Chapter 8: Q. 40 (page 692)

In Exercises in Section 8.2, you were asked to find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of $x_{0}$ . In Exercises give Lagrange’s form for the remainder $R_{4} (x) .$
$\sqrt{x}, 1$

Short Answer

Expert verified

The Value is $R_{4} (x) = \frac{7}{256} c^{\frac{9}{2}} (x - 1)^{5}$

Step by step solution

Given information

The function is $\sqrt{x}, 1$

Simplification

The function's derivatives are $f (x) = \sqrt{x}$ is

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

That is,

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

similarly,

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

Similarly,

$f^{'''} (x) = \frac{d}{dx} [\frac{3}{8} x^{\frac{5}{2}}] = \frac{3}{8} - \frac{5}{2} x^{\frac{7}{2}} = - \frac{15}{16} x^{\frac{7}{2}}$

That is,

$f^{(4)} (x) = - \frac{15}{16} x^{\frac{7}{2}}$

Lastly,

$f^{(3)} (x) = \frac{d}{dx} [- \frac{15}{16} x^{\frac{7}{2}}] = - \frac{15}{16} \frac{d}{dx} [x^{\frac{7}{2}}] = - \frac{15}{16} - \frac{7}{2} x^{- \frac{9}{2}} = \frac{105}{32} x^{\frac{9}{2}}$

Finding Lagrange’s fourth form

If f is a function that may be differentiated $n + 1$ times in some open interval containing the point $x_{0}$ and $R_{n} (x)$ is the nth remainder for $f$ at $x = x_{0}$ , then $R_{n} (x)$ is the $n^{th}$ remainder for f at $x = x_{0}$ . Then at least one c exists between $x_{0}$ and x such that $R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} {(x - x_{0})}^{n + 1}$

For , $f^{(5)} (x) = \frac{105}{32} x^{\frac{9}{2}}$ and $x_{0} = 1$ is role="math" localid="1650350756752" $R_{4} (x) = \frac{\frac{105}{32} c^{\frac{9}{2}}}{5!} (x - 1)^{5}$

Finally, $R_{4} (x) = \frac{7}{256} c^{\frac{9}{2}} (x - 1)^{5}$

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In Exercises in Section 8.2, you were asked to find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of $x_{0}$ . In Exercises give Lagrange’s form for the remainder $R_{4} (x) .$
$\sqrt{x}, 1$

Short Answer

Step by step solution

Given information

Simplification

Finding Lagrange’s fourth form

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In Exercises in Section 8.2, you were asked to find the fourth Taylor polynomial P4(x)for the specified function and the given value of x0. In Exercises give Lagrange’s form for the remainder R4(x).x,1

Short Answer

Step by step solution

Given information

Simplification

Finding Lagrange’s fourth form

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Logic and Functions

Calculus

Probability and Statistics

Mechanics Maths

Pure Maths

Study anywhere. Anytime. Across all devices.

In Exercises in Section 8.2, you were asked to find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of $x_{0}$ . In Exercises give Lagrange’s form for the remainder $R_{4} (x) .$
$\sqrt{x}, 1$