Pick any exercise in 8.8 of the calculus textbook. Solve and post your solution. If you have issues doing so, discuss them.
I chose the Problem 8.8.10:
In Exercises 7–12, find a formula for the nth 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.)
Function and Taylor Series Equality:
\(f(x) = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x - c)^n\) on \(i\)
First four derivatives:
\(f'(x) = \frac{1}{x+1}\) \(f''(x) = -\frac{1}{(x+1)^2}\) \(f^{(3)}(x) = \frac{2}{(x+1)^3}\) \(f^{(4)}(x) = -\frac{6}{(x+1)^4}\)
First term (c = 0, n = 0) \(= \frac{\ln(1 + 0)}{0!}(x - 0)^0 = \ln(1) = 0\)
Second term (c = 0, n = 1) \(= \frac{\frac{1}{0+1}}{1!}(x - 0)^1 = x\)
Third term (c = 0, n = 2) \(= \frac{-\frac{1}{(0+1)^2}}{2!}(x - 0)^2 = -\frac{1}{2}x^2\)
Fourth term (c = 0, n = 3) \(= \frac{\frac{1}{(0+1)^3}}{3!}(x - 0)^3 = \frac{1}{6}x^3\)
Fifth term (c = 0, n = 4) \(= \frac{-\frac{1}{(0+1)^4}}{4!}(x - 0)^4 = -\frac{1}{24}x^4\)
Using the first 5 terms, the Taylor series is: \(0 + x - \frac{1}{2}x^2 + \frac{1}{6}x^3 - \frac{1}{24}x^4 + ...\)
Or in summation form:
\(\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}(-1)^{n+1}\)
Though this doesn’t work for the first term (?).