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Dec 6, 2019 at 10:26 vote accept ThunderBiggi
Dec 5, 2019 at 0:18 answer added Carl Woll timeline score: 3
Dec 4, 2019 at 23:56 history edited ThunderBiggi CC BY-SA 4.0
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Dec 4, 2019 at 23:55 comment added ThunderBiggi @DanielLichtblau Ok, I didn't tell you what all the objects in the general formula are, but they are in the worst case scenario real numbers, hence nothing has changed. The old formula was $\sum_{n=0}^{N}\,a_n(x)\,y^{n}+y^{\alpha}\,\sum_{k=0}^{K}b_k(x)\,y^{k}+y^{2\,\alpha}\,\sum_{j=0}^{J}c_j(x)\,y^{j}+...$, which is equivalent to the new one. I also edited my post with a possible solution that you can copy paste to see exactly what I want.
Dec 4, 2019 at 23:50 history edited ThunderBiggi CC BY-SA 4.0
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Dec 4, 2019 at 23:23 comment added Daniel Lichtblau I think the requirement changedfor the general case. the last time I looked it was as I described in prior comments. Now it is not. It is unrealistic to expect people to foresee such changes. For what it is worth, the earlier requirement would be the easier to meet.
Dec 4, 2019 at 22:33 comment added ThunderBiggi @DanielLichtblau I am not sure I really understand what you are saying. I rewrote the general formula. The simple example is what I want. The general formula is a mathematical way of writing it - that is obviously not unique due to commutativity, distributivity and so on.
Dec 4, 2019 at 22:19 comment added Daniel Lichtblau The general case has powers in y^a with coefficients that have nested powers in y. There is no mixing of the integer powers and integer multiple-of-a powers.
Dec 4, 2019 at 22:05 history edited ThunderBiggi CC BY-SA 4.0
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Dec 4, 2019 at 21:36 comment added ThunderBiggi @DanielLichtblau It is late here, so I might be missing something, but doesn't the simple example correspond to $a_0=-2$, $a_1=2-x$, $a_2=-(1-x)x$, $b_0=-1$ and $b_1=-(1-x)$. As for the comment before the note - I know that in the simple example case I can just write the explicit powers, but my real life application has very complicated expressions with hundreds of terms, so I need something automated
Dec 4, 2019 at 21:14 comment added Daniel Lichtblau I should note that the desired general form is in conflict with the specific example, in terms of treating "mixed" powers such as y^(1+a).
Dec 4, 2019 at 21:09 comment added Daniel Lichtblau Could do that but only if the forms you are after are explicitly in the input. Check for example In[57]:= expr = (-1 + (-1 + x) y) (2 + x y + y^a); Collect[Expand@expr, {y^a, y^(a + 1), y}, Factor] Out[58]= -2 + (-2 + x) y + (-1 + x) x y^2 - y^a + (-1 + x) y^(1 + a)
Dec 4, 2019 at 21:04 comment added ThunderBiggi @DanielLichtblau I tried that. Is there a way to specify a general form, like when one does a replacement someExpr/.{x^(1+k_):>x^k} for example.
Dec 4, 2019 at 20:23 comment added Daniel Lichtblau Comes close: In[42]:= expr = (-1 + (-1 + x) y) (2 + x y + y^a); Collect[expr, {y^a, y}, Factor] Out[43]= -2 + (-2 + x) y + (-1 + x) x y^2 + y^a (-1 + (-1 + x) y)
Dec 4, 2019 at 19:58 comment added ThunderBiggi @MarcoB I added a simpler example
Dec 4, 2019 at 19:58 history edited ThunderBiggi CC BY-SA 4.0
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Dec 4, 2019 at 17:55 comment added MarcoB Could you perhaps come up with a simpler expression to work on, and show the exact form (not the general formula) you would like it transformed to?
Dec 4, 2019 at 17:50 history edited MarcoB CC BY-SA 4.0
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Dec 4, 2019 at 17:42 history asked ThunderBiggi CC BY-SA 4.0