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.

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.

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.

The substitution we made leads us to a very interesting formula:
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))]

Unizor - Limits - Number e as a Limit

We have introduced a number e, an extremely important number in calculus and analysis, as a base of an exponential function y=a^x with a steepness of 1 at the argument value x=0. We have also indicated without sufficient rigorousness that this number is the limit of an infinite sequence:
e = lim [n→∞] (1+1/n)^n.
In this lecture we will rigorously prove that the above limit does exist and can be used as a definition of a number e.

In the lecture Problems 3 of this chapter about limits we have proved that an infinite monotonously increasing sequence that is bounded from above has a limit. This theorem is a foundation of this lecture. We will prove that a sequence (1+1/n)^n is monotonously increasing and has an upper bound. Therefore, it has a limit as n→∞, and that limit is a definition of a number e.

The fact that our sequence (1+1/n)^n is bounded from above has already been proven in the previous lecture about a function F(x)=e^x where we have proved that for any natural number n
2 ≤ (1+1/n)n ≤ 3
So, all we have to prove now is the monotonic character of this sequence.

So, let's prove that for all n greater than 1
(1+1/n)^n ≥ [1+1/(n−1)]^(n−1)
Direct method to prove this is to use the Binomial formula by Newton.
The expression on the left is a sum of n positive terms, the ith of them being
n!/[(n−i)!·i!·n^i]
The expression on the right is a sum of n−1 positive terms, the i-th of them being
(n−1)!/[(n−i−1)!·i!·(n−1)^i]
The expression on the left has one more positive term in the sum. Now we will prove that every i-th term of the left expression is greater than the corresponding i-th term of the right expression.
Indeed, we can get rid of i! since it is the same for both left and right common terms.
What remains from the left term can be written as
n·(n−1)·...·(n−i+1)/n^i =
= 1·[1−1/n]·...·[1−(i−1)/n]
What remains from the right term can be written as
(n−1)·(n−2)·...·(n−i)/(n−1)^i =
= 1·[1−1/(n−1)]·...·[1−(i−1)/(n−1)]
The term on the left is greater because each member of a product in the left term is greater than corresponding member of a product in the right term since
1-k/n ≥ 1-k/(n−1)
for each k from 1 to i−1.

This completes the proof that a sequence (1+1/n)^n is monotonically increasing. Together with the fact that it is less than 3 for any positive integer number n, proven in the previous lecture about a function ex (that is, 3 is the upper bound for this sequence) it proves that a sequence (1+1/n)^n tends to a limit, which is some real number between 2 and 3. This number is the number e introduced in the previous lecture from a different angle, as a base of an exponential function that has a steepness of 1 at point x=0.

Based on this, let's mention one more equality related to limits.
Let x be any real number. Consider an exponential function F(x)=e^x.
We can state now that
lim[n→∞](1+x/n)^n = e^x

Intuitively, it's obvious since
(1+x/n)^n = [(1+x/n)^(n/x)]^ x
When n goes to infinity, n/x goes to infinity as well and the expression in square brackets tends to e, so the whole expression tents to e^x. Although it's not an absolutely rigorous proof, this consideration should suffice for now.

Unizor - Trigonometry - Exponentiation of Complex Numbers

Continuing on trigonometric representation of complex numbers, let's research how to raise a complex number in trigonometric form to some power (the process called exponentiation). It's important to understand that the rules we will be dealing with are not theorems that we prove, but definitions of new operations. We just prove that these definitions are reasonable in a sense that they preserve important properties of new operations, similar to properties we already know for real numbers, like
a^0 = 1 and
a^(P+Q) = a^P·a^Q and
a^(-P) = 1/(a^P) and
a^(P·Q) = (a^P)^Q

Since we know how to multiply two complex numbers in trigonometric form, we can easily derive a formula for raising a complex number to a positive integer power since this process is just a multiplication by itself certain number of times.
Indeed, using the property of multiplication addressed in the previous lecture,
[r·cos(φ)+i·r·sin(φ)]^2 =
= [r·cos(φ)+i·r·sin(φ)] ·
· [r·cos(φ)+i·r·sin(φ)] =
=r·r·cos(φ+φ)+i·r·r·sin(φ+φ)=
= r^2·cos(2·φ)+i·r^2·sin(2·φ) =
= r^2·[cos(2·φ)+i·sin(2·φ)]

Similarly, by induction, for any natural N,
[r·cos(φ)+i·r·sin(φ)]^N =
= r^N·cos(N·φ)+i·r^N·sin(N·φ) =
= r^N·[cos(N·φ)+i·sin(N·φ)]

Expanding this to negative integer numbers, we will use the main property of exponentiation we would certainly want to preserve:
a^(M+N) = a^M · a^N
and, derived from it for N=−M
1 = a^0 = a^(M−M) = a^[M+(−M)] = a^M · a^(−M)
from which follows:
a^(−M) = 1 / (a^M)

Therefore, it's reasonable to define an exponentiation of complex numbers with negative integer power −N (where N is a positive integer) as
[r·cos(φ)+i·r·sin(φ)]^(−N) =
= 1/[r^N·cos(N·φ)+i·r^N·sin(N·φ)]

Now it's easy to notice that for any argument (phase) ψ of a complex number in polar form
[cos(ψ)+i·sin(ψ)]·[cos(−ψ)+i·sin(−ψ)] =
= cos(ψ−ψ)+i·sin(ψ−ψ) =
= cos(0) + i·sin(0) = 1
Therefore,
1 / [cos(ψ)+i·sin(ψ)] =
= cos(−ψ)+i·sin(−ψ)

Using the above equality for exponentiation to a negative integer power and similar equality for real numbers 1/(r^N)=r^(−N), we derive
[r·cos(φ)+i·r·sin(φ)]^(−N) =
= r^(−N)·[cos(−N·φ)+i·sin(−N·φ)]
Notice the universality of the formula for natural exponent (a consequence of plain multiplication)
{r·[cos(φ)+i·sin(φ)]}^N =
= r^N·[cos(N·φ)+i·sin(N·φ)]
which has exactly the same form if N is a negative integer.

Let's proceed to rational exponent of a complex number in trigonometric form.
The simple approach to define exponentiation with rational exponent is to use the above formula for integer N.
Let's define a new absolute value (modulus) q=r^N and a new argument ψ=N·φ.
Then r=q^(1/N), φ=ψ/N and the formula would look like this:
{q^(1/N)·[cos(ψ/N)+i·sin(ψ/N)]}^N =
= q·[cos(ψ)+i·sin(ψ)]
This is a justification for the general definition of raising a complex number to a power equal to a rational number 1/N (where N is an integer):
{q·[cos(ψ)+i·sin(ψ)]}1/N =
= q^(1/N)·[cos(ψ/N)+i·sin(ψ/N)]
Notice, again, the general format of the formula is exactly the same as for positive integer exponents, that is the absolute value is raised to a power and an argument is multiplied by this power.

From here it's just one little step to derive a formula for any rational exponent. We just combine the rules for z^M and z^(1/N) into one rule for z^(M/N):
{r·[cos(φ)+i·sin(φ)]}^(M/N) =
= r^(M/N)·[cos((M/N)·φ)+i·sin((M/N)·φ)]

Imagine how cumbersome the formula for raising a complex number into a rational power would look in the traditional representation of a complex number as z=a+b·i.

To make this lecture complete, it's necessary to say a few words about irrational exponents. Here, as with raising real numbers to irrational power, exact rigorous mathematics is quite involved. Sufficient to say that the theory of limits is used and, as irrational numbers can be considered as limits of sequences of rational numbers, the definition of irrational exponentiation is based on the same limit principle and is defined as a limit of corresponding rational exponentiation.

Example
Let's see what (−1)^1/2 is if we use the trigonometry.
(−1)^1/2 = [cos(π)+i·sin(π)]^1/2 = cos(π/2)+i·sin(π/2) = 0+i·1 = i
as expected, because this is a definition of a complex i.

Unizor - Limits - F(x)=e^x

In the chapter describing exponential functions we introduced the concept of steepness β of an exponential function F(x)=a^x at point x=0 as a limit of a ratio between increment of a function between points x=0 and x=1/n (that is F(1/n)−F(0)=a^(1/n)−a^0) and a corresponding increment of an argument (that is 1/n−0) when n→∞, so a point x=1/n is getting closer and closer to a point x=0.
After a trivial simplification (since a^0=1) the steepness of a function F(x)=a^x at point x=0 is defined as
β = lim[n→∞](a^(1/n)−1)/(1/n) = lim[n→∞]n·(a^(1/n)−1)

At this point we would like you to recall the material presented in the lecture Algebra - Exponential Functions - Problems 2 about a steepness of exponential functions 2^x and 3^x at a point x=0. Refer to that lecture for refreshing.

We have proved there that the steepness of a function 2^x at point x=0 is less than 1, while the steepness of a function 3^x at point x=0 is greater than 1.
Assuming continuity of steepness as it depends on the base of the exponential function, it is reasonable to expect that somewhere between 2 and 3 there is a value of a base of an exponential function with the steepness equaled exactly to 1. This value of a base is denoted as e in mathematics and its approximate value is 2.71.
Therefore, we can write:
lim[n→∞] (e^(1/n)−1)/(1/n) = 1

Slightly deviating from the rigorousness we used to apply to our statements, we can derive from the last equality that there is an approximation
(e^(1/n)−1)/(1/n) ≅ 1
which becomes better and better as n→∞.
Or, resolving this for e,
e^(1/n)−1 ≅ 1/n or
e^(1/n) ≅ 1+1/n or
e ≅ (1+1/n)^n
Returning to exact equality, we can write the value of e as the following limit:
e = lim[n→∞] (1+1/n)^n

Summarizing the above logic, we have determined a specific value of the base of the exponential function with the steepness of 1 at point x=0. This value is designated a symbol e and is equal to
e = lim[n→∞] (1+1/n)^n
Thus, we have defined an exponential function
F(x) = e^x
Obviously, all properties of exponential functions are preserved in this particular function, including but not limited to:
e^0 = 1
e^1 = e
e^(x+y) = e^x·e^y
e^(x·y) = (e^x)^y

Another interesting fact is that steepness, as we introduced it for function F(x)=e^x at point x=0, can be defined at any other point using exactly the same methodology.
Consider any point x=x0. Now step forward from this point by a value 1/n to a point x=x0+1/n. Construct a ratio between the difference in the values of a function in these two points and the difference in the values of an argument:
β = [F(x0+1/n) − F(x0)] / (1/n)
As n→∞, the limit of this ratio characterizes the steepness of a function F(x) at point x0.

In case of the function we have introduced in this lecture, F(x)=e^x, the steepness at any point x0 equals to a limit of the following expression, as n→∞:
[e^(x0+1/n) − e^x0] / (1/n) =
= [e^(x0)·e^(1/n) − e^(x0)] / (1/n) =
= e^(x0)·[e^(1/n) − 1] / (1/n)
As n→∞, e^(x0) remains a constant while the expression
[e^(1/n) − 1] / (1/n), which represents the steepness of function ex at point x=0, has a limit of 1.
Therefore, the steepness of function e^x at any point x0 equals to a value of this function at this same point, that is e^(x0). This is quite a remarkable and unique property of function F(x)=e^x.