Find a formula for the \(n^{th}\) term of the Taylor series of \(f(x)\) centered at \(c\), by finding the coefficients of the first few powers of \(x\) and looking for a pattern. \(f(x)=cosx\) at \(c=\frac{\pi}{2}\)
Using the formula for a Taylor Series centered at c, I can generate the first several terms of the expansion. I will also use the fact that \(cos(\frac{\pi}{2})=0\) and \(sin(\frac{\pi}{2})=1\):
At \(c=0\):
\(\frac{cos(\frac{\pi}{2})}{0!}(x-\frac{\pi}{2})^0 = 0\)
At \(c=1\):
\(\frac{-sin(\frac{\pi}{2})}{1!}(x-\frac{\pi}{2})^1\) = \(\frac{-1}{1!}(x-\frac{\pi}{2})^1\)
At \(c=2\):
\(\frac{-cos(\frac{\pi}{2})}{2!}(x-\frac{\pi}{2})^2 = 0\)
At \(c=3\):
\(\frac{sin(\frac{\pi}{2})}{3!}(x-\frac{\pi}{2})^3\) = \(\frac{1}{3!}(x-\frac{\pi}{2})^3\)
At \(c=4\):
\(\frac{cos(\frac{\pi}{2})}{4!}(x-\frac{\pi}{2})^4 = 0\)
At \(c=5\):
\(\frac{-sin(\frac{\pi}{2})}{5!}(x-\frac{\pi}{2})^5\) = \(\frac{-1}{5!}(x-\frac{\pi}{2})^5\)
The pattern at this point is clear. The expansion can be simplified and summarized:
\[0 -(x-\frac{\pi}{2})+0x^2+\frac{1}{6}(x-\frac{\pi}{2})^3+0x^4-\frac{1}{120}(x-\frac{\pi}{2})^5...\]
\[-(x-\frac{\pi}{2})+\frac{1}{6}(x-\frac{\pi}{2})^3-\frac{1}{120}(x-\frac{\pi}{2})^5...\] Finally, these expansion can be summarized as:
\[\sum_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\frac{\pi}{2})^{2n+1}}{(2n+1)!}\]