## Friday, December 26, 2014

### 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.

## Friday, December 19, 2014

### 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.

## Saturday, December 13, 2014

### 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.

## Thursday, December 11, 2014

### Unizor - Trigonometry - Representation of Complex Numbers

We strongly advise students to refresh the knowledge of Complex Numbers. The lectures dedicated to these numbers are presented in the Algebra part of this course. Special attention should be dedicated to graphical representation of complex numbers.

Here is the main concept of the graphical representation of complex numbers.
We can always consider a complex number z = a + b·i as a pair of two real numbers (a, b) and each such pair (i.e. each complex number) we can put into a correspondence with a point on a coordinate plane with Cartesian coordinates (a, b).

But every point on a plane can be identified not only by its Cartesian coordinates, but a pair of polar coordinates - the distance from the origin r and a polar angle φ.
The conversion from the polar coordinates into Cartesian is a simple trigonometric identities:
a = r·cos(φ)
b = r·sin(φ)
The conversion from Cartesian coordinates to polar is also fully defined by the following equalities:
r = √(a^2+b^2)
cos(φ) = a/√(a^2+b^2)
sin(φ) = b/√(a^2+b^2)

Hence, each complex number a+b·i can be represented in the polar form as
r·cos(φ)+r·sin(φ)·i = r·[cos(φ)+i·sin(φ)]
In this form the distance from the origin r is called magnitude or modulus, or absolute value of a complex number, while the polar angle φ is called an argument or a phase.

Incidentally, real numbers can be considered as a subset of complex numbers with an imaginary component (that is, coefficient at i) equaled to zero. In polar form it means that the argument equals to 0 for positive real numbers or π for negative real numbers. In both cases the imaginary part equals to 0 since sin(0)=sin(π)=0.

While it's easy to add two complex numbers in their traditional form, their product looks much more complicated.

Here is a sum:
(a1+b1·i) + (a2+b2·i) =
= (a1+a2) + (b1+b2)·i

And here is a product:
(a1+b1·i) · (a2+b2·i) =
=a1·a2+a1·b2·i+a2·b1·i+b1·b2·i^2=
= (a1·a2−b1·b2)+(a1·b2+a2·b1)·i
because i^2 = −1

The situation with product is much simpler in the polar form of representation of complex numbers:
r1·[cos(φ1)+i·sin(φ1)] ·
r2·[cos(φ2)+i·sin(φ2)] =
= r1·r2·[cos(φ1)·cos(φ2) +
+ i·cos(φ1)·sin(φ2) +
+ i·sin(φ1)·cos(φ2) +
+ i^2·sin(φ1)·sin(φ2)] =
= r1·r2·{[cos(φ1)·cos(φ2) − sin(φ1)·sin(φ2)] +
+ i·[cos(φ1)·sin(φ2) + sin(φ1)·cos(φ2)]} =
= r1·r2·[cos(φ1+φ2)+sin(φ1+φ2)]
So, the magnitudes are multiplied but arguments are added. The final formula is relatively simple.

There is a very clear geometric meaning of multiplication of complex numbers in polar form.
Consider only complex numbers that have a magnitude of 1, that is all numbers of a form
cos(φ)+i·sin(φ).
They all lie on a unit circle around the origin of coordinates.

When one such number with an argument φ1 is multiplied by another with an argument φ2, the result will be a new complex number with a magnitude of 1, still on the same unit circle, and an argument equal to φ1+φ2.
It means that algebraic operation of multiplication is, geometrically, a rotation.
Remarkable, is not it!

If we consider multiplication by a complex number with a magnitude not equal to 1, geometrically, we still deal with a rotation, but also deal with a stretching by a factor equal to a magnitude of a multiplier.
Multiplication by a positive real number is only a stretching since positive real numbers, considered as a subset of complex numbers, have argument equaled to zero in polar form and, therefore, there is no rotation.
Multiplication by a negative real number is a stretching with a change in the direction to an opposite since negative real numbers, considered as a subset of complex numbers, have argument equaled to π in polar form, which, if added, changes the direction to an opposite.

Multiplication by i is a rotation by π/2=90° because, in polar form,
i = 0+1·i = cos(π/2)+i·sin(π/2),
that is, i is a complex number with magnitude 1 and argument π/2 in polar form.
Therefore, multiplication by i is a rotation by an angle π/2.
Incidentally, this is obvious in a traditional representation of complex numbers since
(a+b·i)·i = −b+a·i
and a segment connecting the origin of coordinates and a point (a,b) should be rotated by an angle π/2 around the origin of coordinates to coincide with a segment from the origin of coordinates to a point (−b,a).