# What is the problem in happily using the MacLaurin expansion of $e^x$ with $e^{ix}$?

I would like some mathematician confirming this very simple argument, or telling me otherwise what am I missing:

I want to have a linear approximation to this function: $$e^{iA}$$ around $A=0$, with $A$ real.

I have seen in Lagrangian Field Theory how this is happily used as that approximation:

$$1+iA$$

That is, simply using the MacLaurin expansion expression and changing $x$ into $iA$

Please confirm that this makes sense only because:

• the Laurent series for $e^{Z}$ (Z complex) is formally the same as the usual MacLaurin expansion, AND
• because the function is analytical around $A=0+0i$ AND
• because $A$ is restricted to be real,

But in any other case ($A$ not generally real and therefore 2-dimensional, the real function well behaved but not the corresponding complex function in that point, etc) it would be a complete nonsense to blindly use the MacLaurin expansion of the corresponding real function. Is that right?

• I believe the correct spelling is Maclaurin, not McLaurin... – gt6989b Feb 19 '13 at 17:56
• @Jyrki Lahtonen, I encourage you to post your comment as an answer. The ability of expanding matrix operators in series has some importance in Quantum Mechanics, where additionally, the fact you mention, that $AB$ equals or not $BA$, has a physical meaning. – Mephisto Feb 19 '13 at 21:22
• @gt6989b, thanks. I've corrected the spelling of MacLaurin – Mephisto Feb 19 '13 at 21:35

As pointed out by joriki the exponential functin as well as all functions represented by a converging power series make perfect sense in the complex domain. Even more so, we can plug in a square matrix in to a power series and it still makes sense as long as we identify the zeroth power as the identity matrix.

A standard example is the 2x2 matrix $$A=\left(\begin{array}{rr}0&−1\\1&0\end{array}\right).$$

We easily get by series expansion (using the Maclaurin series of trig functions) that $$e^{xA}=\left(\begin{array}{rr}\cos x&−\sin x\\\sin x&\cos x\end{array}\right).$$ There is a caveat though. The familiar rule $$e^{B+C}=e^Be^C$$ may not hold for matrices $B$ and $C$ unless $BC=CB$. Observe that with the above matrix $A$ the matrices $B=xA$ and $C=yA$ do commute. Therefore the equation $e^{B+C}=e^Be^C$ turns into the matrix relation $$\left(\begin{array}{rr}\cos (x+y)&−\sin (x+y)\\\sin (x+y)&\cos (x+y)\end{array}\right)=\left(\begin{array}{rr}\cos x&−\sin x\\\sin x&\cos x\end{array}\right)\left(\begin{array}{rr}\cos y&−\sin y\\\sin y&\cos y\end{array}\right),$$ which has the hopefully familiar interpretation that rotation by angle $y$ followed by rotation by angle $x$ gives rotation by angle $x+y$. I believe that in Lie group language this is expressed by saying that $A$ is an infinitesimal generator for the group of rotations. This also shows in the derivative rule $$\frac{d}{dx}e^{Ax}=Ae^{xA}$$ that I invite you to verify as an exercise. Basically it means that if we start rotating the point $P=(x_1,x_2)$ about the origin, it will start its motion into the direction $A(x_1,x_2)^T$.

Caveat: in higher dimensions the infinitesimal generators for rotations about different axes do not commute, so we can only combinet rotations about the same axis using this language.

What makes this tick? The key ingredient is that we have a notion of an absolute value in all theses spaces: reals, complex numbers, square matrices. Presumably you know what that is in the cases of the number systems, for square matrices $A$ it is given by $$\left\Vert A\right\Vert=\max_{x\in\mathbb{C}^n,\left\Vert x \right\Vert=1}\left\Vert Ax\right\Vert.$$ Furthermore, the estimate $\left\Vert A^n\right\Vert\le\left\Vert A\right\Vert^n$ holds (a similar estimate holds for the product of any two square matrices of same size). In the case of reals and complex numbers and the usual absolute value we have equality here, but inequality is good enough to show that the series $$e^A=\sum_{n=0}^\infty \frac1{n!}A^n$$ converges entrywise. The argument for convergence is the same as in the more familiar cases.

• Thanks very much for this impressive answer. It may be interesting for you to know that the non-commutative property of matrices makes them interesting for representing operators in Quantum Mechanics. Observing a quantity affects the result, and so doing a second measure is affected by the first one, therefore the order of the measures matters, and matrices enter the model naturally for operators that yield physical quantities. – Mephisto Feb 24 '13 at 3:20

No, that's not right. $\mathrm e^x$ is a perfectly well-behaved entire function on the complex plane, arguably one of the most well-behaved that there are. There's no problem at all in plugging in any value you like, and there's no such thing as separate real and complex Maclaurin series for it. There's nothing blind about using the series for complex arguments; in fact one could argue to the contrary that one would be blind to many important aspects of this function and its Maclaurin series if one considered it solely for real arguments.

• I am wondering up to which point I can assume that for other functions as well, if it is more or less safe or if it only makes sense here because of the good behaviour of the exponential. Was not there something more general called Laurent series and path-dependent... well, path-dependent "things" in the complex plane that demand special attention? (I am sorry, I am an experimental physicist and all this staff is buried a bit deep among funny memories of undergraduate days) – Mephisto Feb 19 '13 at 21:30

There's no reason to worry about $A$ being real. $$e^{iz}=1+iz+(iz)^2g(z),$$ where $g$ is analytic everywhere, and non-zero at $z=0$. In particular, $g(z)$ is bounded away from zero in some neighborhood of $z=0$, meaning that there exist $m,M>0$ such that $m\leq|g(z)|\leq M$ whenever $z$ is sufficiently close to $0$--say with $|z|<\delta_0$.

Hence, if we wish to ensure that $e^{iz}$ is "close enough" to $1+iz$, we need only ensure that $(iz)^2g(z)$ is "close enough" to zero. Since $$\left|(iz)^2g(z)\right|=|iz|^2|g(z)|=|z|^2|g(z)|,$$ then we can make this as small as we like, simply by taking $|z|$ sufficiently small--in particular, $$|z|<\min\left\{\delta_0,\sqrt{\frac{\epsilon}{M}}\right\}$$ works.

The exponential is really well behaved, even (or especially) for complex numbers. If you wish to only deal with $e^{ia}$ for real $a$ you can reduce the complex case to this case with the identity:

$$e^{i(a+ib)} = e^{ia}e^{-b}$$