I am going to present a new feature to all my 0 readers - the "Gem of the Week". This is a reprise of a sometime feature on my old private blog, "Chemical of the Day". I am expanding the topic somewhat from chemicals to anything I find interesting. None of these are necessarily news, but they might be.
This week, it's Euler's Identity.
The mathematicians treat this one as almost holy.
\[e^{i\pi}+1=0\]
They will go on and on about the five most important constants, the amazing, almost miraculous nature, etc, blah, blah, blah. However, this is just one facet of the gem. if we dig a bit deeper, we find the whole gem underneath.
To start off with, we have complex numbers. Aggressively ignoring history, we present the concept in basically its modern form. The product of any two negative numbers is positive, so no real number is the square root of a negative number. So, we invent the concept of imaginary numbers. The imaginary unit \(i\) is created out of thin air, and to begin with, it's most important property is \(i^2=-1\) or taking the square root of both sides, \(i=\sqrt{-1}\). Any time you have a negative number \(-a\) under a square root you can always factor it \(\sqrt{-a}=\sqrt{(-1)a}=\sqrt{-1}\sqrt{a}=i\sqrt{a}\).
It's amazing how far we can run with this simple definition.
First, real numbers and imaginary numbers can be added to produce complex numbers, but you can't simplify the addition. If you add the real number 3 to the imaginary number 4\(i\), the best you can do is 3+4\(i\). You can treat \(i\) as an unknown variable and do math with it, and thereby obtain the addition and multiplication rules. The interesting one is multiplication as we will see here:
\[\begin{eqnarray*}(a+bi)(c+di)&=&ac+adi+bic+bdi^2\\
&=&ac-bd+i(ad+bc)\end{eqnarray*}\]
We used the property \(i^2=-1\) to simplify \(bdi^2\) to \(-bd\).
Real numbers are visualized as points on the number line. Complex numbers are visualized as points on the complex plane, where the x coordinate is the coefficient of the real part, and the y coordinate is the coefficient of the imaginary part.
Now the fun begins. If you have a point on a plane, you can express it in rectangular coordinates like we have above, or in polar coordinates, where instead of real and imaginary parts, you have the magnitude, the distance from the origin, and the argument, the angle from the +x axis to the segment from the origin to the point.
So, if you have a complex number \(a+bi\), you can figure out the magnitude from the Pythagorean theorem:
\[r=\sqrt{a^2+b^2}\]
and the argument from trigonometry:
\[\theta=\tan^{-1}\frac{b}{a}\]
If we are calculating this on a computer, we use theta=atan2(b,a)to make sure that the quadrant is right.
So, any complex number \(a+bi\) can be expressed in polar form as \(r\angle\theta\). The reverse is also true, and can be expressed in one formula:
\[\begin{eqnarray*}r\angle\theta&=&r\cos\theta+ri\sin\theta\\
&=&r(cos\theta+i\sin\theta)\end{eqnarray*}\]
When this is first presented in algebra textbooks, the part in parentheses is often called the \(\operatorname{cis}\) function (cosine i sine), so we have \(r\angle\theta=r\operatorname{cis}(\theta )\). However, you never see this in advanced or professional works. We'll see why presently.
Here comes the gem - Look at the Taylor series expansion of cosine and sine:
\[\sin \theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...=\sum_{i=0}^\infty\frac{-1^{n}\theta^{(2n+1)}}{(2n+1)!}\]
\[\cos \theta=1-\frac{\theta^2}{2}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}...=\sum_{i=0}^\infty\frac{-1^{n}\theta^{(2n)}}{(2n)!}\]
The general terms are correct but not particularly important, just look at the first few specific terms. Now the number we get when we add cosine and the imaginary unit times sine is as follows:
\[\cos \theta+i\sin \theta=1+i\theta-\frac{\theta^2}{2!}-i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+i\frac{\theta^5}{5!}+\frac{\theta^6}{6!}-i\frac{\theta^7}{7!}...\]
Now just for fun and because I know this will work, let's look at the Taylor series for the natural exponential function. The odd thing is that we will use an imaginary argument \(x=i\theta\)
\[e^x=1+x+\frac{x^2}{2!} +\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}...=\sum_{i=0}^\infty\frac{x^n}{n!}\]
The Taylor series expansion was derived in real number space, so there is no guarantee that it will work with complex numbers, but let's define the imaginary exponent to be whatever the formula gives. We accept as an extension of the definition of exponentiation the following formula:
\[\begin{eqnarray*}e^{i\theta}&=&1&+&i\theta&+&\frac{(i\theta)^2}{2!}&+&\frac{(i\theta)^3}{3!}&+&\frac{(i\theta)^4}{4!}&+&\frac{(i\theta)^5}{5!}&+&\frac{(i\theta)^6}{6!}&+&\frac{(i\theta)^7}{7!}&...\\ &=&1&+&i\theta&+&\frac{i^2\theta^2}{2!}&+&\frac{i^3\theta^3}{3!}&+&\frac{i^4\theta^4}{4!}&+&\frac{i^5\theta^5}{5!}&+&\frac{i^6\theta^6}{6!}&+&\frac{i^7\theta^7}{7!}&...\\\end{eqnarray*}\]
We can clean this up a bit by noting that \(i^3=ii^2=i(-1)=-i\), \(i^4=(i^2)^2=(-1)^2=1\), and so on in a pattern repeating every increment
of 4 in the exponent.
\[\begin{eqnarray*}e^{i\theta} &=&1&+&i\theta&+&\frac{(-1)\theta^2}{2!}&+&\frac{(-i)\theta^3}{3!}&+&\frac{(1)\theta^4}{4!}&+&\frac{(i)\theta^5}{5!}&+&\frac{(-1)\theta^6}{6!}&+&\frac{(-i)\theta^7}{7!}&...\\
&=&1&+&i\theta&-&\frac{\theta^2}{2!}&-&i\frac{\theta^3}{3!}&+&\frac{\theta^4}{4!}&+&i\frac{\theta^5}{5!}&-&\frac{\theta^6}{6!}&-&i\frac{\theta^7}{7!}&...\\\end{eqnarray*}\]
Or in other words, exactly the same as \(cos\theta+i\sin\theta\) above. The \(\operatorname{cis}(\theta)\) function is exactly the exponential function with imaginary argument \(e^{i\theta}\). That's the gem. That's the miracle -- the Taylor series of the trig sum and the exponential match. Once we have \(\cos\theta+i\sin\theta=\operatorname{cis}(\theta)=e^{i\theta}\), it is easy to plug in the circle constant \(\pi\) and get
\[\begin{eqnarray*}e^{i\pi}&=&\cos\pi+i\sin\pi\\
&=&-1+i(0)\\
&=&-1\\
e^{i\pi}+1&=&0\end{eqnarray*}\]
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