Mixing
Proof: Legendre Polynomials Solving the Corresponding Differential Equation
$begingroup$
In a homework question, we are asked to show that the Legendre polynomials do indeed solve the Legendre Differential Equation:
$$
frac{d}{dx} left[ (1 - x^2) frac{d}{dx} P_n(x) right] + n (n + 1) P_n(x)
= 0.
$$
According to Wikipedia, it is sufficient to show that after deriving it n+1 times the following equation must hold true:
$$
(x^2 - 1) frac{d}{dx} (x^2 - 1)^n
= 2 n x (x^2 - 1)^n.
$$
This was also given as a hint in the problem. Since I don't want to blindly accept the hint, I was trying to figure out how it was deduced, but didn't manage to.
My question therefore is, how do we arrive at the equation above (the "hint equation")?
I can see, that the left part of the equation is nearly equal to the first part of the Legendre Differential Equation, except for a missing outer derivative.
However, I am somehow missing the steps taken to arrive at the right side: Where does the $2nx$ come from?
real-analysis analysis ordinary-differential-equations legendre-polynomials
edited Jan 18 at 15:15
Viktor Glombik
9841527
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
$endgroup$
add a comment |
$begingroup$
In a homework question, we are asked to show that the Legendre polynomials do indeed solve the Legendre Differential Equation:
$$
frac{d}{dx} left[ (1 - x^2) frac{d}{dx} P_n(x) right] + n (n + 1) P_n(x)
= 0.
$$
According to Wikipedia, it is sufficient to show that after deriving it n+1 times the following equation must hold true:
$$
(x^2 - 1) frac{d}{dx} (x^2 - 1)^n
= 2 n x (x^2 - 1)^n.
$$
This was also given as a hint in the problem. Since I don't want to blindly accept the hint, I was trying to figure out how it was deduced, but didn't manage to.
My question therefore is, how do we arrive at the equation above (the "hint equation")?
I can see, that the left part of the equation is nearly equal to the first part of the Legendre Differential Equation, except for a missing outer derivative.
However, I am somehow missing the steps taken to arrive at the right side: Where does the $2nx$ come from?
real-analysis analysis ordinary-differential-equations legendre-polynomials
edited Jan 18 at 15:15
Viktor Glombik
9841527
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
$endgroup$
$begingroup$
What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction.
$endgroup$
– Raghav
Nov 26 '13 at 21:44
add a comment |
$begingroup$
In a homework question, we are asked to show that the Legendre polynomials do indeed solve the Legendre Differential Equation:
$$
frac{d}{dx} left[ (1 - x^2) frac{d}{dx} P_n(x) right] + n (n + 1) P_n(x)
= 0.
$$
According to Wikipedia, it is sufficient to show that after deriving it n+1 times the following equation must hold true:
$$
(x^2 - 1) frac{d}{dx} (x^2 - 1)^n
= 2 n x (x^2 - 1)^n.
$$
This was also given as a hint in the problem. Since I don't want to blindly accept the hint, I was trying to figure out how it was deduced, but didn't manage to.
My question therefore is, how do we arrive at the equation above (the "hint equation")?
I can see, that the left part of the equation is nearly equal to the first part of the Legendre Differential Equation, except for a missing outer derivative.
However, I am somehow missing the steps taken to arrive at the right side: Where does the $2nx$ come from?
real-analysis analysis ordinary-differential-equations legendre-polynomials
edited Jan 18 at 15:15
Viktor Glombik
9841527
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
$endgroup$
In a homework question, we are asked to show that the Legendre polynomials do indeed solve the Legendre Differential Equation:
$$
frac{d}{dx} left[ (1 - x^2) frac{d}{dx} P_n(x) right] + n (n + 1) P_n(x)
= 0.
$$
According to Wikipedia, it is sufficient to show that after deriving it n+1 times the following equation must hold true:
$$
(x^2 - 1) frac{d}{dx} (x^2 - 1)^n
= 2 n x (x^2 - 1)^n.
$$
This was also given as a hint in the problem. Since I don't want to blindly accept the hint, I was trying to figure out how it was deduced, but didn't manage to.
My question therefore is, how do we arrive at the equation above (the "hint equation")?
I can see, that the left part of the equation is nearly equal to the first part of the Legendre Differential Equation, except for a missing outer derivative.
However, I am somehow missing the steps taken to arrive at the right side: Where does the $2nx$ come from?
real-analysis analysis ordinary-differential-equations legendre-polynomials
real-analysis analysis ordinary-differential-equations legendre-polynomials
edited Jan 18 at 15:15
Viktor Glombik
9841527
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
edited Jan 18 at 15:15
Viktor Glombik
9841527
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
edited Jan 18 at 15:15
Viktor Glombik
9841527
edited Jan 18 at 15:15
Viktor Glombik
9841527
edited Jan 18 at 15:15
Viktor Glombik
9841527
9841527
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
asked Nov 25 '13 at 18:45
R.G.R.G.
359210
359210
$begingroup$
What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction.
$endgroup$
– Raghav
Nov 26 '13 at 21:44
add a comment |
$begingroup$
What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction.
$endgroup$
– Raghav
Nov 26 '13 at 21:44
$begingroup$
What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction.
$endgroup$
– Raghav
Nov 26 '13 at 21:44
$begingroup$
What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction.
$endgroup$
– Raghav
Nov 26 '13 at 21:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is the other way around: if you wish to prove Rodrigues' formula
$$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
where $f^{(n)}$ is symbolic for the $n$th derivative of $f$, then the first step is to prove that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the differential equation.
When $$[(1-x^2)y']' + n(n+1)y = 0$$
then obviously also $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$
So we wish to somehow evaluate $$(1-x^2)[(x^2-1)^n]^{(n+2)} - 2x[(x^2-1)^n]^{(n+1)}$$
Now starting with $$(x^2-1) cdot [(x^2-1)^n]^{(1)} = (x^2-1) cdot n(x^2-1)^{(n-1)} cdot 2x = 2nx(x^2-1)^{n}$$
and taking left and right the $(n+1)$st derivative, we find that
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + binom{n+1}{1}cdot 2x cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{2}cdot 2 cdot [(x^2-1)^n]^{(n)} \ = 2nx cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{1}cdot 2n cdot [(x^2-1)^n]^{(n)}$$
or
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$
which leads to
$$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$
or, equivalently
$$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$
which shows that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the Legendre differential equation.
Normalizing $P_n = alpha cdot F$ by requiring $P_n(1) = 1$, we then find the factor $alpha = 1/2^nn!$
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
It is the other way around: if you wish to prove Rodrigues' formula
$$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
where $f^{(n)}$ is symbolic for the $n$th derivative of $f$, then the first step is to prove that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the differential equation.
When $$[(1-x^2)y']' + n(n+1)y = 0$$
then obviously also $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$
So we wish to somehow evaluate $$(1-x^2)[(x^2-1)^n]^{(n+2)} - 2x[(x^2-1)^n]^{(n+1)}$$
Now starting with $$(x^2-1) cdot [(x^2-1)^n]^{(1)} = (x^2-1) cdot n(x^2-1)^{(n-1)} cdot 2x = 2nx(x^2-1)^{n}$$
and taking left and right the $(n+1)$st derivative, we find that
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + binom{n+1}{1}cdot 2x cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{2}cdot 2 cdot [(x^2-1)^n]^{(n)} \ = 2nx cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{1}cdot 2n cdot [(x^2-1)^n]^{(n)}$$
or
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$
which leads to
$$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$
or, equivalently
$$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$
which shows that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the Legendre differential equation.
Normalizing $P_n = alpha cdot F$ by requiring $P_n(1) = 1$, we then find the factor $alpha = 1/2^nn!$
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
It is the other way around: if you wish to prove Rodrigues' formula
$$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
where $f^{(n)}$ is symbolic for the $n$th derivative of $f$, then the first step is to prove that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the differential equation.
When $$[(1-x^2)y']' + n(n+1)y = 0$$
then obviously also $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$
So we wish to somehow evaluate $$(1-x^2)[(x^2-1)^n]^{(n+2)} - 2x[(x^2-1)^n]^{(n+1)}$$
Now starting with $$(x^2-1) cdot [(x^2-1)^n]^{(1)} = (x^2-1) cdot n(x^2-1)^{(n-1)} cdot 2x = 2nx(x^2-1)^{n}$$
and taking left and right the $(n+1)$st derivative, we find that
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + binom{n+1}{1}cdot 2x cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{2}cdot 2 cdot [(x^2-1)^n]^{(n)} \ = 2nx cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{1}cdot 2n cdot [(x^2-1)^n]^{(n)}$$
or
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$
which leads to
$$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$
or, equivalently
$$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$
which shows that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the Legendre differential equation.
Normalizing $P_n = alpha cdot F$ by requiring $P_n(1) = 1$, we then find the factor $alpha = 1/2^nn!$
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
It is the other way around: if you wish to prove Rodrigues' formula
$$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
where $f^{(n)}$ is symbolic for the $n$th derivative of $f$, then the first step is to prove that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the differential equation.
When $$[(1-x^2)y']' + n(n+1)y = 0$$
then obviously also $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$
So we wish to somehow evaluate $$(1-x^2)[(x^2-1)^n]^{(n+2)} - 2x[(x^2-1)^n]^{(n+1)}$$
Now starting with $$(x^2-1) cdot [(x^2-1)^n]^{(1)} = (x^2-1) cdot n(x^2-1)^{(n-1)} cdot 2x = 2nx(x^2-1)^{n}$$
and taking left and right the $(n+1)$st derivative, we find that
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + binom{n+1}{1}cdot 2x cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{2}cdot 2 cdot [(x^2-1)^n]^{(n)} \ = 2nx cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{1}cdot 2n cdot [(x^2-1)^n]^{(n)}$$
or
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$
which leads to
$$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$
or, equivalently
$$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$
which shows that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the Legendre differential equation.
Normalizing $P_n = alpha cdot F$ by requiring $P_n(1) = 1$, we then find the factor $alpha = 1/2^nn!$
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
$endgroup$
It is the other way around: if you wish to prove Rodrigues' formula
$$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
where $f^{(n)}$ is symbolic for the $n$th derivative of $f$, then the first step is to prove that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the differential equation.
When $$[(1-x^2)y']' + n(n+1)y = 0$$
then obviously also $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$
So we wish to somehow evaluate $$(1-x^2)[(x^2-1)^n]^{(n+2)} - 2x[(x^2-1)^n]^{(n+1)}$$
Now starting with $$(x^2-1) cdot [(x^2-1)^n]^{(1)} = (x^2-1) cdot n(x^2-1)^{(n-1)} cdot 2x = 2nx(x^2-1)^{n}$$
and taking left and right the $(n+1)$st derivative, we find that
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + binom{n+1}{1}cdot 2x cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{2}cdot 2 cdot [(x^2-1)^n]^{(n)} \ = 2nx cdot [(x^2-1)^n]^{(n+1)} + binom{n+1}{1}cdot 2n cdot [(x^2-1)^n]^{(n)}$$
or
$$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$
which leads to
$$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$
or, equivalently
$$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$
which shows that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the Legendre differential equation.
Normalizing $P_n = alpha cdot F$ by requiring $P_n(1) = 1$, we then find the factor $alpha = 1/2^nn!$
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
edited Apr 23 '14 at 5:55
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
answered Apr 23 '14 at 5:49
Maestro13Maestro13
1,081724
1,081724
add a comment |
add a comment |
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$begingroup$
What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction.
$endgroup$
– Raghav
Nov 26 '13 at 21:44