Question

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

Short answer

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

Step-by-step solution

Explanation of solution

The equation is provided $e^{xy}$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 $\mathbf{f}\left(x\right)$input values close to zero. In many practical applications, it is equivalent to the function it represents.

Calculation

The following equation is provided.

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

Since,

${\mathrm{e}}^{\mathrm{u}}=1+\mathrm{u}+\frac{{\mathrm{u}}^2}{2}+\frac{{\mathrm{u}}^3}{6}+\frac{{\mathrm{u}}^4}{24}........-----\left(1\right)$

And,$f(x,y)=e^{xy}-------\left(2\right)$

Substitute for $u=xy$in (1)

$$\begin{gathered} f\left(x,y\right)=e^y=1+y++\frac{{\left(xy\right)}^2}{2}\frac{{\left(xy\right)}^3}{6}+\frac{{\left(xy\right)}^4}{24} \\ =1+xy+\frac{x^2{y}^2}{2}+\frac{x^3{y}^3}{6}+\frac{x^4{y}^4}{24}+....... \end{gathered}$$

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