Chapter 1: Q44P (page 32)

$f (x) = \sqrt x, a = 25 .$

Short Answer

Expert verified

The first few terms of Taylor’s series expansion are:

$\sqrt{x} = 5 + (\frac{x - 25}{10}) - \frac{(x - 25)^{2}}{1000} + \frac{(x - 25)^{3}}{50000} + \dots$

Step by step solution

Given Information

The given function.

$f (x) = \sqrt x$

Definition of Taylor’s series

The Taylor’s series is a power series that yields the expansion of a function in the vicinity of a point if the function is continuous, all of its derivatives exist, and the series converges.

Find the Taylor’s series for x

Rewrite $\sqrt{x}$ ,

$\sqrt{x} = \sqrt{25 + (x - 25)} \sqrt{x} = 5 \sqrt{1 + \frac{(x - 25)}{25}}$

Use the Maclaurin series of $\sqrt{1 + x}$ and replace $x$ by $\frac{x - 25}{25}$ ,

localid="1657421695758" $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{8} + \frac{x^{3}}{16} - \frac{5 x^{4}}{128} + \dots \sqrt{1 + \frac{(x - 25)}{25}} = 1 + \frac{(\frac{x - 25}{25})}{2} - \frac{{(\frac{x - 25}{25})}^{2}}{8} + \frac{{(\frac{x - 25}{25})}^{3}}{16} + \dots \sqrt{x} = 5 \times \sqrt{1 + \frac{(x - 25)}{25}}$

$= 5 + (\frac{x - 25}{10}) - \frac{(x - 25)^{2}}{1000} + \frac{(x - 25)^{3}}{50000} + \dots$

Hence, first few terms of Taylor’s series expansion about $a = 25$ are:

$\sqrt{x} = 5 + (\frac{x - 25}{10}) - \frac{(x - 25)^{2}}{1000} + \frac{(x - 25)^{3}}{50000} + \dots$

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$f (x) = \sqrt x, a = 25 .$

Short Answer

Step by step solution

Given Information

Definition of Taylor’s series

Find the Taylor’s series for x

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f(x)=√x,a=25.

Short Answer

Step by step solution

Given Information

Definition of Taylor’s series

Find the Taylor’s series for x

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Measurements

Waves Physics

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Famous Physicists

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Study anywhere. Anytime. Across all devices.

$f (x) = \sqrt x, a = 25 .$