200!

d0

dx0(x2 − 1)0

= (x2 − 1)0 = 1

When l = 1

P1(x) =1

211!

d1

dx1(x2 − 1)1

=1

2(2x) = x

When l = 2

P2(x) =1

222!

d2

dx2(x2 − 1)2

=1

8(12x2 − 4) =

1

2(3x2 − 1)

Rodrigues’ Formula: Proof

Let v = (x2 − 1)l . Differentiate once

dv

dx= l(x2 − 1)l−1(2x)

(x2 − 1)dv

dx= l(x2 − 1)(2x) = l ⋅ v ⋅ 2x

(x2 − 1)dv

dx= 2xlv (1)

Rodrigues’ Formula: Proof Cont’d

Differentiate Eqn. (1) l + 1 times using Leibniz rule

dl+1

dx l+1

((x2 − 1)

dv

dx

)=

dl+1

dx l+1(2xlv)

(x2 − 1)dl+2v

dx l+2+ (l + 1)(2x)

dl+1v

dx l+1+

(l + 1)l

2!⋅ 2 ⋅ d

lv

dx l

= 2lxdl+1v

dx l+1+ 2l(l + 1)

dlv

dx l(2)

Rodrigues’ Formula: Proof Cont’d

Rearrange Eqn. (2), we get

(1− x2)dl+2v

dx l+2− 2x

dl+1v

dx l+1+ l(l + 1)

dlv

dx l= 0

(1− x2)

(dlv

dx l

)′′

− 2x

(dlv

dx l

)′

+ l(l + 1)dlv

dx l= 0

Which show that the Rodrigues’ formula satisfied the Legendreequation.

Rodrigues’ Formula: Proof Cont’d

Here we see that

C ⋅ dlv

dx l= Pl(x)

The derivative for x2n, when v = (x2 − 1)l = (x + 1)l(x − 1)l is

dlv

dx l= (x + 1)l

dl

dx l+l C1 ⋅ l(x + 1)l−1 ⋅ dl−1

dx l−1(x − 1)l

+ ⋅ ⋅ ⋅+ (x − 1)ldl

dx l(x + 1)l = 0

When x = 1,dlv

dx l= 2l ⋅ l!

Rodrigues’ Formula: Proof Cont’d

Thus, when x = 1

C ⋅ dlv

dx l= Pl(1) = 1

Which give C as

C =1

2l ⋅ l!So,

Pl(x) =1

2l ⋅ l!dlv

dx l

=1

2l ⋅ l!dl

dx l(x2 − 1)l

Proved

Generating Function for Legendre Polynomials

Many properties of Legendre Polynomials can be derive by usingGenerating function

Φ(x , h) =1√

1− 2xh + h2, ∣h∣ < 1, (3)

Generating Function for Legendre Polynomials: Cont’d

Maclaurin series for (1− y)−1/2 for −1 < y < 1 is

1√1− y

= 1 +1

2y +

3

8y2 +

15

48y3 +

105

384y4 +

945

3840y5 + ⋅ ⋅ ⋅

Setting y = 2xh − h2, we obtain

1√1− 2xh − h2

= 1 +1

2(2xh − h2) +

3

8(2xh − h2)2

+15

48(2xh − h2)3 +

105

384(2xh − h2)4

+945

3840(2xh − h2)5 + ⋅ ⋅ ⋅

Generating Function for Legendre Polynomials: Cont’d

Expand the previous equation and collecting h term, we get

1√1− 2xh − h2

= 1 + xh +

(−1

2+

3

2

)h2 +

(−3

2x +

5

2x3)h3

+

(3

8− 15

4x2 +

35

8x4)h4

+

(15

8x − 35

4+

63

8x5)h5 + ⋅ ⋅ ⋅

= P0(x) + P1(x)h + P2(x)h2 + P3(x)h3

+ P4(x)h4 + P5(x)h5 + ⋅ ⋅ ⋅

=∞∑l=0

hlPl(x)

Generating Function for Legendre Polynomials: Cont’d

When x = 1

Φ(1, h) = (1− 2h + h2)−1/2

=1

1− h

= 1 + h + h2 + ⋅≡ P0(1) + P1(1)h + P2(1)h2 + ⋅ ⋅ ⋅

Thus Pl(1) = 1.

Generating Function for Legendre Polynomials: Cont’d

To show that generating function satisfy Legendre equation, weuse this equation

(1− x2)∂2Φ

∂x2− 2x

∂Φ

∂x+ h

∂2

∂h2(hΦ) = 0

Where

∂Φ

∂x=

h

(1− 2xh + h2)3/2

∂2Φ

∂x2=

3h2

(1− 2xh + h2)5/2

∂2

∂h2(hΦ) = − −2x + hx2 − 2xh2 + 3h

(−1 + 2xh − h2)√

1− 2xh + h2

Generating Function for Legendre Polynomials: Cont’d

Inserting

Φ(x , h) =∞∑l=0

hlPl(x)

into the previous equation, we get

(1− x2)∞∑l=0

hlP′′l (x)− 2x

∞∑l=0

hlP′l (x) +

∞∑l=0

l(l + 1)hlPl(x) = 0

∞∑l=0

hl(

(1− x2)P′′l (x)− 2xP

′l (x) + l(l + 1)Pl(x)

)= 0

Which is Legendre Equations.

Recursion Relations for Legendre Polynomials

1. lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)

2. xP′l (x)− P

′l−1(x) = lPl(x)

3. P′l (x)− xP

′l−1(x) = lPl−1(x)

4. (1− x2)P′l (x) = lPl−1(x)− lxPl(x)

5. (2l + 1)Pl(x) = P′l+1(x)− P

′l−1(x)

6. (1− x2)P′l−1(x) = lxPl−1(x)− lPl(x)

Proof for lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)

Given

Φ(x , h) = (1− 2xh + h2)−1/2 =∞∑l=0

hlPl(x)

We differentiate with respect to h,

∂Φ

∂h= − 1

2 3√

1− 2xh + h2(−2x + 2h)

(1− 2xh + h2)∂Φ

∂h= (x − h)Φ

Or

(1− 2xh + h2)∞∑l=1

lhl−1Pl(x) = (x − h)∞∑l=0

hlPl(x)

Proof for lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x):Cont’d

Expanding the previous equation, we get,

∞∑l=1

lhl−1Pl(x)− 2x∞∑l=1

lhlPl(x) +∞∑l=1

lhl+1Pl(x)

= x∞∑l=0

hlPl(x)−∞∑l=0

hl+1Pl(x)

Adjusting h indices so that all become hl−1, we get

∞∑l=1

lhl−1Pl(x)− 2x∞∑l=2

(l − 1)hl−1Pl−1(x) +∞∑l=3

(l − 2)hl−1Pl−2(x)

= x∞∑l=1

hl−1Pl−1(x)−∞∑l=2

hl−1Pl−2(x)

Proof for lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x):Cont’d

We can write the equation from the previous slide as

lPl(x)− 2x(l − 1)Pl−1(x) + (l − 2)Pl−2(x) = xPl−1(x)− Pl−2(x)

When we rearrange we will get the first recursion formula.

lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)

Proof for xP′

l (x)− P′

l−1(x) = lPl(x)

Given

Φ(x , h) = (1− 2xh + h2)−1/2 =∞∑l=0

hlPl(x)

Differentiate with respect to h, we get

∂Φ

∂h=

(x − h)3√

1− 2xh + h2=∞∑l=1

lhl−1Pl(x) (4)

Differentiate with respect to x , we get

∂Φ

∂x= − 1

2 3√

1− 2xh + h2(−2h)

=h

3√

1− 2xh + h2=∞∑l=0

hlP′l (x) (5)

Proof for xP′

l (x)− P′

l−1(x) = lPl(x): Cont’d

Dividing Eqn. (4) with Eqn. (5), we get

x − h

h=

∑∞l=1 lh

l−1Pl(x)∑∞l=0 h

lP′l (x)

(x − h)∞∑l=0

hlP′l (x) = x

∞∑l=0

hlP′l (x)−

∞∑l=0

hl+1P′l (x)

=∞∑l=1

lhlPl(x)

Proof for xP′

l (x)− P′

l−1(x) = lPl(x): Cont’d

Adjusting h indices so that all become hl , we get

x∞∑l=0

hlP′l (x)−

∞∑l=0

hl+1P′l (x)

= x∞∑l=0

hlP′l (x)−

∞∑l=1

hlP′l−1(x)

=∞∑l=1

lhlPl(x)

So we get the second recursion relation

xP′l (x)− P

′l−1(x) = lPl(x)

Proof for P′

l (x)− xP′

l−1(x) = lPl−1(x)

The first recursion relation is

lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)

Differentiate with respect to x , we get

lP′l = (2l − 1)Pl−1 + (2l − 1)xP

′l−1 − (l − 1)P

′l−2

Rearrange, using second recursion relation, we get

l [P′l − xP

′l−1]− (l − 1)[xP

′l−1 − P

′l−2]

= l [P′l − xP

′l−1]− (l − 1)[(l − 1)Pl−1]

= (2l − 1)Pl−1

Proof for P′

l (x)− xP′

l−1(x) = lPl−1(x): Cont’d

Then

l [P′l − xP

′l−1] = (l − 1)(l − 1)Pl−1 + (2l − 1)Pl−1

= l2Pl−1

So we get the third recursion relation.

P′l − xP

′l−1 = lPl−1

Proof for (1− x2)P′

l (x) = lPl−1(x)− lxPl(x)

Using

P′l (x)− xP

′l−1(x) = lPl−1(x) (6)

xP′l (x)− P

′l−1(x) = lPl(x) (7)

Multiply Eqn. (7) with x and subtracting from Eqn. (6), we get

(1− x2)P′l (x) = lPl−1(x)− lxPl(x)

Proof for (2l + 1)Pl(x) = P′

l+1(x)− P′

l−1(x)

Using first recursion relation

lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)

Replacing l with l + 1, we get

(l + 1)Pl+1 = (2(l + 1)− 1)xPl − (l + 1− 1)Pl−1

= (2l + 1)xPl − lPl−1

Differentiate with respect to x , we get

(l + 1)P′l+1 = (2l + 1)Pl + (2l + 1)xP

′l − lP

′l−1

Proof for (2l + 1)Pl(x) = P′

l+1(x)− P′

l−1(x): Cont’d

Inserting the second recursion formula xP′l − P

′l−1 = lPl into the

previous equation, we get

(l + 1)P′l+1 = (2l + 1)Pl + (2l + 1)(lPl + P

′l−1)− lP

′l−1

(l + 1)P′l+1 − (l + 1)P

′l−1 = (2l + 1)(1 + l)Pl

So we getP

′l+1 − P

′l−1 = (2l + 1)Pl

Proof for (1− x2)P′

l−1(x) = lxPl−1 − lPl(x)

Try to prove this recursion relation on your own.