Taylor Series of e^x
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Joseph Melo Manalo
on 21 Mar 2020
Answered: Mohamed Hakim
on 21 May 2021
The program calculates e^x by adding terms of the series and stopping when the absolute value of the term that was added last is smaller than 0.0001. Use a while-end loop, but limit the number of passes to 30. If in the 30th pass the value of the term that is added is not smaller than 0.0001, the program stops and displays a message that more than 30 terms are needed.
clear
clc
fprintf('Solving e^x using Taylor Series\n');
x=input('x = ');
n=30;
i=0;
f=1;
while i<n || i==n
f=f+(x.^i)/factorial(i);
i=i+1;
end
y=f-1
This is where I am stuck, I don't know how to limit the n to 30 and how can I make a condition that if the value of the term is smaller than 0.0001 it will stop.
Any help would be appreciated, Thank you!
1 Comment
Walter Roberson
on 21 Mar 2020
while term is in range && iterations is in range
Calculate a term
accumulate term into total
increment iterations
Accepted Answer
Subhamoy Saha
on 21 Mar 2020
clear
clc
fprintf('Solving e^x using Taylor Series\n');
x=input('x = ');
n=30;
i=0;
f=1;
while i<=n
last_term=(x.^i)/factorial(i);
if (x.^i)/factorial(i)<0.0001
msgbox('Last term is smaller than 0.0001 and hence stopped')
break
elseif i==n && last_term>0.0001
msgbox('More steps needed')
break
end
f=f+last_term;
i=i+1;
end
y=f-1
4 Comments
More Answers (2)
Mohamed Hakim
on 20 May 2021
function [ts]=newton(x,n)
i=1;
ts=1;
while i<n || i==n
ts=ts+(x.^i)/factorial(i);
i=1+i;
end
end
1 Comment
Walter Roberson
on 20 May 2021
Is there a reason to write
i<n || i==n
instead of
i<=n
?
Mohamed Hakim
on 21 May 2021
y= @(x) 2*x^2-5*x+3;
x1=input("enterfirst number");
x2=input("enterfirst number");
if f(x1)*f(x2)==0
disp("no");
end
0 Comments
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