Blog Post Maths: Outline

Convergent and Divergent series.

What is \(i\)?

What is trigonometry (unit circle)?
What is the sine function?
What is the cosine function?
[Explanation of angles in the unit circle. The angle is the arc length of the circle measured from (1,0) to the (x,y) co-ordinate of the point on the circle we are interested in. Now, it is hard, very very hard, to work out the actual length of the curve. Indeed it was thought impossible before calculus. Essentially what you have to do is split the x distance between the x=1 and x=x into equal parts, and work out the y co-ordinate for each x. You can then pythagoras the distance between each (x,y) and (x+1,y+1). Add up all the distances and you get an approximation to the arc length. To get the REAL arc length rather than an approximation you need to use an infinite amount of parts. This is where calculus comes in

A bit more about the calculus. The length of an arc between points a and b on the curve of a "rectangular function" \[f(x)\] is the same as:
\[\int_a^b \sqrt{1+\left(f`(x) \right)^2} dx\]


The \(f(x)\) bit for the circle in the positive x,y quadrant just means how high up from the x-axis do you draw the dot for a given x value. Or in other words find y, if given x. Now, we know that \(x^2 + y^2 = 1\) so we also know that \(y^2 = 1 - x^2\) because we have taken \(x^2\) away from both sides, and if you do the same thing to both sides, then the equation remains true. In a similar vein we also know that \(y = \sqrt{1-x^2}\), and that is our \(f(x)\). That can also be written as \[f(x) = (1-x^2)^\frac{1}{2}\]. That makes (magic wand waved) \[f`(x)=\frac{-x}{\sqrt{1-x}\sqrt{x+1}}\]. So the full formula for the length of and arc on a circle is \[\int_a^b \sqrt{1+\left(\frac{-x}{\sqrt{1-x}\sqrt{x+1}} \right)^2} dx\]. And, the version for working out the value for \(\theta\) (the angle at the centre of the unit circle) is \[\int_1^\alpha \sqrt{1+\left(\frac{-x}{\sqrt{1-x}\sqrt{x+1}} \right)^2} dx\], where \(\alpha\) is a value on the x-axis between 0 and 1.

It is claimed that the indefinite integral of that equation is \[\sqrt{\frac{1}{1 - x^2}} \sqrt{1 - x^2} \arcsin(x)\].

However, a different site alleges that \[f`(x)\] is actually \[-\frac{x}{\sqrt{1-x^2}}\] . This makes the length equation \[\int_a^b \sqrt{1+\left(-\frac{x}{\sqrt{1-x^2}} \right)^2} dx\].

integral(sqrt(x^2/(1-x^2)+1), x)

History
Algebra - Alkwarizmi
Functions - Euler
Symbolic notation - descartes and fermat
imaginary - bombelli
plane decartes (2d graphing of functions) wessel (plotting complex numbers on the plane)