What is the difference between a Taylor series and a Maclaurin series?
[answer]
A Taylor series of \(f(x)\) is
\[f(x-a) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \dots,\] while the Maclaurin series is the special case for \(a= 0\).
Write the first three nonzero terms of the Maclaurin series representation of \(e^x \), \(\sin x \), \(\cos x \), and \(\tan x \).
[answer]
\begin{align*}
e^x&\simeq 1 + x + \frac{x^2}{2!}\\
\sin x &\simeq x - \frac{x^3}{3!} + \frac{x^5}{5!}\\
\cos x &\simeq 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\\
\tan x & \simeq x + \frac{x^3}{3} \tag*{Sorry, I will go no higher.}
\end{align*}
Show that \( \cos x = (e^{jx} + e^{-jx})/2\), \( \sin x = (e^{jx} - e^{-jx})/2j\).
Write the first three terms of the Maclaurin series representation of \(\ln (1+x)\).
[answer]
\[\ln (1+x) \simeq x - \frac{x^2}{2} + \frac{x^3}{3} \text{ near } x = 0\]
Write the first three terms of the Maclaurin series representation of \(\ln (1-x)\)
[answer]
\[\ln (1-x) \simeq -x - \frac{x^2}{2} - \frac{x^3}{3} \text{ near } x = 0\]
Write the first three terms of the Taylor series representation of \(\ln (-x)\) about \(x= -1\). Verify your answer for by checking that it can be obtained from the previous expansion for \(\ln (1-x)\) by shifting the latter one unit in the negative \(x\) direction.
[answer]
\[\ln (-x) \simeq -(x+1) - \frac{(x+1)^2}{2} - \frac{(x+1)^3}{3} \text{ near } x = -1\] This matches the previous result when the previous result is shifted one unit in the \(-x\) direction.