#math101

Proof: Let ℇ > 0 given let |x - 0| < δ, ie x |x| < δ with δ to be determined.

 We want to prove à

|x + 1

____ - 1               < ℇ, whenever |x| < δ

1 – x      |

We have to find the δ

 x + 1                      1 + x – (1 – x)

____ - 1  =            ___________

1 – x                     1 – x

                =             2x

                             ____

                             1 – x

 So,

|x + 1                                  |2x

____ - 1               =             ____

1 – x      |                            1 – x|

                                              =            2|x|                      2δ

                                                           ______  <             ______

                                                           |1 + x|                 |1 + x|

 |1 - x| > |1| - |x|

 For δ < 1 (so |x| < 1), we have (1 – x)  > 1 - |x| > 1 - |x| > 1 – δ

                 1                            1

∴           _______   <            _____

              |1 – x|                  1 – δ

    ∴

|1 + x                                  2δ                          2δ

_______ - 1        <            _____    <             ____

1 – x           |                       |1 – x|                 1 - δ

                               2δ                                         ℇ

Now choose at _____     = ℇ, ie δ =            ____

                             1 – c                                    2 + ℇ

 Then we have

|1 + x                                  2δ          

_______ - 1        <            _____    = ℇ

1 – x           |                       1 – ℇ      

                                              ℇ

Whenever |x| < δ =        ____

                                            2 + ℇ

 |x – L| < δ

i.e. – δ < x – c < δ

 i.e.

c – δ < x < δ + c

 Our limit, so far, are two sided. We take x à c+ (approach from above c) and x à c- (approach from below c)

This will not work for √x as x à 0

We need one sided limits in this case.

eg Discuss the limit x à 1 for

              x2, x > 1

f(x) = {

              1 – x, x < 1

  Two sided limits at x = 1

From the left x à 1- , f(x) = 1 – x à 0 but 0 is not in the range of f from the right, x à x+ f(x) = x2 = 1

  Limits as x à + ∞

Are one sided.

 x à ∞ is from below.

 x à -∞ is from above

 Formally: - we say f(x) à L as x à ∞, OR               lim          f(x) = L, if given any ℇ > 0 there exists some x

                                                                                         x à ∞

such that |f(x) - L| < 3, whenever x > X.

Prove that 1/x à 0 as x à ∞

 Proof: Let ℇ > 0 be given

              Let x > X, X to be determined

              [We want to prove that |1/x| < whenever x > X]

              We can take x > 0, as x > 0 and |1/x| = 1/x < 1/X

              Choose X = 1/ ℇ (as 1/x = ℇ)

So,  |1/x| < ℇ whenever x > 1/ ℇ             e.g. prove that 1/√(x 2+ 1) à 0 as à ∞

 Proof: Let ℇ > 0 be given. Let x > X, x to be determined |1/√(x 2+ 1) – 0| = 1/√(x 2+ 1) < 1/x2 <

                                                                                                                       Limit -->

1/X, choose x = 1/ ℇ

Then |1/√(x 2+ 1)| < ℇ for x > X

 eg Prove that (√(x + 1) -√(x - 1))/x à as x à ∞

 Proof: Let ℇ > 0 be given Let x > X, X to be determined

Firstly (√(x + 1) -√(x - 1))/x           =  ((√(x + 1) -√(x - 1))/x) * ((√(x + 1) + √(x + 1))/(√(x + 1) + √(x + 1))

                                                           = ((x + 1) – (x – 1))/(x(√(x + 1) + √(x + 1))

                                                           = 2/(x(√(x + 1) + √(x + 1)) < 2/x

Since √(x + 1) + √(x + 1) > 1

But 2/x < 2/X, as x > X

∴ (√(x + 1) -√(x - 1))/x < 2/x

Choose x = 2/ℇ (as 2/x = ℇ)

∴ (√(x + 1) -√(x - 1))/x < ℇ

http://u.cubeupload.com/Brenorenz/surjectiveinjective1.png

 Bijective

 (2) f: |R -> {0, 1)

 f(x) =      {o, x irrational

              {I, x rational

Not injective

Is surjective

(3)

 http://u.cubeupload.com/Brenorenz/surjectiveinjective2.png

 f: |N à |R

f(x) = x2 – 1

 not surjective

is injective

 -//-

 Limits

Notions f(x) approaches some value  /_ as x approaches some value c € Domain f

              Indicate by f(x) à L as x à c

 Or          lim f(x) = L

              X ->

              The closer I get to x = L the closer x gets to L

 Approximation

              Given a degree of approximation |f(x) – L ) < 0.01

                             Can you find the x that gives this approximation?

                             eg f(x) = x2 à 1 as x à 1

                             Given the approximation |x2 - 1| < 0.01

                             Can we find the possible x’s?

                             Solution |x2 - 1| < 0.01

                             Means -0.01 < x2 – 1 < 0.01

                             Ie 0.99 < x2 < 1.01

                                            ∴ √0.99 < x < √1.01

                                            ∴ |x2 - 1| < 0.01 whenever √0.99 < |x| < √1.01

 Formal Definition of a Limit

We say f(x) à L as x à c or

Lim

xà c

  If given any ℇ > 0 there is a δ > 0 such that |f(x) = L| < ℇ whenever |x - L| < ℇ

Means given the approximation ℇ : - |f(x) - L| < ℇ

We have to find the δ (depend on          ) to make |f(x) - L| < ℇ whenever |x - c| < δ

eg (a) Prove x2 à 1 as x à 1

 Solution: Let ℇ > 0 be given we must find a δ such that |x2 - 1| < ℇ whenever |x – 1| < δ

We must find the approximate δ

Now |x2 - 1|       = |(x - 1)(x + 1)|

                             = |x - 1||x + 1| < |x + 1|

              Also (x - 1) < δ, - δ < x – 1 < δ

              ie 1 – δ < x < 2 + δ

              ∴ 2 – δ < x + 1 < 2 + δ

              Choose δ < 1, then |x + 1| < 3 and 3 δ

              |x2 – 1| < δ |x + 1| < 3 δ Now choose δ ≤ 1/3 ℇ

De Moivres Theorem

Note zn = rn (cosθ + isinθ)n , How do we calculate (cosθ + isinθ)n ?

 Theorem (De Moivre) If n is a rational number then (cosθ + isinθ)n = cos(nθ) + isin(nθ)

Proof Exercise

              Cos(A+B) = cosAcosB – SinAsinB

              Sin(A+B) = SinAcosB + SinBcosA

e.g. Calculate: (i) (1 + i)4 (ii) (√3/2 + I 1/2)3

Solutions

(i)               Polar form for 1 + I : -

Now |1 + i| = √(i2 + i2)

                      = √2

∴ (1 + i)/√2 = cosθ + isinθ

 i.e.  cosθ = 1/√2

                                     } ∴ θ       = π/4

                                                    = 45°

       sinθ = 1/√2

 ∴ √2(cos(π/4) + isin(π/4)) – the polar form.

Then (1 + i)4        = (√2)4 x (cos(π/4) + isin(π/4))4

                             = (√2)4  x (cosπ + isinπ)

                             = 4(cosπ + isinπ)

                             = -4, as cosπ = -1 sinπ = 0

 (ii)              Polar form of √3/2 + i/2

|√3/2 + i/2| = √((√3/2)2 + (i/2)2)

                      = √(3/4 + 1/4)

                      = √1

                      = 1

(√3/2 + i/2) = cosθ + isinθ

With      cos θ = √3/2

                                            }

              sin θ = ½

 Polar form

√3/2 + i/2 = cos(π/6) + isin(π/6)

∴ (√3/2 + i/2)3   = (cos(π/6) + isin(π/6))3

                             = cos(π/2) + isin(π/2)

                             = i, cos(π/2) = 0, sin(π/2) = 1

 Functions

f: A à B, sets A, B

f: a|à b, f(a) = b, a € A, b € B

g: B à C

              g: b <--> c, g(b) = c, b € B , c € C

                             g0 f(a) = g(f(a))

 Injective (1 – 1, one to one)

              For function       f:A -> B

                                            F: a|àB

Given any b € B there is at most one a € A each that f(a) = b, the function is called injective

Surjective (onto)

              A function of f:AàB is called surjective if for every b € B there is at least one a € A such that f(a) = b.

 --//--

 R – valued functions on |R

  http://cubeupload.com/im/Brenorenz/Rvaluedfunction1.png

 http://cubeupload.com/im/Brenorenz/Rvaluedfunction2.png

 f: |R è |R

f(x) = - 1

Not injective and not surjective as a function from |R +-> |R

 What about f:|R à {y € |R : u ≥ 0}

                                            f(x) = x2

 http://cubeupload.com/im/Brenorenz/Rvaluedfunction3.png

 Not injective (two x’s for every y)

Is surjective.

 What about f: {x € |R: x ≥ 0} à { y € |R: y ≥ 0

f(x) = x2

  http://cubeupload.com/im/Brenorenz/Rvaluedfunction4.png

 Both surjective and injective.

|z1 + z2|, |z1| + |z2| are both non-negative number so we just need to prove |z1 + z2| ≤ (|z1| + |z2|)

Now, |z1 + z2| - (|z1| + |z2|)2

                 ______

= (z1 + z2)( z1 + z2) – (|z1|2 + |z2|2 + z|z1z1|)

                 __   __

= (z1 + z2)(z1 + zz) – (z1 zz + z1 zz +z)

     _         _                     _            _           _         _

= z1z1 + z1z1 + z1 z2 + z2 z1 + z2 + z2 – (z1 z1 + z2 z2 + z|z1 z2|)

      _         _

= z1 z2 + z2 z1 - z|z1 z2|

Now ζ = z1 z2 is a complex number and put = ζ x + iy

                                   _

              We have z1 z2 = ζ

                                   _       _____

                                                 _

                               z1 z2 =  (z1 z2) = ζ

                                              _____

              | z1 z2|  = √(z1 z2) (z1 z2)

                                        _        _

                             = √(z1 z2) (z2 z1)

                                                       _          _

Ie |z1 + z2|2 – (|zz + zz|)2 = ζ + ζ - z√ ζ ζ

                                            = (x + iy) + (x – iy) - z√(x2 + y2)

                                            = z[x - √(x2 + y2)] ≤ 0

 -//-

 Complex Conjugation

The complex conjugation of a complex of 2 = x + y. It is denoted by z-.

i.e. if z = x + iy then z- = - iy

Note: z z- = x2 + y2 (real and positive) Find the complex conjugates of:

(1)   2 + 3i

(2)   |2| - i

(3)   1/(1 + i)

 Solution

.        .

2 + 3i = 2 – 3i

 .   .

|2| - i

 .            .

1/(1 + i)  = 1/(1 – i)

 Theorem

If 21 and 22 are complex numbers then

 (i)               .       .      . .    . .

21 + 22 = 21 + 22

 (ii)              .       .               . . . .

Supermum x k for x S

           And K - x for some x S and > 0

 Infinum k x, for all x S and k +> x, for some x S and any > 0.

Eg        1 s1 = { x Q x ½ }

2 s2 = { x Q x2 < 2 }

           3 s3 = { ½, 2/3, ¾, ... n/(n + 1), ... }

           4 s4 = { 1, ½, ¼, ... 2-n }

 Which sets are bounded? What are the supermum and infinmum?

Solutions

1 Bounded                   Supremum is ½ and is contained in S.

                                   Infinmum is 0 and is contained in S.

2 Boundednote              S2 = { x Q : - √2 < x √2}

                                   Supremum is √2, note √2 is not in S2

                                   Infinmum is -√2, note √2 is not in S2

3 Bounded                   Supremum is 1, note 1 is not in S3

                                   Infinmum is ½, note ½ is in S3

4 Bounded                   Supremum is 1, note 1 is in S4

                                   Infinmum is 0, note 0 is not in

 -//-

 Complex numbers

Any equation x2 = a, a R

Cannot be solved for x R, if a < 0

Extend R so that x2 = -1 has a solution

Yes can solve it in C (Complex numbers)

 C is R with an element I = √-1 (and its multiples etc)

Note R ⸦ C

Any complex number z C, can be put into the form z = a +ib; a, b R

a = Real part of z  = a + ib, Re(z) = a

b = Imaginary part of z, a + ib, Im(z) = b

eg 2 – 3i C

Re(2-3i) = 2

Im(2-3i) = -3

Addition of elements C

Eg (2-3i) + (1+i) = 3 – 2i

Note: Re(z1 + z2) = x1 + x2

Im (z1 + z2) = y1 + y2

Multiplication

Eg (2 – 3i)(1 + i)         = 2 + 2i – 3i – 3i2

                                   = 2 – i – 3i2, but i2 = -1

                                   = 2 – i – 3 – 1

                                   = 5 – i

General           z1 = x1 + iy1

                       Z2 = x2 + iy2

 z1z2 = (x1 + iy1)( x2 + iy2)       = x1x2 + ix1y2 + iy1x2 + i2y1y2

                                               = x1x2 – y1y2 + i(x1y2 + x2y1)

(a)   (2 + 3i)(10 – i)             = 20 – 2i + 30i – 3i2

= 23 + 28i

(b)   (1 + i)2             = (1 + i)(1 = i)             (c) (√2 - i)2      = (√2)2 + (-i)2 - 2√2i

= 1 + i + i + i2                                     = 2 – 1 - 2√2i

= 2i                                                      = 1 - 2√2i

 (d) i3 = i2i = -i

(e) i4 = (i2) = (-1)2 = i

Math 101 – Lecture 8

Basic Arithmetic

The set of real numbers R has the following properties a+b = b+a R for a, b R  a+ 0 = 0 + a, the additive identity.

For each a ( 0) R, there exists b R such that ab = 1 where 1 is the multiplicative identity a*1 = 1*a = a.

Order for any two real numbers a, b R  only one of the following is true:

 a > b or

a = b or

a < b

 If a, b, c R and a > c, c > b then a > b

           If a, b, c R and a > b then a + c > b + c

If a, b, c R and a > b, if c > 0 then ac > bc

 From these properties, together with the deankind axiom all the familiar results for real numbers.

Eg Index Laws a^m a . a . a .... a, m IN

                       {          m times  }

           Extend to a^-m = 1/(am) = 1(a . a . a .... a) =               1/a . 1/a . 1/a .... 1/a

                                                           {          m times  }        {          m times  }

           Extend to a^(p/q), p, q Z

Eventually extend to a^-σ , σ R

 Absolute value – a ‘distance measure’ on R

|a| = {a, a 0

           -a, for a < 0

Theorem (properties of absolute values)

1.      |a| 0 for all a R, note  |a| = 0 if any only if a = 0

2.      |ab| = |a| * |b|, for all a, b R

Cases

(i)                 |ab| = ab = |a| |b|     ie ab > 0 for [(i) a, b > 0

(ii) a, b < 0

(ii)               |ab| = ab = (-a)(-b)

3.      |a|^2 = a^2

4.      |a+b| |a| + |b|, a, b R

Proof: If a+b<0 then |a + b| = -(a + b)

                                               = (-a) b),

Now – a |a|, for any a R

 If a + b > 0 then |a + b| = a + b

But a |a| and b |b|

|a + b| |a| + |b|

5.      |a – b| |a| - |b|, for any a, b  R

-||-

Bounded Sets

Infinite sets : 1) Q

                 2) Prime numbers

                 3) Z

     4) {x  Q: -1 x 1}

4) is a subset of an interval of the real number line.

[a, b] = {x^(R): a x b} – a closed interval.

[a, b] = { x^(R): a x b }

Also have [a, b] and [a, b]

 Bounded sets: - A set S R is bounded above if thre is K R, such that K for all x  S

     S is said to be bounded below.

                 If there is k  R such that x  k for all x    S

                       S is bounded if it bounded above and below.

*The least upper bound (supremum) for a bounded set S R is K R such that K x for some x    S and any > 0

*The greatest lower bound (infinum) for is K R such that [ k x, x S

                                                                                                     K + > x for some x S and any > 0