Mathematics: Find the taylor series for f(x centered at the given value of a. [assume that f has a power series expansion. do not show that rn(x ? 0. ] f(x = ln x, a = 5

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

Ответ:

bremarie102013

17.01.2022

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

$f(x) = ln(x) \\ f^{1}(x) = \frac{1}{x} \\ f^{2}(x) = -\frac{1}{x^{2}} \\ f^{3}(x) = \frac{2}{x^{3}} \\ f^{4}(x) = \frac{-6}{x^{4}}$
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Since we need to have it centered at 5, we must take the value of f(5), and so on.

$f(5) = ln(5) \\ f^{1}(5) = \frac{1}{5} \\ f^{2}(5) = \frac{-1}{5^{2}} \\ f^{3}(5) = \frac{1(2)}{5^{3}} \\ f^{4}(5) = \frac{-1(2)(3)}{5^{4}}$
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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

$f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}$

$f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}$

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Ответ:

darriennichole

13.05.2021

Mathematics

(1) Animals must expend energy to get food, they must actively obtain food from the environment.

(2) Clams use their valves to get food.

(3) Cats, tigers and lions use their claws in getting food.

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Find the taylor series for f(x centered at the given value of
a. [assume that f has a power series expansion. do not show that rn(x ? 0. ] f(x = ln x, a = 5

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