Calculating Remainders for Taylor Series of Sine Function

[stukbv]

Usually to do the remainder we take Rn(x) = (f differentiated n+1 times at a ).(x-c)ⁿ⁺¹/(n+1)!,
but when my function is sin(x) do i take (f differentiated 2n+2 times at a ).(x-c)²ⁿ⁺²/(2n+2)!?

Thanks

 

[micromass]

No, the remainer remains

R_n(x)=\frac{\sin^{(n+1)}(a)(x-c)^{n+1}}{(n+1)!}

So, for example, we have

\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+R_6(x)

or, we can also have

\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+R_7(x)

which is better since the remainder is smaller...

 

[stukbv]

Ok cool thanks a lot - conflicting lecturers grr!