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I am having some trouble translating a pseudocode for Horner's algorithm into a proper code in MatLab. I think my confusion stems from the fact that the code assumes that the first vector entry can be referred to by 0, whereas in MatLab, this has to be 1. I have tried to modify my code accordingly, but I don't get it to work properly.

The pseducode is as follows: input n, (a_i,: 0 ≤ i ≤ n), z_0 for k = 0 to n-1 do for j = n-1 to k step -1 do a_j = a_j + z_0*a_(j+1) end do end do output (a_i: 0 ≤ i ≤ n) Here is my attempt at writing this in MatLab, where a is an input vector representing coefficients in a polynomial: function x = horner(a,z_0) n = length(a); for k = 1:n-1 for j = n-1:-1:k a(j) = a(j) + (z_0)*a(j+1); end end x = a; I tried this on the vector a = [1 -4 7 -5 -2] which represents coefficients in a polynomial. I also set z_0 = 3. According to my book, I should have received the output vecor a = [1 8 25 37 19], but my code gives the output vector a = [-245 -313 -146 -29 -2]. Installing Air Conditioner In Crank Window. If anyone can help me clear up this code, I would be very grateful!

Try this - here a is the vector of polynomial coefficients listed with a(1) as the coefficient of the highest degree term in your polynomial. If your vector is the opposite way round, simply set b = fliplr(a) and call the function using vector b. This function will evaluate the polynomial using Horners algorithm. Note that this assumes z_0 is the value that you want the polynomial evaluated at, hence a single value is returned (not a vector) function x = horner(a,z_0) n = length(a); result = a(1); for j = 2:n result = result*z_0 + a(j); end x = result; If you want to pass in a vector of values z to evaluate so you can evaluate multiple points (the elements of z) at the same time, you can pass them in via a vector: function x = horner(a,z) n = length(a); m = length(z); result = a(1)*ones(1,m); for j = 2:n result = result.*z + a(j); end x = result; now the returned x will be your vector of results.