I'm having significant issues with a problem and would appreciate any help at all with it. It is regarding proving the orthogonality of Legendre polynomials using a specific recursion formula and property of Legendre polynomials.

I have been given the following relation $$x P'_n(x) -P'_{n-1}(x) = n P_n (x) $$ And that $$ \int^{1}_{-1} f(x) P_n (x) dx = 0 $$ for any polynomial $f(x)$ of degree less than $n$.

I must then show that for $ n \geq 1$

$$ \int^{1}_{-1} P_n (x)^2 dx = \frac{1}{2n} \int^{1}_{-1} x \frac{d}{dx} (P_n (x)^2)dx $$

And then determine the value of the integral, that being $\frac{2}{2n+1}$. I must use the two stated facts, I'd be grateful for any help with this, thank you guys.

  • $\begingroup$ Have you tried integration by parts yet? $\endgroup$ – Somos Nov 14 '20 at 0:31
  • $\begingroup$ My first thought too but I decided its a no-go. Id like to see your approach, @Somos. $\endgroup$ – CogitoErgoCogitoSum Nov 14 '20 at 1:31

$$\int\limits_{-1}^{1} P_n(x)^2 \;dx = $$ $$\int\limits_{-1}^{1} P_n(x)P_n(x) \;dx = $$ $$\frac{1}{2n}\int\limits_{-1}^{1} 2n P_n(x)P_n(x) \;dx = $$

Substituting your first relation in: $$\frac{1}{2n}\int\limits_{-1}^{1} 2 \left[x P'_n(x) - P'_{n-1}(x)\right] P_n(x)\;dx = $$

$$\frac{1}{2n}\int\limits_{-1}^{1} 2 x P'_n(x) P_n(x) \; dx - \frac{1}{2n}\int\limits_{-1}^{1} 2 P'_{n-1}(x) P_n(x)\;dx = $$

Given your second property defining orthogonality (since after all $P'$ is of lesser degree than $P$) the entire second integral drops out.

$$\frac{1}{2n}\int\limits_{-1}^{1} x 2 P_n(x) P'_n(x) \; dx = $$

Reverse of the chain rule:

$$\frac{1}{2n}\int\limits_{-1}^{1} x \frac{d}{dx} P^2_n(x) \; dx $$


And for anyone who asks/wonders, or is inclined to criticize... I answered this question because it was asked, and this is a Q&A forum. I shouldnt have to justify answering questions on a Q&A forum though, but this is the culture here.

  • $\begingroup$ Did you read the part in the question stating: "And then determine the value of the integral, that being 2/(2n+1)."? You seem to have not done that part yet. $\endgroup$ – Somos Nov 14 '20 at 21:55
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    $\begingroup$ Youre right, I didnt. I dont feel the need to do all 100% of the work when OP is hindered only by the first 10%. Dont know what your point is. I dont believe in "handing out answers". OP should be able to do some of it, dont you think? $\endgroup$ – CogitoErgoCogitoSum Nov 14 '20 at 23:20

I immediately thought of integration by parts. Here is my work:

By the product rule of derivative we have $$ \frac{d}{dx} x\, P_n(x)^2 = P_n(x)^2 + x\, \frac{d}{dx}(P_n(x)^2) \tag{1} $$

Note that Legendre polynomials have the property $$ P_n(1) = 1,\quad \text{ and } \quad P_n(-1)=(-1)^n. \tag{2} $$

By integrating equation $(1)$ and using equation $(2)$ we get $$ 2 = 1 - (-1) = \int_{-1}^1 P_n(x)^2 dx + \int_{-1}^1 x\, \frac{d}{dx}(P_n(x)^2) dx. \tag{3} $$

Define $\, a_n := \int_{-1}^1 P_n(x)^2 dx.\,$ Equation $(3)$ implies that $$ 2 - a_n = \int_{-1}^1 x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{4} $$ We are given the property that $$ x P'_n(x) -P'_{n-1}(x) = n P_n (x). \tag{5} $$ Multiply this equation $(5)$ by $\,P_n(x)\,$ and integrate to get $$ \int_{-1}^1\!P_n(x)(xP'_n(x)\!-\!P'_{n-1}(x))dx \!=\!\int_{-1}^1\!P_n(x)nP_n(x)dx. \tag{6} $$ We are also given the property that $$ \int_{-1}^1 P_n(x)\,f(x)\,dx = 0 \tag{7} $$ for any polynomial $\,f(x)\,$ of degree less than $\,n.\,$ Now notice that $$ x\, \frac{d}{dx}(P_n(x)^2) = 2\,x\,P_n(x)P'_n(x). \tag{8} $$ Combining equations $(6)$, $(7)$, and $(8)$ we get that $$ \frac12(2-a_n) = n\,a_n. \tag{9}$$ Dividing this equation $(9)$ through by $\,n\,$ and using the definition of $\,a_n\,$ and equation $(4)$ gives the result we wanted: $$ \int_{-1}^1 P_n(x)^2 dx = \frac{2-a_n}{2n} = \frac1{2n} \int_{-1}^1 x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{10}$$ Also, solving for $\,a_n\,$ in equation $(9)$ gives $$ a_n = \frac2{2n+1} \tag{11}$$ which was also requested.

NOTE: The property in equation $(2)$ was not given to us. It would have to be proven using equations $(5)$ and $(7)$ which are given to us. It can be done, and would take some effort, but I have not given the steps required.

  • $\begingroup$ A few points/criticisms, if you dont mind. For one thing, it isnt obvious how your suggestion to use "integration by parts" is being utilized. Frankly I dont see I by P in your work. Secondly, your work doesnt flow all that well... you make a few leaps. The logic is sound but effort must be used to comprehend the somewhat convoluted steps. Its just a bit scattered and disorderly. Thirdly, property (2), although true, was not one of the properties OP provided; Im not sure its appropriate to use it. Just some things to consider... $\endgroup$ – CogitoErgoCogitoSum Nov 14 '20 at 20:27
  • $\begingroup$ In the Wikipedia article on Integration by parts is stated "The rule can be thought of as an integral version of the product rule of differentiation." That is what I used. You say "leaps". I say "steps". I don't see anything convoluted. In my NOTE at the end I admit the status of property (2), but the original quesition did not explicitly state than no other properties of Legendre polynomials can be used. If you have a better proof, you can answer the question yourself, which, of course, you have done. Thanks for commenting. $\endgroup$ – Somos Nov 14 '20 at 20:45

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