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taylor series using forward method

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Ronald Aono
Ronald Aono on 21 Oct 2019
Answered: Yuan Li on 21 Oct 2019

% question 3 part b
clc
clear all
real_val=14301/400;
h=[10^-4 10^-3 10^-2 10^-1 10^0 10^1];%step sizes
%defining the function f(x)
n=length(h);
x = 2.1;
f=@(x) x^5-2*x^4+3*x^2-1;
df=@(x) 3*x^4 - 8*x^3 + 6*x;
df=cell(1:n);
error=[];
for i=1:n
f(i)=(f(x + h(i)) - f(x))/ (h(i));% evaluating the derivative at h(i)
error(i)=abs((df(i)-real_val)/(real_val)); % evaluating the absolute error at h(i)
end

% creating a log log plot of the error against
figure(1)
loglog(h,error)
grid on
xlabel('step size')

keep getting this error
Conversion to function_handle from double is not possible.

Error in Tay_ser (line 25)
f(i)=(f(x + h(i)) - f(x))/ (h(i));% evaluating the derivative at h(i)

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Accepted Answer

Yuan Li
Yuan Li on 21 Oct 2019
I can't figure out the relationship between the following codes:
f(i)=(f(x + h(i)) - f(x))/ (h(i));% evaluating the derivative at h(i)
error(i)=abs((df(i)-real_val)/(real_val)); % evaluating the absolute error at h(i)
how about reviewing this way?
temp(i)=(f(x + h(i)) - f(x))/ (h(i));% evaluating the derivative at h(i)
error(i)=abs((temp(i)-real_val)/(real_val)); % evaluating the absolute error at h(i)

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