91Ó°ÊÓ

The equation is provided and to find the two variables 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.

The following equation is provided:

cos x sinh y.

Since,

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

And,

$\sin hy=y+\frac{y^3}{2}+\frac{y^5}{120}+\frac{y^7}{5070}+....$

suggestthat,

role="math" localid="1659091871698" $$\begin{gathered} f(x,y)=\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+....\right)\left(y+\frac{y^3}{6}+\frac{y^5}{120}+\frac{y^7}{5070}+...\right) \\ =y+\frac{y^3}{6}-\frac{x^{2}y}{2}+\frac{x^{4}y}{24}-\frac{x^2{y}^3}{12}+\frac{y^5}{120}.... \end{gathered}$$

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:

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

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

Hence, role="math" localid="1659091908725" $f(x,y)=y+\frac{y^3}{6}-\frac{x^{2}y}{2}+\frac{x^{4}y}{24}-\frac{x^2{y}^3}{12}+\frac{y^5}{120}....$