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Say I start with an expression with potentially an arbitrary number of variables (input-dependent), for example Exp[x]Sin[y]z^(-1)w, and I want to expand in all variables to a certain power. I could do this

Series[Exp[x]Sin[y]z^(-1)w,{x,0,5},{y,0,5},{z,0,5},{w,0,5}]

But is there a more compact way to achieve this? Not to mention that this method fails when the number of variables depend on the inputs. I tried something like

inputs={x,y,z,w}
Thread[{inputs,0,5}]

This gives

{{x, 0, 5}, {y, 0, 5}, {z, 0, 5}, {w, 0, 5}}

as an output. This is as far as I got. I tried to extract the elements using Row and StringRiffle but it doesn't work. Any help is appreciated, thanks!

Edit: Changed my example from all exponentials to a more general one.

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1 Answer 1

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There are two answers to your question. The simple answer is as follows. Since you only have the exponents, the expression Exp[x]Exp[y]Exp[z]Exp[w] is equal to Exp[a], where a=x+y+z+w. Therefore, one can operate as follows:

(Series[Exp[a], {a, 0, 5}] // Normal) /. 
  a -> x + y + z + w // Expand

I do not post the result since it is a bit too large, but you easily get it by evaluating this code.

However, I guess that your question is about a more general situation, in which the functions are not necessarily all exponents and may differ from one another. To be specific, let us expand exprA = Exp[x] Sin[y] up to the 5th order both in x and y.

To do this, let us first replace x and y by t x and t y:

exprB = exprA /. {x -> t*x, y -> t*y}

(*  E^(t x) Sin[t y]  *)

Now let us expand the result in terms of t up to t^5 and then put t=1:

 exprC = (Series[exprB, {t, 0, 5}] // Normal) /. t -> 1

(* y + x y + (x^2 y)/2 - y^3/6 + 1/6 (x^3 y - x y^3) + 
 1/120 (5 x^4 y - 10 x^2 y^3 + y^5)  *)

Now to bring this to a better visible form one can use Collect or Expand:

Expand[exprC]

(*  y + x y + (x^2 y)/2 + (x^3 y)/6 + (x^4 y)/24 - y^3/6 - (x y^3)/6 - (
 x^2 y^3)/12 + y^5/120   *)

Done. Have fun!

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  • $\begingroup$ Yes I used a very bad example, thank you! Your second method seems to be a very smart way to get around things although I'm a bit concerned what if I have polynomials like x^n/y^m, then some powers of t will cancel. $\endgroup$
    – Lepnak
    Commented Jun 16, 2021 at 9:49
  • $\begingroup$ @Lepnak It is no problem, since in the end you get rid of t and get a pure polynomial in terms of x and y. It becomes independent of t and starting from this point it "forgets" to which powers of t belonged this or that term. Another story that you may then want to regroup the result somehow. But this is a different task. $\endgroup$ Commented Jun 16, 2021 at 9:55
  • $\begingroup$ Ah yes you're right. Although I notice I might need to expand to a higher order. What I mean is for example let exprA=1/(1-x)(1-y^2), then if I expand to {t,0,5} I do not see x^2y^4, x^3y^4 etc. I need to expand to {t,0,10} to see all the y^5 terms (although now I also see redundant x^10 terms). Whereas if I do {x,0,5},{y,0,5} then I see all terms up to x^5, y^5. I think if x and y are both small numbers and I want to expand in powers of xy, then yours is definitely the way to go! Although I'm working on more of a character formula where I want to expand x and y individually. $\endgroup$
    – Lepnak
    Commented Jun 16, 2021 at 10:26

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