Mod1: Finding Roots

Question 1

more advanced techniques to solve mathematical problems to obtain an exact solution

Answer 1

Analytical Solution

Question 2

Math problem is reformulated so it can be solved by arithmetic operations to obtain its approx. solutions

Answer 2

Numerical Solution

Question 3

A source of error rooting from limited significant figures to represent exact num

Answer 3

Round-off error

Question 4

A source of error from truncating or approximating a mathematical procedure

Answer 4

truncation Error

Question 5

A source of error through mathematical functions

Answer 5

Propagation error

Question 6

A source of error through mathematical functions

Answer 6

Propagation error

Question 6

A source of error through mathematical functions

Answer 6

Propagation error

Question 6

A source of error through mathematical functions

Answer 6

Propagation error

Question 7

Converting from Binary to decimal

Answer 7

1. Get sign bit by:
   1- Negative
   0- positive
2. Get mantissa by
   Adding 0. to the leftmost mantissa
   Adding 1 to mantissa
3. Converting mantissa from binary to decimal
   Get exponent by
   Converting exponent to decimal
4. Get e bias
   2^(exponent bits-1) - 1
5. Plug in to the formula and compute
   s*m base 10 * 2^ (e base 10- e bias)

Question 8

Converting from Decimal to binary

Answer 8

1. Get sign bit
   + = 0
   
   • = 1
2. Convert Decimal to binary
3. Normalize to scientific notation to get the initial mantissa and the unbiased exponent
   Get the exponent of the base two to get unbiased exponent
4. Get the mantissa by
   subtracting 1. to the equation
   Convert decimal to binary
5. Get the exponent of base 2 by:
   Adding 127 to unbiased exponent
   Convert decimal to binary

Question 9

Two Categories of Roots of an Equation

Answer 9

Bracketing Methods
Open Methods

Question 10

This method is where the root is located within an interval prescribed by a lower and an upper bound. Repeated application of these methods always results in closer estimate of the true value of the root

Answer 10

Bracketing Methods

Question 11

Bracketing methods includes ___ and __ methods

Answer 11

bisection and false-position

Question 12

This method of finding roots of an equation require only a single starting value of x or two starting values that do not necessarily bracket the root

Answer 12

Open Methods

Question 13

Open methods includes __ and ___

Answer 13

Newton-Rapson and Secant methods

Question 14

___ is a bracketing method for finding roots of an equation that can be done by getting the midpoint of the boundary continuously until the midpoint will approximately or equal to zero when plugin to the function f(x)

Answer 14

Bisection method

Question 15

How to do Bisection method:

Answer 15

1. Check whether point A and point b have opposite values when plugged in to f(x)
2. If yes, then get the midpoint between the two points
3. Plugin the midpoint to f(x). If the answer is closer to zero, use that as the new boundary.
   4 Verify the boundary by using comparisons (<>=)
4. Repeat process from 1 to 4 until the value of f(x) is approx or equal to zero or to the specified value

Question 16

How to to False-Position Method

Answer 16

1. Get the f(x)’s of the roots a and b
2. Subs the values to the formula to get bn or the next point
3. Subs f(bn)
4. If the f(bn) is positive, change a to bn
   If f(bn) is negative, change b to bn

Question 17

How to do newton-raphson Method

Answer 17

1. Guess a point and sub f(guess)
   If f(guess) is not = 0, continue
2. Derive f(x) then f’(guess)
3. Subs to formula
4. Repeat Process

Question 18

How to do secant method

Answer 18

1. Guess two points
   2.Compute f(guess1) and f(guess 2)
2. Subs to formula
3. Repeat until the new point is approx zero

Question 19

Reason for roots to fail:
The point give a __ slope

Answer 19

Zero slope

Question 20

Reason for roots to fail:
The consecutive points is in ___ of a function

Answer 20

Symmetry