Taylor Series of e^x
98 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Joseph Melo Manalo
le 21 Mar 2020
Réponse apportée : Mohamed Hakim
le 21 Mai 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 commentaire
Walter Roberson
le 21 Mar 2020
while term is in range && iterations is in range
Calculate a term
accumulate term into total
increment iterations
Réponse acceptée
Subhamoy Saha
le 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 commentaires
Plus de réponses (2)
Mohamed Hakim
le 20 Mai 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 commentaire
Walter Roberson
le 20 Mai 2021
Is there a reason to write
i<n || i==n
instead of
i<=n
?
Mohamed Hakim
le 21 Mai 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
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