Question

1. Give pseudocode for the algorithm for evaluating (for a particular value v of x) a polynomial of degree n using Horner's rule. Illustrate the actions for the following: 5x5 + 6x4 - 17x3 + x +9. What is the count of multiplications and addition operations?

Answer 1

) Answer :- for a Orner's Rule: This method is used to simpliby the process of evaluating the polynomial of value of a. degree consider a polynomial a a50 +0424+03 x² + 0 x 10,9, 40 can be expressed as ((ceas 1+ Qu)* +ag) 2 +a2)x + ap)a+ao In above representation, we can evaluate expression early, compared to the original representation Pseudo code for evaluating polynomial using Homer's Rule. 1. Input degree of the polynomial as n (ne. -degree of polynomial) 2. Create an array of size n+1 ise., a[n+] 3. Input coefficients in to array, starting from coefficient of higher degree monomial. 4. create variable named value, initialize with a co]. a [o] 5. for loop with ia until is less than or equal ton and increment i by 1. value = value*x ta[i] 6. End loop 7 print the value. value can

( 2 ) n Time complexity : 0(0) Give to illutrate with example 52S+6x4_151°4x +9 n = degree of polynomial = 5 a [6] = {5,6, -17,0,1,9} let & be vie. xav value = a[0] for (int i=1; irani) value = value * x +CI] Illustration of loop Ist iteration value = 5x+6 2nd iteration value = (5x+6) x + (-17) Srd iteration value = [(5x+6) +--17))x +o 4th iteration value=[(5x+6)x - 17 Jan+o)x+) 5th iteration value = ({[(5x+6) * - 17 )a+0)x+1)x+9 After end for loop we got value= ((((54+6) 2-17) 4+0) x + 1) +9 This nothing but representing the polynomial in Homer's way to evaluate easily.

Here generally as 2 * has a numerical value, But for demonstration I used it Each iteration has a multiplication and addition there are 5 iterations for above polynomial. Therebor 5 multiplication and 5 addition operations. Here it we observe 5 is the degree of given polynomial therebore for degree polynomial we have multiplication and n addition operations. Use 1] in the above algorithm That is calculate value of polynomial for x = 1 1st iteration : value = 5x+6 = 11 2nd iteration : value = (11) x + (-17) = -6 3rd iteration: value = (-6)xto =-6. 4 th iteration value=(-6)x+ 1 =-5 5th iteration : value = (-548 + 9 = 4 we got value = 4 substitute 21 in 5x+624_17x3 + x +90 500) + 6(0-1301) +9 4 our value matched with the direct substitution value

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