       Re: Fit with lists instead of functions

• To: mathgroup at smc.vnet.net
• Subject: [mg40297] Re: [mg40283] Fit with lists instead of functions
• From: Dr Bob <majort at cox-internet.com>
• Date: Sun, 30 Mar 2003 04:09:54 -0500 (EST)
• References: <200303291020.FAA04122@smc.vnet.net>
• Reply-to: majort at cox-internet.com
• Sender: owner-wri-mathgroup at wolfram.com

```Here are a couple of ways to do it.  In either case, Regress can be
replaced with Fit.

Known values:

n = 10;
knowns = Sin@Range@n;

Predictor variable lists:

a = Array[Random[] &, n];
b = Range@n;
c = b^2;

The simplest thing:

Regress[Transpose@{a, b, c, knowns}, {1, x, y, z}, {x, y, z}]

A more flexible method:

f = Interpolation[a];
g = Interpolation[b];
h = Interpolation[c];
Regress[knowns, {1, f@x, g@x, h@x}, x]

The second method can be easily modified to include other predictor
functions:

Regress[knowns, {1, f@x, g@x, h@x, Cos@x}, x]

Bobby

On Sat, 29 Mar 2003 05:20:26 -0500 (EST), Martin <martin_rommel at yahoo.com>
wrote:

> It ought to be simple, I just don't know where to find it:
> I want to do a (linear, least square) fit but instead of functions in
> a variable I want to use lists of the same length as the data list.
>
> And by the way, I am still running 4.0
>
> Thanks.
>
>

--
majort at cox-internet.com
Bobby R. Treat

```

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