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nirvana33 [79]
3 years ago
9

Find the taylor series for f(x centered at the given value of

Mathematics
1 answer:
Lady_Fox [76]3 years ago
3 0
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

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}}
.
.
.
Since we need to have it centered at 5, we must take the value of f(5), and so on. 

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}}
.
.
.
Following the pattern, we can see that for f^{n}(x),

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}}

This applies for n\geq 1. Expressing f(x) in summation, we have 

\sum_{n=0}^{\infty} \frac{f^{n}(5)}{n!} (x-5)^{n}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}
<span>
Answer: </span>f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}

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