Answer:
[tex]\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}[/tex]
The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval
[-1.113826815, 1.113826815]
Step-by-step explanation:
According to Taylor's theorem
[tex]\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)[/tex]
with
[tex]\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}[/tex]
for some c in the interval (-x, x)
In the particular case f
f(x)=cos(x)
we have
[tex]\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)[/tex]
therefore
[tex]\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0[/tex]
and the polynomial approximation of T5(x) of cos(x) would be
[tex]\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}[/tex]
In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that
[tex]\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}[/tex]
for some c in (-x,x). So
[tex]\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}[/tex]
and we must find the values of x for which
[tex]\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652[/tex]
Working this inequality out, we find
[tex]\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815[/tex]
Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval
[-1.113826815, 1.113826815]
