# What is the MacLaurin formula of higher orders for multivariable functions? Example: f(x,y)=cosx cosy

What is the MacLaurin formula of higher orders for multivariable functions? Example: $f\left(x,y\right)=\mathrm{cos}x\mathrm{cos}y$
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Duncan Kaufman
You can expand separately
- $\mathrm{cos}x=1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{4!}+o\left({x}^{4}\right)$
- $\mathrm{cos}y=1-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{4!}+o\left({y}^{4}\right)$
and then multilply taking the terms to the desidered order.
That is for order IV
$f\left(x,y\right)=\mathrm{cos}x\cdot \mathrm{cos}y=\left(1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{4!}+o\left({x}^{4}\right)\right)\left(1-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{4!}+o\left({y}^{4}\right)\right)=\phantom{\rule{0ex}{0ex}}=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+\frac{{x}^{4}}{4!}+\frac{{y}^{4}}{4!}-\frac{{x}^{2}{y}^{2}}{4}+o\left(|\left(x,y\right){|}^{4}\right)$
Note that
- for order II: $f\left(x,y\right)=\mathrm{cos}x\cdot \mathrm{cos}y=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+o\left(|\left(x,y\right){|}^{2}\right)$
- for order III: $f\left(x,y\right)=\mathrm{cos}x\cdot \mathrm{cos}y=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+o\left(|\left(x,y\right){|}^{3}\right)$
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Gauge Odom
Your example is special in so far as the function is "separated" into two functions of one variable.
$g\left(t\right):=f\left(t\phantom{\rule{thinmathspace}{0ex}}x,t\phantom{\rule{thinmathspace}{0ex}}y\right)$
of one variable $t$, and compute its value at $t=1$, using the Taylor expansion of a function of one variable:

and compute the higher derivatives ${g}^{\left(p\right)}\left(0\right)$ using repeatedly the chain rule. Collecting equal terms you obtain