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This one FLOORs me...

Wilmer

In Memoriam
Mar 19, 2012
376
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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.
 
Last edited:

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Well-known member
Jan 26, 2012
890
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

In Memoriam
Mar 19, 2012
376
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%3D0



These equations I'm told are for buckling of columns, in structural engineering!
Further comments appreciated!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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 $A\to 0.0000149558.$
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

In Memoriam
Mar 19, 2012
376
Thanks.

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

Agree?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

In Memoriam
Mar 19, 2012
376
How are you getting your range?
Sorry...was doing something different...not applicable here...

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