I can't find the mistake in the code
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Jesus Alejandro Rodriguez Morales 2020 年 9 月 20 日
The following code takes a vector of coefficient p, defines a function that returns the value of the polynomial given the scalar input x, and returns a function to handle it. However, when the code is assessed with random polynomials fails. For example, "Incorrect answer for pf = poly_fun([ 0 6 7 5 0 8 1 8 6 -7 -4 ])". Any suggestions?
Thanks in advance,
function fh = poly_fun(p)
function polynomial = poly(x)
n = length(p)-1;
polynomial = sum(p.*x.^(n:-1:0));
fh = @poly;
Steven Lord 2020 年 9 月 20 日
I've seen two different conventions for representing polynomials as vectors of coefficients. The one used by functions like polyfit and polyval has the high order term as the first element.
p = [1 2 3];
x = 4;
v1 = polyval(p, x) % x^2+2*x+3 evaluated at x = 4 is 27
The other has the high order term as the last element.
v2 = p(1)+x*p(2)+x.^2*p(3) % 1+2*x+3*x^2 evaluated at x = 4 is 57
Based on the fact that the sample polynomial used in the grading of your assignment has 0 as its first element, I'm wondering if they're using the latter convention.
Alternately, does your assignment say your function should return a function handle or the numeric result of evaluating that polynomial?
その他の回答 (3 件)
Thiago Henrique Gomes Lobato 2020 年 9 月 20 日
You can't use "abs" since the coefficients/x values can be negative.
Herren Thomas D'Souza 2020 年 9 月 27 日
If you workout the algorithm on a piece of paper or mentally by giving smaller arrays as input co-effcients like,
>> z = poly_fun([-1,2])
>> ans = z(2)
You'll now understand clearly how polynomial = sum(p.*x.^(n:-1:0)); is working.
For your logic to work, you can use, flip(p) before the nested function. Or you can use the same sum function to iterate from 0, without needing to flip.