1 ACCEPTED SOLUTION
Accepted Solutions
As for the check: Both solutions pass, but there is some numeric roundoff for the positive one. Not surprising if you consider the coefficients of your quadratic equations are in the range of 10^12, 10^13 and you provide just 12 or 13 decimals of the constant in the denominator.
Generally numeric evaluation (as opposed to symbolic one) is limited to 16-17 decimals anyway.
Im really not quite sure what is going on in that sheet. I am a beginner and I think this is probably a bit easier than what that looks like. Here is what it can really simplify down to. The only two variables are Ka and x, the other things are just units. I want to solve for x.
Its basically a quadratic equation you want to solve. So there are two solution - given the value of Ka you supplied, one solution for x is negative around -10^13, the other is positive (around 0.666).
I used the symbolic solve and because I assumed you would be interested in the positive solution, I used the command assume, x>0 to omit the negative solution. If you delete this assume statement, you get both solutions in a 2x1 vector - give it try.
I also used the variable M in the equation instead of the copied value - hope thats what you intended.
In fact I first tried to use a solve block like Mike has done, but I wasn't able to find a guess value which would yield the positive solution.
There is a big drawback in using the symbolic solver - it doesn't know anything about units and sees m, L and mol as unknown variables. Its able to cancel mol, but not m^3 and L. Thats why I told the symbolic solver to substitute L by 10^-3 m^3. That way all units can be cancelled and we get the scalar solution as expected.
As you demanded also a symbolic solution, you either have to solve before you define Ka or make Ka unknown to the symbolics by writing Ka:=Ka.
Here you are - the symbolic solution, followed by both solution in their numeric representation:
As for the check: Both solutions pass, but there is some numeric roundoff for the positive one. Not surprising if you consider the coefficients of your quadratic equations are in the range of 10^12, 10^13 and you provide just 12 or 13 decimals of the constant in the denominator.
Generally numeric evaluation (as opposed to symbolic one) is limited to 16-17 decimals anyway.
Mike Armstrong wrote:
Great explanation Werner
Maybe, but at least partially wrong. The number of decimals provided for the constant has nothing to do with the roundoff errors.
And strange enough - forcing the symbolics in exact mode by providing the constant as fraction of integers makes it even worse - at least the numeric eval.
Also note the difference between the numeric and symbolic evaluation of Y.
