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Bad Solver
#1
Posted 19 February 2006 - 12:01 PM
solve((x+1)^2=0) --> {x=-1}
but i want to know how many times it happens...!!
i think OS should upgrade this
#2
Posted 19 February 2006 - 12:32 PM
i think that cp300 solve is very bad because
solve((x+1)^2=0) --> {x=-1}
but i want to know how many times it happens...!!
i think OS should upgrade this
Well the answer is good, no? (-1+1)^2=0^2=0
What do you mean with "how many times it happens"??? You want to konw how many solutions there are? You should just write dim(solve((x+1)^2=0)), as the returned value is a list!
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#3
Posted 20 February 2006 - 03:19 PM
#4
Guest_pabloec20_*
Posted 09 March 2006 - 04:08 PM
usin os 1 dunno if that has anything to do
#5
Posted 10 March 2006 - 05:55 PM
Because the original equation is quadratic it has two solutions, but they are both the same. The solution is x=-1, but this is a repeated root. I think most algebra systems will not give you the multiplicity of a root.i did the dim trick and i get "1" what does it mean? the multiplicity is 2 so its not the multiplicity what i am gettin...
usin os 1 dunno if that has anything to do
#6
Posted 10 March 2006 - 07:16 PM
#7
Posted 10 March 2006 - 10:43 PM
If a is a root of P(x) with multiplicity m, then a is also a root of dP(x)/dx with multiplicity m-1, d?P(x)/dx? with multiplicity m-2, ...
#8
Posted 11 March 2006 - 10:31 AM
you can differentiate (x+1)^2, it gives 2(x+1). As -1 is root of (x+1)^2 and of 2(x+1), you can say the multiplicity is at least 2
If a is a root of P(x) with multiplicity m, then a is also a root of dP(x)/dx with multiplicity m-1, d?P(x)/dx? with multiplicity m-2, ...
i know this idea but there is equations that you can not do this,something like (x+a)^2*(x+b )(x+c)
there could be a x=-a answer 3 times.. at last it is to slow to do this
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