MHB How can the polynomial $L_n(x)$ be used to solve the equation Laguerre?

[evinda]

Hello! (Wave)

The differential equation $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$, that is called equation Laguerre, is given.

Let $L_n$ be the polynomial $L_n(x)=e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x})$ (show that it is a polynomial), $n=1,2,3, \dots$. Show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$.

So we have to substitute $L_n$ in the differential equation and see that it is only satisfied for $a=n$, right? )Do we differentiate the Leguerre polynomial as follows?
$$$$
$$\frac{d}{dx} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+\frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})$$

$$\frac{d^2}{dx^2} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+ e^{x} \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+\frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})$$Or have I done something wrong?

 

[I like Serena]

Hey! (Smile)

I believe you have dropped an $e^x$. (Worried)

 

[evinda]

Hey! (Smile)

I believe you have dropped an $e^x$. (Worried)

At which point? (Thinking)

 

[I like Serena]

At which point? (Thinking)

$\d {}x (e^x f(x))= e^xf(x) + e^xf'(x)$ (Sweating)

 

[evinda]

Oh yes, right... So it is as follows, right? (Thinking)
$$\frac{d}{dx} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+e^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})$$

$$\frac{d^2}{dx^2} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+ e^{x} \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+e^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+e^x \frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})=e^x \frac{d^n}{dx^n} (x^n e^{-x})+ 2e^{x} \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+e^x \frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})$$

Substituting $L_n$ at the equation I got the following:

$xy''+(1-x)y'+ay=0 \\ \Rightarrow x e^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+x e^x \frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})+(a+1) e^x \frac{d^n}{dx^n}(x^n e^{-x})=0 $

Is it right or have I done something wrong? (Worried)

If it is right, how could we show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$? (Thinking)

 

[I like Serena]

Substituting $L_n$ at the equation I got the following:
$xy''+(1-x)y'+ay=0 \\ \Rightarrow x e^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+x e^x \frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})+(a+1) e^x \frac{d^n}{dx^n}(x^n e^{-x})=0 $

Is it right or have I done something wrong? (Worried)

I think you've dropped an $e^x$ in the term with the (n+1)th derivative. (Thinking)

If it is right, how could we show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$? (Thinking)

How about making the derivatives the same?
That is, differentiate once respectively twice where applicable? (Wondering)

 

[evinda]

I think you've dropped an $e^x$ in the term with the (n+1)th derivative. (Thinking)

It should be as follows, right? (Thinking)

$$xe^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+e^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+(a+1)e^x \frac{d^n}{dx^n}(x^n e^{-x})+x e^x \frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})=0$$

How about making the derivatives the same?
That is, differentiate once respectively twice where applicable? (Wondering)

I haven't understood it... Could you explain it further to me? (Thinking)

 

[I like Serena]

It should be as follows, right? (Thinking)

$$xe^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+e^x \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+(a+1)e^x \frac{d^n}{dx^n}(x^n e^{-x})+x e^x \frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})=0$$

Yep. (Nod)

I haven't understood it... Could you explain it further to me? (Thinking)

$$\d {} x \left(\frac {d^n} {dx^n} f(x)\right) = \frac {d^{n+1}} {dx^{n+1}} f(x) = \frac {d^n}{dx^n}\left( \d {} x f(x)\right) = \frac {d^n}{dx^n}f'(x)$$
(Wasntme)

 

[evinda]

$$\d {} x \left(\frac {d^n} {dx^n} f(x)\right) = \frac {d^{n+1}} {dx^{n+1}} f(x) = \frac {d^n}{dx^n}\left( \d {} x f(x)\right) = \frac {d^n}{dx^n}f'(x)$$
(Wasntme)

You mean that we have to write it like that? (Thinking)

$$x e^x \frac{d^n}{dx^n} \left( nx^ne^{-x}-x^ne^{-x} \right)+e^x \frac{d^n}{dx^n} \left( n x^{n-1}e^{-x}-x^n e^{-x} \right)+(a+1) e^x \frac{d^n}{dx^n}(x^n e^{-x})+x e^x \frac{d^n}{dx^n} \left( n(n-1) x^{n-2} e^{-x}-n x^{n-1} e^{-x}- n x^{n-1} e^{-x}+x^n e^{-x}\right)=0$$

 

[Opalg]

The differential equation $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$, that is called equation Laguerre, is given.

Let $L_n$ be the polynomial $L_n(x)=e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x})$ (show that it is a polynomial), $n=1,2,3, \dots$. Show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$.

There does not seem to be any easy way to show that the Rodrigues polynomial $L_n(x)$ satisfies the Laguerre equation with $a=n$. I came across the paper Rodrigues-type formulae for Hermite and Laguerre polynomials by V. Radulescu, which has an ingenious proof of this result. For me, this is a particularly attractive method because it uses the machinery of operators to simplify the algebraic calculations. (Operators, algebra, Opalg, get the connection?)

For a differentiable function $y$, let $D$ denote the operation of differentiation, let $M$ denote the operation of multiplication by $x$, and let $I$ denote the identity operator. So $D(y) = \frac{dy}{dx}$, $M(y) = xy$ and $I(y) = y.$ Using this notation, the product rule for differentiation, $\frac d{dx}(xy) = x\frac{dy}{dx} + y$, becomes $DM(y) = MD(y) + I(y)$, which we can write as $DM = MD + I$.

Laguerre's equation $xy''+(1-x)y'+ny=0$ becomes $MD^2(y) + (I-M)Dy + nI(y) = 0$, or more simply $MD^2 + (I-M)D + nI = 0$. Using the relation $DM = MD + I$, you can re-write that as $(D-I)MD = -nI.$

We want to show that a solution to that equation is the function $L_n(x) = e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x}) = e^xD^nM^n(e^{-x})$. The first step in the solution is to show that $L_n(x) = (D-I)^nM^n(y_0)$, where $y_0$ is the constant function $1$. This is done by using induction on $n$ to show the more general result $e^xD^nM^k(e^{-x}) = (D-I)^nM^k(y_0)$ for any positive integer $k$.

So the aim is to show that if $y_n = (D-I)^nM^n(y_0)$, then $(D-I)MD(y_n) = -ny_n,$ in other words $y_n$ is an eigenfunction for the operator $S = (D-I)MD$, corresponding to the eigenvalue $-n.$

That result is still not easy to prove. There are several intermediate steps, some of which are just algebraic calculations, while others have to be proved by induction. They are

$(1)\qquad (D-I)^nM - M(D-I)^n = n(D-I)^{n-1},$

$(2)\qquad (D-I)^{n+1}M^{n+1} = \bigl((D-I)M + nI\bigr)(D-I)^nM^n,$

$(3)\qquad S(D-I)M - (D-I)MS = S - (D-I)M \quad (\text{where }S = (D-I)MD),$

and finally

$(4)\qquad S(D-I)^nM^n(y_0) = -n(D-I)^nM^n(y_0).$

Details of all these calculations are in Radulescu's paper. The key idea throughout is to use the product rule $DM = MD + I$ in order to shift $M$s past $D$s.

 

[evinda]

There does not seem to be any easy way to show that the Rodrigues polynomial $L_n(x)$ satisfies the Laguerre equation with $a=n$. I came across the paper Rodrigues-type formulae for Hermite and Laguerre polynomials by V. Radulescu, which has an ingenious proof of this result. For me, this is a particularly attractive method because it uses the machinery of operators to simplify the algebraic calculations. (Operators, algebra, Opalg, get the connection?)

For a differentiable function $y$, let $D$ denote the operation of differentiation, let $M$ denote the operation of multiplication by $x$, and let $I$ denote the identity operator. So $D(y) = \frac{dy}{dx}$, $M(y) = xy$ and $I(y) = y.$ Using this notation, the product rule for differentiation, $\frac d{dx}(xy) = x\frac{dy}{dx} + y$, becomes $DM(y) = MD(y) + I(y)$, which we can write as $DM = MD + I$.

Laguerre's equation $xy''+(1-x)y'+ny=0$ becomes $MD^2(y) + (I-M)Dy + nI(y) = 0$, or more simply $MD^2 + (I-M)D + nI = 0$. Using the relation $DM = MD + I$, you can re-write that as $(D-I)MD = -nI.$

We want to show that a solution to that equation is the function $L_n(x) = e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x}) = e^xD^nM^n(e^{-x})$. The first step in the solution is to show that $L_n(x) = (D-I)^nM^n(y_0)$, where $y_0$ is the constant function $1$. This is done by using induction on $n$ to show the more general result $e^xD^nM^k(e^{-x}) = (D-I)^nM^k(y_0)$ for any positive integer $k$.

So the aim is to show that if $y_n = (D-I)^nM^n(y_0)$, then $(D-I)MD(y_n) = -ny_n,$ in other words $y_n$ is an eigenfunction for the operator $S = (D-I)MD$, corresponding to the eigenvalue $-n.$

That result is still not easy to prove. There are several intermediate steps, some of which are just algebraic calculations, while others have to be proved by induction. They are

$(1)\qquad (D-I)^nM - M(D-I)^n = n(D-I)^{n-1},$

$(2)\qquad (D-I)^{n+1}M^{n+1} = \bigl((D-I)M + nI\bigr)(D-I)^nM^n,$

$(3)\qquad S(D-I)M - (D-I)MS = S - (D-I)M \quad (\text{where }S = (D-I)MD),$

and finally

$(4)\qquad S(D-I)^nM^n(y_0) = -n(D-I)^nM^n(y_0).$

Details of all these calculations are in Radulescu's paper. The key idea throughout is to use the product rule $DM = MD + I$ in order to shift $M$s past $D$s.

Interesting... (Thinking)

Do you maybe have also an idea of an other way or is it the only one? (Thinking)

 

[Opalg]

Do you maybe have also an idea of an other way or is it the only one? (Thinking)

There is a closed form for the polynomial $L_n(x)$, namely $$L_n(x) = \sum_{k=0}^n (-1)^k(n-k)!{\textstyle {n\choose k}^2}x^k.$$ I imagine that the original proof must have made use of this, but I do not know for sure.