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I'm trying to write the beginning of the power series expansion at $x = y = 0$, up to and including quadratic terms, for the function:

$$f(x,y) = \frac{\cosh(x)\cosh(y)}{\sqrt{1+2(x-y)} - 0.5}$$

and using this, how can we get $f_{xy}$?

Thanks

start: $\cosh(x) = (e^x+e^{-x})/2$ and inserting that in for $y$, but how to take care of the denominator and put it into a series?

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1 Answer

up vote 2 down vote accepted

The Taylor series for two variables up to quadratic terms is:$$ f(x, y) = f(x_0, y_0) + (x- x_0) \dfrac{\partial f}{\partial x} + (y - y_0) \dfrac{\partial f}{\partial y} + \dfrac{1}{2} \left [ (x - x_0)^2 \dfrac{\partial^2 f}{\partial x^2} + 2 (x - x_0)(y - y_0) \dfrac{\partial^2 f}{\partial x \partial y} + (y - y_0)^2 \dfrac{\partial^2 f}{\partial y^2} \right ] + \cdots$$ The calculations are tedious if done by hand. For example$$ \frac{\partial f}{\partial x}=\frac{\sinh (x) \cosh (y)}{\sqrt{2 (x-y)+1}-0.5}-\frac{\cosh(x) \cosh (y)}{\left(\sqrt{2 (x-y)+1}-0.5\right)^2 \sqrt{2(x-y)+1}}$$which evaluated at (0,0) is -4.

Using software I found that:$$ f(x,y)=2-4x+4y+11x^2-20xy+11y^2+\cdots$$

EDIT

Following the comments here is a more detailed calculation of the expansion of $f(x,y)=\sqrt{2(x-y)+1}$. The derivatives are easy:$$ f_x=-f_y=\frac{1}{\sqrt{2 (x-y)+1}} ; f_{xx}=f_{yy}=-f_{xy}=-\frac{1}{(2 (x-y)+1)^{3/2}}$$

Now evaluate those at (0,0) and you get the expansion:$$f(x,y)=1+x-y-x^2/2-y^2/2+xy+\cdots$$

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This is practical, but unfortunately I need to get practice by hand –  mary Sep 26 '12 at 6:21
    
In that case you'll have to start calculating those derivatives. –  ivan Sep 26 '12 at 7:07
    
No, its just using the fact that cosh(x) = 1/2(e^x+e^-x) –  mary Sep 26 '12 at 7:08
    
and getting the series expansion for e^x and e^-x and multiplying that, then taking the denominator such as 1/1-x = 1+x+x^2+... and multiplying that with our numerator, but I don't know how to handle with this messy square root –  mary Sep 26 '12 at 7:09
    
Hmm, you are not allowed to expand $\sqrt{2(x-y)+1}-0.5$? –  ivan Sep 26 '12 at 7:21
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