Chapter 4: Q6P (page 192)

Find the two-variable Maclaurin series for the following functions.
$e^{x + y}$

Short Answer

Expert verified

The answer is: $f (x, y) = 1 + x + y + \frac{x^{2} + 2 xy + y^{2}}{2} + \frac{x^{3} + y^{3} 3 x^{2} y + 3 {xy}^{2}}{6} + . . . . .$

Step by step solution

Explanation of solution

The equation is provided $e^{x + y}$ 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.

$f (x, y) = e^{x + y}$

Since, $e^{u} = 1 + u + \frac{u^{2}}{2} + \frac{u^{3}}{6} + \frac{u^{4}}{24} + . . . . . . . . (1)$

And, $f (x, y) = e^{x + y} - - - - - - - (2)$

Substitute for $u = x + y$ in (1)

$f (x, y) = 1 + x + y + \frac{{(x + y)}^{2}}{2} + \frac{{(x + y)}^{3}}{6} + \frac{{(x + y)}^{4}}{24} + . . . . . = 1 + x + y + \frac{x^{2} + 2 xy + y^{2}}{2} + \frac{x^{3} + y^{3} 3 x^{2} y + 3 {xy}^{2}}{6} + . . . . .$

Hence, $f (x, y) = 1 + x + y + \frac{x^{2} + 2 xy + y^{2}}{2} + \frac{x^{3} + y^{3} 3 x^{2} y + 3 {xy}^{2}}{6} + . . . . .$

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Find the two-variable Maclaurin series for the following functions.
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Short Answer

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Explanation of solution

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