# What is the Conceptual meaning of Euler's number 'e'?

1. Apr 25, 2013

### cdux

Since its inception lies on the calculation of interest rates is it a, so to speak, 'Financial' number?

Also why is our Mathematician so determined to use it so much on integrals related to probability distributions? Is it because it is integrated with less lengthy calculations usually? :)

2. Apr 25, 2013

### slider142

The main reason Euler's number shows up in so many places is due to the exponential function "f(x) = Aex" being a fixed point for differentiation operators for any constant A: its derivative is itself. This allows the functions Aekx to be very useful in solving differential equations (most applied equations are derived from statements about differentials), and subsequently the original function occupies a special central place in complex analysis and the theory of integration (which forms the basis for probability and statistics).
The details are quite varied, but you can read more about them in "e: the Story of a Number" by Eli Maor.

3. Apr 25, 2013

### mathman

The density function for the standard normal distribution is e-x2/2/√(2π).

4. Apr 25, 2013

### cdux

I most probably miss something but from that I understood "it's easy to solve". Surely though it can't be the sole reason of spending so much time on it since after all, ease of use can't be the sole factor of choice for research.

5. Apr 25, 2013

### cdux

That is I guess is an important factor for probability and statistics since if I recall correctly N() can be the approximation for many solutions.

6. Apr 25, 2013

### SteamKing

Staff Emeritus
d(e^x)/dx = e^x

ln(e^x) = x

e^(i*pi) - 1 = 0 where i^2 = -1

e may have started out financial, but it didn't stay that way.

7. Apr 26, 2013

### the_wolfman

In my opinion its not the number $e$ that is important. Instead it is the function $e^x$ that is important.

$e^x$ is important because it is the only nonzero solution to the equation $\frac{d}{dx}f(x)=f(x)$ with the condition $f(0)=1$.

It also follows that $e^x$ is a solution to the equation $\frac{d^n}{dx^n}f(x)=f(x)$

This property makes the function $e^x$ useful in manipulating challenging differential equations into a form that is solvable.

8. Apr 26, 2013

### slider142

You could say that. The sole purpose of any mathematical endeavor is to make an object simpler to work with. We certainly don't enter any problem hoping to make the problem's solution more complicated than the problem. In the same way that choosing the basis vectors for real n-dimensional space to be orthogonal makes the equations that represent many objects easy to write. We could choose some terribly complicated looking functions to span functional space if we wanted to, but the family of functions ekx are nicer to express solutions in for many purposes. Just like it is sometimes nicer to use spherical coordinates, sometimes it is nicer to use cosine and sine functions. There are other families of functions useful in the theory of differentiation and integration, but they only make things look simpler in more specialized applications.
The fact that the basic tools of differentiation and integration don't change the form of this particular function allows it to be more widely visible when these tools are applied, which is where you see it the most. Its connection to sine and cosine, caused by its fixed point behavior, make it extremely visible in any field that uses complex analysis. Very specific differentiations and integrations make the other functions more visible.

Last edited: Apr 26, 2013
9. Apr 26, 2013

### Mandelbroth

By asking this question, I'm supposing that you wouldn't get the "we like math for the sake of math" concept. But, in truth, that's where most of our interest is.

Many calculations are actually more difficult due to Euler's constant (meaning e, rather than my personal favorite number, $\gamma$). A good example is in calculating the area under a normal curve. Consider the "normal function" N, with parameters μ and σ.

$$N(x;\mu,\sigma)=\frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma\sqrt{2\pi}}$$
Now, consider the indefinite integral of that. That is to say, $\displaystyle \int \frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma\sqrt{2\pi}}\ dx$. The closest we can get to putting that expression in terms of elementary functions is $\displaystyle \frac{1}{\sqrt{\pi}}\int_{0}^{(\frac{x-\mu}{\sigma\sqrt{2}})}e^{-x^2} \ dx$. This can be difficult to calculate, though we can approximate with numerical integration to an arbitrary degree of accuracy.

The reason it arises in probability is because of the central limit theorem.

We often consider the fact that $e^{ix}=\cos{x}+i\sin{x}$ a phenomenon. It is a fairly incredible formula, and thus deserving of much appreciation.

10. Apr 27, 2013

### TheAbsoluTurk

e = lim x -->infinity (1 + 1/n)^n

But the Swiss are known for their finance, aren't they?

11. Apr 29, 2013

### HallsofIvy

Staff Emeritus
Or, the reverse, "e" is the unique value of a such that

$$\lim_{x\to 0}\frac{a^x- 1}{x}= 1$$

12. May 2, 2013

### bahamagreen

This concept is likely much older than the calculation of interest compounded continually.

If you are planting seed crops, e shows up intuitively at least, where one would like to determine how much of the harvest must be set aside in order to maintain a self sustainable operation.

If you do the calculation for just the next season there is not enough in that next season to fully sustain the one after that, because that next season also needs a portion aside for the subsequent season. One will figure out pretty quickly that the base portion set aside for any next season needs an additive adjustment to ensure the following season, and another little adjustment for the subsequent season after that... ad infinitum.

The adjustments in the series get progressively smaller because they are progressively further in the future, and you typically have a multiplier acting on your crop each season - with corn, one kernel of corn produces x ears, each with y kernels of corn, so you have a xy:1 multiplier (assuming all goes well).

If the portion of seed crop you put aside does not meet the limit sum of the series of additive adjustments, your growing operation will not be self sustainable.

13. May 2, 2013

### thelema418

With regard to probability distributions, you might want to look up a proof or the starting points for deriving the normal probability density function:
$p(x) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$

The VERY condensed reason $e$ shows up in that formula begins with an integration of $\frac{1}{p}$. This will give $\ln p$. Since you want the solution of $p$, you need to exponentiate: $e^{\ln p} = p$. But then, the other side of the equation will also be raised to $e$. Again, this is very condensed and only meant to illustrate why you get an $e$ in those types of problems.

14. May 2, 2013

### micromass

Staff Emeritus
Which proof or derivation are you refering to here?

In my opinion, the only reason we care about normal distributions is because of the Central Limit Theorem. So is it that what you are refering to?

15. May 2, 2013

### thelema418

The question posed on this thread concerned why "e" shows up frequently in probability distributions. I inferred from the author's question a belief that the mathematician forced e into the equation for the sake of a task. Rather, it is a consequence that follows from a set of assumptions about the shape of a normal distribution.

There's probably discussion in some calculus books. There's also this discussion about the topic by D. Teague: http://courses.ncssm.edu/math/Talks/PDFS/normal.pdf

16. May 2, 2013

### cdux

This sounds incredibly intuitive and the answer I need.

I find it difficult to put it into numbers so I can't fully understand it. Can someone help with walking through it for "2-4 of years of harvest"?

Is "how much to put aside" just 1/n? Where is 'e' in this concept exactly?

17. May 2, 2013

### rcgldr

In math classes normally e is introduced as the solution when solving for a for ln(a) = 1:

$$\int_1^e \frac{1}{x} dx = 1$$

For continuously compounding interest, there's a limit:

$$\lim_{n\rightarrow +\infty} (1 + \frac{i}{n})^n = e^i$$

You can do a web search for proofs of this.

18. May 3, 2013

### cdux

By the way, sorry if I have been misunderstood in the OP but I didn't ask for the derivation of e per se but for a conceptual understanding of it (I knew it was (1+1/n)^n when n tends to infinity).

PS. I still wonder if that 'crops/seeds' analogy of bahamagreen is valid and if I can put it into numbers more clearly (that pinpoint where 'e' (and maybe 'n') is exactly in it).

19. May 3, 2013

### chingel

For example, when you do an analysis of the forces on a rope wound around a pole (the capstan equation), you get the result that if you consider a small angle, the difference in tension in the rope in one end compared to the other is $dT=-T\mu d\alpha$.

So after a small turn, the tension in the rope is $T_{1}=T_{0}-T_{0}\mu d\alpha=T_{0}(1-\mu d\alpha)$. After yet another small turn the tension is $T_{2}=T_{1}-T_{1}\mu d\alpha=T_{1}(1-\mu d\alpha)=T_{0}(1-\mu d\alpha)^2$ and after n turns the tension is $T_{n}=T_{0}(1-\mu d\alpha)^n$. If you want to know the tension after a total angle of $\alpha$ and lets suppose you divided it into n small angles, then $d\alpha=\alpha/n$. From there $T_{n}=T_{0}(1-\mu \alpha/n)^n$. Now this should seem very familiar to you as the well-known definition of the exponential function, the second term is the function $e^{-\mu\alpha}$ in the limit of very large n, so you get that $T_{n}=T_{0}e^{-\mu \alpha}$.

The same situation happens for example when you try to calculate the change in pressure with height of an isothermal atmosphere with constant gravity. There are lots and lots of other situations where the derivative of something is proportional to itself and then the solution uses e.

You could have gotten the same result from $\frac{dT}{d\alpha}=-T\mu$ by integrating $\frac{T'}{T}=-\mu$. You would get that $ln(T)=-\mu \alpha+c$, from where the same result follows. You could also figure the solution out directly by being aware of the properties of the exponential function and the chain rule.

So this function with the weird limit definition turns out to be exactly the weird limit that comes up in this sort of problems. That's all there is to it, it is the limit that is needed, it is not just something made up for obscure reasons and fiddled into someplace where it isn't necessary at all.

20. May 4, 2013

### lavinia

Statistical analysis often involves estimating the parameters of a sample distribution. When the distribution is unknown this is impossible. But one can create a new distribution by averaging independent choices from the original distribution. The distribution of the averages is close to the normal distribution. Thus averaging translates a unknown distribution into an approximation of a known distribution. Thus it is possible to estimate the parameters of the distribution of averages.

The normal distribution fortuitously only has two parameters. This simplifies the problem since in general a distribution may have many parameters again making parameter estimation impossible in practice.

This is one reason why the normal distribution is important.

The normal distribution also arises in continuous stochastic processes where change in the process is independent of its history. These processes are important in statistical modelling of numerous real world situations such as stochastic control. They also model processes in physics such as heat diffusion.