# Orthogonality of Legendre polynomials using specific properties

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.

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

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

QED

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.

• 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. – Somos Nov 14 '20 at 21:55
• 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? – 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.

• 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... – CogitoErgoCogitoSum Nov 14 '20 at 20:27
• 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. – Somos Nov 14 '20 at 20:45