Riemann prose, part 6


Instead of treating the x-y graph as being a simple combination of a single-dimensional variable plus the result of the equation, what if we make x into two dimensions, and call the output of the equation “z”?

z = 2x + 3y

We’re no longer drawing a line; now we’re making a sheet, or plane. We want “x” and “y” to be independent of each other, and to have certain properties. We want to keep the scaling factor, so that “x” times x moves us left and right on the horizontal axis, and “x” times y moves us up and down on the vertical axis.

For “y”, we want something connected to that “-1” mentioned in the last blog entry. We want “y” times x to be xy, but “y” times y to be -1.

You probably already know where I’m going with this.

If we calculate the areas of each of the squares in the four quadrants:

50 * 50 = 2,500
50 * (-50) = -2,500
(-50) * 50 = -2,500
(-50) * (-50) = 2,500

We can get the areas in each case, but how do we work backwards from the area to get the length of the side for that square?

y = squareroot(2500) = +50, or -50
y = sqrt(2500) = +/-50

Ok, fine, this works. But,

y = sqrt(-2500)?

Fail. We don’t have a number that, times itself, equals -1.

Yet. But, this is exactly the connection to -1 that we’re looking for in our speculated vertical axis element, “y”. “y” * “y” = -1.

I’ll quit the drama, and say that we’re reached “i”. The purpose of “i” is to rotate a point counterclockwise around the origin (0,0).

Because “i” doesn’t physically exist, it’s a bookkeeping placemarker, numbers that include “i”, in the form of “b*i”, were called “imaginary numbers”, while our existing numbers (“a”) were identified as “real numbers”. Combining them in the form of “a + i*b” creates complex numbers.

“a” and “i*b” are related by being the two sides of a right-angle triangle. We can either write a complex number as (1.3 + 2.7i), or as a vector – magnitude (mag) and angle. Where mag is sqrt(a^2 + b^2), and the angle (theta) is arctan(b/a).

2 * (a + ib) = 2a + 2ib (scaling)
i * (a + ib) = ia – b (rotation)
(a + ib)^2 = a^2 + 2iab – b^2 (both)

Complex numbers are symbolized as “s”. Now, where have we seen “s” before?

Hmm. That looks like the Riemann Zeta function again. 1/n^2 could be rewritten as

1/n^(a+ib), where a = 2 and b = 0

How would that work?

Let’s backtrack and just use real numbers.

n^a = n * n * n *… “a” times
n^2 = n * n
n^3 = n * n * n
n^4 = n * n * n * n

Here’s an interesting thing: n*n*n*n*n = (n*n)*(n*n*n) = (n^2) * (n^3) = n^5

Exponents (the powers following the “^” symbol) of a specific value “n” are additive. So,

n^(a+ib) = (n^a) * (n^ib)

We already have an idea of what “n^a” does, as long as “a” is a positive integer. What if it’s NOT a positive integer?

1/n = n^(-1)

Having a negative real number in the exponent is just the same as taking the reciprocal of “n”.

n^(1/2) = sqrt(n)

If n squared (n^2 = n * n) = 2,500, then the value of n, which squared gives us 2,500, must be + or – 50. That is, n = 2500^(1/2) = +/-50.

A cube is defined as a*a*a = a^3,
so the cube root of a number will be the length of one side of that cube, “a”.

8^(1/3) = 2

We could also write n^1/2 as n^0.5

So, fractional values in the real portion of the exponent just refer to root operations (square root, cube root, fractional roots, etc). That is, for “n^2.5”, n is equal to:

n*n * extra

Where “extra” is whatever number gives you “n” when multiplied by itself 1/0.5 (2) times.

9^2.5 = 9^2 * 9^0.5 = 9*9 * sqrt(9) = 81 * 3 = 271

So, what do imaginary exponents do?

Advertisements
Leave a comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: