Question

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

${\mathit{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}$

Short answer

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

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.

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

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

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

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

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