July 22nd, 2017, 01:40 AM  #1 
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  On Euler's formula
Hello everybody, I am dealing with a text, where I found the following notes about Euler's formula (I am giving literal translation of the whole fragment): а(φ) = cosφ + jsinφ; j^2 = 1 > da(φ)/dφ = d(cosφ + jsinφ)/dφ = sinφ + jcosφ = j(cosφ + jsinφ) = jа(φ) > ∫[da(φ)/а(φ)] = j∫dφ > lnа(φ) = jφ + k > а(φ) = Ke^jφ; K = a(0) = cos0 + jsin0 = 1 > e^jφ = cosφ + jsinφ > A◦(t) = A(t)e^j[ω(t)t+ψ] = A(t)cos[ω(t)t+ψ] + jA(t)sin[ω(t)t+ψ] – a “circulating” vector; φ  angle; ψ – “initial” angle; ω – angular velocity; t  time. This is extremely complex notation for me. I need help from good Mathematician who can review the fragment and tell me whether there are errors in the quoted notation. And eventually.... give me a short, common sense answer to the question "what is meant by this impenetrable notation?" Thank you very much for the efforts... 
July 22nd, 2017, 03:38 AM  #2 
Senior Member Joined: Oct 2009 Posts: 906 Thanks: 354 
Yes, it is correct. It is basically deriving Euler's formula from a first order differential equation. It sweeps a lot under the rug though. So while it is formally correct, it is not rigorous from mathematician's standards. I recommend you to look at Visual Complex analysis by Needham. He uses and explains the same argument in his first chapter. 
July 22nd, 2017, 03:53 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326 
A complex number, z, has the form z = a + bi, where a and b are real numbers and i is the "imaginary" unit, which satisfies the equation i² = 1. On occasion (especially in engineering contexts), the letter j is used instead of i. The real numbers a and b are called the real part and imaginary part of the complex number, and the point (a, b) in Cartesian coordinates represents the complex number, the plane for which the Cartesian coordinates are defined being known as an Argand diagram. If the point (a, b) is given in polar coordinates as (r, θ), where r $\small\geqslant$ 0 and $\pi$ < θ $\leqslant$ $\pi$, r is called the modulus of the complex z, also denoted by z, and θ is called the argument of the complex number z, occasionally denoted by arg(z) or arg z. By elementary trigonometry, a = r*cos(θ) and b = r*sin(θ), so z = r(cos(θ) + sin(θ)i). If r = 1, z = cos(θ) + sin(θ)i and can be regarded as a function of θ, occasionally denoted by z(θ). You need to understand from the context that this notation doesn't mean the product of z and θ. The notation dz/dθ (or sometimes z'(θ)) is used for the derivative of z with respect to θ. Loosely speaking, dz/dθ is the rate of change of z with respect to θ. The process of finding dz/dθ as a function of θ is called differentiation. The derivative of a constant is zero, so θ is usually being treated as a variable. It can be shown that the derivative of cos(θ) with respect to θ is sin(θ) and the derivative of sin(θ) with respect to θ is cos(θ). Hence if r = 1, z'(θ) = sin(θ) + cos(θ)i = i(cos(θ) + sin(θ)i) = iz. The derivative of the function $e^{i\theta}$ with respect to $\theta$ is $ie^{i\theta}$. It can be shown that $e^{iθ} = \cos(\theta) + \sin(\theta)i$ for all real values of $\theta$, which is Euler's formula. In what you posted, the letter a was used instead of the letter z, and the letter φ was used instead of the letter θ. Also, the imaginary unit was denoted by j instead of i. Do you understand the above so far? Also, do you understand what is meant by the inverse of a function and know that for real $x$ the function $\ln(x)$ is the inverse of the function $e^x$ and vice versa? Last edited by skipjack; July 22nd, 2017 at 06:21 AM. 
July 22nd, 2017, 04:03 AM  #4  
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  Quote:
In layman's terms, I want to know one thing. You said this notation expresses sweeping derivation of Euler's formula. In other words, the author is skipping certain parts of the rigorous mathematical derivation. But does the author "discover" something new in the derivation? Does he add something new? Or this is just the standard derivation which is written in the textbooks?  
July 22nd, 2017, 05:02 AM  #5  
Senior Member Joined: Oct 2009 Posts: 906 Thanks: 354  Quote:
But the author's argument is definitely not new. It's very well known. So the author didn't really discover anything new. However, he does give a convincing argument what $e^{i\theta}$ SHOULD be.  
July 22nd, 2017, 06:24 AM  #6 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
The following might help. The exponential function can be defined by a series. So for a real number, x $\displaystyle {e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + .........$ This also works for functions of x So if $\displaystyle f\left( x \right) =  2{x^2}$ substituting into the above series gives $\displaystyle {e^{  2{x^2}}} = 1 + \frac{{\left( {  2{x^2}} \right)}}{{1!}} + \frac{{{{\left( {  2{x^2}} \right)}^2}}}{{2!}} + \frac{{{{\left( {  2{x^2}} \right)}^3}}}{{3!}} + .........$ $\displaystyle = 1  2{x^2} + 2{x^4}  \frac{4}{3}{x^6} + \frac{2}{3}{x^8}  ...........$ This also works for complex numbers z = a+jb $\displaystyle {e^z} = 1 + \frac{z}{{1!}} + \frac{{{z^2}}}{{2!}} + \frac{{{z^3}}}{{3!}} + .........$ Now if a is zero we have an imaginary number thus $\displaystyle z = \left( {0 + j\theta } \right) = \left( {j\theta } \right)$ substituting $\displaystyle {e^{j\theta }} = 1 + \frac{{\left( {j\theta } \right)}}{{1!}} + \frac{{{{\left( {j\theta } \right)}^2}}}{{2!}} + \frac{{{{\left( {j\theta } \right)}^3}}}{{3!}} + .........$ If we multiply all these out and collect we find that we have two collections, those which are real and those which are multiplied by j or imaginary $\displaystyle {e^{j\theta }} = \left( {1  \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^2}}}{{4!}}  \frac{{{\theta ^2}}}{{6!}} + ....} \right) + j\left( {\theta  \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^5}}}{{5!}}  \frac{{{\theta ^7}}}{{7!}} + ....} \right)$ these two collections are the power series for cos and sine thus $\displaystyle {e^{j\theta }} = \cos \theta + j\sin \theta $ This is the easiest way to do this and, of course, ties in with differential equations because you can usually find a power series solution to a differential equation. So there are often two ways to define a mathematical function by power series or by differential equation. Final point your translation of "circulating vector" is incorrect. The correct term is rotating vector. I say this because the term circulation has a particular meaning in mathematics which is not this one. Last edited by skipjack; July 22nd, 2017 at 06:59 AM. 
July 22nd, 2017, 09:30 AM  #7 
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0 
Thanks a lot for the extensive reply! I am taking my time to "absorb" and comprehend the information.... I am getting so much help!! I feel so excited! And I am so eager to learn!! By the way: "circulating vector" was my error in the translation. I am translating the text into English. Now I know the proper English phrase is "rotating vector". 
July 22nd, 2017, 09:48 AM  #8  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
You will find pretty well everything you want there, well into the future.  

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