ppval and polyval giving different results

Question

cr 2022-4-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1699785-ppval-and-polyval-giving-different-results

评论： Torsten 2022-4-19

I created a piecewise polynomial using mkpp. But when I evaluate a set of points I get incorrect results using ppval, although polyval() gives correct result. Looks like I'm missing something. Any idea what that is?

Thanks!

>> ppm
ppm = 
struct with fields:
form: 'pp'
breaks: [749.5 16747 32768 48771 64784]
coefs: [4×2 double]
pieces: 4
order: 2
dim: 1
K>> ppm.coefs
ans =
0.00031269      -10.246
0.00031268      -10.246
0.00031274      -10.247
0.00031265      -10.245
>> ppval(ppm,ppm.breaks(1))
ans =
      -10.246
>> polyval(ppm.coefs(1,:),ppm.breaks(1))
ans =
      -10.012
      
>> polyval(ppm.coefs(1,:),8750)
ans =
        -7.51
>> ppval(ppm,8750)
ans =
      -7.7444      

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Answer 1

Bruno Luong 2022-4-19

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https://ww2.mathworks.cn/matlabcentral/answers/1699785-ppval-and-polyval-giving-different-results#answer_945805

编辑：Bruno Luong 2022-4-19

在 MATLAB Online 中打开

Here is a correct way to know how ppval works

pp=spline(cumsum(rand(1,10)),rand(1,10));
x=3;
ppval(pp,x)
ans = 2.1433
% 
i=discretize(x,pp.breaks);
polyval(pp.coefs(i,:),x-pp.breaks(i))
ans = 2.1433

Your evaluation

polyval(ppm.coefs(1,:),8750)

is wrong unless pp.breaks(1) is 0.

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Answer 2

Torsten 2022-4-19

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https://ww2.mathworks.cn/matlabcentral/answers/1699785-ppval-and-polyval-giving-different-results#answer_945855

I think you fed "polyval" with the coefficients cut to a certain number of digits.

You must use the coefficients in full precision to get equal results from ppval and polyval.

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Bruno Luong 2022-4-19

"And I don't think you must recalculate them for the shifted polynomial."

You are wrong here is the doc of mkpp

"Polynomial coefficients, specified as an L-by-k matrix with the ith row coefs(i,:) containing the local coefficients of an order k polynomial on the ith interval, [breaks(i), breaks(i+1)]. In other words, the polynomial is coefs(i,1)*(X-breaks(i))^(k-1) + coefs(i,2)*(X-breaks(i))^(k-2) + ... + coefs(i,k-1)*(X-breaks(i)) + coefs(i,k)."