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Qiuyi Wei on 2 Apr 2019
Answered: Torsten on 2 Apr 2019
I am trying to write a code that output a scalor apporximation for function "sin(x)" that have an error less than 0,05. I try to write Taylor series myself(I cannot use "taylor" code) but my error keep getting larger and larger( the wrror supposed to be smaller and smaller). My math is really bad and I don't know why I keep getting wrong answer. Can anyone help me figure out what's wrong for my code and how can I fix it?
Thanks

Qiuyi Wei on 2 Apr 2019
x=randi([-20 20]);
a=x+randi([2 20]);
k = sin(x); % true value
y = zeros(1,201); % approximation
y(1) = sin(a);
d = zeros(1,200) % derivative
for n = 1:200
e = mod(n,4);
if e==1;
d(n) = sin(a);
elseif e==2;
d(n) = cos(a);
elseif e==3;
d(n) = -sin(a);
elseif e==0;
d(n) = -cos(a);
end
y(n+1) = y(n)+d(n).*(x-a).^(n)./factorial(n);
end
error = abs((y-k)./k);
check = error<=0.05
t = find(check==1);
w = y(t);
est = k(1);
Qiuyi Wei on 2 Apr 2019
Here is my code

Torsten on 2 Apr 2019
d(1) = cos(a), d(2) = -sin(a), d(3) = -cos(a), d(4) = sin(a)
So your if-statement to determine f^(n)(a) is wrong.
Further, "error" must be defined as
error = abs((y(end)-k)/k)
Further, you should not always explicitly calculate (x-a)^n/n!. Note that
(x-a)^(n+1)/(n+1)! = (x-a)^n/n! * (x-a)/(n+1).
So you can use a recursion to determine (x-a)^(n+1)/(n+1)! from (x-a)^n/n!.