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

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

Explanation

Solution

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)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...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⁡ xf(x) = \ln \,x . The formula for the Taylor’s series expansion is given by f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...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)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}
Now consider the function f(x)=ln⁡ xf(x) = \ln \,x , the value of f at x=1 is f(1)=0f(1) = 0
Let we find the derivatives of the function
The first derivative of the function is f′(x)=1xf'(x) = \dfrac{1}{x} , The first derivative can be written as f′(x)=x−1f'(x) = {x^{ - 1}}
The second derivative of the function is f′′(x)=−x−2f''(x) = - {x^{ - 2}} ,
The third derivative of the function is f′′′(x)=2x−3f'''(x) = 2{x^{ - 3}} ,
The fourth derivative of the function is f′′′′(x)=−6x−4f''''(x) = - 6{x^{ - 4}} ,
The fifth derivative of the function is fv(x)=24x−5{f^v}(x) = 24{x^{ - 5}} ,
Likewise the nth derivative is given by
f(n)(x)=(−1)n−1(n−1)!x−n{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)!{f^{(n)}}(x) = {( - 1)^{n - 1}}(n - 1)!
Therefore the Taylor’s series for the function is written in generally as
⇒ln⁡x=∑n=1∞(−1)n−1(n−1)!n!(x−1)n\Rightarrow \ln x = \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{n!}}{{(x - 1)}^n}}
On simplifying we get
⇒ln⁡x=∑n=1∞(−1)n−1n(x−1)n\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
⇒ln⁡1=0\Rightarrow \ln 1 = 0
When the value of x = 2 the Taylor’s series expansion is given by
⇒ln⁡2=∑n=1∞(−1)n−1n\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=∑n=1∞(−1)n−1n(x−1)n\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.