Chapter 8: Q 14. (page 704)

Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.
$\sqrt{x}, x_{0} = 1$

Short Answer

Expert verified

$P_{3} (x) = 1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3}$

Step by step solution

Given information

$x_{0} = 1$

Calculation

Consider the function $f (x) = \sqrt{x}$

Since for any function $f$ with a derivative of order 3 at $x = 0$ the third-order Taylor polynomial at $x_{0} = 1$ is given by

P_{3} (x) = f (\frac{π}{2}) + f^{'} (1) (x - 1) + \frac{f^{''} (1)}{2!} (x - 1)^{2} + \frac{f^{'''} (1)}{3!} (x - 1)^{3}

Therefore, first find the value of the function along with $f^{'} (x), f^{''} (x)$ and $f^{''} (x)$ at $x_{0} = 1$

Calculation

Thus, the value of the function at $x = 1$ is

$f (1) = \sqrt{1} = 1$

The derivatives of the function $f (x) = \sqrt{x}$ are

$f^{'} (x) = \frac{d}{d x} [\sqrt{x}] = \frac{1}{2 \sqrt{x}}$

So, at $x = 1$

$f^{'} (1) = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$

Also

$f^{''} (x) = \frac{d}{d x} (\frac{1}{2 \sqrt{x}}) = \frac{1}{2} \frac{d}{d x} (x)^{- \frac{1}{2}} = \frac{1}{2} \cdot - \frac{1}{2} (x)^{- \frac{3}{2}} = - \frac{1}{4} x^{- \frac{3}{2}}$

So, at $x = 1$

$f^{''} (1) = - \frac{1}{4} (1)^{- \frac{3}{2}} = - \frac{1}{4}$

Again

$f^{'''} (x) = \frac{d}{d x} [- \frac{1}{4} x^{- \frac{3}{2}}] = - \frac{1}{4} \frac{d}{d x} (x)^{- \frac{3}{2}} = - \frac{1}{4} \cdot - \frac{3}{2} x^{\frac{5}{2}} = \frac{3}{8} x^{- \frac{5}{2}}$

So, at $x = 1$

$f^{'''} (1) = \frac{3}{8} (1)^{- \frac{5}{2}} = \frac{3}{8}$

Therefore, the third-order Taylor polynomial for the function $f (x) = \sqrt{x}$ at $x = 1$ is

P_{3} (x) = 1 + \frac{1}{2} \cdot (x - 1) + \frac{(- \frac{1}{4})}{2!} (x - 1)^{2} + \frac{\frac{3}{8}}{3!} (x - 1)^{3}

Implies that

P_{3} (x) = 1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3}

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.
$\sqrt{x}, x_{0} = 1$

Short Answer

Step by step solution

Given information

Calculation

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Probability and Statistics

Theoretical and Mathematical Physics

Geometry

Decision Maths

Pure Maths

Study anywhere. Anytime. Across all devices.

Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.x,x0=1

Short Answer

Step by step solution

Given information

Calculation

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Probability and Statistics

Theoretical and Mathematical Physics

Geometry

Decision Maths

Pure Maths

Study anywhere. Anytime. Across all devices.

Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.
$\sqrt{x}, x_{0} = 1$