Let f(X)=cos2x. Find a quadratic polynomial px so that p0=f0, p'0=f'0 and p''0=f''0.
px=

Question

Let f(X)=cos2x. Find a quadratic  polynomial px  so that p0=f0, p'0=f'0 and p''0=f''0.
 px=

ash2326 · Answer

Let's find f(0)= cos 0=1= p(0)
P(x) must have a constant term let it be 1
Let's find f'(x)
$$f'(x)= -2\sin 2x$$
x=0
$$f'(x)=0$$
 p'(0)=0
Let's find f''(x)
$$ f''(x)= -4 cos 2x$$
$$ f''(0)=-4$$
so p''(0)=-4
p"(x)= -4
We can p'(x)= -4x+c
p(x)=-2x^2+c1x+c2
p(0)=1
c2=1
p'(0)=0
c2=0
p''(0)=-4
so
$$p(x)= 1-2x^2$$

anonymous · Answer

thanks

ash2326 · Answer

ddx27 did you understand?

anonymous · Answer

yea i just had a problem with findin f''(0)

anonymous · Answer

$$f(0) = 1 $$
 $$f \prime(x) = -2\sin2x $$
$$f \prime(0) = 0$$
$$f \prime \prime(x) = -4 \cos 2x$$
$$f \prime \prime(0) = -4$$
$$p(x)=ax ^{2}+bx+c$$
$$p(0) = c$$
$$p \prime(x)=2ax+b$$
$$p \prime(0) = b$$
$$p \prime \prime(x) = 2a$$
$$p \prime \prime(0) = 2a$$
$$Equating$$
$$c = 1$$
$$a = -2$$
$$b = 0$$
$$p(x) = -2x ^{2} + 1$$