What is the MacLaurin formula of higher orders for multivariable functions? Example: $f(x,y)=\mathrm{cos}x\mathrm{cos}y$

cubanwongux
2022-09-13
Answered

What is the MacLaurin formula of higher orders for multivariable functions? Example: $f(x,y)=\mathrm{cos}x\mathrm{cos}y$

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Duncan Kaufman

Answered 2022-09-14
Author has **17** answers

You can expand separately

- $\mathrm{cos}x=1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{4!}+o({x}^{4})$

- $\mathrm{cos}y=1-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{4!}+o({y}^{4})$

and then multilply taking the terms to the desidered order.

That is for order IV

$$f(x,y)=\mathrm{cos}x\cdot \mathrm{cos}y=(1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{4!}+o({x}^{4}))(1-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{4!}+o({y}^{4}))=\phantom{\rule{0ex}{0ex}}=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+\frac{{x}^{4}}{4!}+\frac{{y}^{4}}{4!}-\frac{{x}^{2}{y}^{2}}{4}+o(|(x,y){|}^{4})$$

Note that

- for order II: $f(x,y)=\mathrm{cos}x\cdot \mathrm{cos}y=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+o(|(x,y){|}^{2})$

- for order III: $f(x,y)=\mathrm{cos}x\cdot \mathrm{cos}y=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+o(|(x,y){|}^{3})$

- $\mathrm{cos}x=1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{4!}+o({x}^{4})$

- $\mathrm{cos}y=1-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{4!}+o({y}^{4})$

and then multilply taking the terms to the desidered order.

That is for order IV

$$f(x,y)=\mathrm{cos}x\cdot \mathrm{cos}y=(1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{4!}+o({x}^{4}))(1-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{4!}+o({y}^{4}))=\phantom{\rule{0ex}{0ex}}=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+\frac{{x}^{4}}{4!}+\frac{{y}^{4}}{4!}-\frac{{x}^{2}{y}^{2}}{4}+o(|(x,y){|}^{4})$$

Note that

- for order II: $f(x,y)=\mathrm{cos}x\cdot \mathrm{cos}y=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+o(|(x,y){|}^{2})$

- for order III: $f(x,y)=\mathrm{cos}x\cdot \mathrm{cos}y=1-\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}+o(|(x,y){|}^{3})$

Gauge Odom

Answered 2022-09-15
Author has **4** answers

Your example is special in so far as the function is "separated" into two functions of one variable.

$$g(t):=f(t\phantom{\rule{thinmathspace}{0ex}}x,t\phantom{\rule{thinmathspace}{0ex}}y)$$

of one variable $t$, and compute its value at $t=1$, using the Taylor expansion of a function of one variable:

$$f(x,y)=g(1)=\sum _{p=0}^{n}\frac{1}{p!}{g}^{(p)}(0)+{R}_{n}\text{},$$

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

$${g}^{(p)}(0)=\sum _{k=0}^{p}{\textstyle (}\genfrac{}{}{0ex}{}{p}{k}{\textstyle )}{f}_{{x}^{p-k}{y}^{k}}(0,0)\phantom{\rule{mediummathspace}{0ex}}{x}^{p-k}\phantom{\rule{thinmathspace}{0ex}}{y}^{k}\text{}.$$

$$g(t):=f(t\phantom{\rule{thinmathspace}{0ex}}x,t\phantom{\rule{thinmathspace}{0ex}}y)$$

of one variable $t$, and compute its value at $t=1$, using the Taylor expansion of a function of one variable:

$$f(x,y)=g(1)=\sum _{p=0}^{n}\frac{1}{p!}{g}^{(p)}(0)+{R}_{n}\text{},$$

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

$${g}^{(p)}(0)=\sum _{k=0}^{p}{\textstyle (}\genfrac{}{}{0ex}{}{p}{k}{\textstyle )}{f}_{{x}^{p-k}{y}^{k}}(0,0)\phantom{\rule{mediummathspace}{0ex}}{x}^{p-k}\phantom{\rule{thinmathspace}{0ex}}{y}^{k}\text{}.$$

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