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From: Alessio Giberti on 7 Apr 2010 07:26
Hi all. A vector function of one variable, defined as follows,

Y=expA[z].(Y0+Integrate[expA[-s].b[s],{s,0,z}])

give me different results if I use ReplaceAll instead of writing
directly the value of z:

Y /. z -> 0
Out[41]=
{0.207031, 0.408203, 0.50293, 0.40625, 0.206543, 0., -0.00195313,
0.00195313, 0., 0.}

expA[0].(Y0 + Integrate[expA[-s].b[s], {s, 0, 0}])
Out[42]=
{0.206431, 0.408134, 0.503139, 0.405617, 0.206019, 0., 0., 0., 0., 0.}


Where expA[z] is a matrix (10x10 in this case), Y0 is a vector of
constants and b is a vector of functions. Since the last 5 components of
Y0 are all zero, the first result cannot be correct!
Why these 2 different results?

From: dh on 8 Apr 2010 08:04
On 07.04.2010 13:26, Alessio Giberti wrote:
> Hi all. A vector function of one variable, defined as follows,
>
> Y=expA[z].(Y0+Integrate[expA[-s].b[s],{s,0,z}])
>
> give me different results if I use ReplaceAll instead of writing
> directly the value of z:
>
> Y /. z -> 0
> Out[41]=
> {0.207031, 0.408203, 0.50293, 0.40625, 0.206543, 0., -0.00195313,
> 0.00195313, 0., 0.}
>
> expA[0].(Y0 + Integrate[expA[-s].b[s], {s, 0, 0}])
> Out[42]=
> {0.206431, 0.408134, 0.503139, 0.405617, 0.206019, 0., 0., 0., 0., 0.}
>
>
> Where expA[z] is a matrix (10x10 in this case), Y0 is a vector of
> constants and b is a vector of functions. Since the last 5 components of
> Y0 are all zero, the first result cannot be correct!
> Why these 2 different results?
>

Hi Alessio,
in the first case, the integral is calculated symbolically and
afterwards numerical values are substituted. My guess is that during the
evaluation of this expression we will have truncation errors.
On the other hand, the integral is evaluated numerically and the
algorithmus cares about the precision.
cheers, Daniel

--

Daniel Huber
Metrohm Ltd.
Oberdorfstr. 68
CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh(a)metrohm.com>
Internet:<http://www.metrohm.com>


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