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Calc - How to produce the reverse of a row?

asked 2015-07-06 18:42:57 +0200

Weux082690 gravatar image

updated 2016-03-09 21:11:07 +0200

Alex Kemp gravatar image

Essentially, I have two sequences of numbers in Row 2 and Row 3, and I want to produce a sequence of numbers in Row 4 such that:
A4 = A2A3
B4 = A2
B3 + B2A3
C4 = A2
C3 + B2B3 + C2A3
D4 = A2D3 + B2C3 + C2B3 + D2A3
etc. (This is the Cauchy product for the mathematically minded)

I want to use the function SUMPRODUCT which will greatly simplify the formulas, but I need some way to get the reverse of Row 2 or Row 3 as an array. Is there some function or extension that I am missing? Or is this simply not possible in LibreOffice Calc?

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answered 2015-07-06 18:55:45 +0200

Lupp gravatar image

updated 2015-07-06 18:57:32 +0200

There is not a specific function but you can get the effect using OFFSET like in

{=OFFSET($A$2;0;1024-COLUMN($A$1:$AMJ$1))}

as an array formula. Of course, you will replace the constants by appropriate expressions. Be cautious with zeros. They may represent originally blank cells.

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Comments

(OT) Brave Cauchy product revisited.

@Weux082690 Out of curiosity: For what reason do you want to calculate the terms (sums) being the addends of the Cauchy product? If you have, say, 20 addends of the first and of the second (infinite) series, the product of the two partial sums should approximate the product of the two series better than the partial sum over 20 terms of the Cauchy product.

Lupp gravatar imageLupp ( 2015-07-07 01:15:02 +0200 )edit

As an example I just calculated 20 terms of the standard series for e^2 and for e^3 as well. I got 148.4131591.for the product of the two partial sums and 148.41310787 adding 20 terms of the Cauchy product. The first value is hardly distinguishable from the true value of e^5. The second one is easily.

I think the Cauchy product can sometimes help in generalizations ...

Lupp gravatar imageLupp ( 2015-07-07 01:21:48 +0200 )edit

I was computing the terms of a series whose generator function was the solution of G(x) = 1+x(G(x))^2+x^2(G(x))^3. The closed form for G(x) is complex and difficult to differentiate, so I was computing the terms recursively using Cauchy products.

Weux082690 gravatar imageWeux082690 ( 2015-07-07 21:12:43 +0200 )edit

Thanks for comforting me.

Lupp gravatar imageLupp ( 2015-07-07 21:30:51 +0200 )edit
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Asked: 2015-07-06 18:42:57 +0200

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Last updated: Jul 06 '15