关于使用勒让德节点给出错误的多项式插值的 Matlab 代码
Matlab code about polynomial interpolation using legendre nodes giving error
我正在尝试通过求解线性方程组（给出 Vandermonde 矩阵）使用勒让德节点绘制多项式插值。我这样做是为了 j = 5, 7, 9,...,35. 这是我的代码：
clear all;
clc;
for j=5:2:35
syms z;
xnodes_3 = vpasolve(legendreP(j,z) == 0);
fnodes_3 = 1./(1+25*xnodes_3.^2);
M = 1001;
x = linspace(-1,1,M)';
coeff_vector_3 = fliplr(vander(xnodes_3))\fnodes_3;
p_3 = polyval(fliplr(coeff_vector_3'), x');
plot(x,p_3);
hold on
end
这给我以下错误：
Error using polyval (line 74)
Inputs must be floats, namely single or double.
Error in trial (line 15)
p_3 = polyval(fliplr(coeff_vector_3'), x');
谁能帮忙。我整天都坚持这个。非常感谢。
编辑： 在给出错误之前，这些是 coeff_vector_3.
的值
coeff_vector_3 =
1.0
3.4109834835328701522238688633873e-40
-4.0512155881301394351090454057919
-2.2893329650954534538375610295397e-39
3.519395781194136574901680371827
coeff_vector_3 =
1.0
-5.4307705791518293802399784884843e-39
-7.0462203744169433121822078413245
1.6720348984736620597732967039872e-38
14.571492730176072614195066605034
-8.9896434936753099542494371570515e-39
-8.802481667857292936084380384082
coeff_vector_3 =
1.0
1.1895297197316568231506041907424e-38
-10.150118913095781981168545036265
-1.656929497262701054447685901442e-37
35.95471688613268525301895310519
4.5856583163339969237259878878631e-37
-49.454723982395860648314603704114
-3.2722088392672170042736433493855e-37
22.920673601437558097637629851961
coeff_vector_3 =
1.0
-3.4070795021123689133565334809069e-39
-13.071031040337698565017684899241
6.8318012404587502749750786583848e-38
68.314781882425933034158628631267
-3.3980135924018244407798124502157e-37
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6.3588750161147154164385284410183e-37
161.4436432060590296122021924084
-3.9427101228610381581618919044928e-37
-60.714837165602428758170595282528
coeff_vector_3 =
1.0
6.7743419377835896476400888375877e-38
-15.641391643053141648840522708178
-1.9475752167281198208034560655283e-36
110.27654036792279068622874894977
1.4804067399849635653881337742821e-35
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625.05088500456434046698274321406
5.4368319346540047502133835673377e-35
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-2.3508099236679182700026469488146e-35
162.06317362181401093418125128426
coeff_vector_3 =
1.0
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4.4054014482556972567553468342051e-34
1758.2722399177410497794450154309
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8.4191487946456620598105592526566e-34
1589.1382218176969571732232005908
-2.9452686618521626935986359589406e-34
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coeff_vector_3 =
1.0
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8.6232679831839388915776313632354e-34
1164.7454439082133880485584414644
coeff_vector_3 =
1.0
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0.000000000000000000000000000000073483825464302587081023771309299
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coeff_vector_3 =
1.0
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coeff_vector_3 =
1.0
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coeff_vector_3 =
1.0
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0.0000000000000000000000000000002984388487867261972619424655104
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0.000000000000000000000000000033568649770568733394781883417097
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coeff_vector_3 =
1.0
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coeff_vector_3 =
1.0
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0.000000000000000000000000000015551600880358959919486980706785
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0.00000000000000000000000060331315400572171509211045884152
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0.00000000000000000000000083453523141059274280186623566073
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coeff_vector_3 =
1.0
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0.00000000000000000000000000000089479127468902153227416406121596
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0.00000000000000000000000000095657556607957131683371121860286
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0.000000000000000000000000092472458408994114348875083201744
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0.000000000000000000000001678115966841866698521838475859
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0.0000000000000000000000013743750544997383390940925209134
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coeff_vector_3 =
1.0
-1.247835296084725146240039248023e-33
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0.00000000000000000000000000000023712779410607480588238778794733
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coeff_vector_3 =
1.0
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你的coeff_vector_3仍然是象征性的，因为你使用了vpasolve。要将它们传递给 polyval，请先将它们转换为双打。
coeff_vector_3 = double(fliplr(vander(xnodes_3))\fnodes_3); % convert to doubles here
p_3 = polyval(fliplr(coeff_vector_3'), x');
我正在尝试通过求解线性方程组（给出 Vandermonde 矩阵）使用勒让德节点绘制多项式插值。我这样做是为了 j = 5, 7, 9,...,35. 这是我的代码：
clear all;
clc;
for j=5:2:35
syms z;
xnodes_3 = vpasolve(legendreP(j,z) == 0);
fnodes_3 = 1./(1+25*xnodes_3.^2);
M = 1001;
x = linspace(-1,1,M)';
coeff_vector_3 = fliplr(vander(xnodes_3))\fnodes_3;
p_3 = polyval(fliplr(coeff_vector_3'), x');
plot(x,p_3);
hold on
end
这给我以下错误：
Error using polyval (line 74)
Inputs must be floats, namely single or double.
Error in trial (line 15)
p_3 = polyval(fliplr(coeff_vector_3'), x');
谁能帮忙。我整天都坚持这个。非常感谢。
编辑： 在给出错误之前，这些是 coeff_vector_3.
的值coeff_vector_3 =
1.0
3.4109834835328701522238688633873e-40
-4.0512155881301394351090454057919
-2.2893329650954534538375610295397e-39
3.519395781194136574901680371827
coeff_vector_3 =
1.0
-5.4307705791518293802399784884843e-39
-7.0462203744169433121822078413245
1.6720348984736620597732967039872e-38
14.571492730176072614195066605034
-8.9896434936753099542494371570515e-39
-8.802481667857292936084380384082
coeff_vector_3 =
1.0
1.1895297197316568231506041907424e-38
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-1.656929497262701054447685901442e-37
35.95471688613268525301895310519
4.5856583163339969237259878878631e-37
-49.454723982395860648314603704114
-3.2722088392672170042736433493855e-37
22.920673601437558097637629851961
coeff_vector_3 =
1.0
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coeff_vector_3 =
1.0
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coeff_vector_3 =
1.0
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coeff_vector_3 =
1.0
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你的coeff_vector_3仍然是象征性的，因为你使用了vpasolve。要将它们传递给 polyval，请先将它们转换为双打。
coeff_vector_3 = double(fliplr(vander(xnodes_3))\fnodes_3); % convert to doubles here
p_3 = polyval(fliplr(coeff_vector_3'), x');