We all have studied about the Euler's Number \( e \). We know it is the natural base. We know it is the value to which \( (1 + 1/n)^{n} \) approaches as \( n \) tends to infinity. We also know its Taylor expansion.

We learnt these things from our textbooks. But I never found one which explains all the point clearly. These textbooks simply teach these as facts.

I want to know why Euler mentioned this number? By what purpose did he invented (or discovered) it? I was told that its convenient to use \( e \) as the base. Why this irrational is so convenient (actually, I haven't been convenient) ? How do we know its value and its taylor expansion?

Now when we know that

\[ e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ... = 2.7182...\]

how did we discovered that

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... \]?

Also, how Euler proved that

\[ e^{ix} = cos\theta + isin\theta \]

## Comments

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TopNewestIf you write out the taylor/maclauren series expansion of e^x, cos, and sine, euler's formula is immediate.

As to why e is so famous, i'd say its because of the fact d/dx e^x = e^x, a property which no other function satisfies. This is why taking logarithms in base e are helpful.

In proving that d/dx e^x = e^x, you can apply the power rule on that very same expansion you have written

d/dx e^x = 1 + (2

x)/2! + (3x^2)/3! + (4x3)/4! ...when you cancel of the common factors, you will see that you end up with the same series you had started with, ie e^x

As to why its taylor series expansion is like this, its mainly due to the above fact, that d/dx e^x = e^x Just plug that in in the formula you have for taylor/maclauren series expansion - link – Harshit Kapur · 4 years, 7 months ago

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I remember being told that Euler created e as a number to satisfy the equation \(\frac{d}{dx}\)e\(^x\) = e\(^x\) – Harel Dor · 4 years, 7 months ago

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– Harshit Kapur · 4 years, 7 months ago

i'm unsure as to how he could 'create e as a number' to me it seems more likely that he was seeing if the limit if 1/(n!) converges or not, and it did, since factorials get large pretty quicklyLog in to reply

The proof that \(e = 1 + 1 + \frac{1}{2!} +\frac{1}{3!}\frac{1}{4!} + ... \) is a very beautiful one, which comes from the binomial expansion of \(e = (1 + \frac{1}{N} )^N \). The process for \(e^x\) is very similar. However, I don't know where the \(e^{ix} = cosx+isinx\) comes from. – Esteban Gomezllata · 4 years, 7 months ago

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– Calvin Lin Staff · 4 years, 7 months ago

What do you mean by \( e = ( 1 + \frac {1}{N})^N\)? Is \(N=2, 3\)? Otherwise, how do you take the binomial expansion?Log in to reply

Although the original problem is solved now (almost), I have new series of questions on the same topic.

i) How did Euler knew (with reference to Harel D.) that an exponential function would be the one whose derivative is the function itself? ii) Aren't there other functions such that \( \frac{d}{dx} f(x) = f(x) \) ? iii) For what values of x, the expansion of \( e^x \) (except x=0) is rational? iv) Proof of Taylor's Theorem? – Avinash Pandey · 4 years, 7 months ago

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At last, after rigorous search on internet and libraries, I found answers of most of them. Actually, we use a theorem called Taylor's Theorem which can be used to represent any function as a series of polynomial terms.

\[ f(x)= \sum_{k=0}^{\infty} \frac{x^k}{k!} f^{(k)} (0) \]

Now using the fact told by Harel D., Euler knew that "zeroth derivative of \( e^x \) evaluated at zero is one and kth derivative, where k is natural number, of \( e^x \) is \( e^x \) itself". Now apply Taylor's Theorem to get the expansion. Now when we know \( e^x \), then we can evaluate \( e=e^1 \).

Again using Taylor's Theorem, we know the expansion of sinx and cosx. And thus Euler's Theorem is immediate. Also, we can verify (I had little hassles evaluating it) :

\[ \lim_{x \to \infty} (1 + \frac{1}{x})^x = e \].

But still, the important one remains. Taking this irrational as the base, we can write the expansion easily. But still the expansion is infinite going. It never ends. So it doesn't really helps.

Also, I would now like to know for what values of x, the expansion of \( e^x \) (except x=0) is rational. – Avinash Pandey · 4 years, 7 months ago

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– Harshit Kapur · 4 years, 7 months ago

it won't, if you're looking for real-valued x. extend your search to complex numbers, and you will be amazed :) google euler's identityLog in to reply

I found this page interesting: http://www-history.mcs.st-and.ac.uk/HistTopics/e.html – Arndt Jonasson · 4 years, 7 months ago

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