Answer:
Step-by-step explanation:
By the general expresion:
T4(x) = f(a) + (f'(a)/1!)*(x-a) + (f''(a)/2!)*(x-a)² + (f'''(a)/3!)*(x-a)³ + (f''''(a)/4!)*(x-a)^4
Then:
f(x) = sinx ; f'(x) = cosx ; f''(x) = -sinx ; f'''(x) = -cosx ; f''''(x) = sinx
Since: sin(π/6) = 1/2 ; cos(π/6) = [tex]\sqrt{3}/2[/tex]
Thus, in the expresion for T4(x):
[tex]T4(x) = 1/2 + (\sqrt{3}/2 )*(x-\pi/6) - (1/4)*(x-\pi/6)^{2} - (\sqrt{3}/12)*(x-\pi/6)^{3} + (1/48)*(x-\pi/6)^{4}[/tex]
Inequality, by the Taylor's theorem (jpg adjunt):
We need to find M in the interval indicate. Of the analysis from the graf of the function (jpg adjunt), we see a good candidate is:
M = f(π/3) = sin(π/3) = [tex]\sqrt{3}/2[/tex]
Then, in the Taylor's theorem:
[tex]|R(x)| \leq (\sqrt{3}/2)* \frac{|x-\pi/6 |^{5} }{5!}[/tex]
Answer:
(a) [tex]T_4(\frac{\pi}{6})=\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi}{6})-\frac{1}{2*2!}(x-\frac{\pi}{6})^2-\frac{\sqrt{3}}{2*3!}(x-\frac{\pi}{6})^3+\frac{1}{2*4!}(x-\frac{1}{2})^4[/tex]
(b) [tex]R_4(x)=\frac{f^{(5)}(\alpha)}{(5!)}(x-\frac{\pi}{6})^5=\frac{cos\alpha}{5!}(x-\frac{\pi}{6})^5[/tex]
Step-by-step explanation:
(a) We have that Taylor's polynomial is given by
[tex]T_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+...+\frac{f^{(n)}(a)}{n!}(x-a)^n[/tex]
In this case we have to compute T4, with a = pi\6. Hence we have to calculate the first derivative until the fourth derivative of f(x):
[tex]f(x)=sinx \ ; \ f(\frac{\pi}{6})=sin(\frac{\pi}{6})=0.5\\\\f'(x)=cosx \ ; \ f'(\frac{\pi}{6})=cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\\\\f''(x)=-sinx \ ; \ f''(\frac{\pi}{6})=-sin(\frac{\pi}{6})=-0.5\\\\f'''(x)=-cosx \ ; \ f'''(\frac{\pi}{6})=-cos(\frac{\pi}{6})=-\frac{\sqrt{3}}{2}\\\\f^{(4)}=sinx \ ; \ f^{(4)}(\frac{\pi}{6})=sin(\frac{\pi}{6})=0.5[/tex]
By replacing we have
[tex]T_4(\frac{\pi}{6})=\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi}{6})-\frac{1}{2*2!}(x-\frac{\pi}{6})^2-\frac{\sqrt{3}}{2*3!}(x-\frac{\pi}{6})^3+\frac{1}{2*4!}(x-\frac{1}{2})^4[/tex]
(b) the Taylor's inequality is given by
[tex]|R_n(x)|\leq \frac{M}{(n+1)!}|x-a|^{n+1}\\\\R_(x)=\frac{f^{(n+1)}(\alpha)}{(n+1)!}(x-a)^n+1[/tex]
where f(x)<=M, Hence we have
[tex]R_4(x)=\frac{f^{(5)}(\alpha)}{(5!)}(x-\frac{\pi}{6})^5=\frac{cos\alpha}{5!}(x-\frac{\pi}{6})^5[/tex]
HOPE THIS HELPS!
