Find the Taylor polynomial for Natural Log (x) at 1

This video calculates a Taylor Polynomial for Natural Log of x ln(x) at point 1 to 5 terms.
Take the Derivative of Ln(x) 5 times , up to and including the 5th Derivative.
Then take there respective values for x=1.
There is a shortcut to calculate all the derivatives to save time .
Also when going through the stages it is important not to substitute everything in all in 1 go as things can get confusing

#taylorseries
#taylorserrano
#polynomials
#polynomial
#maclaurinseries
#maclaurin
#derivatives
#triginometry
#calc1
#calculus

The graphing program used is found here

The latex used to make this video is as follows

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\begin{document}

\large

Calculate the Taylor Polynomial for 5 terms at point $1$ for the function $f(x)=\ln(x)$.\\
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The formula for a Taylor series of degree n at a point a for any function is as follows.\\
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\[
T_n(x)=f(a)+\cfrac{f^{'}(a)}{1!}(x-a)^1+\cfrac{f^2(a)}{2!}(x-a)^2+\cfrac{f^3(a)}{3!}(x-a)^3+\cfrac{f^4(a)}{4!}(x-a)^4+\cfrac{f^5(a)}{5!}(x-a)^5......\cfrac{f^n(a)}{n!}(x-a)^n
\]
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Differentiating the function $\ln(x)\ $ 5 times and the values of each derivative at $1$.\\
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\begin{align*}
& f(x)=\ln(x) & f(1)=0 \\
& f^{'}(x)=\cfrac{1}{(x)} \ =\ x^{-1} & f^{'}(1) = 1 \\
& f^2(x)=\cfrac{-1}{(x)^2} \ = \ -1(x)^{-2} & f^2(1)=-1 \\
& f^3(x)=\cfrac{2}{(x)^3}\ = \ 2(x)^{-3} & f^3(1)=2 \\
& f^4(x)=\cfrac{-6}{(x)^4}\ = \ -6(x)^{-4} & f^4(1)=-6 \\
& f^5(x)=\cfrac{24}{(x)^5} \ = \ 24(x)^{-5} & f^5(1)=24 \\
\end{align*}
\[
T_5(x)=f(1)+\cfrac{f^{'}(1)}{1!}(x-1)+\cfrac{f^2(1)}{2!}(x-1)^2+\cfrac{f^3(1)}{3!}(x-1)^3+\cfrac{f^4(1)}{4!}(x-1)^4+\cfrac{f^5(1)}{5!}(x-1)^5
\]
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So the Taylor Polynomial of degree 5 at point $1$ for $\ln(x)$ is\\
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\[
T_5(x)=0+\cfrac{1}{1}(x-1)+\cfrac{-1}{2!}(x-1)^2+\cfrac{2}{3!}(x-3)^3+\cfrac{-6}{4!}(x-1)^4+\cfrac{24}{5!}(x-1)^5
\]
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this simplifies to\\
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\[
T_5(x)=-1+x-\cfrac{(x-1)^2}{2}+\cfrac{(x-1)^3}{3}-\cfrac{(x-1)^4}{4}+\cfrac{(x-1)^5}{5}
\]

\clearpage

\end{document}

[Ensign_Cthulhu]

9:28 To keep things consistent and in pattern, the leading -1 + x can be written as (x-1)/1, then alternate with the "even" terms being negative.