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Re: Getting crude approximation to a function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg58855] Re: Getting crude approximation to a function
*From*: Peter Pein <petsie at dordos.net>
*Date*: Thu, 21 Jul 2005 03:08:01 -0400 (EDT)
*References*: <dbkkpu$s6a$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Mukhtar Bekkali schrieb:
> Assume I have a function f[x], x is some variable, given below (my real
> function is much more complex). I would like to obtain its crude
> approximation. I used command Series, first order expansion. The
> resulting function has coefficients that have high precision. I do not
> need that since my expansion is very crude anyway. I need coefficients
> that are rational number approximations to these coefficients. How do I
> obtain this? It seems to me that command Chop takes care of
> coefficients that are not product with variable x but cannot handle
> coefficients that are not standalone. For instance, in this example
>
> \!\(\(\(Normal[
> Series[0.71 p\
> x + \(1\/3\)
> x\^2 - 4, {x, 1, 1}]] // Expand\) // Chop\) //
> Rationalize\)
>
> I would like to obtain output of the form (-13/10)+(29/10)x.
> Mathematica gives me (-13/10)+2.8972x instead, where it keeps
> 2.8971974507154195` in the memory. I need this because I use
> InequalitySolve package and it refuses to function unless all numbers
> are rational.
>
> Mukhtar Bekkali
>
Hi Mukhtar,
I don't find the definition of p, so I have to guess:
In[1]:= p = N[Expand[Pi*(1 + 34/(25*x))]];
In[2]:= appr = Expand[Normal[
Series[(p/Sqrt[2])*x + (1/3)*x^2 - 4, {x, 1, 1}]]]
Out[2]= -1.3121729353856435 + 2.8881081357458496*x
If you prefer the 10 in the denominator, try
In[3]:= appr /. a_Real :> Round[10*a]/10
Out[3]= -(13/10) + (29*x)/10
If you just want /any/ approximation, then
In[4]:= Rationalize[appr, 0.0125]
Out[4]= -(13/10) + (26*x)/9
does the job too.
--
Peter Pein
Berlin
http://people.freenet.de/Peter_Berlin/
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