Solving Nonlinear Equations with Numeric Methods

[Wilmer]

A = [1.25 * 10^(-9)] / [(27/2) * ((B + 2)*(C + 2))^(-2)]

B = FLOOR[1.29 * A^(-1/4)] - 2

C = FLOOR[A^(-1/4) * 4.476 * ((B + 2) / (11*B + 43))^(1/2)] - 2

3 equations, 3 unknowns. Solve for A,B,C.

I can only see numeric methods here.
Any insights? Thanks.

 

[Ackbach]

Several thoughts:

1. You could quite easily eliminate one variable, as all three equations are explicit. [EDIT]: See Bacterius below for a better answer.

2. You're most definitely not guaranteed a unique solution. The floor function is not 1-1, so inverting it can be problematic. You're not guaranteed that a solution exists, and if one does exist, you're not guaranteed that it is unique.

3. For a highly nonlinear system like this, I'm thinking numerical is your only option. The Excel Solver routine might give you good results, or maybe MATLAB or a MATLAB clone like GNU Octave if you can't afford MATLAB.

 

[Nono713]

You can actually eliminate two variables to solve an equation in A (express all B's in terms of A, then substitute the C in the first equation by the third equation). You can then obtain expressions for B (and then C) trivially. But this equation which has lots of floor functions and exponents is probably going to be difficult/impossible to rearrange for A, so I'd say numerical is probably easiest here.

 

[Ackbach]

You can actually eliminate two variables to solve an equation in A (express all B's in terms of A, then substitute the C in the first equation by the third equation). You can then obtain expressions for B (and then C) trivially. But this equation which has lots of floor functions and exponents is probably going to be difficult/impossible to rearrange for A, so I'd say numerical is probably easiest here.

Good catch!

 

[CaptainBlack]

I believe there is no solution.

Change the variables to a=1/A, B and C and the solution is a=0, B=-2, C=-2

(or my algebra could have gone horribly wrong, you could also help yourself by simplifying the equations)

CB

 

[Wilmer]

Thanks everyone...
I tried it, REMOVING the floor function, this way:

Let u = 1.25 * 10^(-9) / (27/2) , v = 1.29 , w = 4.476

Then the equations can be rewritten:
A = u * [(B + 2)(C + 2)]^2

B = v / A^(1/4) - 2

C = w / A^(1/4) * SQRT[(B + 2) / (11B + 43)] - 2

Manipulations/contortions(!) of above lead to:
11vA^2 + 21A^(9/4) - uv^3w^2 = 0 ; (I'm 99.9% sure that's correct!)

And that also seems to require numeric solving; you were correct in "many solutions";
Wolfram hands out 8 of them; with one real where A = .0000160191 :

http://www.wolframalpha.com/input/?i=11*1.29*a%5E2%2B21*a%5E%289%2F4%29-%281.25*10%5E%28-9%29%2F13.5%29*1.29%5E3*4.476%5E2%3D0These equations I'm told are for buckling of columns, in structural engineering!
Further comments appreciated!

 

[Ackbach]

We can simplify the expressions as follows:
\begin{align*}
A&=\frac{5}{54}\,\times 10^{-9}(B+2)^{2}(C+2)^{2}\\
B&=\left\lfloor \frac{1.29}{\sqrt[4]{A}}\right\rfloor-2\\
C&=\left\lfloor\frac{4.476}{\sqrt[4]{A}}\sqrt{\frac{B+2}{11B+43}}\right\rfloor
-2
\end{align*}
Now we follow Bacterius's suggestion to obtain the following single equation
for $A$:
$$A=\frac{5\times 10^{-9}}{54}\left\lfloor\frac{1.29}{\sqrt[4]{A}}\right\rfloor^{2}
\times\left\lfloor\frac{4.476}{\sqrt[4]{A}}\sqrt{\frac{\left\lfloor\frac{1.29}{\sqrt[4]{A}}\right\rfloor}{11\left\lfloor\frac{1.29}{\sqrt[4]{A}}\right\rfloor+21}}\right\rfloor^{2}.$$
If you plot the LHS and the RHS on the same axis, you do get an intersection
on a horizontal portion of the RHS's graph at around $A=0.000015$. Using the
FindRoot command in Mathematica yielded the solution
$$A\to 0.0000150371,\quad B\to 18.,\quad C\to 18.$$
Naturally, $B$ and $C$ must be integers.
If I use the FindRoot command just on the $A$ equation, I get a more refined value of $A\to 0.0000149558.$

 

[Ackbach]

On even more reflection, I believe you can come up with an exact answer. Let $B=C=18$. Then $A=1.48\overline{148}\times 10^{-5}$ satisfies all equations.

 

[Wilmer]

Thanks.

However, B=18 and C=18 results in range A = 1.43000 * 10^(-5) to 1.72000 * 10^(-5)

Agree?

 

[Ackbach]

Thanks.

However, B=18 and C=18 results in range A = 1.43000 * 10^(-5) to 1.72000 * 10^(-5)

Agree?

You get an exact value from plugging $B=18$ and $C=18$ into the first equation. There's no range. I get $A=1/67500$. How are you getting your range?

 

[Wilmer]

How are you getting your range?

Sorry...was doing something different...not applicable here...

Thanks everyone for the responses. All's fine now.