Taylor Series in Multiple Variables

Click For Summary
SUMMARY

The discussion focuses on deriving the leading order terms in the Taylor series for the function f(x,y) = Sqrt(a*x^8 + b*x^4*y^4 + c*y^8), centered at (0,0). The method involves rewriting the function using the binomial series expansion, specifically applying the formula (1+x)^p = Σ (p choose k) x^k. The expansion is contingent on the values of x and y, with particular attention to the condition |x| < 1 for the validity of the expansion.

PREREQUISITES
  • Understanding of Taylor series and power series expansions
  • Familiarity with binomial coefficients and the binomial theorem
  • Knowledge of multivariable calculus, specifically functions of two variables
  • Basic algebraic manipulation of polynomial expressions
NEXT STEPS
  • Study the derivation of Taylor series for multivariable functions
  • Learn about the convergence criteria for power series expansions
  • Explore alternative methods for function expansion, such as using polar coordinates
  • Investigate the implications of the condition |x| < 1 on the series expansion
USEFUL FOR

Mathematicians, physics students, and anyone involved in advanced calculus or numerical analysis who seeks to understand multivariable Taylor series expansions.

gschran
Messages
1
Reaction score
0
Can anyone help me for the leading order terms in the taylor series for the function
f(x,y) = Sqrt(a*x^8+b*x^4*y^4+y^8),
centered at x=0,y = 0 and a,b,c constants?
 
Physics news on Phys.org
Your question really depends on the values of x and y. And your expression doesn't have c. I assumed your function to be:

[tex]f(x,y)=\sqrt{a x^8+b x^4y^4+c y^8}[/tex]

I only outline the method to obtain a power series here.

[tex]f(x,y)=c y^8\sqrt{1+\left(\frac{a x^8+b x^4y^4}{c y^8}\right)}[/tex]

By

[tex](1+x)^p=\sum _{k=0}^{\infty } \binom{p}{k}x^k[/tex]

,we have:

[tex]f(x,y)=c y^8\sum _{k=0}^{\infty } \binom{\frac{1}{2}}{k}\left(\frac{a x^8+b x^4 y^4}{c y^8}\right)^k[/tex]

There is another expansion method as well, in which you take out the power x instead of y. It depends on the values of x and y. Since the expansion of [tex](1+x)^p[/tex] is valid for |x| < 1 (|x| = 1 is more complicated).
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
17K