This is really amazing how J does it, but
in fact the logic is simple:
(1.10 * _50 + ])^:3 b. _1
(50 + 0.90909090909090906 * ])^:3
- power remains power (^:)
- 1.1 times (_50 plus o) becomes 50 plus (%1.1 times o)
This could be solved by hand like this
f=: (1.1 * _50 + ])^:3
2 -~/\f i.10 NB. linear with such factor
1.331 1.331 1.331 1.331 1.331 1.331 1.331 1.331 1.331
f 0 NB. and such free element
_182.05
lsfit=: {:@] %. [EMAIL PROTECTED] ^/ i.@>:@[ NB. also
]C=: 1 lsfit (,: f) 0 1
_182.05 1.331
+/(f +/@- C p. ]) i.1000 NB. very close
6.24262e_8
NB. 1000 = f x <=> 0 = _1000 + f x
p. _1000 0+C
+-----+-------+
|1.331|888.092|
+-----+-------+
1{::p. _1000 0+C NB. sought value
888.092
C p. 888.092
1000
f 888.092
1000
--- Scott Russell <[EMAIL PROTECTED]> wrote:
> Thanks Henry.
>
> Can you give me the original equation and then how do you make an inverse of
> the equation ?
>
> Scott
>
> ----- Original Message -----
> From: Henry Rich<mailto:[EMAIL PROTECTED]>
> To: 'Programming forum'<mailto:[email protected]>
> Sent: Monday, May 08, 2006 2:59 PM
> Subject: RE: [Jprogramming] J question
>
>
> I would never have thought of doing it this way, but you can
> solve it in J just as you would in Excel. Defines the function,
> and use the inverse to ask what value would give you 1000:
>
> (1.10 * 50 -~ ])^:3 ^:_1 (1000)
> 888.092
>
> Henry Rich
>
> > -----Original Message-----
> > From: [EMAIL PROTECTED]<mailto:[EMAIL PROTECTED]>
> > [mailto:[EMAIL PROTECTED] On Behalf Of Scott Russell
> > Sent: Monday, May 08, 2006 2:34 PM
> > To: [email protected]<mailto:[email protected]>
> > Subject: [Jprogramming] J question
> >
> > Hello,
> >
> > I'm a newbie to J and I have a general programming question.
> >
> > Say I wanted to find a solution to a problem that would
> > normally require an iterative technique. For a very
> > simplified example:
> >
> > Say you purchased a CD or some other financial instrument
> > where you wanted to accumulate $1,000 at the end of 3 years.
> > This CD had an annual fee of $50 (payable at the beginning of
> > the year) and it guaranteed an interest rate of 10%. How
> > much would you need to invest at the beginning of the 3 years
> > ? In other words, solve for X:
> >
> > (((((X-50) * 1.10) - 50) * 1.10) - 50) * 1.10 = $1,000
> >
> > In Excel, I could set up a macro that searched for the
> > solution through the bisection algorithm or I could use the
> > "Goal Seek" function if I really wanted to take a short cut.
> >
> > But I wouldn't know where to start in J. Any ideas ? Thanks
> > ----------------------------------------------------------------------
> > For information about J forums see
> > http://www.jsoftware.com/forums.htm<http://www.jsoftware.com/forums.htm>
>
> ----------------------------------------------------------------------
> For information about J forums see
> http://www.jsoftware.com/forums.htm<http://www.jsoftware.com/forums.htm>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
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