Find the linearization, say L(x)=ax+b, of fx=sin^(2)x at the point x=pi/4

Question

Find the linearization, say L(x)=ax+b, of fx=sin^(2)x  at the point x=pi/4

anonymous · Answer

Find L(x)

anonymous · Answer

Hi Lifesaver,

You have to take a Taylor series expansion about pi/4 for sin^2(x) and truncate after the first two terms.  So, 
$$f(x)=\sum_{k=0}^{\infty}[f^k(x_0)/k!](x-x_0)^k$$
Set $$x_0=\pi/4$$
and truncate after k=1 (i.e. sum up only the first two terms, k=0 and k=1).

In this instance, 
$$\sin^2(x)=\sin^2(x_0)+2\sin(x)\cos(x)(x-x_0)+...$$
so that,
$$\sin^2(x)=\sin^2(\pi/4)+2\sin(\pi/4)\cos(\pi/4)(x-\pi/4)+...$$
After truncating the above (i.e. dropping everything after the term 2sin(x)cos(x)(x-x_0), and finding values for sin(pi/4), cos(pi/4), etc., you have what you're looking for,
$$L(x)=x+(1/2-\pi/4)$$ as a linear approximation to sin^2(x) about the point pi/4.

I hope I've interpreted your question correctly.