# Tag Info

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There isn't a unique answer, especially if (infinite) series are used. But if, as the example formula indicates, you want a polynomial in 1/z, then the result is unique up to multiplying the numerator and denominator by a scalar factor. Divide @@ Map[ Collect[Factor @ #, u] &, (1 + u)^2 Through[{Numerator, Denominator}[f /. z -> 1/u]] ] /. ...

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Since Numerator@f == (Denominator@f /. {h2 -> h1, w2 -> w1}) it should be suficient to have only the series decomposition of the numerator, no? Or are you aiming for something else? Series[Numerator@f, {z, 0, 2}]

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One way would be like the following. Let us define the function rule as follows: Clear[rule]; rule[expr_] := ReplaceAll[ expr, {Sin[2 γ_] -> 2*Sin[γ]*Cos[γ], Cos[2 γ_] -> Cos[γ]^2 - Sin[γ]^2}]; and map this function on your expression. For the sake of shortness I take here only a small part of your otherwise a too ...

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If you want the term $\frac{\partial f}{\partial r}$ to disappear you need to introduce new function which would be: w2 = f[r, θ] r which means that you have to make a substitution f -> w2/r, this way: lapla1 /. f -> (w2[#, #2]/# &) // Simplify // ExpandAll If you once used f or w, don't change theirs definitions, use a new one, you ...

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Basing on the following thread: Change variables in differential expressions and using great code by Jens for visualisation purposes (I have replaced part [vars__Symbol] with [vars__] because you are using Substripted names which are not Symbols, but that's only about visualisation). You can do the following in the first step: M2 = M /. f -> (f[#, #2, ...

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The simplest way to get the output you want is to enter the objects representing the vectors as String like this: "{1,0,1}" + "{2,0,1}" + 3 "{1,0,1}" - "{4,4,7}" - 3 "{2,0,1}" 4 {1,0,1} - 2 {2,0,1} - {4,4,7} This is just a particularly simple version of what @Szabolcs mentioned in the comment: you don't want the lists representing the vectors to be ...

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