Chapter 4: Q4P (page 192)

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

Short Answer

Expert verified

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

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 $f (x)$ 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)

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

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

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

Short Answer

Step by step solution

Explanation of solution

Calculation

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