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Laguerre Polynomial L3(x)
5.666667
value at the first x; 2.333333 at the last x · 51 rows
x L3(x)
-1 5.666667
-0.9 5.0365
-0.8 4.445333
-0.7 3.892167
-0.6 3.376
-0.5 2.895833
-0.4 2.450667
-0.3 2.0395
-0.2 1.661333
-0.1 1.315167
0 1
0.1 0.714833
0.2 0.458667
0.3 0.2305
0.4 0.029333
0.5 -0.145833
0.6 -0.296
0.7 -0.422167
0.8 -0.525333
0.9 -0.6065
1 -0.666667
1.1 -0.706833
1.2 -0.728
1.3 -0.731167
1.4 -0.717333
1.5 -0.6875
1.6 -0.642667
1.7 -0.583833
1.8 -0.512
1.9 -0.428167
2 -0.333333
2.1 -0.2285
2.2 -0.114667
2.3 0.007167
2.4 0.136
2.5 0.270833
2.6 0.410667
2.7 0.5545
2.8 0.701333
2.9 0.850167
3 1
3.1 1.149833
3.2 1.298667
3.3 1.4455
3.4 1.589333
3.5 1.729167
3.6 1.864
3.7 1.992833
3.8 2.114667
3.9 2.2285
4 2.333333

What is the Laguerre Polynomial Table Calculator?

This tool tabulates and plots the Laguerre polynomial \(L_n(x)\) over a sequence of x values. The Laguerre polynomials are the orthogonal-polynomial solutions of the differential equation \(x\cdot y'' + (1 - x)\cdot y' + n\cdot y = 0\), and they appear throughout quantum mechanics (the radial part of the hydrogen atom), numerical integration (Gauss-Laguerre quadrature), and signal processing. This calculator uses the standard normalization with \(L_n(0) = 1\).

Graph of the first four Laguerre polynomials plotted over a range of x
Curves of the first few Laguerre polynomials L_n(x) over a range of x.

How to use it

Enter four numbers: the order n (a non-negative integer), the initial value of x, the increment (step) between successive x values, and the number of rows. The calculator generates \(x = \text{startX},\ \text{startX} + \text{stepX},\ \text{startX} + 2\cdot\text{stepX},\ \ldots\) and evaluates \(L_n(x)\) at each, returning a two-column table and a line graph.

The formula explained

Rather than expanding the polynomial, the calculator uses the numerically stable three-term recurrence: \(L_0(x) = 1\), \(L_1(x) = 1 - x\), and for \(k \ge 1\), $$L_{k+1}(x) = \frac{(2k + 1 - x)\cdot L_k(x) - k\cdot L_{k-1}(x)}{k + 1}.$$ This requires only \(O(n)\) work per point. The first few polynomials are \(L_2(x) = 1 - 2x + \frac{x^2}{2}\) and \(L_3(x) = 1 - 3x + 1.5x^2 - \frac{x^3}{6}\).

Diagram showing two previous polynomials combining to produce the next via recurrence
The recurrence builds L_{k+1}(x) from the two previous polynomials.

Worked example

For \(n = 3\) and \(x = -1\): $$L_3(-1) = 1 + 3 + 1.5 + 0.16667 = 5.66667.$$ Checking with the recurrence: \(L_0 = 1\), \(L_1 = 2\), \(L_2 = 3.5\), $$L_3 = \frac{6\cdot 3.5 - 2\cdot 2}{3} = \frac{17}{3} = 5.66667.$$ At \(x = 0\), \(L_3(0) = 1\); at \(x = 1\), \(L_3(1) = -0.66667\).

FAQ

What normalization is used? The standard form with \(L_n(0) = 1\), not the unnormalized \(n!\cdot L_n(x)\) seen in some references.

What if n = 0? \(L_0(x) = 1\) everywhere, a flat horizontal line. For \(n = 1\) you get the straight line \(1 - x\).

How large can n be? The recurrence is stable for moderate \(n\). For very large \(n\) or large \(|x|\) the values grow rapidly and floating-point overflow may eventually occur.

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