4

Let $f : \mathbb{R} \to \mathbb{C}$ be defined by

$$f(t) = \exp(I t^2/2),$$

where $I$ is the imaginary unit. Then $f$ is the first derivative of the Euler spiral.

Problem

Provide a non-recursive formula for the n:th derivative $f^{(n)}(t)$ where $n \in \mathbb{N}$.

Instances

Sagemath will happily compute an arbitrary derivative. Here are some of the first, where I have divided away the common factor $f(t)$:

$$ \begin{aligned} f^{(1)}(t) / f(t) & = I*t \\ f^{(2)}(t) / f(t) & = -t^2 + I \\ f^{(3)}(t) / f(t) & = -I*t^3 - 3*t \\ f^{(4)}(t) / f(t) & = t^4 - 6*I*t^2 - 3 \\ f^{(5)}(t) / f(t) & = I*t^5 + 10*t^3 - 15*I*t \\ f^{(6)}(t) / f(t) & = -t^6 + 15*I*t^4 + 45*t^2 - 15*I \\ f^{(7)}(t) / f(t) & = -I*t^7 - 21*t^5 + 105*I*t^3 + 105*t \\ f^{(8)}(t) / f(t) & = t^8 - 28*I*t^6 - 210*t^4 + 420*I*t^2 + 105 \\ \end{aligned} $$

These were computed with the following Sagemath script:

t, k = var('t', 'k')
df = exp(I * (1/2) * t^2)
with assuming(t, 'real'):
    print('\\begin{aligned}')
    for n in range(1, 9):
        print(f'f^{{({n})}}(t) / f(t) & = {(diff(df, t, n) / df).simplify_full()} \\\\')
    print('\\end{aligned}')
kaba
  • 2,881

2 Answers2

4

The derivatives of $g(z) = \exp(-x^2/2)$ can be expressed with the help of the probabilist's Hermite polynomials: $$ g^{(n)}(x) = (-1)^n \operatorname{He}_n(x) g(x) \, , $$ see also Higher derivatives of an exponential function. The Hermite polynomials have the explicit expression $$ \operatorname{He}_n(x) = n! \sum_{m=0}^{\lfloor n/2\rfloor} \frac{(-1)^{m}}{m!(n-2m)!} \frac {x^{n-2m}}{2^{m}} \,. $$ Then $f(t) = \exp(i t^2/2) = g(\omega t)$ where $\omega$ is chosen such that $\omega^2 = -i$. It follows that $$ f^{(n)}(t) = \omega^n (-1)^n \operatorname{He}_n( \omega t) g(\omega t) = P_n(t) f(t) $$ with $P_n(t) = \omega^n (-1)^{n} \operatorname{He}_n( \omega t)$. Using $\omega^2 = -i$ this simplifies to $$ P_n(t) = \sum_{m=0}^{\lfloor n/2\rfloor} \frac{n! i^{n-m}}{2^m m!(n-2m)!} t^{n-2m} \,. $$

The (absolute values of the) coefficients of the Hermite polynomials are also known as Bessel numbers, compare A100861 in The On-Line Encyclopedia of Integer Sequences®.

Check the result with Maxima:

(%i1) P(n, t) := sum(n! * (%i)^(n-m)/(m! * (n-2*m)!) * t^(n-2*m)/2^m, m , 0, floor(n/2))$

(%i2) P(1, t);
(%o2)                                %i t


(%i3) P(2, t);
                                          2
(%o3)                               %i - t


(%i4) P(3, t);
                                       3
(%o4)                            - %i t  - 3 t


(%i5) P(4, t);
                                4         2
(%o5)                          t  - 6 %i t  - 3


(%i6) P(8, t);
                    8          6        4           2
(%o6)              t  - 28 %i t  - 210 t  + 420 %i t  + 105
Martin R
  • 128,226
1

Let the OP's polynomials be $\operatorname{Hi}_n(t)=\frac{f^{(n)}(t)}{f(t)}$, $n\geq0$. By Cauchy's integral formula it can be shown that $$e^{i(\frac{z^2}2+tz)}=\sum_{n=0}^\infty\operatorname{Hi}_n(t)\frac{z^n}{n!}\tag1$$ is their generating function. On the other hand, $$e^{2tz-z^2}=\sum_{n=0}^\infty H_{n}(t)\frac{z^n}{n!}\tag2$$ is the generating function for the physicist's Hermite polynomials. From $(1)$ and $(2)$, $$\operatorname{Hi}_n(t)=\frac{w^{3n}}{(\sqrt2)^n}H_n\left(\frac{t}{\sqrt2\,w}\right)$$ where $w=\exp(\tfrac{\pi i}4)$.

Bob Dobbs
  • 15,712