# 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?
function fh = poly_fun(p)
function polynomial = poly(x)
n = length(p)-1;
polynomial = sum(p.*x.^(n:-1:0));
end
fh = @poly;
end

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### 採用された回答

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?
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Jesus Alejandro Rodriguez Morales 2020 年 9 月 21 日
Thank you for your answer, Steven. Yeah, the function should return a function handle and after we put the value of x should show the numeric result. For example, if I run the following values works perfectly.
pf = poly_fun(1:5);
pf(1)
pf = poly_fun(ones(1,10));
pf(2)
But, when the grader run something like pf = poly_fun([ -10 5 0 -4 -4 2 -2 10 -6 ]) my function fail the test.

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### 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.
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Thiago Henrique Gomes Lobato 2020 年 9 月 20 日
Depending of your inputs the direction of the sum may be wrong. This code is basically the same as yours but using the "(... ,2)" in the sum so the dimensions are right:
n = length(p)-1;
polynomial = @(x,p)(sum(p.*x.^((length(p)-1):-1:0),2 ));
A = polynomial( [-1:0.01:1]',[ 0 6 7 5 0 8 1 8 6 -7 -4 ] ); % Note the array dimensions
B = polyval([ 0 6 7 5 0 8 1 8 6 -7 -4 ],[-1:0.01:1]');
norm(A-B) % Compare with matlab polyval
ans =
1.9746e-14

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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.
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Alberto Zekry 2020 年 10 月 14 日
function fh = poly_fun(p)
polynomial=0
function polynomial = poly(x)
polynomial=polyval(p(end:-1:1),x);
end
fh = @poly;
end

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R2020b

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