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Re: Re: Uniform design
*To*: mathgroup at smc.vnet.net
*Subject*: [mg48387] Re: [mg48365] Re: Uniform design
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 26 May 2004 02:41:56 -0400 (EDT)
*References*: <c8pu2d$kq9$1@smc.vnet.net> <200405251118.HAA03729@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
As I do not have the time to discuss or even to consider these examples
in detail I will just comment on one, which I selected at random.
> z + O[n] /. z -> 1/n + 1
I have no idea why you think what you are doing is legitimate. It seems
to me that quite obviously it is not.
Let's first of all just look at the documentation:
In[6]:=
?SeriesData
SeriesData[x, x0, {a0, a1, ... }, nmin, nmax, den] represents \
a power series in the variable x about the point x0. The ai are the \
coefficients in the power series. The powers of (x-x0) that appear are \
nmin/den, (nmin+1)/den, ... , nmax/den.
Now look at:
FullForm[z+O[n]]
SeriesData[n,0,List[z],0,1,1]
To my mind this tell's you clearly that z is now considered as a
independent of n (coefficient of n^0) and obviously a substitution z->
1+ 1/n will break things.
Consider a somewhat different case:
In[10]:=
f=z[n] + O[n]
Out[10]=
SeriesData[n, 0, {z[0]}, 0, 1, 1]
Obviously this assumes that z has no singularities at 0. If you now
define:
z[n_]:=1/n
In[13]:=
f
Power::"infy" Infinite expression
encountered.
SeriesData[n, 0, {ComplexInfinity}, 0, 1, 1]
Is this a bug? Not to my mind. While it is probably not clearly stated
anywhere this sort of thing is not legitimate I think anyone who makes
even a slight effort to investigate how O[n] works and looks at the
SeriesData output would realize that doing any of the above is simply
asking for trouble (nonsense in -> nonsense out).
Since it seems that some people did not understand the intention of my
comments on Leibniz and Newton I feel I have to put it more bluntly.
I do not think Mathematica was ever meant to be "fool proof". A fool
can easily break it. It is of course even easier to do it for someone
who is not a fool but only pretending to be one. Mathematica offers you
a great deal more freedom than other programs of this kind and any kind
of freedom has costs, particularly for those who choose to turn off
their brains (not to mention those who do not have any).
Andrzej Kozlowski
On 25 May 2004, at 20:18, Maxim wrote:
> Continuing on the topic of uniform design, let's see how well various
> functions handle SeriesData. For instance, series expansions can be
> integrated just like other expressions:
>
> In[1]:=
> Integrate[Log[a] + x + O[a], {x, 0, 1}]
>
> Out[1]=
> a + O[a]^2
>
> The result is incorrect, probably because Integrate overlooks the
> possibility that SeriesData can contain logarithmic terms. Next, we
> can substitute values into series:
>
> In[2]:=
> z + O[n] /. z -> 1/n + 1
> z + O[n]^2 /. z -> 1/n + 1
>
> Out[2]=
> 1 + O[n]
>
> Out[3]=
> 1/n + 1 + O[n]^3
>
> Only in the first case we lose the 1/n term, in the second case lose
> O[n]^2 term. Let's look at Limit:
>
> In[4]:=
> Limit[O[n]/n^2, n -> 0]
>
> Out[4]=
> ComplexInfinity
>
> Strictly speaking, this is incorrect too, because n^2 is also O[n].
> Another issue is that Mathematica outputs O[n]/n^2 as 1/O[n], which I
> suppose can be acceptable as a conventional notation, but it's
> questionable in the strict mathematical sense: O(n)/n^2 is O(1/n) and
> not 1/O(n). Limit of 1/O(n) (not Mathematica's 1/O[n]!) would indeed
> be equal to ComplexInfinity. Another example:
>
> In[5]:=
> Limit[z/n + O[n], n -> 0]
>
> Out[5]=
> ComplexInfinity
>
> At least this one is good? Not exactly, because standard Mathematica
> behaviour in such cases is to return z*Infinity.
>
> In[6]:=
> HarmonicNumber[n, 2] + O[n, Infinity]
> -HarmonicNumber[n, 2] + O[n, Infinity]
>
> Out[6]=
> Pi^2/6 + O[1/n]^1
>
> Out[7]=
> -HarmonicNumber[n, 2] + O[1/n]^1
>
> The last example demonstrates a different issue: transformation rules
> catch only some of all the possible cases; here changing the sign
> prevents the application of the rule, sometimes f[n]+O[n] and
> Series[f[n],{n,0,0}] work slightly differently.
>
> Lastly, there are some other problems with notation: O[x]^0 is allowed
> in the output, but not as the input; O[1/x] will be output exactly as
> O[x,Infinity] but means something different.
>
> Maxim Rytin
> m.r at inbox.ru
> (don't use the prontomail address anymore please)
>
>
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