Bounds (बंध )

Upper Bound and Bounded Above:-
Let S is a subset of R. And a∈R. Then a is called upper bound of S if x≤a for all x∈S.
And then S is called bounded above.

Lower Bound and Bounded Below:- 
Let S is a subset of R. And a∈R. Then a is called lower bound of S if a≤x for all x∈S.
And then S is called bounded below.

Least Upper Bound:-
Let S is a subset of R. And a∈R. Then a is called least upper bound of S if it is satisfied two axioms.
1) x≤a for all x∈S.
2) there are exist a Upper Bound t such that a≤t.
I.e. lub is less or equal all other upper bounds.


Greatest lower Bound:-
Let S is a subset of R. And a∈R. Then a is called least upper bound of S if it is satisfied two axioms.
1) a≤x for all x∈S.
2) there are exist a lower Bound w such that w≤a.
I.e. glb is greater or equal all other lower bounds. 

Mathspics

Mathspics

Complex numbers

A complex number is a number that can be expressed in the form a + bi, where a and bare real numbers and i is the imaginary unit, that satisfies the equation i² = −1.In this expression, a is the real part and b is the imaginarypart of the complex number.

   
  Complex numbers extend the concept of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. The complex number a + bi can be identified with the point (a, b) in the complex plane. A complex number whose real part is zero is said to be purely imaginary, whereas a complex number whose imaginary part is zero is a real number. In this way, the complex numbers contain the ordinary real numbers while extending them in order to solve problems that cannot be solved with real numbers alone.

As well as their use within mathematics, complex numbers have practical applications in many fields, including physics, chemistry, biology, economics,electrical engineering, and statistics. The Italian mathematician Gerolamo Cardano is the first known to have introduced complex numbers. He called them "fictitious" during his attempts to find solutions to cubic equations in the 16th century.

Complex numbers allow solutions to certain equations that have no solutions in real numbers. For example, the equation

has no real solution, since the square of a real number cannot be negative. Complex numbers provide a solution to this problem. The idea is to extend the real numbers with the imaginary unit i where i² = −1, so that solutions to equations like the preceding one can be found. In this case the solutions are−3 + 3i and −3 − 3i, as can be verified using the fact that i² = −1:

According to the fundamental theorem of algebra, all polynomial equational with real or complex coefficients in a single variable have a solution in complex numbers.

Definition:-

A complex number is a number of the form a+ bi, where a and b are real numbers and i is the imaginary unit, satisfying i² = −1. For example, −3.5 + 2i is a complex number.

The real number a is called the real part of the complex number a + bi; the real numberb is called the imaginary part of a + bi. By this convention the imaginary part does not include the imaginary unit: hence b, not bi, is the imaginary part.The real part of a complex number z is denoted by Re(z) orℜ(z); the imaginary part of a complex number z is denoted by Im(z) or ℑ(z). For example,

Hence, in terms of its real and imaginary parts, a complex number z is equal to . This expression is sometimes known as the Cartesian form of z.

A real number a can be regarded as a complex number a + 0i whose imaginary part is 0. A purely imaginary number bi is a complex number 0 + bi whose real part is zero. It is common to write a for a + 0i and bifor 0 + bi. Moreover, when the imaginary part is negative, it is common to write a − bi withb > 0 instead of a + (−b)i, for example 3 − 4iinstead of 3 + (−4)i.

The set of all complex numbers is denoted by ℂ, 

Addition:-

To add two complex numbers we add each part separately:

(a+bi) + (c+di) = (a+c) + (b+d)i

Example: (3 + 2i) + (1 + 7i) = (4 + 9i)

Multiplying

To multiply complex numbers:

Each part of the first complex number gets multiplied by
each part of the second complex number

Just use "FOIL", which stands for "Firsts, Outers,Inners, Lasts"

Firsts:a × c Outers:a × di Inners:bi × c Lasts:bi × di

(a+bi)(c+di) = ac + adi + bci + bdi2

Like this:

Example: (3 + 2i)(1 + 7i)

(3 + 2i)(1 + 7i)		= 3×1 + 3×7i + 2i×1+ 2i×7i
		= 3 + 21i + 2i + 14i2
		= 3 + 21i + 2i − 14	(because i2 = −1)
		= −11 + 23i

But There is a Quicker Way!

Use this rule:

(a+bi)(c+di) = (ac−bd) + (ad+bc)i

Example: (3 + 2i)(1 + 7i) = (3×1 − 2×7) + (3×7 + 2×1)i = −11 + 23i

1. Add: (7 + 5i) + (8 - 3i)



2. Add: (2 + 3i) + (-8 - 6i)



3. Express the sum of and
in the form .



4. Add and .


Example: i²

i can also be written with a real and imaginary part as 0 + i

Complex Plane

We can also put complex numbers on a Complex Plane.

• The Real part goes left-right
• The Imaginary part goes up-down

i2 = (0 + i)2		= (0 + i)(0 + i)
		= (0×0 − 1×1) + (0×1 + 1×0)i
		= −1 + 0i
		= −1

Prove that  for any integer n.

Corollary 1.2.15: DeMoivre's Formula
	For any integer n and any real number twe have (cos(t) + i sin(t))n = cos(nt) + i sin(nt) Proof

DeMoivre's Formula is quite something. It says that if you take a number on the unit circle (i.e. with lenght 1) with initial argument (angle) t and multiply it by itself, it simply rotates around the unit circle by that angle t. Each time you multiply the number by itself, the vector rotates another tdegrees. In other words, in this case the power operator results in a simple rotation.

Powers of a vector z with |z|=1

Two interesting questions related to this rotation, taken from the field of Complex Dynamics, are: suppose z is a complex number with |z|=1. Then:

• find conditions for Arg(z) such that zⁿ = z for some integer n. Such a point, incidentally, is called periodic of order n.
• if Arg(z)/ is irrational, what can you say about the sequence {z, z², z³, z⁴, ...}? Does it, for example, converge? Such a sequence, incidentally, is called the orbit of z.

Polar coordinates can be especially helpful for finding roots, in particular for complex numbers of lenght 1.

Proposition 1.2.16: Finding Roots
	For any positive integer n and any non-zero complex number a = r cis(t) the equation zn = a has exactly n distinct roots given by: z = where k = 0, 1, 2, ... n-1. Proof

In a previous example we found the two square roots of i, which turned out to be a fair amount of work. The above proposition allows us to dispense with such a question quickly. For example, the two solutions for

z² = i = cis(/2)

are:

z₁ = cis(/2/2) = cis(/4)
z₂ = cis((/2 + 2)/2) = cis(5/4)

which you can quickly check using DeMoivre's Formula. Here is a geometric interpretation of this proposition: let's find, for example, the three third-roots of i, i.e. we want to find all solutions toz³ = i.

1: Draw the vector i	2: Divide angle by 3 for first root	3: Draw 3 equally spaced segments, starting at the first root

Note, in particular, that the third root of i turned out to be -i, which indeed checks out:

(-i)(-i)(-i) = i*i*(-i) = (-1)*(-i) = i

This proposition is very satisfying: it says that at least every simple polynomial equation of degreen has n solutions. Later we will see that this is true in general: every n-th degree polynomial hasn roots, no "if's" and "but's". This, in fact, is a sign of things to come: many theorems in complex analysis will turn out to be very "satisfying" and nicely structured, which is one reason that the study of complex analysis is a lot of fun (I think -:). But first a few more 'profane' examples.

Example 1.2.17: Finding roots geometrically
	Find the cube roots of 8 and iand draw them. Find all 4 fourth-roots of -1 and draw them geometrically Find all 5 fifth-roots of 1 and draw them geometrically Find both square-roots of 3i-2 by (a) using polar coordinates and (b) using rectangular coordinates and aformula from the previous section. Confirm that both methods result in the same answers.

Let's conclude this chapter with a result that illustrates what a 'nicely structured' theorem in complex analysis can look like.

Proposition 1.2.18: Roots of Unity
	The n n-th roots of unity are given bywnk, where k = 0, 1, 2, ... n-1 and wn = cis(2/n) They form the vertices of a regular polygon and add up to zero, i.e. they satisfy the equation: 1 + wn + wn2 + ... + wnn-1 = 0 Proof

This is neat: not only does the equation zⁿ = 1have exactly n solutions (one of which is, of course, z=1), but the solutions have this really pretty geometric structure of forming a regular polygon, which implies algebraically that they add up to zero as vectors. Here are, for example, the eight roots of z⁸=1:

You can see their regular structure. When you add them all as vectors, you indeed get the zero vector.


(.Ref from Wikipedia and Mathisfun private.Shu.edu)

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differentibily

We present a new proof of the differentiability of exponential functions. It is based
entirely on methods of differential calculus. No current or recent calculus text gives or
cites a proof of the differentiability that depends only on such elementary tools. Our
proof makes it possible to give a comprehensive treatment of the derivative properties
of exponential and logarithmic functions in that order in differential calculus, building
on the standard introduction to these topics in precalculus courses. This is the logical
order and has considerable pedagogical merit.
Most calculus books defer the treatment of exponential and logarithmic functions to
integral calculus in order to prove differentiability. A few texts introduce these topics in
differential calculus under the heading of “early transcendentals” but defer the proof of
differentiability to integral calculus. Both approaches have serious pedagogical faults,
which are discussed later in this paper.
Our proof that exponential functions are differentiable provides the missing link
that legitimizes the “early transcendentals” presentation.
Preliminaries
We assume that ar has been defined for a > 0 and r rational in a precalculus course
and that the familiar rules of exponents are known to hold for rational exponents. It is
natural to define ax for a > 0 and x irrational as the limit of ar as r → x through the
rationals. In this way, ax is defined for all real x.

Basic properties of ax for real x are inherited by limit passages from corresponding
properties of ar for r rational. These properties include the rules of exponents with
real exponents and
ax is positive and continuous,
ax is increasing if a > 1,
ax is decreasing if a < 1.
It is not especially difficult to justify the definition of ax for x irrational and to
derive the foregoing properties of ax for x real, but there are a lot of small steps. A
program along these lines is carried out by Courant in [2, pp. 69–70]. The general idea
of each step is well within the grasp of students in typical calculus classes. However,
just as properties of ar with r rational are routinely stated without proof, it is better to
give just an overview of the basic properties of ax with x real, illustrated with graphs,
and move on to the question of differentiability, which is more central to differential
calculus.
A more complete development, beginning with the derivation of properties of ar
with r rational, might be given in an honors class. The properties can be extended to
ax with x real with the aid of the density of the rationals in the reals and the squeeze
laws for limits. The conclusion that ax with a > 1 is increasing also relies on the
following proposition which should seem evident from graphical considerations:
If f is a continuous function on a real interval I
and f is increasing on the rational numbers in I,
then f is increasing on I.
The same proposition will provide a key step in the proof that ax is differentiable.
Henceforth, we restrict our attention to properties of ax with a > 1. Corresponding
properties of ax with 0 < a < 1 follow from ax = (1/a)
−x .
The differentiability of ax
Consider an exponential function ax with any a > 1. In order to prove that ax is differentiable
for all x, the main task is to prove that it is differentiable at x = 0. Our proof
of this depends only on methods of differential calculus. It is motivated by the fact
that the graph of ax (see Figure 1) is concave up, even though this fact is not assumed
a priori

figure 1

Graph of ax with B = (x, ax ) and C = −x, a−x for x > 0

In Figure 1, imagine that x → 0 with x > 0 and x decreasing. Then B and C slide
along the curve toward A. The upward bending of the curve seems to imply that
slope AB decreases, slope AC increases,
and slope AB − slope AC → 0.
It follows that the slopes of AB and AC approach a common limit, which is the slope of
the tangent line T in Figure 1 and the derivative of f (x) = ax at x = 0. This geometric
argument will be made rigorous.
The curve in Figure 1 is actually the graph of f (x) = 2x . The following table gives
values of the slopes of AB and AC rounded off to two decimal places. It appears that
the slopes of AB and AC approach a common limit, which is f
(0) = slope T ≈ 0.7.


x                    1               1/2         1/4          1/8         1/16        1/32
slope AB        1              .83         .76          .72 .       71           .70
slope AC       .50            .59        .64 .         66         .67            .69


With this preparation, we are ready to prove that f (x) = ax is differentiable at
x = 0. The foregoing geometric description of the proof and the numerical evidence
should be informative and persuasive to students, even if they do not follow all the
details of the argument.

Theorem 1. Let f (x) = ax with any a > 1. Then f is differentiable at x = 0 and
f
(0) > 0.
Proof. To express our geometric observations in analytic terms, let
m(x) = f (x) − f (0)/x − 0
= ax − 1/x
.
In Figure 1, x > 0 and
slope AB = m(x),
slope AC = m(−x).
We shall prove that, as x → 0 with x > 0 and x decreasing, m(x) and m(−x) approach
a common limit, which is f
(0).
To begin with, m(x) is continuous because ax is continuous. The crux of the proof,
and the only tricky part, is to show that
m(x) is increasing on (0,∞) and (−∞, 0).
We give the proof only for (0,∞) since the proof for (−∞, 0) is essentially the same.
We show first that m is increasing on the rationals in (0,∞). Fix rational numbers r
and s with 0 < r < s and let a vary with a ≥ 1. Define
g(a) = m(s) − m(r ) =a^s -1/s = a^r-1/r

.
Then g(a) is continuous for a ≥ 1 and
g
(a) = a^s -1-a^r-1>0 for a>0

Thus, g(a) increases as a increases and g(a) > g(1) = 0 for a > 1, so
m(r) < m(s) for 0 < r < s.
Thus, m(x) is continuous on (0,∞) and m(x) increases on the rational numbers in
(0,∞). As noted earlier, this implies that m(x) increases on (0,∞). The argument for
the interval (−∞, 0) is similar.
For x > 0,
                         m(−x) = m(x)a−x ,
                        0 < m(−x) < m(x),
                    0 < m(x) − m(−x) = m(x) 1 − a−x .

Let x → 0 with x decreasing. Then
m(x) decreases, m(−x) increases, m(x) − m(−x) → 0.
It follows that m(x) and m(−x) approach a common limit as x → 0, which is f
(0).
Furthermore, 0 < m(−x) < f '(0) < m(x) for x > 0, which implies that f ' (0) > 0.
We believe that this proof is new. We have been unable to find any other proof
that depends only on methods of differential calculus. Theorem 1 and familiar reasoning give the principal result on the differentiability


special refrence to www.maa.org