Question 5d213

Sep 28, 2015

See the explanation.

Explanation:

f(x)=sum_(k=0)^n f^((k))(0) x^k/(k!) + f^((n+1))(epsilon)x^(n+1)/((n+1)!)

$f \left(x\right) = {\cos}^{2} x$

$f \left(0\right) = 1$
$f ' \left(x\right) = - 2 \cos x \sin x = - \sin 2 x \implies f ' \left(0\right) = 0$
$f ' ' \left(x\right) = - 2 \cos 2 x \implies f ' ' \left(0\right) = - 2$
${f}^{\left(3\right)} \left(x\right) = 4 \sin 2 x \implies {f}^{\left(3\right)} \left(0\right) = 0$
${f}^{\left(4\right)} \left(x\right) = 8 \cos 2 x \implies {f}^{\left(4\right)} \left(0\right) = 8$
${f}^{\left(5\right)} \left(x\right) = - 16 \sin 2 x \implies {f}^{\left(5\right)} \left(0\right) = 0$
${f}^{\left(6\right)} \left(x\right) = - 32 \cos 2 x \implies {f}^{\left(6\right)} \left(0\right) = - 32$

and so on...

f(x)=f(0)+f'(0) x^1/(1!)+f''(0) x^2/(2!)+f^((3))(0) x^3/(3!)+...

f(x)=1+0-2x^2/(2!)+0+8x^4/(4!)+0-32x^6/(6!)+....
f(x)=1+0-2^1x^2/(2!)+0+2^3x^4/(4!)+0-2^5x^6/(6!)+....
f(x)=1+0-2^(2-1)x^2/(2!)+0+2^(4-1)x^4/(4!)+0-2^(6-1)x^6/(6!)+....

f(x)=1+sum_(k=1)^oo (-1)^k 2^(2k-1) x^(2k)/((2k)!)#