Question

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

$\sqrt{\mathbf{1}\mathbf{+}\mathbf{xy}}$

Short answer

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

Step-by-step solution

Explanation of solution

The equation is provided $\sqrt{1+xy}$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,$\sqrt{1+u}=1+\frac{u}{2}-\frac{u^2}{8}+\frac{u^3}{16}-\frac{5u^4}{128}+........\left(1\right)$

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

Substitute for $u=xy$in equation (1),

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

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