Not a bug.
take a 5 term taylor series and add/subtract a 7 term taylor series and
the computation is done to 5 terms only.
This is right because expanding the 5 term taylor series to
+0*x^6+0*x^7 is unwarranted.
If you do something that takes the expression out of the taylor series
representation, so you have a 5 term
polynomial and a 7 term polynomial, then their difference can be
computed as a polynomial. (or a number).
It's a feature.
RJF
Sheldon Newhouse wrote:
> Jer?nimo Alaminos Prats wrote:
>
>> Hi all,
>> I've using maxima for about a month in a calculus class. I'm writing
>> some notes to help the students and now I'm having some trouble with
>> the behavior of Taylor series.
>> If I define a function and its Taylor polinomial, I have no problem
>> plotting both or calculating the difference
>>
>> (%i3) taylor(log(1+x),x,0,5);
>> (%o3) x-x^2/2+x^3/3-x^4/4+x^5/5+...
>> (%i5) define(g(x),taylor(f(x),x,0,5));
>> (%o5) g(x):=x-x^2/2+x^3/3-x^4/4+x^5/5+...
>> (%i6) g(2);
>> (%o6) 76/15
>> (%i8) f(2)-g(2);
>> (%o8) (15*log(3)-76)/15
>> but
>>
>> plot2d[f(x)-g(x),[x,0.1,5])
>>
>> plots the zero function. I can obtain the expected (at least for me)
>> behavior using
>>
>> plot2d(f(x)-trunc(g(x)),[x,0.1,5])
>>
>> Is this the right thing to do? I do not understand when it's mandatory
>> to use trunc with taylor and when it is optional? I mean, what's the
>> difference between taylor(...) and trunc(taylor(...))?
>>
>> If someone it's interested the notes (in spanish) are available as a
>> work in progress at
>>
>> http://www.ugr.es/~alaminos/docencia_2/calculo_telecomunicaciones/practicas_de_ordenador/
>>
>> Of course any comments are welcome.
>>
>> Thank you,
>> Jer?nimo.
>>
>>
>>
>> _______________________________________________
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>> Maxima at math.utexas.edu
>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>
>>
>>
> This looks like a bug to me.
> What do others think?
>
> For instance,
> (%i37) display2d: false;
>
> (%o37) false
> (%i38) f(x):= taylor(log(1+x),x,0,5);
>
> (%o38) f(x):=taylor(log(1+x),x,0,5)
> (%i39) g(x):= taylor(log(1+x),x,0,7);
>
> (%o39) g(x):=taylor(log(1+x),x,0,7)
> (%i40) f(x);
>
> (%o40) x-x^2/2+x^3/3-x^4/4+x^5/5
> (%i41) g(x);
>
> (%o41) x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/6+x^7/7
> (%i42) g(x)-f(x);
>
> (%o42) +0
> (%i43) ratexpand(g(x)) - ratexpand(f(x));
>
> (%o43) x^7/7-x^6/6
>
> Using 'ratexpand' seems to fix the problem, but I don't think that
> should be necessary.
>
> -sen
>
>
>
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