Chapter 4: Q5P (page 192)

Find the two-variable Maclaurin series for the following functions.
$\sqrt{1 + xy}$

Short Answer

Expert verified

$f (x, y) = 1 + \frac{xy}{2} - \frac{x^{2} y^{2}}{8} - \frac{x^{3} y^{3}}{16} - \frac{5 x^{4} y^{4}}{24} . . . . . . . .$

Step by step solution

Explanation of solution

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

Maclaurin Series:

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

Calculation

The following equation is provided:

$f (x, y) = \sqrt{1 + u}$

Since, $\sqrt{1 + u} = 1 + \frac{u}{2} - \frac{u^{2}}{8} + \frac{u^{3}}{16} - \frac{5 u^{4}}{128} + . . . . . . . . (1)$

And, $f (x, y) = \sqrt{1 + u} - - - - - (2)$

Substitute for $u = x y$ in equation (1),

$f (x, y) = 1 + \frac{xy}{2} - \frac{{(xy)}^{2}}{8} + \frac{{(xy)}^{3}}{16} - \frac{5 {(xy)}^{4}}{128} + . . . . . . . . = 1 + \frac{xy}{2} - \frac{x^{2} y^{2}}{8} + \frac{x^{3} y^{3}}{16} - \frac{5 x^{4} y^{4}}{24} + . . . . . . . .$

Hence,role="math" localid="1659082954746" $f (x, y) = 1 + \frac{xy}{2} - \frac{x^{2} y^{2}}{8} - \frac{x^{3} y^{3}}{16} - \frac{5 x^{4} y^{4}}{24} . . . . . . . .$

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Find the two-variable Maclaurin series for the following functions.
$\sqrt{1 + xy}$

Short Answer

Step by step solution

Explanation of solution

Calculation

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Find the two-variable Maclaurin series for the following functions.1+xy

Short Answer

Step by step solution

Explanation of solution

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Find the two-variable Maclaurin series for the following functions.
$\sqrt{1 + xy}$