taylor series using forward method

2 Ansichten (letzte 30 Tage)
Ronald Aono
Ronald Aono am 21 Okt. 2019
Beantwortet: Yuan Li am 21 Okt. 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)

Akzeptierte Antwort

Yuan Li
Yuan Li am 21 Okt. 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)

Weitere Antworten (0)

Kategorien

Mehr zu Log Plots finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by