- #1

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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?

- Thread starter gschran
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- #1

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- 0

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?

- #2

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[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).

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