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Question: How do you find the Taylor series for \(\ln (x)\) about the point \(x = 1\)?...

How do you find the Taylor series for ln⁡(x)\ln (x) about the point x=1x = 1?

Explanation

Solution

Here we need to apply the formula of Taylor series about the point aa which is:
f(x)=∑n=0∞fn(a)n!(x−a)nf(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}(a)}}{{n!}}{{(x - a)}^n}} and here fn(a){f^n}(a) is the derivative of the function at aa which is done nn times. So we just need to apply this formula and get the series for the function f(x)=ln⁡(x)f(x) = \ln (x)

Complete step-by-step answer:
Here we are given to find the Taylor series or expansion of the function ln⁡(x)\ln (x) about the pointx=1x = 1
So we must know that the Taylor series is the sum of the series till infinity which is expressed in the form of the function’s derivative at a single point. Here that point is a=1a = 1
Now we know that formula to find the Taylor series for the function f(x)f(x) is given by:
Value off(x)=∑n=0∞fn(a)n!(x−a)nf(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}(a)}}{{n!}}{{(x - a)}^n}} and here fn(a){f^n}(a) is the derivative of the function at aa which is done nn times. So we just need to apply this formula and get the series for the function f(x)=ln⁡(x)f(x) = \ln (x)
f′(x)=ddx(ln⁡(x))=1xf'(x) = \dfrac{d}{{dx}}(\ln (x)) = \dfrac{1}{x}
f′′(x)=ddx(1x)f''(x) = \dfrac{d}{{dx}}(\dfrac{1}{x})
We know that ddx(1xn)=−nxn+1\dfrac{d}{{dx}}\left( {\dfrac{1}{{{x^n}}}} \right) = - \dfrac{n}{{{x^{n + 1}}}}
So we get f′′(x)=ddx(1x)=−1x2f''(x) = \dfrac{d}{{dx}}(\dfrac{1}{x}) = - \dfrac{1}{{{x^2}}}
f′′′(x)=ddx(−1x2)=2x3f'''(x) = \dfrac{d}{{dx}}( - \dfrac{1}{{{x^2}}}) = \dfrac{2}{{{x^3}}}
f4(x)=ddx(2x3)=−3.2x4{f^4}(x) = \dfrac{d}{{dx}}(\dfrac{2}{{{x^3}}}) = - \dfrac{{3.2}}{{{x^4}}}
Now we need to find these all values at x=a=1x = a = 1 as we know that in the formula we need to find the values of the derivatives at the point aa which is 11 over here.
So we get:
f0(1)=f(1)=ln⁡(1)=0{f^0}(1) = f(1) = \ln (1) = 0
f′(1)=11=1f'(1) = \dfrac{1}{1} = 1
f′′(1)=−112=−1f''(1) = - \dfrac{1}{{{1^2}}} = - 1
f′′′(1)=213=2f'''(1) = \dfrac{2}{{{1^3}}} = 2
f4(x)=−3.214=−(3)(2){f^4}(x) = - \dfrac{{3.2}}{{{1^4}}} = - (3)(2)
Now we know the pattern which is given by:
f(x)=∑n=0∞fn(a)n!(x−a)nf(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}(a)}}{{n!}}{{(x - a)}^n}}
f(x)=f0(1)0!(x−1)0+f1(1)1!(x−1)1+f2(1)2!(x−1)2+f3(1)3!(x−1)3+.............f(x) = \dfrac{{{f^0}(1)}}{{0!}}{(x - 1)^0} + \dfrac{{{f^1}(1)}}{{1!}}{(x - 1)^1} + \dfrac{{{f^2}(1)}}{{2!}}{(x - 1)^2} + \dfrac{{{f^3}(1)}}{{3!}}{(x - 1)^3} + .............
Now we can substitute the values from above in this above equation to get the series:
So we get:
⇒\Rightarrow f(x)=0+1(x−1)−1(1)2!(x−1)2+23!(x−1)3+.............f(x) = 0 + 1(x - 1) - \dfrac{{1(1)}}{{2!}}{(x - 1)^2} + \dfrac{2}{{3!}}{(x - 1)^3} + .............
So we get the Taylor series as:
⇒\Rightarrow ln⁡(x)=0+1(x−1)−1(1)2!(x−1)2+23!(x−1)3−............\ln (x) = 0 + 1(x - 1) - \dfrac{{1(1)}}{{2!}}{(x - 1)^2} + \dfrac{2}{{3!}}{(x - 1)^3} - ............
We can write this in the general form as:
⇒\Rightarrow f(x)=ln⁡(x)=∑n=0∞(−1)n−1(x−1)nnf(x) = \ln (x) = \sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}\dfrac{{{{(x - 1)}^n}}}{n}}

Note: Here in these types of problems the student must remember the formula of the Taylor series and must try to find the derivatives unless we are clear with the next terms upon seeing the previous terms. So we need to find at least 44 terms in order to get its series.