Question

Question #5d213

Answer 1

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(x)=cos^2x`

`f(0)=1`
`f'(x)=-2cosxsinx=-sin2x => f'(0)=0`
`f''(x)=-2cos2x => f''(0)=-2`
`f^((3))(x)=4sin2x => f^((3))(0)=0`
`f^((4))(x)=8cos2x => f^((4))(0)=8`
`f^((5))(x)=-16sin2x => f^((5))(0)=0`
`f^((6))(x)=-32cos2x => f^((6))(0)=-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)!)`