Respuesta :
Taylor series is [tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]
To find the Taylor series for f(x) = ln(x) centering at 9, 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)
[tex]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 9, we must take the value of f(9), and so on.
f(9) = ln(9)
[tex]f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{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{1.2.3.4.5...........(n-1)}{9^{n} } \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}[/tex]
This applies for n ≥ 1, Expressing f(x) in summation, we have
[tex]\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}[/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!(9)^{n}(x9)^{2} }[/tex]
Taylor series is [tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]
Find out more information about taylor series here
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