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The derivative of the exponential map is given by (wiki):

$$ \frac{d}{dt} e^{X(t)} = e^{X(t)} \frac{1 - e^{-ad_{X(t)}}}{ad_{X(t)}} \frac{d}{dt}X(t) $$

Is there a reasonable formula for higher order derivatives: $$ \frac{d^n}{dt^n} e^{X(t)} = ?? $$

Or more ideally: $$ \frac{d}{dt_1}\dots\frac{d}{dt_n} e^{X(t_1, \dots, t_n)} = ?? $$

I have tried the direct formal argument from wiki for $t_1, t_2$ but got lost in it pretty quickly.

tom
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  • I am familiar with ${d \over dt}e^{X(t)}=\int _0^1 ds e^{(1-s)X(t)}{d \over dt}X(t) e^{sX(t)}$ which is easy to generalize – user619894 May 19 '22 at 13:27
  • That is exactly what I was trying but I would like, at least, reduce it to one integral. Direct application results in n integrals for n-th derivative. – tom May 19 '22 at 13:44
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    of course. It is the Dyson Series. – user619894 May 19 '22 at 13:45
  • Never heard of Dyson series, by looking at the wiki articles it is not obvious to me at all that it is directly related. – tom May 19 '22 at 13:53
  • It is a special case, where $A(t)=D+B(t)$ but even there you typically need the multiple integrals. – user619894 May 19 '22 at 14:25

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