Chapter 4: Q3P (page 192)

Find the two-variable Maclaurin series for the following functions.
$\frac{In (1 + x)}{1 + y}$

Short Answer

Expert verified

$f (x, y) = x - \frac{x^{2}}{2} - xy + \frac{x^{3}}{3} + \frac{x^{2} y}{2} + {xy}^{2} . . . . . .$

Step by step solution

Explanation of solution

The equation is provided $\frac{In (1 + x)}{1 + y}$ and to find the two variable Maclaurin for the following function.

McLaren Series:

If the values of the function's consecutive derivatives at zero are known, a Maclaurin series can be used to approximate a function f(x) with input values close to zero. In many practical applications, it is equivalent to the function it represents.

Calculation

The following equation is provided:

$f (x, y) = \frac{In (1 + x)}{1 + y} - - - - - - (1)$

Since, $(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + . . . - - - - (2)$

And, ${(1 + y)}^{- 1} = 1 - y + y^{2} - y^{3} + y^{4} - y^{5} + . . . - - - - - (3)$

Substitute (2) and (3) in (1),

$f (x, y) = I n (1 + x) {(1 + y)}^{- 1} = x - \frac{x^{2}}{2} - x y + \frac{x^{3}}{3} + \frac{x^{2} y}{2} + x y^{2} . . . . . . = x - \frac{x^{2}}{2} - x y + \frac{x^{3}}{3} + \frac{x^{2} y}{2} + x y^{2} . . . . . .$

Hence, $f (x, y) = x - \frac{x^{2}}{2} - x y + \frac{x^{3}}{3} + \frac{x^{2} y}{2} + x y^{2} . . . . . . .$

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Find the two-variable Maclaurin series for the following functions.
$\frac{In (1 + x)}{1 + y}$

Short Answer

Step by step solution

Explanation of solution

Calculation

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Find the two-variable Maclaurin series for the following functions.In(1+x)1+y

Short Answer

Step by step solution

Explanation of solution

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Solid State Physics

Fields in Physics

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Study anywhere. Anytime. Across all devices.

Find the two-variable Maclaurin series for the following functions.
$\frac{In (1 + x)}{1 + y}$