|
Hello John:
Thanks for answering this question. I did not think that this was a
frivolous question. It is part of a Fibonacci speed computation
programming problem.
Since I am new to Matlab, I had no idea what vpa was/is. Subsequently,
I did find an example using vpa. In case anyone else needs this, here
is the code to do what I asked:
>> A=[0,1;1,1]
>> vA = vpa(A,30)
>> vA^10000
ans =
[ .20793608237133498072112648988638e2090, .
33644764876431783266621612005100e2090]
[ .33644764876431783266621612005100e2090, .
54438373113565281338734260993738e2090]
Finally, it must be related to precision, since the "fix" is to use
variable precision arithmetic.
Thanks again.
Regards..
--------------------------------------------------------------------------------------------
On Jul 4, 5:08 pm, "John D'Errico" <woodch...@rochester.rr.com> wrote:
>
> (This is not a precision question.)
>
> Could you do this, then someone else
> would be asking why they cannot compute
>
> A^10000000000000
>
> or some arbitrarily higher power.
>
> If you truly need this, there is
> always the symbolic toolbox, using
> vpa.
>
> John- Hide quoted text -
>
> - Show quoted text -
|