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UA - MATH 300 - Class Notes - Week 1

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Reviews

1/14/16 Introduction To Numerical Analysis

- Design / implement algorithms / methods
- Analyze when / why they work / fail
- Classify competing methods

Don't forget about the age old question of What question is being answered by Epistomology?

Example 1:

Find the area under the unit circle shifted up by 1. (in the first quadrant)

We also discuss several other topics like a hydrated lithium iodide sample, lii·xh2o, weighing 4.450 g is dried in an oven. when the anhydrous salt is removed from the oven, its mass is 3.170 g. what is the value of x?

Equation of circle: x2 + y2 = r2 = 1

Solve for y → y = If you want to learn more check out What is the distance light travels in one year?

When you shift it by 1: y =

- Geometrically: S = (area of unit circle) + area of unit square

= (1)2 + 1 = 1.7854

- Analytically (by a closed - form formula)

S = f(x)dx = = + arcsin x + xIf you want to learn more check out How does our nervous system work?

Evaluate in matlab

- Define a function f(x) = + arcsin x + x by using F = @(x) (x/2)*sqr + (1 - x2) - (½) arcsin x + x

Then to evaluateDon't forget about the age old question of f 288 scantron

F(1) - F(0)

Answer = 1.7854

- Riemann Sum

We also discuss several other topics like What refers to the study of microscopic structures and tissues?

A = A1 + A2 + A3 + A4

= f(x1)x + f(x2)x + f(x3)x + f(x4)

f(xi)

Riemann Sums continued → in Matlab

a =0 b =1 n = 4 (all separate names)

Dx = (b - a)/n

Dx = ¼ = 2500

→ x = [1.n] x dx

X = 0.2500 0.5000 0.7500 1.0000

Format rat

X

X = ¼ ½ ¾ 1

F = @(x) sqrt(1 - x2) + 1

A = sum (fx)) . dx

A = 3217 / 1981

Format

A = 1.6239

Then you can easily redo everything for a larger n, thus a closer approximate

Say you use n = 10,000

Dx =

X =

A =

A = 1.7853 → off by only .0001

Method 3 using Matlab is the best for complicate, real - life problems

Example 2: Find ln(2)

- Riemann Sum

Lnx = dt

Ln2 = dx

F = @(x) 1. /x x1 = a + ix

Dx = (b.a) / n i = 0 = n

X = a + [1:n] = dx

A = sum(f(x) x dx)

A = .9950

H = 1000000

X = a + [1:n] x dx

A = sum(f(x)x dx)

A = .6931

log(2) = .6931

- Series - ln(1 - x) = ∫dx = ∫(1 + x + x2 +,...)dx

= x + x+ … then x = -1 -ln(2) = -(1 ---)

Note: Review of Geometric series

Sn - xSn = 1 = xn+1

(1 - x) Sn = 1 - xn+1 → Sn =

= x < 1 → = x < 1

= 1 + x + x2

Note - he used octet, after done of Matlab

Example: Evaluate ln(2)

**Riemann Sum

ln(x) = ∫1x dt

ln(2) = ∫12 dt

A = f(x1) ==← with right endpoint

X1 = a + ix

A = f(xi-1) aka. For n = 1 you use X0,etc ← with left endpoints

Or A = f(xi)

Octave 1 > a = 1

Octave 2 > b = 2

Octave 3 > n = 10,dx = (b-a)/n