[Date Index]
[Thread Index]
[Author Index]
Re: please solve
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52942] Re: please solve
*From*: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
*Date*: Fri, 17 Dec 2004 05:19:25 -0500 (EST)
*References*: <cprk0m$r87$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Your matrix x'x is singular because x is rank 1 (all rows the same) so it
can't be uniquely inverted. You could try replacing it by x'x + eps I where
eps is a small positive number and I is the identity matrix. This has the
effect of regularising the inverse matrix.
x = Table[t^i, {j, 3}, {i, 0, 2}]
gives
{{1, t, t^2}, {1, t, t^2}, {1, t, t^2}}
and
Simplify[Inverse[x.Transpose[x]+\[Epsilon] IdentityMatrix[3]]]
gives
{{(2 + 2*t^2 + 2*t^4 + \[Epsilon])/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 +
\[Epsilon])),
-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))),
-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon])))},
{-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))),
(2 + 2*t^2 + 2*t^4 + \[Epsilon])/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 +
\[Epsilon])),
-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon])))},
{-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))),
-((1 + t^2 + t^4)/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 + \[Epsilon]))),
(2 + 2*t^2 + 2*t^4 + \[Epsilon])/(\[Epsilon]*(3 + 3*t^2 + 3*t^4 +
\[Epsilon]))}}
and
Simplify[Inverse[x.Transpose[x]+\[Epsilon] IdentityMatrix[3]].x]
gives
{{1/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t/(3 + 3*t^2 + 3*t^4 + \[Epsilon]),
t^2/(3 + 3*t^2 + 3*t^4 + \[Epsilon])}, {1/(3 + 3*t^2 + 3*t^4 +
\[Epsilon]),
t/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t^2/(3 + 3*t^2 + 3*t^4 +
\[Epsilon])},
{1/(3 + 3*t^2 + 3*t^4 + \[Epsilon]), t/(3 + 3*t^2 + 3*t^4 + \[Epsilon]),
t^2/(3 + 3*t^2 + 3*t^4 + \[Epsilon])}}
Steve Luttrell
<rcmcll at yahoo.com> wrote in message news:cprk0m$r87$1 at smc.vnet.net...
> Greetings:
>
> Could someone please solve this symbolically?
> This is just the ols formula for beta-hat but I need a symbolic
> solution for this special case.
>
> b = inv(x'x)x'y
>
> where
>
> x = 1 t t^2
> 1 t t^2
> 1 t t^2
>
>
> and simplify simplify simplify!!
>
> Thanks,
>
> Bob
>
Prev by Date:
**Re: Help on a recursive function**
Next by Date:
**Bug in 5.1 FE?**
Previous by thread:
**Re: please solve**
Next by thread:
**Re: please solve**
| |