3
$\begingroup$

How to find $n $ th derivative of complex exponential \begin{align*} \frac{d^{n}}{dx^{n}}e^{ix^2/(2a)} \end{align*} One method is series soultion. I want a formula which works faster in my programme. Please help me.

  • 1
    The chain rule. – 2017-01-18
  • 0
    @Alqatrkapa Nope, you need more than that. – 2017-01-18
  • 0
    @SimpleArt It gives a recursive algorithm for computing the derivative. It doesn't give a closed form, sure. – 2017-01-18
  • 1
    https://en.wikipedia.org/wiki/Hermite_polynomials – 2017-01-18
  • 0
    @Alqatrkapa Yes it does, there is a closed form... – 2017-01-18
  • 0
    can you say something to the variables? – 2017-01-18
  • 2
    @Dr.SonnhardGraubner You mean like talking to them? – 2017-01-18
  • 0
    no variables can not talk to you – 2017-01-18
  • 0
    i mean about the variables – 2017-01-18
  • 1
    But you asked if we could say something to the variables. :D – 2017-01-18
  • 0
    @skorpion: Could you please explain why a series solution (which gives a closed formula in terms of factorials) isn't fast enough for your purposes? – 2017-01-18
  • 0
    @AndrewD.Hwang I am thinking he had an infinite series in mind... – 2017-01-18
  • 0
    @skorpion: Without more information on what exactly you need (an algebraic formula for the $n$th derivative, a series expansion, an efficient means of evaluating the $n$th derivative at an arbitrary point...) it's difficult to answer your question in a way likely to be useful to you. I've voted to close, pending additional context. – 2017-01-18
  • 0
    @AndrewD.Hwang: thanks.....I got my answer. I am evaluation a complex function. It has eight summations already........Thats why its slow. – 2017-01-18

1 Answers 1

2

One may use Faà di Bruno's formula, with $f(x)=e^x$ and $g(x)=ix^2/2a$, whereupon we get

$$\frac{d^n}{dx^n}f(g(x))=\sum_{k=1}^n f(g(x))\cdot B_{n,k}\left(ix/2a,i/2a,0,0,\dots,0\right)$$

where $B_{n,k}$ are the bell polynomials. (closed form, though probably not good for numerical computations. As per WolframAlpha,

$$\frac{d^n}{dx^n}e^{cx^2}=2^ne^{cx^2}(cx)^nn!\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-4cx^2)^{-k}}{k!(n-2k)!}$$

  • 0
    This will likely be only a starting point for OP if he is using an ordinary programming language which doesn't handle complex $c$. He'll have to split it out into real and imaginary parts in that case, so there is more work to be done here. I don't know what language he's using, of course, just thinking about what I would be doing in this case (using C++ with native types, say). But +1 for you, nonetheless. – 2017-01-18
  • 0
    D: I wonder why you would need that so badly... – 2017-01-18
  • 0
    Yes, especially in such generality. I would think that in any specific application OP might need some specific derivative, not all of them. But who knows, he hasn't posted much about why he wants this. – 2017-01-18