Question

How do you find the nth Taylor series for \[\ln \,x\] in powers of \[x - 1\] ?

Answer 1

Hint: Here in this question we have to find the Taylor’s series. By using the formula of Taylor's series \[f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...\] Substituting the value of a to the formula hence we can determine the solution for the given question.

Complete step-by-step answer:
We have to determine the nth Taylor’s series by using the Taylor’s series expansion. Here the function is \[f(x) = \ln \,x\] . The formula for the Taylor’s series expansion is given by \[f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...\] the value of a is 1. The general form of Taylor’s series expansion is written as \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}\]
Now consider the function \[f(x) = \ln \,x\] , the value of f at x=1 is \[f(1) = 0\]
Let we find the derivatives of the function
The first derivative of the function is \[f'(x) = \dfrac{1}{x}\] , The first derivative can be written as \[f'(x) = {x^{ - 1}}\]
The second derivative of the function is \[f''(x) = - {x^{ - 2}}\] ,
The third derivative of the function is \[f'''(x) = 2{x^{ - 3}}\] ,
The fourth derivative of the function is \[f''''(x) = - 6{x^{ - 4}}\] ,
The fifth derivative of the function is \[{f^v}(x) = 24{x^{ - 5}}\] ,
Likewise the nth derivative is given by
 \[{f^{(n)}}(x) = {( - 1)^{n - 1}}(n - 1)!{x^{ - n}}\]
Suppose when the value of x = 1 we have
 \[{f^{(n)}}(x) = {( - 1)^{n - 1}}(n - 1)!\]
Therefore the Taylor’s series for the function is written in generally as
 \[ \Rightarrow \ln x = \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{n!}}{{(x - 1)}^n}} \]
On simplifying we get
 \[ \Rightarrow \ln x = \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}}{n}{{(x - 1)}^n}} \]
When the value of x = 1 the Taylor’s series expansion is given by
 \[ \Rightarrow \ln 1 = 0\]
When the value of x = 2 the Taylor’s series expansion is given by
 \[ \Rightarrow \ln 2 = \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}}{n}} \]
Hence we have determined the nth Taylor series for the given function.
So, the correct answer is “\[ \ln x = \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}}{n}{{(x - 1)}^n}} \]”.

Note: In Taylor’s series expansion we have the derivatives, so we have to apply the differentiation to the function. If the function is a logarithmic function then we use standard differentiation formula and then we determine the solution. While solving the above function we use simple arithmetic operations.