On 7/19/12 4:32 AM, Rupert Swarbrick wrote:
> Michele Minelli <micheleminelli1 at gmail.com> writes:
>> Hi everyone,
>> First of all I am italian so sorry for my English...
>> Anyway, my problem is quite simple (I think) but I'm getting mad on it!
>> I have to solve a quite simple equation where the unknown value is the
>> second bound of the integral.
>> I wrote this:
>>
>> solve(0.1=integrate(1/(2.9*(3.3-u)-(3.3-u)^2/2), u, 2.26, x), x);
>>
>> and this is the result:
> <snip>
>> (%o38)
>> [log(2*x+5)=(565040900*log(10*x-33)+565040900*log(-1)-1775128338*%i+113906055)/565040900]
>>
>> I answered positive because I know that the result should be 2.48 but
>> in the end it comes out with imaginary values...
>> I also tried to set logabs to true and solveradcan to true but nothing
>> good happened.
>> How can I solve that equation?
>> Thank you very much.
> Here's a quick attempt. First note that for symbolic calculations like
> integration etc., Maxima automatically rationalizes its arguments. Since
> the numbers then come out horrible, let's do it by hand:
I think this is a nicely done example of how to get maxima to do what
you want when maxima can't do it by itself.
I think having examples like this on the wiki would be very nice too. I
always learn a lot from examples like this, and such examples are easier
for me to find on a wiki than on a mailing list with lots of other
discussions.
Ray
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