Click here to purchase the entire book in PDF format. Euler's Identity So far, we've looked at logarithms with a base of 10. As we've seen, a logarithm is just an exponent backwards, so if $A^{B} = C$ then $\log_{A} C = B$. Therefore $\log_{10} 100 = 2$. There is a beast in mathematics known as a natural logarithm which is just like a regular old everyday logarithm except that the base is a very specific number - it's e. ``What's e?'' I hear you cry... Well, just like $\pi$ is an irrational number close to 3.14159, e is an irrational number that is pretty close to 2.718281828459045235360287 but it keeps on going after the decimal place forever and ever. (If it didn't, it wouldn't be irrational, would it?) How someone arrived at that number is pretty much inconsequential - particular if we want to avoid using calculus, but if you'd like to calculate it, and if you've got a lot of time on your hands, the math is: $$e = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...$$ (2.60) (If you're not familiar with the mathematical expression ``!'' you don't have to panic! It's short for factorial and it means that you multiply all the whole numbers up to and including the number. For example, $5! = 1*2*3*4*5$.) How is this e useful to us? Well, there are a number of reasons, but one in particular. It turns out that if we raise e to an exponent x, we get the following. $$e^{x} = \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + ...$$ (2.61) Unfortunately, this isn't really useful to us. However, if we raise e to an exponent that is an imaginary number, something different happens. $$e^{j\theta} = \cos(\theta) + j \sin(\theta)$$ (2.62) This is known as Euler's identity or Euler's formula. Notice now that, by putting an i up there in the exponent, we have an equation that links trigonometric functions to an algebraic function. This identity, first proved by a man named Leonhard Euler2.4, is really useful to us. There's just a couple of extra things to take note of: Since $\cos(\pi) = -1$ and $\sin(\pi) = 0$ then: $$e^{j\pi} = \cos(\pi) + j \sin(\pi)$$ (2.63) $$e^{j\pi} = -1 + 0$$ (2.64) therefore $$e^{j\pi} + 1 = 0$$ (2.65)