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Taylor Series of e^x

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Joseph Melo Manalo
Joseph Melo Manalo 2020-3-21
回答： Mohamed Hakim 2021-5-21
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 个评论
Walter Roberson
Walter Roberson 2020-3-21
while term is in range && iterations is in range
Calculate a term
accumulate term into total
increment iterations

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采纳的回答

Subhamoy Saha
Subhamoy Saha 2020-3-21
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 个评论
显示 2更早的评论
Subhamoy Saha
Subhamoy Saha 2020-3-25
I didn't realised that @Walter.
Joseph Melo Manalo
Joseph Melo Manalo 2020-3-25
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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更多回答（2 个）

Mohamed Hakim
Mohamed Hakim 2021-5-20
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
Mohamed Hakim
Mohamed Hakim 2021-5-21
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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