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For homework I have to find the derivative of $\text {exp}(6x^5+4x^3)$ but I am not sure if this is equivalent to $e^{6x^5+4x^3}$ If there is a difference, what do I do to calculate the derivative of it?

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    $\begingroup$ There is no difference; they are alternative notations for the same thing. $\endgroup$ – Brian M. Scott Apr 8 '15 at 21:33
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    $\begingroup$ It is just a matter of notation. For instance, if the power is a long polynomial we use $\exp$ so we will not get confused. This notation helps avoiding mistake. $\endgroup$ – FreeMind Apr 8 '15 at 22:07
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    $\begingroup$ Why didn't you ask the person who set the homework? $\endgroup$ – OrangeDog Apr 9 '15 at 16:41
  • $\begingroup$ Possible duplicate of math.stackexchange.com/q/137499/37903 $\endgroup$ – OrangeDog Apr 9 '15 at 16:42
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    $\begingroup$ why so many upvotes? $\endgroup$ – zed111 Apr 9 '15 at 17:20
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Yes. They are the same thing.

When exponents get really really complicated, mathematicians tend to start using $\exp(\mathrm{stuff})$ instead of $e^{\mathrm{stuff}}$.

For example: $e^{x^5+2^x-7}$ is kind of hard to read. So instead one might write: $\exp(x^5+2^x-7)$.

Note: For those who use Maple or other computer algebra systems, e^x is not usually the same as exp(x). In Maple, e^x is the variable $e$ raised to the variable $x$ power whereas exp(x) is Euler's number $e$ raised to the $x$ power.

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    $\begingroup$ Agreed, they are two different ways of looking at the same thing... one way $e^x$ kind of the "mathy" way to view it, and $\text{exp}(x)$ is more the "programatical" way to view it. $\endgroup$ – TravisJ Apr 8 '15 at 21:39
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    $\begingroup$ @TravisJ It has nothing to do with programming and everything to do with the clarity of written mathematics, as the answer you're commenting on makes clear. $\endgroup$ – David Richerby Apr 9 '15 at 10:38
  • $\begingroup$ @DavidRicherby, by "programatical" I meant that $\text{exp}(x)$ emphasizes that you have a function, named exp() and the input to that function is $x$ (or whatever else...). It is a convenient side effect that functional notation is sometimes easier to read than other notation. $\endgroup$ – TravisJ Apr 9 '15 at 12:45
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    $\begingroup$ @TravisJ: Functions are about the "mathiest" thing there is; it's bizarre to log onto a math site and tell mathematicians that functions are "programatical" instead of "mathy". $\endgroup$ – ruakh Apr 9 '15 at 14:43
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    $\begingroup$ @TravisJ: No one is objecting to your statement that "they are two different ways of looking at the same thing"; you don't need to defend it. The problem is with your contrast between "mathy" and "programatical"; they are both mathy. It's true that only one is of particular CS interest, but that's not relevant to mathematicians. (Your claim is like saying that 'good' and 'OK' are different perspectives, 'good' being kind of "Englishy" and 'OK' being more "Portuguesical", because Portuguese has borrowed only the latter.) $\endgroup$ – ruakh Apr 9 '15 at 15:33
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Yes. The purpose for the notation $\exp$ is twofold:

  • It allows one to talk about the exponentiation function itself, without specifying a particular input. For example, one can write that $\exp$ is a homomorphism from the additive group on $\mathbb{R}$ to the multiplicative group on $\mathbb{R}$. One may also say that $\exp$ and $\log$ are inverses.

  • It allows you to write exponentiation without pushing the body of exponentiation into a superscript. For example, one may write the following, which is unwieldy to write without $\exp$ notation:

$$\prod_i e^{x_i} = \exp \sum_i x_i$$

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    $\begingroup$ Allow me to add to your last line that most physicists do not come across situations where $e$ might be ambiguous and therefore they use $e$ for exponentials (except when the argument gets too long). $\endgroup$ – Hrodelbert Apr 9 '15 at 11:25
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    $\begingroup$ Yes, I'm with @Hrodelbert: the statement that "exp" is preferred in physics seems unjustified. $\endgroup$ – David Z Apr 9 '15 at 11:50
  • $\begingroup$ Watch out, log(x) might resolve to log10(x) [or even log2(x)], not ln(x). $\endgroup$ – Joshua Apr 9 '15 at 16:59
  • $\begingroup$ @Hrodelbert That last line was edited in. Removed it $\endgroup$ – Solomonoff's Secret Apr 9 '15 at 17:02
  • $\begingroup$ Exp is a homomorphiam from the additive to the multiplicative. You've written it backwards. $\endgroup$ – Thomas Andrews Apr 9 '15 at 19:36
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As other answers say, in your homework (and, indeed, in most places in mathematics) there is no difference.

I have seen a beginning textbook first defining a certain function $\exp(x)$, then proving certain properties of it, and finally using those properties to motivate calling it $e^x$.

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    $\begingroup$ That's how I "learned" about the exponential function in our maths lectures. We first introduced exp(x) as the power series then proved that this equals some constant to the power of x and than defined that constant as e. $\endgroup$ – Roman Reiner Apr 9 '15 at 6:27
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I agree with these two answers, but I want to add one thing: Well defines.

$e$ is some (positive) number, so (without knowing the function $\exp$), you can compute $e^n$ for $n \in \mathbb{N}$ – just multiply $e$ $n$ times with itself. You can also compute $e^{-n} = \frac{1}{e^n}$ and even $e^\frac{p}{q} = \sqrt[q] e^p$ (for $n, q \in \mathbb{N}, p \in \mathbb{Z}$). One can prove that the $\exp$ function yields the same numbers with these arguments. This justifies the notation.

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Although I agree with the answers already provided that in this situation (and indeed in most other ones in mathematics) there is no difference between the two notations, I would like to add the following for completeness:

In manifold theory (most particularly Lie Group theory or Riemannian geometry), the exponential map $\exp$ is a map from a tangent space to the manifold itself. For Lie groups, it expresses the local group structure and allows to lift many problems from the group to the tangent space (the Lie algebra). It also defines integral curves on the manifold and is therefore related to geodesics (which is more obvious from the viewpoint of Riemannian geometry).

This exponential $\exp$ coincides with the usual exponential for the case of the Lie group $\mathbb{R}$. It also coincides with the definition of the matrix exponential $$ e^A = \sum_{n=0}^\infty\frac{A^n}{n!}. $$ However, I believe this cannot be done in general, although I do not have an example available.

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  • $\begingroup$ I am a bit surprised be the final sentence. Can you give an example of situation where $e^t$ is defined but different from $\exp(t)$? $\endgroup$ – Marc van Leeuwen Apr 9 '15 at 13:45
  • $\begingroup$ @MarcvanLeeuwen No I cannot. I guess what I remembered is that for disconnected Lie groups, $\exp$ does not cover the entire group, implying that not all group elements have an exponential representation. This, as you rightly question, has nothing to do with the mapping $t\mapsto e^t$. Nevertheless, I presume there do exist Lie groups for which the exponential mapping cannot be interpreted as a version of the map $t\mapsto e^t$, are there not? $\endgroup$ – Hrodelbert Apr 9 '15 at 15:39
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    $\begingroup$ Yes, there are non-linear Lie groups, groups that cannot be embedded in any $GL(n,\Bbb R)$, and they still have an $\exp$ map, but writing $e^t$ is problematic as it cannot be defined as a matrix exponential. By the way I think the notion of $\exp$ is just more general, and one could do away with the notation $e^x$ (and the cumbersome constant $e$) altogether by always writing $\exp(x)$ instead; for instance in case of a matrix exponential $\exp(A)$ seems much less confusing to me than $e^A$. $\endgroup$ – Marc van Leeuwen Apr 9 '15 at 15:44
  • $\begingroup$ I agree with your statement that $\exp$ is simply more general. I guess that was the main point I was trying to convey here. Thank you for your comments, they were enlightening. $\endgroup$ – Hrodelbert Apr 9 '15 at 19:31
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While both expressions are generally the same, $\exp(x)$ is well-defined for a really large slurry of argument domains via its series: $x$ can be complex, imaginary, or even quadratic matrices.

The basic operation of exponentiation implicated by writing $e^x$ tends to have ickier definitions, like having to think about branches when writing $e^{1\over2}$ or at least generally $a^b$.

Exponentiation can be replaced by using $\exp$ and $\ln$ together via $a^b=\exp(b\ln a)$, and the ambiguities arise from the $\ln$ part of the replacement. So it can be expedient to work with just $\exp$ when the task does not require anything else. Informally, $e^x$ is used equivalently to $\exp(x)$ anyway but the latter is really more fundamental and well-defined.

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    $\begingroup$ Are you saying $e^x$ is defined over a smaller range than $\exp(x)$, for instance that the former is not defined when $x$ is complex? If so, have you evidence for that? I thought $e^{i\pi} = -1$ was a reasonably well known identity. $\endgroup$ – abligh Apr 9 '15 at 8:01
  • $\begingroup$ @abligh Carl Mummert points out that if you interpret $e^x$ as $\exp(x\log e)$ and are not careful with the branch of log, you can get undesirable results such as $e^{i\pi}=-\exp(2n\pi^2)$ for any $n\in\Bbb Z$. Although "multivalued functions" can sometimes be useful, we don't want that sort of thing to happen in the definition of $\exp$ since it is already analytic. $\endgroup$ – Mario Carneiro Apr 14 '15 at 21:02
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Another reason we use $\exp(x)$ is when defining it in terms of its power series. At that point, we don't know that $\exp(x)=e^x$ when $x$ is real.

Also, there are general problems using the notation $a^z$ when $z$ is complex. $a^z$ is actually a multi-valued function. $e^z$ is thus sometimes ambiguous, so we will, in those cases, prefer $\exp(z)$ to clarify that we are talking about the single-valued function.

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  • $\begingroup$ Why would $a^z$ be multivalued? If $a>0$, we define $a^z = e^{z \cdot \log a}$. $\endgroup$ – goblin Apr 9 '15 at 20:28
  • $\begingroup$ What if $a$ isn't real? @goblin To write $x^y$ in general, you require multivalued functions or some horrible branch cut. $\endgroup$ – Thomas Andrews Apr 9 '15 at 20:29
  • $\begingroup$ Being real isn't strong enough; we want $a \in \mathbb{R}_{>0}$. Otherwise, you don't write $a^z$ without explaining to your reader explicitly what your notation means. If $a$ isn't in $\mathbb{R}_{>0},$ I would advocate writing $a^z_\mathbf{C}$ for whatever version of the exponential you need, where $a^z_\mathbf{C}$ is interpreted as $\mathbf{C}(a,z)$ and $\mathbf{C}$ is understood to be a 2-place function. $\endgroup$ – goblin Apr 9 '15 at 20:43
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    $\begingroup$ But that's my point, $a^z$ is ambiguous, in general, depending on which branch of $\log a$ you take. As I said, it is a multi-valued function. $\endgroup$ – Thomas Andrews Apr 9 '15 at 20:44
  • $\begingroup$ The optimal convention surely is that $a^z$ is unambiguous, we only write $a^z$ when $a>0$, and we decorate the notation in some way whenever we want to extend so that $a$ can be a negative or complex number. $\endgroup$ – goblin Apr 9 '15 at 20:47
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We use exp(Θ) for discrete values. Matlab can only use exp() because everything is discrete in it. In reality, you can use e^Θ in your algebra because continuous.

You can still consider (exp^(jΘ)-e^(jΘ)) / 2j to be a sin if you want discrete values.

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