206.11.1.12 quadratic approximating polynomial

[karush]

$\tiny{206.11.1.12}$
$\textsf{a.Find the linear approximating polynomial for} \\$
$$\displaystyle f(x)=\cos{x}
\textsf{ centered at $\displaystyle a=\frac{\pi}{4}$.}
\text{approximate} \cos(0.28\pi)$$
$\textsf{ using}$
$$f(x)=f(a)+f'(a)(x-a)$$
$\textsf{b. Find the quadratic approximating polynomial}\\$.
$\textsf{Assume we plug into this formula
with $x=0.28\pi$ and $\displaystyle a=\frac{\pi}{4}$
with $f'(x)=-\sin\left({x}\right)$
and $f''(x)=-\cos\left({x}\right)$} \\$
$$\displaystyle f(x)\approx P_2(x)
=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2$$
$\textsf{c. Use the polynomials obtained in (a) and (b) to approximate the given quality}\\$

$\textsf{still confused abour this?}$

 

[HallsofIvy]

Exactly what is your difficulty? You are told exactly what to do!

$\tiny{206.11.1.12}$
$\textsf{a.Find the linear approximating polynomial for} \\$
$$\displaystyle f(x)=\cos{x}
\textsf{ centered at $\displaystyle a=\frac{\pi}{4}$.}
\text{approximate} \cos(0.28\pi)$$
$\textsf{ using}$
$$f(x)=f(a)+f'(a)(x-a)$$

Okay, f(x)= cos(x) and f'= -sin(x). At x= $\pi/4$ $cos(\pi/4)= sin(\pi/4)= \frac{\sqrt{2}}{2}$.

So $cos(x)= -\frac{\sqrt{2}}{2}(x-\pi/4)+ \frac{\sqrt{2}}{2}$.

$\textsf{b. Find the quadratic approximating polynomial}\\$.
$\textsf{Assume we plug into this formula
with $x=0.28\pi$ and $\displaystyle a=\frac{\pi}{4}$
with $f'(x)=-\sin\left({x}\right)$
and $f''(x)=-\cos\left({x}\right)$} \\$
$$\displaystyle f(x)\approx P_2(x)
=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2$$
$\textsf{tc. Use the polynomials obtained in (a) and (b) to approximate the given quality}\\$

$\textsf{still confused abour this?}$

cosine and sine of "$0.28\pi$" do not give any thing simple ($sin(0.28\pi)$ is approximately 0.7705 and $cos(0.28\pi)$ is approximately 0.6374) so just enter $sin(0.28\pi)$ and $cos(0.28\pi)$:

$f(a)= cos(0.28\pi)$, $f'(a)=-sin(a)= -sin(0.28\pi)$, and $f''(a)= -cos(a)= -cos(0.28\pi)$ so we have
$cos(x)= cos(0.28\pi)- sin(0.28\pi)(x- 0.28\pi)- \frac{cos(0.28\pi)}{2}(x- 0.28\pi)^2$

For (c), what is the "given quality"? (Did you mean "equality" or "inequality"?)

 

[karush]

ok added image