Back to Riemann, Part 4


Ok, let’s get back on topic for e^bi.

Or, maybe not.
I want to look at one more way to calculate e.

The zeta function is generally written as:

z(x) = 1 + 1/2^x + 1/3^x + 1/4^x + 1/5^x…

on into infinity, where x can have the value of a + bi across the complex plane.

If we set x = 1, we get the geometric series, AKA the Harmonic Series:

z(1) = 1 + 1/2 + 1/3 + 1/4 + 1/5…

Which doesn’t converge. It just goes to infinity kind of slowly.

Setting x = 2, however, gives us the series:

z(2) = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2…
z(2) = pi^2 / 6 = 1.6449…

I ran the above series in Excel, to 1/177^2, and got 1.6393, which is pretty close.

z(3) = 1 + 1/2^3 + 1/3^3 + 1/4^3 + 1/5^3…
Which approaches 1.202041
z(4) = pi^4/90 = 1.0823…

So, what this says is that the zeta function z(x) converges for x > 1, but it doesn’t converge to 0 (it converges to 1 for large values of x).

There’s also a series that converges to e^x:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5!…

Where “!” is the symbol for factorial, or for multiplying the integers from 1 to n together.

2! = 1*2 —– = 2
3! = 1*2*3 — = 6
4! = 1*2*3*4 – = 24
5! = 1*2*3*4*5 = 120


(After only 10 terms, the series is already at 2.71828, and the curve looks flat.)

In the last blog entry, I mentioned that taking the derivative of f(x) = e^x => f'(x) = e^x.
We can prove this by taking the derivative of the series above:

d/dx e^x = d/dx (1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5!…)

Recall that if f(x) = x^b, then
d/dx f(x) = b * x^(b-1)

Note also that d/dx of a constant = 0.
We can take the derivatives of each term in the series separately, as in:

d/dx (1) = 0
d/dx (x) = 1*x^0 = 1
d/dx (x^2) = 2*x
d/dx (x^3) = 3*x^2
d/dx (x^4) = 4*x^3
d/dx (x^5) = 5*x^4
etc.

Plugging the terms back into f'(x),
f'(x) = 0 + 1 + 2*x/(1*2) + 3*x^2/1*2*3 + 4*x^3/1*2*3*4…
f'(x) = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5!…
f'(x) = e^x

What’s interesting about this infinite series approach is that it can also be used to find values for sin(x) and cos(x)

sin(x) = x – x^3/3! + x^5/5! – x^7/7! + x^9/9!…
cos(x) = 1 – x^2/2! + x^4/4! – x^6/6! + x^8/8!…

Now, what happens if we play with e^iy?

e^iy = 1 + iy + (iy)^2/2! + (iy)^3/3! +(iy)^4/4! + (iy)^5/5!…
e^iy = 1 + iy – y^2/2! + iy^3/3! + y^4/4! + iy^5/5! – y^6/6!…

If we group the terms with and without i components together,
e^iy = (1 – y^2/2! + y^4/4! – y^6/6!…) + i(y – y^3/3! + y^5/5!…)

Look at the series expansions for sin() and cos() above.

e^iy = cos(y) + i*sin(y)

And, if we pick y = pi
e^i*pi = cos(pi) + i*sin(pi) = -1 + 0 = -1

What’s this tell us? That e^iy defines the unit circle.

Therefore, e^x, where x = a+bi, gives us e^(a+bi) = (e^a) * (e^bi)
= e^a * (cos(b) + i*sin(b))
(if a=0, e^a  = 1)


(The unit circle as a function of e^bi, which we already saw in Riemann Prose part 7.)

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1 Comment

  1. Back to Riemann, Part 7 | threestepsoverjapan

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