Rounding off and Errors.

Rounding off

If a number has large number of digits then it is necessary to cut them to a usable number of figures. This process of omitting some digits is called rounding off.

Rule for rounding off

The rules for rounding off numbers to the appropriate significant figures are as follows:

a) If the digit to be dropped is less than 5, then the preceding digit is left unchanged and if it is more than 5 then the preceding digit is raised by 1.

b) If the digit to be dropped is 5, then the preceding digit is retained unchanged if it is even, but increased by 1 if it is odd.

Example: The number below are round off to four significant figures.

1.6583 to 1.658

30.0567 to 30.06

0.859378 to 0.8594

3.14159 to 3.142

Errors

Numerical errors arise from the use of approximation to represent exact mathematical operation and quantities.

Truncation errors

Truncation errors result when approximation are used to represent exact mathematical procedures. They are caused, when an infinite process is replaced by a finite one.

For example

Sinx = x - x³/3! + x⁵/5! - x⁷/7! + ..... ∞

If it is replaced by x - x³/3! + x⁵/5! then truncation errors occurs.

Rounding off errors

Round off errors result when numbers having limited significant figures are used to represent exact numbers.

For both types, the relationship between the error, true value and the approximation can be formulated as

True value = approximation + error

=> Error = True value - approximation

Suppose that p* is an approximation to p then

Absolute error (Eₐ)

Absolute error is the modulus of numerical difference between the true value and its approximate value. i.e

Eₐ = | p - p*|

It is also denoted by ∆P.

Relative error (Eᵣ)

Relative error is defined as the ratio of absolute error and true value i.e

Eᵣ = Eₐ/p = | p - p*|/p = ∆p/p

Percentage error (Eₚ)

It is denoted by Eₚ and is defined as

Eₚ = 100 × Eᵣ

= ∆p/p × 100

Questions

Q1) If 0.333 is the approximate value of 1/3, find absolute, relative and percentage errors?

Solution. Here x= 1/3 and xₐ = 0.333

Absolute Error

The absolute Error is given by

Eₐ=|x - xₐ|=|1/3 - 0.333| = |1/3 - 333/1000|

= |1000 -999/3000| = 1/3000 = 0.00033Ans

Relative Error

The relative Error Eᵣ is given by

Eᵣ = Eₐ/x = 0.00033/1/3 = 0.00099Ans

Percentage Error

The percentage Error Eₚ is given by

Eₚ = Eᵣ × 100 = 0.099Ans

Q2) Round off the number 537.261 to four significant digits and then calculate absolute error, relative error and percentage error?

Solution. Given number is 537.261 (=x say)

After rounded off to four significant figures, the given number would be 537.3 (= xₐ say)

Absolute error

Eₐ=|x - xₐ|= |537.261 - 537.300| = 0.039

Relative error

Eᵣ = |x - xₐ|/x = 0.039/537.261 = 7.25 × 10⁻⁵

Percentage error

Eₚ= Eᵣ × 100 = 7.25 × 10⁻³

Comments

Gauss's central difference formula for equal intervals.

Gauss's central difference formula for equal intervals: We shall develop central difference formulae which are best suitable for interpolation near the middle of the tabulated set (table). x: x₋₂ x₋₁ x₀ x₁ x₂ y=f(x): y₋₂ y₋₁ y₀ y₁ y₂ Difference table. Gauss's forward interpolation formula for equal intervals: f(x) = y₀+[u/1!]∆y₀+{[u(u-1)]/2!}∆²y₋₁+{[(u+1)(u)(u-1)]/3!}∆³y₋₁+{[(u+1)u(u-1)(u-2)]/4!}∆⁴y₋₂ +......., Where u= (x-x₀)/h Remark: This formula is applicable when u lies between 0 and 1 i.e (0<u<1). Example: Using Gauss's forward formula to evaluate y₃₀ given that y₂₁=18.4708, y₂₅=17.8144, y₂₉=17.1070, y₃₃=16.3432 and y₃₇=15.5154. Solution. The difference table is Forward difference table. To find y=f(x) at x=30, i.e f(30): Taking x₀ = 29, h=4, x=30, then u= (x-x₀)/h = (30-29)/4 = 0.25 Using Gauss's forward difference formula f(x) = y₀+[u/1!]∆y₀+{[u(u-1)]/2!}∆²y₋₁+{[(u+1)(u)(u-1)]/3!}∆³y₋₁+[(u+1)u(u-1)(u-2)/4!]∆⁴y₋₂ +...... => f(30)

Different operators

Different operators We will study the following operators: 1) The Shifting Operator (E): Ef(x) = f(x+h) i.e Ey₀= y₁ E²f(x) = f(x+2h) i.e E²y₀ = y₂ Eⁿf(x) = f(x+nh) i.e Eⁿy₀ = yₙ Here n takes integral or fractional, positive or negative values. For example: E⁻¹f(x) = f(x-h) i.e E⁻¹ y₂ = y₁ E¹/²f(x) = f[x+(1/2)h] i.e E¹/² y₁ = y₃ₗ₂ Properties of Operator E 1) Operator E is distributive. 2) Operator E is commutative with respect to constant. 3) Operator E obeys laws of indices. 2) Forward difference operator (∆): If x₀, x₁,x₂,......., xₙ are equally spaced with interval of differencing h and if y = f(x), then ∆f(x) = f(x+h) - f(x) i.e ∆yᵢ = yᵢ₊₁ - yᵢ for i = 0,1,2,3,..... n-1, The symbol ∆ is called forward difference operator and ∆yᵢ is called first forward difference. Similarly, the second forward differences are ∆²yᵢ =∆yᵢ₊₁ - ∆yᵢ For example: ∆²y₀ =∆y₁ - ∆y₀ = (y₂ - y₁)-(y₁-y₀) = y₂ -2y₁+y₀ Clearly any higher order differences can easily be expresse

Questions (Forward difference operator, Backward difference operator and Shifting operator)

Questions Example: Given that y₅=4, y₆=3, y₇=4, y₈=10, y₉=24, find the value of ∆⁴y₅: (i) By using the difference table and (ii) Without using the difference table. Solution: (i) Forward difference table. From the difference table ∆⁴y₅=0. Ans. (ii) We have ∆⁴y₅= (E-1)⁴y₅ [∵ ∆=E-1] = (E⁴-4E³+6E²-4E+1)y₅ = E⁴y₅-4E³y₅+6E²y₅-4Ey₅+1y₅ = y₉ -4y₈+6y₇-4y₆+y₅ [Eⁿyₓ=yₓ₊ₙ] = 24-4×10+6×4-4×3+4 = 24-40-24-12+4 =0Ans. Example: Given x: 1 2 3 4 5 y: 2 5 10 17 26 Find the value of ∇²y₅. Solution. (i) Backward difference table. From the difference table, we get ∇²y₅=2 Ans. (ii) Without using the difference table to find ∇²y₅: We have ∇²y₅ = (1-E⁻¹)²y₅ [∵ ∇=1-E⁻¹] = (1+E⁻² -2E⁻¹)y₅ = y₅ + y₃ -2y₄ [E⁻ⁿyₓ=yₓ₋ₙ] = 26+10-2×17 = 36-34 = 2 Ans. Example: F

Numerical analysis

Search This Blog