Q - 35E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: The Taylor series for f(x) =  ln (x) about x₂=0 given in equation (13) can also be obtained as follows:

(a) Starting with the expansion 1/ (1-s) =$\sum _{n=0}^{\infty }s\text{'}\text{'}$  and observing that

 '

obtain the Taylor series for 1/x  about x₀ = 1 .

(b) Since   use the result of part (a) and termwise integration to obtain the Taylor series for f (x) =lnx about x₀ = 1 .

Step-by-Step Solution

Answer

a) The Taylor series $\frac{1}{x}$ for around ^x₀=1  is given by:

b) The Taylor serious about ^x₀=1  for ^{f(X) = lnx}   is given by,

Step 1: Taylor series

For a function ^{$f\left(x\right)$} the Taylor series expansion about a point ^x₀  is given by,$f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).(x-x_0)+f\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+...$

Step 2: Solution for part (a)

Rewrite  $\frac{1}{x}$ by adding and subtracting  1 from the denominator, as follows:

$\frac{1}{x}=\frac{1}{1+(x-1)}$and$\frac{1}{1-s}= \sum _{n=0}^{\infty }s\text{'}\text{'}$

It follows,

$$\begin{gathered} \frac{1}{x}=\frac{1}{1+(x-1)} \\ =\sum _{n=1}^{\infty }{(1-x)}^n \\ =\sum _{n=1}^{\infty }{{(-1)}^n(1-x)}^n \end{gathered}$$

Therefore, the Taylor series for $\frac{1}{x}$ around x₀=1 is given by: .$\frac{1}{x}$ $=\sum _{n=1}^{\infty }{(-1)}^n(1-x)$

Step 3: Solution for part (b)

Substitute $\frac{1}{x}=\sum _{n=1}^{\infty }{(-1)}^n{(1-x)}^n$ in the equation $\ln x ={\int }_1^x\frac{1}{t}dt,$ we get,

$\ln x ={\int }_1^x\sum _{n=1}^{\infty }{(-1)}^n{(t-1)}^{n}dt$

Interchange the integral and the sum,

$\ln x =\sum _{n=1}^{\infty }{\int }_1^x{(-1)}^n{(t-1)}^{n}dt$

$\ln x =\sum _{n=1}^{\infty }{(-1)}^n{\int }_1^x{(t-1)}^{n}dt$

Take integration,

$\ln x =\sum _{n=1}^{\infty }{(-1)}^n{{\overline{)\frac{{(t-1)}^{n+1}}{n+1}}}^x}_1$

 

Compute the boundaries,

$\ln x =\sum _{n=1}^{\infty }{(-1)}^n\frac{{(t-1)}^{n+1}}{n+1}$

Let,

 k = n + 1

 n = k - 1

 So,

 $\ln x =\sum _{\mathrm{k}=1}^{\infty }{(-1)}^{\mathrm{k}-1}\frac{{(\mathrm{x}-1)}^k}{k}$

 

Now substitute k for n in the above equation, it is only a dummy index the name is not important, we need to respect the range from 1 to$\infty$ .