Taylor Series of e^x

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Joseph Melo Manalo
Joseph Melo Manalo el 21 de Mzo. de 2020
Respondida: Mohamed Hakim el 21 de Mayo de 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 comentario
Walter Roberson
Walter Roberson el 21 de Mzo. de 2020
while term is in range && iterations is in range
Calculate a term
accumulate term into total
increment iterations

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Respuesta aceptada

Subhamoy Saha
Subhamoy Saha el 21 de Mzo. de 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 comentarios
Subhamoy Saha
Subhamoy Saha el 25 de Mzo. de 2020
I didn't realised that @Walter.
Joseph Melo Manalo
Joseph Melo Manalo el 25 de Mzo. de 2020
I just read the rules in asking homework and I am really sorry for not tagging "homework". I thank you both also for answering my questions. I have learn that I can learn better using actual example, so even though Subhamoy Saha didn't exactly write the code that I need. I can work from there. Again I am sorry and Thank you both!

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Más respuestas (2)

Mohamed Hakim
Mohamed Hakim el 20 de Mayo de 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 comentario
Walter Roberson
Walter Roberson el 20 de Mayo de 2021
Is there a reason to write
i<n || i==n
instead of
i<=n
?

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Mohamed Hakim
Mohamed Hakim el 21 de Mayo de 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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