taylor series and conditional while loop

Question

beshayer alturaiyef 2019-10-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/486331-taylor-series-and-conditional-while-loop

回答： VBBV 2021-10-23

clc

clear all

% inputs

x0 = input('what is the startup value x(i): ');

x1 = input('what is the the value you want to predict f(x) at x(i+1): ');

syms x;

syms f(x);

f(x) = log(x*x);

%solving

tv = log(x1*x1);

iter = 1; et = 100; sol=log(x0*x0);

es = 100;

while 1

solold(iter)=sol;

sol = taylor(f, x, 'Order', iter);

es=(0.5*10^(2-iter));

et=abs((tv - sol)/tv)*100;

if et<es

break;

end

iter = iter + 1;

end

sol

iter

shows the folowing:

Error using symengine

Unable to compute a Taylor expansion.

Error in sym/taylor (line 128)

tSym = mupadmex('symobj::taylor',f.s,x.s,a.s,options);

Error in Q32 (line 18)

sol = taylor(f, x, 'Order', iter);

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Answer 1

Sourav Ghai 2019-10-22

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/486331-taylor-series-and-conditional-while-loop#answer_397604

编辑：Sourav Ghai 2019-10-22

Hi,

You just need to set the property 'ExpansionPoint' to 1 in taylor function

taylor(f, x, 'ExpansionPoint',1,'Order', iter);

This code will display the Taylor series:

clc
clear all
% inputs
x0 = input('what is the startup value x(i): ');
x1 = input('what is the the value you want to predict f(x) at x(i+1): ');
syms x;
syms f(x);
f(x) = log(x*x);
%solving
tv = log(x1*x1);
iter = 1; et = 100; sol=log(x0*x0);
es = 100;
while 1
    disp(sol)
    sol = taylor(f, x, 'ExpansionPoint',1,'Order', iter);
    es=(0.5*10^(2-iter));
    et=abs((tv - sol)/tv)*100;
    iter = iter + 1;
end

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Answer 2

VBBV 2021-10-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/486331-taylor-series-and-conditional-while-loop#answer_814858

 clc
clear all
% inputs
% x0 = input('what is the startup value x(i): ');
% x1 = input('what is the the value you want to predict f(x) at x(i+1): ');
x0 = 2;
x1 = 4;
syms x;
syms f;
f = log(x*x);
%solving
tv = log(x1*x1);
iter = 1; et = 100; sol=log(x0*x0);
es = 100;
while 1
    solold(iter)=sol;
    sol = taylor(f, x, 6,'Order', iter); % use conditional value 
    es=(0.5*10^(2-iter));
    ssol = vpa(subs(sol,x,1.5*iter),4) % substitute it tolerance check 
    sol = ssol;
    et=vpa(abs((tv - ssol)/tv),4);
if et>es
    break;
end
    iter = iter + 1;
end
ssol = 3.584
ssol = 2.584
ssol = 3.021
vpa(sol,4)
ans = 3.021
iter
iter = 3