Oh,Mike,your solution is so beautiful!
recur=:[ (] * [:+/\ [ % ]) [: */\ ]
EMA2=:([EMAIL PROTECTED] , [ * [EMAIL PROTECTED]) recur (1, (<:@[EMAIL
PROTECTED]) $ [EMAIL PROTECTED])
(0.1&EMA -: 0.1&EMA2) p:i.10
1
6!:2 '0.1&EMA i.1e6'
2.80876
6!:2 '0.1&EMA1 i.1e6'
3.781998
6!:2 '0.1&EMA2 i.1e6'
0.314285
2008/3/9, Mike Powell <[EMAIL PROTECTED]>:
> Xu,
>
> Your EMA question is related to algorithms for linear recurrence
> relations. There is a much loved algorithm (probably because it's
> usually found to be lean and mean) that I first encountered at IP
> Sharp in a newsletter quiz. We used it all over the place,
> particularly wherever a present value or discounted cash flow was in
> play. Where it came from before that, I know not. It goes like this.
>
> If you have a linear recurrence relation of the form:
>
> r[0] is m[0] times a[0]
> r[i] is a[i] + m[i] times r[i-1] (i >0)
>
> then this can be calculated in vector languages in the following way:
>
> a recur m;t
> t is timesscan m
> r is t times plusscan a div t
>
> I'd be most interested to see a J algorithm for this, particularly a
> tacit one. I note that we have a reflexive piece (t) and that there
> may be an opportunity for an inverse (* and %).
>
> As far as the EMA is concerned, I suggest Xu's original requirement is
> solved with:
>
> (y[0] , k times 1 drop y) recur (length y) replicate 1 - k
>
>
> Mike Powell
>
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