Connect via MCP →

Enter Calculation

Formula

Advertisement

Results

Hermite polynomial order n = 3
51 points
first Hn(x) = -95, last Hn(x) = 95
i x Hn(x)
0 -2.5 -95
1 -2.4 -81.792
2 -2.3 -69.736
3 -2.2 -58.784
4 -2.1 -48.888
5 -2 -40
6 -1.9 -32.072
7 -1.8 -25.056
8 -1.7 -18.904
9 -1.6 -13.568
10 -1.5 -9
11 -1.4 -5.152
12 -1.3 -1.976
13 -1.2 0.576
14 -1.1 2.552
15 -1 4
16 -0.9 4.968
17 -0.8 5.504
18 -0.7 5.656
19 -0.6 5.472
20 -0.5 5
21 -0.4 4.288
22 -0.3 3.384
23 -0.2 2.336
24 -0.1 1.192
25 0 -0
26 0.1 -1.192
27 0.2 -2.336
28 0.3 -3.384
29 0.4 -4.288
30 0.5 -5
31 0.6 -5.472
32 0.7 -5.656
33 0.8 -5.504
34 0.9 -4.968
35 1 -4
36 1.1 -2.552
37 1.2 -0.576
38 1.3 1.976
39 1.4 5.152
40 1.5 9
41 1.6 13.568
42 1.7 18.904
43 1.8 25.056
44 1.9 32.072
45 2 40
46 2.1 48.888
47 2.2 58.784
48 2.3 69.736
49 2.4 81.792
50 2.5 95

What this calculator does

This tool evaluates the physicists' Hermite polynomial \(H_n(x)\) for a single fixed order n across a sequence of x values. It returns a table of \((x, H_n(x))\) pairs and draws the curve. Hermite polynomials appear throughout quantum mechanics (the energy eigenstates of the harmonic oscillator), probability theory, and numerical analysis (Gauss-Hermite quadrature).

How to use it

Enter the polynomial order n (a non-negative integer such as 0, 1, 2, 3, ...), the initial value of x, the increment (step between successive x values), and the number of repetitions (how many rows to generate). The i-th x value is $$x_i = \text{startX} + i \cdot \text{stepX}$$ for i = 0 to count-1. A negative increment produces a descending table; a zero increment repeats the same x.

The formula

These are the physicists' Hermite polynomials, satisfying the differential equation \(y'' - 2x \cdot y' + 2n \cdot y = 0\) and generated by \(\exp(2xt - t^2)\). We compute them with the stable three-term recurrence $$H_0(x) = 1, \quad H_1(x) = 2x, \quad H_{k+1}(x) = 2x \cdot H_k(x) - 2k \cdot H_{k-1}(x).$$ This avoids factorial overflow. Note these are not the probabilists' \(He_n(x)\), which use \(He_{k+1} = x \cdot He_k - k \cdot He_{k-1}\).

Recurrence relation tree showing how each Hermite polynomial is built from the two previous ones
The three-term recurrence builds each Hermite polynomial from the two preceding orders.

Worked example

For n = 3 the recurrence gives \(H_2(x) = 4x^2 - 2\) and \(H_3(x) = 8x^3 - 12x\). At \(x = -2.5\): $$8(-15.625) + 30 = -95.$$ At \(x = 0\) it is 0, and at \(x = 2.5\) it is +95. With startX = −2.5, stepX = 0.1 and 51 repetitions the table runs from \((-2.5, -95)\) through \((0, 0)\) to \((2.5, 95)\), tracing an odd-symmetric cubic-shaped curve.

FAQ

Which convention is used? The physicists' convention \(H_n\), where \(H_1(x) = 2x\). First few: \(H_0 = 1\), \(H_1 = 2x\), \(H_2 = 4x^2 - 2\), \(H_4 = 16x^4 - 48x^2 + 12\), \(H_5 = 32x^5 - 160x^3 + 120x\).

What if n = 0? \(H_0(x) = 1\) for every x, so the table and graph are a flat line at height 1.

Why do values explode for big n? Hermite polynomials grow extremely fast for large order and large \(|x|\); double precision can overflow beyond roughly 1e308. Keep n and the x range modest for sensible plots.

Overlaid graphs of the first few physicists' Hermite polynomials
Curves of H1 through H4 over a symmetric x-range, showing growing oscillation with order.
Last updated: