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In several posts around this site, I have encountered the expression $\exp(x)$ where $x$ is an arbitrary expression. What does this notation mean?

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    $\begingroup$ Googling tells you the answer. : goo.gl/eBchVh -1 for no effort. $\endgroup$ – evil999man May 21 '14 at 15:37
  • $\begingroup$ Generally, it means the result of evaluating the function $\exp$ at the value $f$. $\endgroup$ – user14972 May 21 '14 at 15:38
  • $\begingroup$ '+1' to counter @Awesome 's downvote. $\endgroup$ – user103816 May 22 '14 at 3:24
  • $\begingroup$ @user31782 Of course, it's hard to give exactly $\frac{2}{5}$ of an upvote. $\endgroup$ – Aidan F. Pierce May 22 '14 at 12:07
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$\exp(x) = e^x{}{}{}{}{}{}{}{}{}{}{}{}{}$

It's often used when we have $\exp(f(x)) = e^{f(x)}$ and $f(x)$ is complicated function. The focus, then, becomes the function in the exponent of $e$ (not to mention that it helps readers to not have to squint to read, say when $f$ is a rational function, $e^{f(x)}$).

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  • $\begingroup$ Also when the author prefers to think of $e^x$ "as a function" rather than as the exponentiation of a particular constant $e$. $\endgroup$ – Jack M May 21 '14 at 15:52
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It denotes the exponential function with base $e$ defined as:

$$\exp(z):=e^z.$$

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This is the exponential function. In other words, $$\mathrm{exp}\,(x)$$ is another way of writing $$e^x,$$ where $$e=2.7182818284590452353...$$

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Usually $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$, if $x$ is something that can be multiplies by itself, divided by an integer, finite sums are defined, and there is some notion of the sums convergence. It makes sense for all finite-dimensional matrices, some operators in infinite dimensional spaces and also for other things.

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