(a+b)^n is an expression, not an equation. What exactly do you mean by
'solving' it? Solving k=(a+b)^n for a or b is trivial -- solve( k=(a+b)^n,
a) works -- so I assume you mean something else.
-s
On Tue, Jan 4, 2011 at 23:19, razif razali <razif66 at gmail.com> wrote:
> what i means is how to solve those equation in terms of a and b...maybe in
> approximation solution if "n" can goes infinity....
>
> On Wed, Jan 5, 2011 at 10:31 AM, Stavros Macrakis <macrakis at alum.mit.edu>wrote:
>
>> What exactly do you mean by "solve (a+b)^n"?
>>
>> On Tue, Jan 4, 2011 at 21:28, razif razali <razif66 at gmail.com> wrote:
>>
>>> how to solve (a+b)^n in maxima?maybe the best approximation that we can
>>> have in maxima...can someone help with this
>>>
>>> --
>>> Regards,
>>>
>>> RAZIF RAZALI,
>>> Tutor & Master Student,
>>> Physics Department,
>>> Faculty Of Science,
>>> Universiti Teknologi Malaysia(UTM).
>>> +60199393606
>>>
>>>
>>>
>>>
>>> _______________________________________________
>>> Maxima mailing list
>>> Maxima at math.utexas.edu
>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>>
>>>
>>
>
>
> --
> Regards,
>
> RAZIF RAZALI,
> Tutor & Master Student,
> Physics Department,
> Faculty Of Science,
> Universiti Teknologi Malaysia(UTM).
> +60199393606
>
>
>
>