Chapter 1: Q40P (page 32)

$f (x) = \frac{1}{x}, a = 1$

Short Answer

Expert verified

$\frac{1}{x} = 1 - (x - 1) + (x - 1)^{2} - (x - 1)^{3} + \dots$

Step by step solution

Given information

The Maclaurin expansion of the function $f (x) = \frac{1}{x}, a = 1$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$\frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + \dots$

Begin by stating the Maclaurin series

State the Maclaurin series.

$f (x) = \frac{1}{x} = \frac{1}{1 + (x - 1)} \frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + \dots$

Make appropriate substitution in the Maclaurin series and evaluate

Replace $x$ with the expression $(x - 1)$ in the third defined Maclaurin series.

$\frac{1}{1 + (x - 1)} = 1 - (x - 1) + (x - 1)^{2} - (x - 1)^{3} + \dots$

$\frac{1}{x} = 1 - (x - 1) + (x - 1)^{2} - (x - 1)^{3} + \dots$

Thus, the final answer is $\frac{1}{x} = 1 - (x - 1) + (x - 1)^{2} - (x - 1)^{3} + \dots$

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$f (x) = \frac{1}{x}, a = 1$

Short Answer

Step by step solution

Given information

Definition of Maclaurin series

Begin by stating the Maclaurin series

Make appropriate substitution in the Maclaurin series and evaluate

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f(x)=1x,a=1

Short Answer

Step by step solution

Given information

Definition of Maclaurin series

Begin by stating the Maclaurin series

Make appropriate substitution in the Maclaurin series and evaluate

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Measurements

Circular Motion and Gravitation

Astrophysics

Fields in Physics

Fluids

Linear Momentum

Study anywhere. Anytime. Across all devices.

$f (x) = \frac{1}{x}, a = 1$