Q 14. Maclaurin and Taylor polynomials... [FREE SOLUTION]

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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{蟺}{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

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

So, at $x = 1$

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

Again

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

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} 路 (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}

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91影视

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

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91影视

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

Theoretical and Mathematical Physics

Probability and Statistics

Statistics

Geometry

Calculus

Mechanics 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.
$\sqrt{x}, x_{0} = 1$