To find the Taylor series for f(x) = ln(x) centering at 5, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have
[tex] f(x) = ln(x) \\ f^{1}(x) = \frac{1}{x} \\ f^{2}(x) = -\frac{1}{x^{2}} \\ f^{3}(x) = \frac{2}{x^{3}} \\ f^{4}(x) = \frac{-6}{x^{4}} [/tex]
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Since we need to have it centered at 5, we must take the value of f(5), and so on.
[tex] f(5) = ln(5) \\ f^{1}(5) = \frac{1}{5} \\ f^{2}(5) = \frac{-1}{5^{2}} \\ f^{3}(5) = \frac{1(2)}{5^{3}} \\ f^{4}(5) = \frac{-1(2)(3)}{5^{4}} [/tex]
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Following the pattern, we can see that for [tex]f^{n}(x)[/tex],
[tex] f^{n}(x) = (-1)^{n-1} \frac{1(2)(3)...(n-1)}{5^{n}} \\ f^{n}(x) = (-1)^{n-1} \frac{(n-1)!}{5^{n}} [/tex]
This applies for [tex] n\geq 1[/tex]. Expressing f(x) in summation, we have
[tex] \sum_{n=0}^{\infty} \frac{f^{n}(5)}{n!} (x-5)^{n} [/tex]
Combining ln2 with the rest of series, we have
[tex] f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n} [/tex]
Answer: [tex] f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n} [/tex]
