Expanding the Function f(x,y) around (1,-1): What's the Easy Way?

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How do u expand f(x,y) = y^2/x^2 about the point (1,-1) ? up to and including quadratic term?
 
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darkar said:
How do u expand f(x,y) = y^2/x^2 about the point (1,-1) ? up to and including quadratic term?
Have you ever used the taylor series on multi-variable functions before?
 
Not really, how? Have only done one variable so far.
 
The general term in a Taylor's series for f(x,y) about the point (a,b) is
\frac{1}{(m+n)!}\frac{\partial^m f}{\partial x^m}\frac{\partial^n f}{\partial y^n}(x- a)^m(y- b)^n

(edited. Thanks, daveb)
 
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Shouldn't that (x - b)^n be (y - b)^n?
 
Bah, everybody likes to do it the hard way. :frown: f(x, y) is just a product of two things whose Taylor series you already know!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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