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How did Euler come up with his famous formula?

  1. Oct 29, 2011 #1
    [itex] e^{ix}=cos(x)+isin(x) [/itex]

    Simple enough?

    Well I am biting my head off because I don't know how he did it.

    Why is this bothering me?

    ex is not a periodic function. How in the world magic happens when you put an i into exponent, and it gets periodic.

    I want to know this so badly.


    Why is this bothering me, further?

    Because I am doing signal analysis and everything, from complex exponential LTI response to Fourier analysis is based on this 2 inch formula.

    :(
     
  2. jcsd
  3. Oct 29, 2011 #2

    gb7nash

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    Euler's derivation is quite long, and probably too long to fit on here. If you're just looking for a proof though, there are a couple of good ones that aren't too bad.
     
  4. Oct 29, 2011 #3
    Thats the point. I searched the internet, from top to bottom and nothing. Nobody even mentions the derivation of eulers formula.

    Do you have any reliable source, where it shows how he did it?

    And no, don't want the proof. I believe its correct, don't need proving :D
     
  5. Oct 29, 2011 #4
    Well, that is a trick question. It's not clear whether the statement in question is a definition or a theorem, since it's not at all obvious initially that it even makes sense to raise a number to a complex power.

    So, if we do not yet know that we can take a number to an imaginary power (if you study complex analysis, there are other approaches), Euler's formula could be thought of as a definition of it. Of course, definitions are not written in the sky, they have to have some sort of motivation behind them. Alternatively, you could interpret the statement as being an equality of functions that are defined in terms of power series. The power series continues to make sense, even if the variable becomes complex. In that case, it's just something that you check by writing down both sides of the equation as power series and making sure that they are equal.

    Knowing Euler, since he was a wiz with power series, my guess is that he probably had the power series for e^x, sin, and cos at his fingertips, and as a result of having some experience dealing with these function, as well as complex numbers, maybe he noticed the similarity of the power series of sin and cos with the power series of e^x, and then realized that if he substituted ix for x in the e^x power series, it showed exactly what the relationship was, as well as providing a concept of imaginary powers. But that is just a guess about his inspiration. Maybe it's not what he was thinking. I'm no historian.

    Generally speaking, I like to not just know that subject, but know how to come up with it. To at least have the illusion that I could have invented it myself. But, it turns out, a lot of times, the process through which things were invented was convoluted and ugly, and that is why we are often never told about the history. On the other hand, I believe in this (overall good and necessary) process of polishing the math, many nice insights were lost, or are now only found in obscure historical documents. Finding these historical gems can be very difficult, so I find it preferable to try to figure out my own possible path for how someone MIGHT have thought of things, although history is one possible source to look to when that fails.

    This is the way I think of it. The whole point of e^x is that it is a function that is its own derivative. If you want to define e^(ix), you want the derivative to be i e^(ix), from the chain rule.

    So, the question is, what function satisfies that property--that when you take the derivative, it gets multiplied by i?

    The answer is given by Euler's equation. Just check what happens when you differentiate the right hand side of the equation.

    Geometrically, multiplication by i is rotation by 90 degrees in the complex plane. If you picture a particle moving around the unit circle, it's coordinates will be

    cos t + i sin t

    The trajectory is perpendicular to the position vector, so it's clear that the velocity vector will be rotated by 90 degrees from the position vectory, in other words, multiplied by i. So, that's the function that has the properties that we want. So, we call it e^(it).
     
  6. Oct 29, 2011 #5
    You have my way of thinking. I see what you are getting at. With that derivative and all.

    I understand that the true derivation might be lost. But I will find it. And most importantly, understand it.

    Thank you very much for your in-depth reply.
     
  7. Oct 29, 2011 #6
    I believe Euler's original method was as follows

    Consider the series expansion of ep

    [tex]{e^p} = 1 + p + \frac{{{p^2}}}{{2!}} + \frac{{{p^3}}}{{3!}}.....[/tex]

    put p = iq

    [tex]{e^{iq}} = 1 + iq - \frac{{{q^2}}}{{2!}} - i\frac{{{q^3}}}{{3!}} + \frac{{{q^4}}}{{4!}} + i\frac{{{q^5}}}{{5!}} - \frac{{{q^6}}}{{6!}}[/tex]

    Note carefully the alternating double plus and minus (for two terms)

    Collect real and imaginary terms

    [tex]\left( {1 - \frac{{{q^2}}}{{2!}} + \frac{{{q^4}}}{{4!}} - \frac{{{q^6}}}{{6!}} + } \right) + i\left( {q - \frac{{{q^3}}}{{3!}} + \frac{{{q^5}}}{{5!}} - } \right)[/tex]

    This provides two series which can immediately be recognised as the expansions of cosq and sin q
     
  8. Oct 29, 2011 #7
    Aaaaaaaaaaaaaaaaaaaaa. Does make sense !!!

    Beauty of series.

    THANK YOU VERY MUCH.
     
  9. Oct 29, 2011 #8
    Yes, that is what I was guessing earlier, but I'm not good with symbols here.

    But, you never know what Euler was really thinking--could be different from what he wrote. You can probably find his original document somewhere.
     
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