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vector decreasing by a factor

3 messages · andre bedon, (Ted Harding), Uwe Ligges

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On 02-May-11 07:55:27, andre bedon wrote:
One expression which would do what you want is

  rev((1+r)^(0:159))

though there may be more efficient ways to do it. This assumes
that r, hence (1+r), is given. If you are given the
value X1 of the first element, which is to be interpreted
as (1+r)^159, then perhaps take (1+r) as X1^(1/159),
though there is a potential slight inaccuracy in recovering
X0 from (1+r)^159. So check this first.

Hoping this helps,
Ted.

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E-Mail: (Ted Harding) <ted.harding at wlandres.net>
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Date: 02-May-11                                       Time: 09:50:55
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#
On 02.05.2011 09:55, andre bedon wrote:
Yes:
(1+r)^(159:0)

Uwe Ligges