Question about Taylor series

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]

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]

Here is my code

Answer 1

[Torsten]

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!.