Some basic idea of differentiation. Here is a (very) informal introduction.
\(e\) is the unique real constant such that the differential equation \[\frac{dy}{dx} = ky\] has a solution of the type \[y=Ce^{kx}\], which is equivalent to saying that \[e = \lim_{n \to \infty}{\left ( 1+ \frac{x}{n} \right ) ^ {\frac{n}{x}}}\]
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 (\(y\)) as a function of time(\(t\)).
Using what we just pointed out, we have:
\[\frac{dy}{dt} \propto y \\ \frac{dy}{dt} = ky \quad \text{where k is the rate constant}\]
Compare that with the equation in the definition!
So, we have \[y(t) = Ce^{kt} \quad \text{where C was the initial size}\]
Well, you must know of compound interest.
Say, my principle was \(P\), rate \(R\) and I compound anually.
Okay, now what if I compound half yearly? Wouldn't it be better because I'd have more money that is compounding during the midyear?
Yeah, that is correct.
Well, if I compounded \(n\) times an year, I'd be getting \[ A = P(1+\frac{R}{n})^{n t}\]
I could keep compounding continuously too, which would mean the amount is never at rest.
That is \[ A_{cont} = \lim_{n \to \infty} P(1+\frac{R}{n})^{n t} \\ = \lim_{n \to \infty} P(1+\frac{R}{n})^{\frac{R n t}{R}} \\ = P \lim_{n \to \infty} (1+\frac{R}{n})^{\frac{n}{R}{Rt}} \\ = P e^{Rt} \]
Yes! Look at the top of the note!
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Here is one (simple) way:
Recall the second second definition.
\[ e = \lim_{n \to \infty} {\left ( 1+ \frac{x}{n} \right ) ^ {\frac{n}{x}}} \\ \implies e^x = \lim_{n \to \infty}{\left ( 1+ \frac{x}{n} \right ) ^ n } \\ = \lim_{n \to \infty}{\left ( 1+ \frac{1}{n} \right ) ^ n} \\ = \lim_{n \to \infty}\sum^{n}_{k=0} {n \choose k} \frac{1}{n^k} \\ = 1+1+\lim_{n \to \infty} \sum_{k=2}^n\frac{n(n1)(n2)\cdots(n(k1))}{k!\,n^k} \\ = 1+ 1+\frac{1}{2!}\left(1\frac{1}{n}\right)+\frac{1}{3!}\left(1\frac{1}{n}\right)\left(1\frac{2}{n}\right)+\cdots \\ = 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots \]
Here you go:
Proposition Let \(f : \mathbb{R} \to \mathbb{R}\) be a differentiable function such that \(f(0) = C\) and \(f'(x) = f(x)\). Then it must be the case that \(f = Ce^x\).
Proof Let \(g(x) = f(x) \frac{e^{x}}{C}\). Then
\[g'(x) = f(x) \frac{e^{x}}{C} + f'(x) \frac{e^{x}}{C} = (f'(x)  f(x)) \frac{e^{x}}{C} = 0\]
by assumption, so \(g\) is constant. But \(g(0) = 1\), so \(g(x) = 1\) identically.
Hence, \[ 1 = f(x)\frac{e^{x}}{C} \\ \implies f(x) = Ce^x \]
Because it is!
You have seen how \(e\) 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 \(\frac{d}{dx}\ln x = \frac{1}{x}\).
More precisely, our bacteria culture which grew as a function of time as \(y(t) = Ce^{kt}\) is of a specific size \(y\) at time \[ t = \frac{\ln(\frac{y}{C})}{k}\]
What is important is to observe that \(\frac{y}{C}\) is the amount by which it grew and \(k\) is the rate constant.
That is one of the most beautiful equations of all time!
\[e^{i x} = 1 + i x + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots \\ =1 + i x  \frac{x^2}{2!} \frac{ix^3}{3!} +\frac{(x)^4}{4!} + \cdots \\ = \left(1 \frac{x^2}{2!} + \frac{x^4}{4!}+\cdots\right) + i\left( x \frac{x^3}{3!}+\frac{x^5}{5!}\cdots\right) \\ = \cos x + i \sin x \]
So, \(e^{i \theta}\) is actually an unit vector on the complex plane with argument \(\theta\). In general, \(R e^{i \theta}\) would be a complex number of modulus \(R\) and argument \(\theta\).
Clearly, \(e^{i \pi} = 1 \implies e^{i \pi} + 1 = 0\).
In essence, \(e\) forms the basis of exponentiation in the complex plane. Here, we use it to show that \(i^i\) is real!
I probably do not know what your question is. Feel free to ask me in the comments.
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Top NewestVery informative . Helped to brushen up my concepts .
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@megh choksi ,@Amrita Roychowdhury, @Sneha Saha : Please have a look
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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 \(\dfrac{dy}{dx}\) 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?
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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
The solution to \[\frac{dy}{dx} = ky\] can be \[y = Ce^{kx}\] because differentiating this in gives the differential equation.
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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?
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"Proof" that \(e^{\pi i}\approx1\)
\(e^{\pi i}\approx(1.000001^{1000000})^{\pi i}\)
\(=(1.000001^{\pi i})^{1000000}\)
\(\approx(1+.000001\pi i)^{1000000}\)
\(\approx1.000005+.00000000001i\)
\(\approx1\)
Therefore, \(e^{\pi i}\approx1\). 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.")
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Before anyone says that's not a real proof: You don't say?!
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I take very high offense on how you called the first way of calculating \(e\) "lame".
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It's not that lame but there are faster converging methods.
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But \(e^{x}\) is one of the most renowned Taylor series simply because \(e^{ix}= \cos x+ i \sin x\) :)
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True.
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@Abhineet Nayyar
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