Legendre Polynomials Orthogonality Relation

In summary, the conversation is discussing the orthogonality relation for Legendre polynomials, specifically the equation \int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1} and how it can be derived. The conversation covers different equations and approaches, with one person pointing out an error when n=m and the other person defending their approach. The conversation ends with the person expressing frustration and deciding to seek help elsewhere.
  • #1
gulsen
217
0
...and orthogonality relation.

The book says
[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]

So I sat and tried derieving it. First, I gather an inventory that might be useful:

[tex](1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1) = 0[/tex]
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

[tex]P_n(-x) = (-1)^n P_n(x)[/tex]
[tex]P_n'(-x) = -(-1)^n P_n'(x)[/tex]

And started the job. Now...
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]
multiplying both sides with [tex]P_m(x)[/tex] and integrating from -1 to 1
[tex]\int_{-1}^{1} [(1-x^2)P_n'(x)]' P_m(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]
applying a partial to the lefthand side
[tex](1-x^2)P_n'(x) P_m(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_n'(x) P_m'(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]
switching n and m's, ve get another equation like it
[tex](1-x^2)P_m'(x) P_n(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_m'(x) P_n'(x) dx = \int_{-1}^{1} -m(m+1)P_m(x) P_n(x) dx[/tex]

substituing these

[tex][-n(n+1) + m(m+1)]\int_{-1}^{1} P_n(x) P_m(x) dx = [(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))] |_{-1}^{1}[/tex]

now, from the inventory
[tex]P_n(-x)P_m'(-x) = -P_n(x) P_m'(x)[/tex]
the righthand side becomes
[tex]\lim_{x \to 1} 2[(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))[/tex]
and in it's final shape
[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2(1-x^2) \frac {(P_n'(x) P_m(x) - P_m'(x) P_n(x))}{-n(n+1) + m(m+1)}[/tex]

For [tex]n \neq m[/tex], there's no problem, it's straightly 0. But for [tex]n = m[/tex]...
And everything starts going wrong. By stating
[tex](1-x^2)P_n'(x) = \int -n(n+1)P_n(x) dx[/tex]
i'm making a little change

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

Now I plug [tex]P_n(x) = P_m(x)[/tex], [tex]x=1[/tex] and [tex]P_n(1) = 1[/tex] and state

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = 2 (\int P_n(x) dx)_{x=1}[/tex]
which is the nonsense of the day.

Where's the error?
 
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  • #2
In the line after you say

"and in it's final shape",

if you intend to let [itex]n=m[/itex] here, then you have effectively divided your expression by [itex]0[/itex]. This line is only valid for [itex]n\neq m[/itex].
 
  • #3
jpr0 said:
In the line after you say

"and in it's final shape",

if you intend to let [itex]n=m[/itex] here, then you have effectively divided your expression by [itex]0[/itex]. This line is only valid for [itex]n\neq m[/itex].

This's not the error.

If you think it this way, you cannot compute a derivative as well, since you're dividing with 0 when computing it as well. Note that I'm dividing 0 with 0, which is indefinite, and from than line on, all I did was to get rid of this indefinition.
 
  • #4
gulsen said:
This's not the error.

If you think it this way, you cannot compute a derivative as well, since you're dividing with 0 when computing it as well. Note that I'm dividing 0 with 0, which is indefinite, and from than line on, all I did was to get rid of this indefinition.
Complete and utter nonsense. If you don't see that the ALGEBRAIC expression 0/0 is meaningless, you're lost.
 
  • #5
gulsen said:
This's not the error.

If you think it this way, you cannot compute a derivative as well, since you're dividing with 0 when computing it as well. Note that I'm dividing 0 with 0, which is indefinite, and from than line on, all I did was to get rid of this indefinition.

Take your expression from the previous line (or whatever), and set [itex]n=m[/itex], and you will see that you have [tex]0=0[/itex]. I'm not sure what you hope to derive from this expression. For example

[tex]
3\times 0 = 5\times 0
[/tex]

Diving this expression by zero gives [itex]5=3[/itex]. ??
Anyway, if you want to derive the orthogonality relation for Legendre polynomials when [itex]n=m[/itex] then it's easier to derive it from the generating function.
 
  • #6
gulsen said:
This's not the error.

If you think it this way, you cannot compute a derivative as well, since you're dividing with 0 when computing it as well. Note that I'm dividing 0 with 0, which is indefinite, and from than line on, all I did was to get rid of this indefinition.
Nobody who know calculus divides by 0 when finding a derivative!
 
  • #7
Look, I'm not a freshman, and I know my calculus that much. Stop lecturing me on that, it goes very funny.
Derivative:
[tex]\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}[/tex]
If you directly set [tex]\Delta x = 0[/tex] which is the thing you ultimately do, you get a 0/0 which is not a solution, this is where you divide by at term that goes to 0 when finding a derivative.
Since you're too pedantic, you'll probably tell me division by limit of a term that goes to zero is different from division by 0, but that name was coined by jpr0 in the first place for my limit, and none of you give a **** to it.

Tell me how this is limit

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

which is said to be "wrong because I'm dividing by 0" is different than

[tex]\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}[/tex]
I appreciate why most of great physicist hate mathematicians. They seem to know you well. I have asked a simple question, yet what you say is everything far from solution. Most questions I raise at physics department goes smooth, while almost all questions I raise right here goes haywire.

Now go **** yourselves guys. I'll this question at QP forum. Even if I can't get a correctly reply, I know that chances are very low that I'll meet such arrogance and idiocy. Have fun, and lock the topic.
 
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  • #8
There's a huge difference between trying to find the limit

[tex]\lim_{x \rightarrow 0} \frac{\text{stuff}}{x},[/tex]

which you are talking about in post #7, and the limit

[tex]\lim_{x \rightarrow 0} \frac{\text{stuff}}{0},[/tex]

which you are doing in post #1, in the case where m = n.


Exercise: if m = n, then -n(n+1) + m(m+1) = _____.



Another mistake (that isn't what's causing the problem you're seeing) is that

[tex](1 - x^2)P'_n(x) = \int -n(n+1) P_n(x) \, dx[/tex]

is incorrect. The left hand side is a specific function, and the right hand side denotes all of the possible antiderivatives of the integrand. You meant to write something like:

[tex](1 - x^2)P'_n(x) = P'_n(0) + \int_0^x -n(n+1) P_n(t) \, dt[/tex]



I appreciate why most of great physicist hate mathematicians. I have asked a simple question, yet what you say is everything far from solution. Most questions I raise at physics department goes smooth, while almost no question I raise right here goes haywire.
Are you sure its the mathematicians' fault? :grumpy: People are telling you that you are literally performing an arithmetic division by zero, but you keep trying to justify it by talking about limit forms. You accuse people of being "too pedantic" -- but the sloppiness in your understanding of limits is precisely what's at fault here.
 
  • #9
Thanks for pointing this out Hurkly. My reply to you is here.
 
  • #10
The problem is that you're working with Legendre polynomials, which implies here that your n and m are integers. Now if you want to "tend n to m", without simply setting n=m, then you will be working with non-integer Legendre functions where the above recursion relations simply do not work (and they probably even diverge at x=1, x=-1 anyway).

The difference between a derivative and doing what you're doing is that the small parameter (h) in the definition of "derivative" ((f(x+h)-f(x))/h) can be continously taken to zero, when the leading term in f(x+h)-f(x) goes linearly to zero as a function of h. How can you continously let one integer tend to another? You can't.

You asked in your original post what was wrong with what you were doing. I pointed out your error. In another post I also suggested a method of proving the orthogonality relation based on the generating function, which is what you will see in most maths texts - sorry if this wasn't good enough for you.

And finally "Look, I'm not a freshman, and I know my calculus that much." -- clearly you don't.
 
  • #11
If you directly set [itex]\bigtriangleup{x}=0[/itex]which is the thing you ultimately do
No, I don't. Nor does anyone who knows mathematics do so.
We do not set [itex]\bigtriangleup{x}=0[/itex] either "directly" or "ultimately", whatever you mean by those vague terms.
 

1. What are Legendre Polynomials?

Legendre Polynomials are a set of orthogonal polynomials used in mathematical analysis, particularly in solving differential equations and approximating functions. They are named after the French mathematician Adrien-Marie Legendre.

2. What is the Orthogonality Relation for Legendre Polynomials?

The Orthogonality Relation for Legendre Polynomials states that if two different Legendre Polynomials are integrated over the range of -1 to 1, the result will be 0. This means that the polynomials are orthogonal to each other, and their inner product is 0.

3. Why is the Orthogonality Relation important?

The Orthogonality Relation is important because it allows us to use Legendre Polynomials as a basis for approximating other functions. By using the Orthogonality Relation, we can find the coefficients for a polynomial that best fits a given function.

4. How is the Orthogonality Relation derived?

The Orthogonality Relation is derived by using the properties of orthogonality and the generating function of Legendre Polynomials. By applying these properties and integrating over the range of -1 to 1, we can prove that the inner product of two different Legendre Polynomials is 0.

5. What are some applications of the Orthogonality Relation for Legendre Polynomials?

The Orthogonality Relation for Legendre Polynomials is used in many fields of science and engineering, such as in solving physical problems, signal processing, and image reconstruction. It is also used in numerical methods for solving differential equations and in computer graphics for creating smooth curves and surfaces.

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