Question

a) Determine the 2n-th Taylor polynomial \(p_{2n}(x)\) of \(f(x) = sin(x)\) at \(x_{0}=\frac{\pi}{2}\)

Answer 1

oh.. i read that wrong.... you want \(\large P_{2n} \)???

Answer 2

yes...

Answer 3

how can i type it greater.. ?

Answer 4

since the nth term is given by \(\huge \frac{-1^nx^{2n+1}}{(2n+1)!} \),
i guess the 2n-th term is to replace the n with 2n?
\(\huge \frac{-1^{2n}x^{2(2n)+1}}{(2(2n)+1)!} \)

???
idk man... sorry....
i haven't touched mclaurin/taylor for a while.....

Answer 5

ok no problem :)