Question:
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. (The formulas for several of these are found in Key Idea 32; show work verifying these formula.)
\(f\left(x\right) = cos x\quad ;\quad c = \pi/2\)
Answer:
So, let’s get the first few terms centered at \(c = \pi\)
\[f\left(x\right) = cos(\frac{\pi}{2}) = 0\] \[{f}^{\prime}\left(x\right) = -sin(\frac{\pi}{2}) = -1\]
\[{f}^{\prime\prime}\left(x\right) = -cos(\frac{\pi}{2}) = 0\]
\[{f}^{\prime\prime\prime}\left(x\right) = sin(\frac{\pi}{2}) = 1\]
\[{f}^{\prime\prime\prime\prime}\left(x\right) = cos(\frac{\pi}{2}) = 0\]
Well, it’s a pretty simple result of just ones and zeros with varying positive/negative multipliers. Also, since we arrived back at our original equation, we do not need to go any further. Therefore, our series looks like the following:
\[\frac{0}{0!} * {\left(x - \frac{\pi}{2} \right)}^{0} - \frac{1}{1!} * {\left(x - \frac{\pi}{2} \right)}^{1} + \frac{0}{2!} * {\left(x - \frac{\pi}{2} \right)}^{2} + \frac{1}{3!} * {\left(x - \frac{\pi}{2} \right)}^{3} + \frac{0}{4!} * {\left(x - \frac{\pi}{2} \right)}^{4}\]
\[= 0 - \left(x - \frac{\pi}{2} \right) + 0 + \frac{{\left(x - \frac{\pi}{2} \right)}^{3}}{3!} + 0\]
So, we only have two terms that alternate negative and positive signs; thus \({\left(-1\right)}^{n+1}\), can replicate this piece of the formula. Now we can focus on the terms themselves:
\[{\left(x - \frac{\pi}{2} \right)}^{1} ; \frac{{\left(x - \frac{\pi}{2} \right)}^{3}}{3!}\]
With our values being raised to the power of 1 and 3, we can use the term \(2n + 1\) to replicate this piece. Also, the first term is divided by \(1!\) while the second term is divided by \(3!\) so we can use the same formula for the denominator. Putting this all together gives us the following:
\[\sum _{ n=0 }^{ \infty }{ {\left( -1 \right)}^{ n+1 } * \frac{{\left( x - \frac{\pi}{2} \right)}^{\left( 2n+1\right)}}{\left( 2n+1\right)!} } \]