# Elegant way to do series expansion around x+$\epsilon$

Suppose I have a function $p(x+\epsilon,y-\epsilon,t)$, and I want to expand it around $(x,y)$ like $$p(x+\epsilon,y-\epsilon,t)=p(x,y,t)+\dfrac{\partial p}{\partial x}\epsilon+\dfrac{\partial p}{\partial y}(-\epsilon)+\cdots$$

A naive way to do it is

Series[p[x+e,y-e,t],{x+e,x,1},{y-e,y,1}]


which will result in the following error message:

General::ivar: "e+x is not a valid variable."
....


Right now I do the expansion in the following clumsy way:

Replace[
Normal@Series[p[x, y, t], {x, a, 1}, {y, b, 1}],
{x - a -> \[Epsilon], y - b -> -\[Epsilon]},
Infinity]
/. {a -> x, b -> y}


Question: Is there an elegant way to do it? In the end, I want to expand $p(x,y,t)$ in $$-(x+z)p(x,y,t)+(x+\epsilon)p(x+\epsilon,y-\epsilon,t)+(1-x-y+\epsilon)p(x,y-\epsilon,t)$$ and keep only the $\epsilon$ terms.

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What's wrong with Series[p[x+e,y-e,t],{e,0,1},{e,0,1}] or Series[p[xx,yy,t],{xx,x,1},{yy,y,1}]? – sebhofer Aug 12 '14 at 10:25
@sebhofer,Actually Series[p[x+e,y-e,t],{e,0,1},{e,0,1}] is what I wanted. – wdg Aug 12 '14 at 10:46
Ups, I just realized this should have been Series[p[x+ex,y-ey,t],{ex,0,1},{ey,0,1}] or just Series[p[x+e,y-e,t],{e,0,1}] – sebhofer Aug 12 '14 at 10:48
@sebhofer Don't forget to post this as an answer. – C. E. Aug 12 '14 at 14:03
@Pickett If I'm honest, I wasn't (and still not am) really sure if this question would survive, as this is a pretty trivial use of Series – sebhofer Aug 12 '14 at 15:45

Depending on your specific needs the straightforward way to do this is

Series[p[x+e,y-e,t],{e,0,1}]


or

Series[p[x+ex,y-ey,t],{ex,0,1},{ey,0,1}]


If you want to extract the terms proportional to $\epsilon$ you can get the correct coefficients with

SeriesCoefficient[p[x + e, y - e, t], {e, 0, 1}]


or if that suits you better with

CoefficientList[Series[p[x + e, y - e, t], {e, 0, 1}], e][[2]]

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