Q. 14 Why do we need to put linear num... [FREE SOLUTION]

Chapter 5: Q. 14 (page 441)

Why do we need to put linear numerators of the form $B_{i} x + C_{i}$ in every term of a partial-fraction decomposition which involves irreducible quadratics? Think about the example $\frac{p (x)}{q (x)} = \frac{1}{{(x^{2} + 1)}^{2}}$ and figure out what goes wrong if we attempt to make a decomposition of the form $\frac{C_{1}}{x^{2} + 1} + \frac{C_{2}}{{(x^{2} + 1)}^{2}}$ .

Short Answer

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The $q (x)$ $= \frac{1}{x^{2} + 1}$ , whichis a reducible quadratic then $q (x)$ is a product of $l_{1} (x)$ and $l_{2} (x)$ of two linear functions. So, $l_{1} (x) = x - 1, l_{2} (x) = x - 2$ .

Step by step solution

Step 1. Given information

$\frac{p (x)}{q (x)} = \frac{1}{{(x^{2} + 1)}^{2}}$ .

Step 2. Let us see some concepts about qx.

If $q (x)$ is an irreducible quadratic then $\frac{p (x)}{q (x)}$ is its own partial fractions decomposition, there is nothing further we can decompose.

If $q (x)$ is a reducible quadratic then $q (x)$ is a product of $l_{1} (x) . l_{2} (x)$ of two linear functions and the obtain a partial function decomposition of the form $\frac{A_{1}}{l_{1} (x)} + \frac{A_{2}}{l_{2} (x)}$ .

The given function is, $q (x) = \frac{1}{x^{2} + 1}$ .

So, $l_{1} (x) = x - 1, l_{2} (x) = x - 2$ .

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Why do we need to put linear numerators of the form Bix+Ciin every term of a partial-fraction decomposition which involves irreducible quadratics? Think about the example pxqx=1x2+12and figure out what goes wrong if we attempt to make a decomposition of the formC1x2+1+C2x2+12.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Let us see some concepts about qx.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Geometry

Mechanics Maths

Calculus

Applied Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.