Question

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

cos ( x + y )

Short answer

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

Step-by-step solution

Explanation of solution

The equation is provided cos ( x + y ) and to find the two variable Maclaurin for the following function.

McLaren Series:

If the values of the function's consecutive derivatives at zero are known, a McLaren 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:

cos ( x + y )

Since,

$\cos u=1-\frac{u^2}{2}+\frac{u^4}{24}+\frac{u^6}{720}+.....$

And,

f ( x,y) = cos (x + y)

Therefore,

$$\begin{gathered} f(x,y)=\left(1+\frac{{(x+y)}^2}{2}+\frac{{(x+y)}^4}{24}+\frac{{(x+y)}^6}{720}+...\right) \\ \left(1+\frac{x^2+2xy+y^2}{2}+\frac{x^{4}4x^{3}y+6x^2{y}^2+4x{y}^3+y^4}{24}+....\right) \end{gathered}$$

Because the Maclaurin series has now expanded around the origin, the values of a and b will be zero. As a result, the generic formula f ( x,y ) is as follows:

localid="1659093776557" $$\begin{gathered} f(x,y)=f(0,0)+f_x(0,0)x+f_y(0,0)y \\ +\frac{1}{2}\left(f_{xx}\right(0,0)x^2+2f_{xy}(0,0)xy+f_{yy}(0,0\left)y^2\right) \\ +\frac{1}{6}+\left(f_{xxx}\right(0,0)x^3+3f_{xxy}(0,0)x{y}^2+f_{yyy}(0,0\left)y^3\right)+... \end{gathered}$$

After calculate this equation, the result will be the same.

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