Euler's Number (e)(e) - Infrequently Asked Questions

  • What do I need to know to read this?

Some basic idea of differentiation. Here is a (very) informal introduction.


  • So, What is ee?

ee is the unique real constant such that the differential equation dydx=ky\frac{dy}{dx} = ky has a solution of the type y=Cekxy=Ce^{kx}, which is equivalent to saying that e=limn(1+xn)nxe = \lim_{n \to \infty}{\left ( 1+ \frac{x}{n} \right ) ^ {\frac{n}{x}}}


  • Why should I care?

Well, these functions pop up all the time. Here is a classical example:

Say, I have a culture of bacteria which is continuously replicating. It is obvious(why?) that the rate of growth of the culture is proportional to the size of the culture itself.

Let us now try to express the size of the culture (yy) as a function of time(tt).

Using what we just pointed out, we have:

dydtydydt=kywhere k is the rate constant\frac{dy}{dt} \propto y \\ \frac{dy}{dt} = ky \quad \text{where k is the rate constant}

Compare that with the equation in the definition!

So, we have y(t)=Cektwhere C was the initial sizey(t) = Ce^{kt} \quad \text{where C was the initial size}


  • I am not very fond of bacteria. Nor do I understand why you emphasized continuously over there.

Well, you must know of compound interest.

Say, my principle was PP, rate RR and I compound anually.

  • At the end of the first year, I have (P+PR)=P(1+R)(P + PR) = P (1+R)
  • At the end of the second year, I have P(1+R)+R(P(1+R))=P(1+R)2P(1+R) + R(P(1+R)) = P(1+R)^2
  • At the end of the tth year, I have P(1+R)tP(1+R)^t

Okay, now what if I compound half yearly? Wouldn't it be better because I'd have more money that is compounding during the mid-year?

Yeah, that is correct.

  • At the end of the first half-year, I have (P+PR2=P(1+R2)(P + P\frac{R}{2} = P (1+\frac{R}{2})
  • At the end of the year, I have P(1+R2)+R2(P(1+R2))=P(1+R2)2P(1+\frac{R}{2}) + \frac{R}{2}(P(1+\frac{R}{2})) = P(1+\frac{R}{2})^{2}
  • At the end of the tth year, I have P(1+R2)2tP(1+\frac{R}{2})^{2t}

Well, if I compounded nn times an year, I'd be getting A=P(1+Rn)nt A = P(1+\frac{R}{n})^{n t}

I could keep compounding continuously too, which would mean the amount is never at rest.

That is Acont=limnP(1+Rn)nt=limnP(1+Rn)RntR=Plimn(1+Rn)nRRt=PeRt A_{cont} = \lim_{n \to \infty} P(1+\frac{R}{n})^{n t} \\ = \lim_{n \to \infty} P(1+\frac{R}{n})^{\frac{R n t}{R}} \\ = P \lim_{n \to \infty} (1+\frac{R}{n})^{\frac{n}{R}{Rt}} \\ = P e^{Rt}


  • Pretty Cool! Could you show this graphically?

Yes! Look at the top of the note!

Sage Sage Code:

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a = lambda n,t: (1+(1/n))^(n*(floor(n*t)/n))
contd = [plot(lambda t: a(n,t), (0,1),ymin=0,ymax=2.8) for n in xrange(1,31)]
anim = animate(contd)
anim.show()


  • You said ee is a real number. How do I compute it's value then?

Here is one (simple) way:

Recall the second second definition.

e=limn(1+xn)nx    ex=limn(1+xn)n=limn(1+1n)n=limnk=0n(nk)1nk=1+1+limnk=2nn(n1)(n2)(n(k1))k!nk=1+1+12!(11n)+13!(11n)(12n)+=2+12!+13!+14!+ e = \lim_{n \to \infty} {\left ( 1+ \frac{x}{n} \right ) ^ {\frac{n}{x}}} \\ \implies e^x = \lim_{n \to \infty}{\left ( 1+ \frac{x}{n} \right ) ^ n } \\ = \lim_{n \to \infty}{\left ( 1+ \frac{1}{n} \right ) ^ n} \\ = \lim_{n \to \infty}\sum^{n}_{k=0} {n \choose k} \frac{1}{n^k} \\ = 1+1+\lim_{n \to \infty} \sum_{k=2}^n\frac{n(n-1)(n-2)\cdots(n-(k-1))}{k!\,n^k} \\ = 1+ 1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots \\ = 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots


  • I still do not see why that is the unique number as promised in the first definition.

Here you go:

Proposition Let f:RRf : \mathbb{R} \to \mathbb{R} be a differentiable function such that f(0)=Cf(0) = C and f(x)=f(x)f'(x) = f(x). Then it must be the case that f=Cexf = Ce^x.


Proof Let g(x)=f(x)exCg(x) = f(x) \frac{e^{-x}}{C}. Then

g(x)=f(x)exC+f(x)exC=(f(x)f(x))exC=0g'(x) = -f(x) \frac{e^{-x}}{C} + f'(x) \frac{e^{-x}}{C} = (f'(x) - f(x)) \frac{e^{-x}}{C} = 0

by assumption, so gg is constant. But g(0)=1g(0) = 1, so g(x)=1g(x) = 1 identically.

Hence, 1=f(x)exC    f(x)=Cex 1 = f(x)\frac{e^{-x}}{C} \\ \implies f(x) = Ce^x


  • Why is the logarithm base ee termed natural?

Because it is!

You have seen how ee relates to every exponential growth scenario, now it is obvious why the natural logarithm relates to time.

The growth velocity of bacteria population varies with it's size. So, the time taken to grow by an amount would vary inversely with it. Hence, we have ddxlnx=1x\frac{d}{dx}\ln x = \frac{1}{x}.

More precisely, our bacteria culture which grew as a function of time as y(t)=Cekty(t) = Ce^{kt} is of a specific size yy at time t=ln(yC)k t = \frac{\ln(\frac{y}{C})}{k}

What is important is to observe that yC\frac{y}{C} is the amount by which it grew and kk is the rate constant.


  • What does it mean to say that eiπ+1=0e^{i \pi} + 1 = 0 ?

That is one of the most beautiful equations of all time!

eix=1+ix+(ix)22!+(ix)33!+(ix)44!+=1+ixx22!ix33!+(x)44!+=(1x22!+x44!+)+i(xx33!+x55!)=cosx+isinxe^{i x} = 1 + i x + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots \\ =1 + i x - \frac{x^2}{2!} -\frac{ix^3}{3!} +\frac{(x)^4}{4!} + \cdots \\ = \left(1- \frac{x^2}{2!} + \frac{x^4}{4!}+\cdots\right) + i\left( x- \frac{x^3}{3!}+\frac{x^5}{5!}\cdots\right) \\ = \cos x + i \sin x

So, eiθe^{i \theta} is actually an unit vector on the complex plane with argument θ\theta. In general, ReiθR e^{i \theta} would be a complex number of modulus RR and argument θ\theta.

Clearly, eiπ=1    eiπ+1=0e^{i \pi} = -1 \implies e^{i \pi} + 1 = 0.

In essence, ee forms the basis of exponentiation in the complex plane. Here, we use it to show that iii^i is real!


  • You still did not answer my question!

I probably do not know what your question is. Feel free to ask me in the comments.

Note by Agnishom Chattopadhyay
4 years, 5 months ago

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@megh choksi ,@Amrita Roychowdhury, @Sneha Saha : Please have a look

Agnishom Chattopadhyay Staff - 4 years, 5 months ago

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solution means a root of an equation according to me , do you agree?

If yes , then you explained in the earlier note that dydx\dfrac{dy}{dx} means rate of change of a quantity with respect to other or a tangent to the curve.

Now how can a magnitude( rate of change of a quantity with respect to other) have a root - seems really odd!

or how can a tangent(which represents a direction) can have root?

U Z - 4 years, 5 months ago

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Solution is a very general term. It does not have a strict definition.

No, by solution of a differential equation, a root, i.e, a number that satisfies the equation is not what is meant.

The solution to a differential equation, is an analytical equation which upon differentiating gives back the equation.

Here is what comprises an analytical expression:

analytic analytic

The solution to dydx=ky\frac{dy}{dx} = ky can be y=Cekxy = Ce^{kx} because differentiating this in gives the differential equation.

Agnishom Chattopadhyay Staff - 4 years, 5 months ago

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@Agnishom Chattopadhyay Thanks sir

U Z - 4 years, 5 months ago

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@Agnishom Chattopadhyay There's a typo in your comment. It should be y=Cekxy=Ce^{kx}

Prasun Biswas - 4 years, 4 months ago

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@Prasun Biswas Fixed. I missed out the {}

Agnishom Chattopadhyay Staff - 4 years, 4 months ago

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@Agnishom Chattopadhyay I thought so. Typing raw LaTeX\LaTeX code quickly sometimes makes you miss the brackets. :D

Prasun Biswas - 4 years, 4 months ago

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Very informative . Helped to brushen up my concepts .

Azhaghu Roopesh M - 4 years, 5 months ago

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I take very high offense on how you called the first way of calculating ee "lame".

tytan le nguyen - 4 years, 5 months ago

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It's not that lame but there are faster converging methods.

Agnishom Chattopadhyay Staff - 4 years, 5 months ago

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True.

tytan le nguyen - 4 years, 5 months ago

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But exe^{x} is one of the most renowned Taylor series simply because eix=cosx+isinxe^{ix}= \cos x+ i \sin x :)

Marc Vince Casimiro - 4 years, 5 months ago

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"Proof" that eπi1e^{\pi i}\approx-1

e1.0000011000000e\approx1.000001^{1000000}

1.000001x1+.000001x1.000001^x\approx1+.000001x for small x (try it)

eπi(1.0000011000000)πie^{\pi i}\approx(1.000001^{1000000})^{\pi i}

=(1.000001πi)1000000=(1.000001^{\pi i})^{1000000}

(1+.000001πi)1000000\approx(1+.000001\pi i)^{1000000}

1.000005+.00000000001i\approx-1.000005+.00000000001i

1\approx-1

Therefore, eπi1e^{\pi i}\approx-1. Somehow. I guess.

(From a correspondence: "Ooh, nice! How did you get from (1+.01πi)^100 to -1 + junk?" "I multiplied it by itself 1000000 times.")

Akiva Weinberger - 4 years, 1 month ago

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Before anyone says that's not a real proof: You don't say?!

Akiva Weinberger - 4 years, 1 month ago

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I think the question is, If this zigzag function is continualy shrinks by a factor k, what will be the its length at infinity?

Magdi Ragheb - 3 years, 8 months ago

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@Abhineet Nayyar

Sanchit Ahuja - 4 years, 5 months ago

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