Suggested languages for you:

Americas

Europe

Q-30E

Expert-verified
Found in: Page 435

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

Question: In Problems 29–34, determine the Taylor series about the point X0 for the given functions and values of X0.30. $f\left(x\right)={x}^{-4}{x}_{0}=1$

The required expression is .$\sum _{n-0}^{\infty }{\left(-1\right)}^{n}\left(x-1\right)$

See the step by step solution

Step 1: Taylor series

For a function the Taylor series expansion about a point is given by,$f\left(x-{x}_{0}\right)=f\left({x}_{0}\right)+f\text{'}\left({x}_{0\right)}.\left(x-{x}_{0}\right)+f\text{'}\text{'}\left({x}_{0}\right).\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+f\text{'}\text{'}\text{'}\left({x}_{0}\right)\frac{{\left(x-{x}_{0}\right)}^{3}}{3!}+....$

Step 2: Derivatives of function at x0.

We have to calculate the Taylor series expansion for, f (x) = x-1 at x0=1.

Calculating the derivatives of function at x0 .

$f\left(x\right)={x}^{-1}thenf\left({x}_{0}\right)=1$

$f\text{'}\left(x\right)={x}^{-2}thenf\text{'}\left({x}_{0}\right)=-1$

$f\text{'}\text{'}\left(x\right)=2{x}^{-3}thenf\text{'}\text{'}\left({x}_{0}\right)=2$

$f\text{'}\text{'}\text{'}\left(x\right)=6{x}^{-4}thenf\text{'}\text{'}\text{'}\left({x}_{0}\right)=-6$

$f\text{'}\text{'}\text{'}\text{'}\left(x\right)=24{x}^{-5}thenf\text{'}\text{'}\text{'}\text{'}\left({x}_{0}\right)=24$

Step 3: Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at x0=1, then,

${\left(x-1\right)}^{-1}=1-1\left(x-1\right)+2.\frac{{\left(x-1\right)}^{2}}{2!}-6\frac{{\left(x-1\right)}^{3}}{3!}+24.\frac{{\left(x-1\right)}^{4}}{4!}+....\phantom{\rule{0ex}{0ex}}=1-\left(x-1\right)+{\left(x-1\right)}^{2}-{\left(x-1\right)}^{3}+{\left(x-1\right)}^{4}+....\phantom{\rule{0ex}{0ex}}=\sum _{n-0}^{\infty }{\left(-1\right)}^{n}.{\left(x-1\right)}^{n}$

Hence, the required expression is $=\sum _{n-0}^{\infty }{\left(-1\right)}^{n}.{\left(x-1\right)}^{n}$