## Tuesday, January 6, 2015

### Unizor - Trigonometry and Complex Numbers - Problems 2

Evaluate the following sums:
PN = cos(x) + cos(2·x) +...+ cos((N−1)·x) + cos(N·x)
and
RN = sin(x) + sin(2·x) +...+ sin((N−1)·x) + sin(N·x)

Solution:
The creative part of this problem is to realize that switching to complex numbers greatly simplifies it, basically converting a trigonometric problem we don't know how to approach to an algebraic problem of summarizing a geometric progression.
Unfortunately, non-creative part of a solution requires some tedious trigonometric transformations (which from pedagogical standpoint play their positive role).
Let's use the Euler's formula
e^(i·φ) = cos(φ)+i·sin(φ).
Consider a series:
SN = e^(i·x) + e^(2·i·x) +...+ e^[(N−1)·i·x]+ e^(N·i·x)
From the Euler's formula for φ=x, φ=2·x,..., φ=(N−1)·x and φ=N·x follows that
SN = PN + i·RN
Since e^(k·i·x) = [e^(i·x)]k, we can transform the above sum SN into
SN = e^(i·x)+[e^(i·x)]^2+...+ [e^(i·x)]^(N−1) + [e^(i·x)]^N
The latter is a sum of a geometric progression.
Without bringing up a formula, which might be hard to remember, we will calculate the sum using the following logic:
SN·e^(i·x) = SN − e^(i·x) + [e^(i·x)](N+1)
Simplifying and resolving it for SN, we obtain:
SN = [e^(i·x) − e^(i·(N+1))·x] / [1−e^(i·x)]
All we have to do now is to convert this complex number into its canonical representation a+b·i and obtain the solutions:
PN = a and RN = b
This is a purely technical exercise with complex numbers.

## Monday, January 5, 2015

### Unizor - Trigonometry and Complex Numbers - Problems 1

Using the Euler's formula
e^(i·x) = cos(x)+i·sin(x),
prove the following equalities for any real numbers x and y.

1. e^(i·0) = 1

2. e^(i·x)·e^(i·y) = e^[i·(x+y)]

3. 1/[e^(i·x)]= e^(−i·x)

4. [e^(i·x)]^y = e^(i·x·y)

5. Absolute value (modulus) of e^(i·x) equals to 1

6. Multiplication of any complex number represented on a coordinate plane by e^(i·x) results in its rotation by an angle of x radians.

## Saturday, January 3, 2015

### Unizor - Trigonometry - Lim (SIN(x)/x)

This lecture combines three different subjects, geometry, trigonometry and a theory of limits, into one extremely important theorem.

Theorem:
Consider an angle x, measured in radians, tends to zero, while remaining positive (staying positive is not really necessary for a theorem but makes it easier to explain the proof). Then the ratio of its sine to its value tends to 1, that is
lim[x→0] sin(x)/x = 1

Proof:
Consider a unit circle around the origin of coordinates O intersecting a positive direction of the X-axis at point P.
Since we are talking in this theorem about an angle decreasing to zero while remaining positive, we need to consider only angles in the first quadrant.

Choose a point A on a unit circle, connect it with a center O and let angle ∠POA measure x radians. Connect point A with point P. Consider a triangle ΔAOP.

Let's compare the area of this triangle and the area of a sector of our unit circle between radius OP and radius OA. Obviously, the area of a sector, that includes the area of a triangle and an additional piece between a chord AP and an arc of a circle, is greater than the area of a triangle. Let's compare them quantitatively.

To determine an area of a triangle ΔAOP, drop a perpendicular from point A to OP, its base on OP being point Q. Obviously, AQ is an altitude in a triangle ΔAOP. The length of this altitude is an ordinate of point A and, therefore, by definition of an sine of an angle, equals to sin(x). The base of our triangle is a radius OP which is equal to 1. Therefore,
area (ΔAOP) = sin(x)/2

Area of a sector AOP that measures x radians is smaller than the area of an entire unit circle, which can be considered as a sector of 2π radians, by a factor of x/2π. The area of a unit circle equals to π·12=π. Therefore,
area (sector AOP) = (x/2π)·π = x/2
So, sin(x) is less than x.

Let's consider another pair of geometric figures. Extend radius OA beyond point A and draw a perpendicular to radius OP at point P. Point of intersection of this perpendicular with an extension to OA is point R. Consider a triangle ΔROP. Obviously, its area is greater than the area of a sector AOP we considered before because it completely includes it in itself. Let's quantify this inequality.

Triangle ΔROP is a right triangle with one cathetus OP equal to 1 and a ratio of another cathetus PR to this OP being a tangent of an angle measured x radians. Therefore, the area of triangle ΔROP equals to tan(x)/2 and x less than tan(x).

Putting together both inequalities, we come up with
sin(x) less than x less than tan(x).

Since we consider angle ∠POA to be a small positive angle measured x radians, all components of these two inequalities are positive. Let's divide all of them by a positive value of sin(x). The result is
1 less than x/sin(x) less than tan(x)/sin(x).
Since, by definition, tan(x)=sin(x)/cos(x), this inequality can be written as
1 less than x/sin(x) less than 1/cos(x)
Inverting these,
1 greater than sin(x)/x greater than cos(x)

The last remaining step of the proof is to use a theorem from the theory of limits that, if a sequence is bounded from below by one sequence and from above by another, and both bounding sequences tend to the same limit, then the sequence "squeezed in-between them" tends to the same limit (see Algebra - Limits - Problems 3 of this course). Now, as angle ∠POA decreases to zero (that is, x→0), the upper bound of our ratio sin(x)/x remains constant 1, the lower bound of this ratio is cos(x), which at point x=0 also equals to 1 and, therefore, tends to 1 as x decreases to zero. Therefore, our ratio, "squeezed between" a constant 1 from above and variable cos(x) that also tends to 1, has a limit of 1:
lim[x→0] sin(x)/x) = 1
End of Proof.

Graphical interpretation of this limit is that the graph of function y=sin(x), as positive x gets closer and closer to zero, becomes more and more like the graph of function y=x, a straight line bisecting the first and third quadrant, while remaining just below this straight line.

Incidentally, as x approaches zero, ratio tan(x)/x also tends to 1 because it's not much different from sin(x)/x since tan(x)=sin(x)/cos(x) and cos(x) is almost equal to 1 for small x. Graphically, this limit means that the graph of function y=tan(x) behaves like the graph of function y=x in the positive vicinity of point x=0, while remaining just above it.

### Unizor - Trigonometry and Complex Exponent - Euler's Formula

The remarkable formula that carries Euler's name looks like this (for any real x):
e^(i·x) = cos(x)+i·sin(x)

This formula can be considered as a foundation for a definition of this concept. However, a lot of considerations were made to present it in this way and this is the only way to define complex exponentiation to preserve all the properties of exponentiation we know from using only real numbers as exponents.

Let's recall a few facts from the material presented earlier.
In Algebra - Limits chapter we discussed a number e and derived a formula that represented this number as a limit:
e = lim[n→∞] (1+1/n)^n.
We have also derived there a more general formula
e^x = lim[n→∞] (1+x/n)^n.

This is not a rigorous proof of the Euler's formula, but reasonable considerations that help us to properly define the operation of raising a real number into complex power.

Let's look at the representation of e^x as a limit above and use it formally for x=i.
e^i = lim[n→∞] (1+i/n)^n.

There is nothing undefined in this formula - we know how to add, subtract, multiply, divide and raise into power complex numbers.
In theory, if we skip the proof that this limit exists, which is beyond the scope of this lecture, this might be a proper definition of a complex exponent. But we will go further to arrive at more constructive Euler's formula.

Let's consider a base of the above expression for ei as a limit - the term (1+i/n) - and represent it as a point A on a complex plane with origin at point O where real part of a number is an abscissa and imaginary part is an ordinate of this point. Abscissa of our point equals to 1, its ordinate equals to 1/n, which in case of large n puts our point A(1,1/n) just slightly above the point P(1,0) lying on the X-axis. This "slightly above" is diminishing as n→∞, so point A moves down closer and closer to point P.

Let's use the representation of complex numbers in polar form as r·[cos(φ)+i·sin(φ)].
In this case
r=OA,
φ = ∠POA.

As n→∞ and, as we mentioned, point A moving down closer and closer to point P, r→1 (the length of OP) and φ→0.

We know that for angles tending to 0 there is a very important limit
lim[φ→0] sin(φ)/φ = 1
Also cos(φ) gets closer and closer to 1 as φ→0.
The above considerations prompt us to replace point (1,1/n) with a point (cos(1/n),sin(1/n)), which is infinitely close to it.

The reason for such substitution is the fact that we have to raise an expression (1+1/n) to n-th power, and this is much easier to do in the polar form. So, we substitute (1+i/n)^n with [cos(1/n)+i·sin(1/n)]^n.

e^i = lim[n→∞] (1+i/n)^n ≅ [cos(1/n)+i·sin(1/n)]^n

Using the properties of polar representation of complex numbers, we can simplify the last expression:
[cos(1/n)+i·sin(1/n)]^n = cos(n·1/n)+i·sin(n·1/n) = cos(1)+i·sin(1)
(just a reminder, angles are in radians, so angle of 1 means 1 radian).

Our final formula is:
e^i = cos(1)+i·sin(1)
Let us emphasize again, we have not proven this formula, since there was no definition of complex exponentiation, but we came up with it using reasonable transformation of properties of complex numbers and exponentiation of real numbers onto the field of complex numbers.

Immediate consequence of the above formula, again transforming known properties of exponentiation of real numbers onto complex numbers and properties of polar representation of complex numbers, is for any real x:
e^(i·x) = (e^i)^x = [cos(1)+i·sin(1)]^x = cos(1·x)+i·sin(1·x) =
= cos(x)+i·sin(x)
which is the Euler's formula we presented above.

The beauty of this formula can be compared with the beauty of a famous Einstein's formula for full energy E=m·c². Euler has brought together complex analysis, trigonometry and limits, seemingly unrelated concepts, and Einstein has brought together energy, mass and speed of light, also quite unrelated from the surface concepts.
Everything is reasonably interrelated in our Universe. We just have to learn to see these relations.

Let's now define complex exponentiation using the above formula. If our definition is correct, all the properties of exponentiation that involves only real numbers must be held in complex case, and that is something that we can prove using our definition and known properties.

We define raising of one particular real number e to any complex power a+i·b (a and b are any real numbers) as
e^(a+i·b) = e^a·e^(i·b) = e^a·[cos(b)+i·sin(b)]

Then, using the logarithms with a base e (natural logarithms), we can define this operation for any other real number. Assume we want to raise a real number d to complex power a+i·b.
We can always represent
d = e^ln(d)
where ln(d)=log[e](d) - natural (based e) logarithm of d.
Therefore,
d^(a+i·b) = d^a·d^(i·b) = d^a·[e^ln(d)]i·b = d^a·e^[i·b·ln(d)]
The last expression in polar form is
d^a·[cos(b·ln(d))+i·sin(b·ln(d))]