Find the nth-order Taylor polynomials centered at 0, for n=0, 1, 2.

[karush]

$\tiny{206.11.1.15-T}$
$\textsf{Find the nth-order Taylor polynomials
centered at 0, for $n=0, 1, 2.$}$ \\
$$\displaystyle f(x)=cos(3x)$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}x^k$$
$\textsf{n=0}\\$
$$P_0\left(x\right)\approx\frac{1}{0!}x^{0}\approx 1$$
$\textsf{n=1}\\$
$$P_1\left(x\right)\approx \frac{1}{0!}x^{0}+\frac{0}{1!}x^{1}= 1+0 = 1$$
$\textsf{n=2}\\$
$$P_2\displaystyle\left(x\right)\approx\frac{1}{0!}x^{0}+\frac{0}{1!}x^{1}+\frac{-9}{2!}x^{2}f\left(x\right)\approx 1- \frac{9}{2}x^{2}$$
$\textsf{first time to try this so kinda ?}$

 

[HallsofIvy]

cos(0)= 1. The derivative of cos(3x) is -3sin(3x) which is 0 at x= 0. The second derivatve is -9 cos(3x) which is -9 at x= 0.

So, yes, those are correct. (But there should be no "f(x)" in that last expression.)

 

[karush]

cos(0)= 1. The derivative of cos(3x) is -3sin(3x) which is 0 at x= 0. The second derivatve is -9 cos(3x) which is -9 at x= 0.

So, yes, those are correct. (But there should be no "f(x)" in that last expression.)

thamk you, I'll post a couple more. the examples i look at are really hard to follow.

 

[karush]

$\tiny{206.11.1.15-T}$
$\textsf{Find the nth-order Taylor polynomials
centered at 0, for $n=0, 1, 2.$}\\$
$$\displaystyle f(x)=cos(3x)$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}x^k$$
$\textsf{n=0}\\$
\begin{align}
f^0(x)&=cos(3x)\therefore f^0(0)=1 \\
P_0\left(x\right)&\approx\frac{1}{0!}x^{0}\approx 1
\end{align}
$\textsf{n=1}\\$
\begin{align}
f^1(x)&=-3\sin(3x)\therefore f^1(0)=0 \\
P_1\left(x\right)&\approx \frac{1}{0!}x^{0}+\frac{0}{1!}x^{1}= 1+0 \approx 1
\end{align}
$\textsf{n=2}\\$
\begin{align}
f^2(x)&=-9\cos(3x)\therefore f^2(0)=-9 \\
P_2 \left(x\right)&\approx\frac{1}{0!}x^{0}+\frac{0}{1!}x^{1}+\frac{-9}{2!}x^{2}
\approx 1- \frac{9}{2}x^{2}
\end{align}