The Taylor series expansion for sin about z_0 = (pi/2)

[laura_a]

Homework Statement

Expand cos z into a Taylor series about the point z_0 = (pi)/2

With the aid of the identity

cos(z) = -sin(z - pi/2)

Homework Equations

Taylor series expansion for sin

sinu = \sum^{infty}_{n=0} (-1)^n * \frac{u^{2n+1}}{(2n+1)!}

and the identity as given above

The Attempt at a Solution

I've subbed in -sin(z- pi/2) into the identity my first prob was how to deal with the negative in front of the sin so I've done

u = z-pi/2

-sin(z-(pi/2)) = -(z-(pi/2)) + 1/3! * (z-(pi/2))^3 - 1/5! * (z-(pi/2))^5

So if that is even correct (because I'm not sure about where to put the negative signs... then what does it mean when it says "about the point" z_0 = (pi/2)

How do I sub that into my answer?

Any suggestions will be much appreciated

Thanks

 

[HallsofIvy]

The Taylor's series of a function 'about' z_0= \pi/2 is the series
\sum_{n=0}^\infty \frac{d^n f}{d x^n}(\pi/2)(z- \pi/2)^n.
That is, you have powers of z- \pi/2 as well as evaluating the derivatives at \pi/2. One method of determining the derivative is to use that definition directly: all derivatives of cos(x) are cos(x), sin(x), -cos(x), -sin(x) and those have values of 0, 1, and -1 at \pi/2.

However, you are completely correct that cos(z)= -sin(z-\pi/2). That means all you need to do is write the Taylor's series for sin(z), which is
z- (1/3!)z^3+ \cdot\cdot\cdot+ ((-1)^n/(2n+1)!)z^{2n+1}+\cdot\cdot\dot
Multiply by -1 to get -sin(z):
-z+ (1/3!)z^3+ \cdot\cdot\cdot+ ((-1)^{n+1}/(2n+1)!)z^{2n+1}+\cdot\cdot\dot
and finally replace z by z-\pi/2:
-(z-\pi/2)+ (1/3!)(z-\pi/2)^3+\cdot\cdot\cdot+ ((-1)^{n+1}/(2n+1)!)(z- \pi/2)^{2n+1}+\cdot\cdot\cdot