# Math Help - Maclaurin expansion

1. ## Maclaurin expansion

I have to expand sinx by using maclaurin theorem which I did but before that I have to prove that certain conditions are being fulfilled by the sinx for which the maclaurin theorem can be applied like all the derivatives exist.But,I also have to fulfill the condition that the maclaurin series that I am going to get is convergent before expanding it.How can I show that?pls explain..........

2. Originally Posted by roshanhero
I have to expand sinx by using maclaurin theorem which I did but before that I have to prove that certain conditions are being fulfilled by the sinx for which the maclaurin theorem can be applied like all the derivatives exist.But,I also have to fulfill the condition that the maclaurin series that I am going to get is convergent before expanding it.How can I show that?pls explain..........
here is what you need to know

If $f(x) = T_n(x) + R_n(x)$ and $\lim_{n \to \infty}R_n(x) = 0$ for $|x - a| < R$, then the Taylor series for $f$ at $a$ converges to $f$ on the interval $|x - a|< R$

( $T_n(x)$ is the nth degree Taylor polynomial of $f$ at $a$).

it also helps to know that if $|f^{(n + 1)}(x)| \le M$ for $|x - a| \le d$, then the remainder $R_n(x)$ satisfies the inequality

$|R_n(x)| \le \frac M{(n + 1)!} |x - a|^{n + 1}$ for $|x - a| \le d$

(the last inequality is called Taylor's inequality)

can you continue?

3. Originally Posted by roshanhero
I have to expand sinx by using maclaurin theorem which I did but before that I have to prove that certain conditions are being fulfilled by the sinx for which the maclaurin theorem can be applied like all the derivatives exist.But,I also have to fulfill the condition that the maclaurin series that I am going to get is convergent before expanding it.How can I show that?pls explain..........
If $T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)x^k}{k!}$ (Taylor polynomial at zero of degree $n$) then for a point $x\in \mathbb{R} - \{0\}$ we know $\sin x - T_n(x) = \frac{f^{(n+1)}(y)}{(n+1)!}x^{n+1}$

Where $y$ is between $0$ and $x$.
Notice that $|f^{(n+1)}(y)| \leq 1$ therefore,
$\left| \sin x - \sum_{k=0}^n \frac{f^{(k)}(0)x^k}{k!} \right| \leq \frac{|x|^{n+1}}{(n+1)!} \to 0$

Thus, we have,
$\sin x = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)x^k}{k!}$