This is a subject of CSE as well as ECE student

1. NUMERICAL
METHODS
NAME – PAYEL DALAL
ROLL NO –
14400120030
STREAM – CSE
SUBJECT –
NUMERICAL METHODS
SUBJECT CODE –
OEC-IT601A

2. Presentation title 2
TOPIC:-
 APPROXIMATION IN NUMERICAL
COMPUTATION
 Truncation and rounding errors
 Fixed and floating point arithmetic
 Propagation of errors

3. APPROXIMATION IN
NUMERICAL
COMPUTATION

4. 4
 Error Definitions:
 True error (Et ): the difference between the true value and
the approximation.
 Absolute error (|Et |): the absolute difference between the
true value and the approximation.
 True fractional relative error: the true error divided by
the true value.
 Relative error (et): the true fractional relative error
expressed as a percentage.
• True fractional relative error = (true value – approximation)/true value
• Et = True value – approximation
• et= true fractional relative error * 100%

5. 5
 We don’t know the true value, so we can’t calculate
the true error - approximate the error
 Relative approximate error – an approximation of
the error relative to the approximation itself
o Relative Approximate Error:
ea =
Approximate error
Approximation
*100%
o Iterative Approach:
ea =
Current Approximation – Previous Approximation
Current Approximation

6. 6
 For iterative approximations, continue to iterate
until the relative approximate error magnitude is
less than a specified stopping criterion
 For accuracy to at least n significant figures set the
stopping criterion to
o Stopping Criterion:
|ea | = es
es = (0.5*10(2-n))%

7. 7
 ROUND-OFF ERROR:
Roundoff errors arise because digital computers cannot represent some quantities
exactly.
There are two major facets of roundoff errors involved in numerical calculations:
• Digital computers have size and precision limits on their
ability to represent numbers.
• Certain numerical manipulations are highly sensitive to
roundoff errors.
Errors
Truncation Errors Round-off Errors

8. 8
 Base-10 computer with a 5 bit word
S1d1.d2*10S0d0
 2 -5= 0.03125  3.1 x 10 -2
 roundoff error =
(0.03125-0.031)/0.03125 = 0.008 = 0.8%
 Because of the limited number of bits for significand and exponent,
Roundoff errors is occur.
π= 3.141593for 16-bit word computer
π= 3.14159265358979 for 32-bit word computer
 Although adding significand digits can improve the approximation, such
quantities will always have some roundoff error when stored in a computer

9. 9
 Truncation errors are those that result from using an approximation in place of an
exact mathematical procedure.
 Example 1: approximation to a derivative using a finite-difference equation:
~
 Example 2: The Taylor Series
 TRUNCATION ERROR:
=
dv v v(ti+1) – v(ti)
dt t ti+1 - ti
=
f(x)= f(x0)+ f' (x0)+ f'' (x0) + ….. + f(n) (x0) + Rn
x – x0
1!
(x – x0)2 (x – x0)n
n!
2!

10. 10
o The Taylor Series:

11. 11
 In general, the nth order Taylor series expansion will be exact for an nth order polynomial.
 In other cases, the remainder term Rn is of the order of h n+1, meaning:
o Truncation Error:
• The more terms are used, the smaller the error, and
• The smaller the spacing, the smaller the error for a given number of terms.
 The total numerical error is the summation of the
truncation and roundoff errors.
 The truncation error generally increases as the
step size increases, while the roundoff error
decreases as the step size increases - this leads to a
point of diminishing returns for step size.
o Total Numerical Error:

12. 12
Figure: Parts of fixed-point representation
 Sign bit:- The fixed-point number representation in binary uses a sign bit. The
negative number has a sign bit 1, while a positive number has a bit 0.
 Integral Part:- The integral part in fixed-point numbers is of different lengths at
different places. It depends on the register's size; for an 8-bit register, the
integral part is 4 bits.
 Fractional part:- The Fractional part is of different lengths at different places. It
depends on the registers; for an 8-bit register, the fractional part is 3 bits.
There are three parts of the fixed-point number representation: Sign bit, Integral
part, and Fractional part.
 Fixed Point Representation:

13. 13
Register Sign Bit Integer Part Fraction Part
8-bit register 1 bit 4 bits 3 bits
16-bit register 1 bit 9 bits 6 bits
32-bit register 1 bit 15 bits 9 bits
Size of Sign Bit, Integer Part, and Fractional Part for different registers are displayed
below:
Now that we have learned about fixed-point number representation, let's see how to represent
it.
The number considered is 4.5
Step 1: We will convert the number 4.5 to binary form. 4.5 = 100.1
Step 2: Represent the binary number in fixed-point notation with the following format.
Figure: Fixed Point Notation of 4.5
o How to write numbers in Fixed-point notation:

14. 14
 Floating Point Representation:
A floating-point representation has three parts: Sign bit, Exponent Part, and Mantissa.
Figure: Parts of floating-point representation
 Sign bit:- The floating-point numbers in binary uses a sign bit. A negative
number has a sign bit 1, while a positive number has a sign bit 0. The sign of
any number depends on mantissa, not on exponent.
 Mantissa Part:- The mantissa part is of different lengths at different places. It
depends on registers like for a 16-bit register, and mantissa part is of 8 bits.
 Exponent Part:- It is the power of the number. It depends on the size of the
register. For example, in the 16-bit register, the exponent part is of 7 bits.

15. 15
The number considered is 53.5
Step 1: We will convert the number 53.5 to binary form. 53.5 =
110101.1
Step 2: Normalize the number ( base is 2) = (1.101011) * 25.
Step 2: Represent the binary number in floating-point notation with the
following format.
Figure: Floating Point Notation of 53.5
o How to write numbers in Floating-point notation:

16. 16
 Error Propagation:
 Let x refer to the floating point representation of the real number x.
 Since computer has fixed word length, there is a difference between x and x (round-off
error)
and we would like to estimate the error in the calculation of f(x):
f(xfl) = |f(x) - f(xfl)|
f(x) - f(xfl) = f'(xfl)(x - xfl)
f(xfl) = f'(xfl) * x
Result: If f'(x) and x are known, then we can estimate the error using this formula
~
• Both x and f(x) are unknown.
• If x, is close to x, then we can use first order Taylor expansion and compute:
~

17. THANK YOU