Simplifying equations in Maple - math

I am using maple (version 12) for the first time.
I want to invert the equations using any maths tool. I tried it using MAPLE.
I have given two equations in this form:
a=x*b^r + y*(b^s)*(c^t) + z*c^u
Now I want to flip this equation (invert this equation)
I want two equations for a and b in terms of c
e.g
a= 0.98989c^2 + 78c + 87
b= 56c + 89
These are not my results. I just gave an example.
Any advice?
My two equations are:
Equation#1 is:
a= 1.093*(b^0.002939) - 0.1887*(b^0.7637)*(c^0.2306) - 0.04425 *(c^0.9143)
Equation#2 is:
a= 1.088*(b^0.003058) -0.1813*(b^0.7627)*(c^0.2501) -0.03958 *(c^0.9365)
From Equation#1, i can find b equation. I put the bequation in Equation#2 . In this way equation#2 has only a and c variables. i can get a equation. and then by putting this equation can get b equation ( with only c variables)
But these equations are very complex and it takes a lot of time to evaluate in MAPLE. How to simplify it? I tried to use natural log, but it doesnot give any answer. It just repeat my equation in command window.
ln(1.093*b^0.2939e-2-.1887*b^.7637*c^.2306-0.4425e-1*c^.9143)
Please guide me.

Related

Solving system of ODEs in vector/matrix form in R (with deSolve?)

So I want to ask whether there's any way to define and solve a system of differential equations in R using matrix notation.
I know usually you do something like
lotka-volterra <- function(t,a,b,c,d,x,y){
dx <- ax + bxy
dy <- dxy - cy
return(list(c(dx,dy)))
}
But I want to do
lotka-volterra <- function(t,M,v,x){
dx <- x * M%*% x + v * x
return(list(dx))
}
where x is a vector of length 2, M is a 2*2 matrix and v is a vector of length 2. I.e. I want to define the system of differential equations using matrix/vector notation.
This is important because my system is significantly more complex, and I don't want to define 11 different differential equations with 100+ parameters rather than 1 differential equation with 1 matrix of interaction parameters and 1 vector of growth parameters.
I can define the function as above, but when it comes to using ode function from deSolve, there is an expectation of parms which should be passed as a named vector of parameters, which of course does not accept non-scalar values.
Is this at all possible in R with deSolve, or another package? If not I'll look into perhaps using MATLAB or Python, though I don't know how it's done in either of those languages either at present.
Many thanks,
H

With my low reputation (points), I apologize for posting this as an answer which supposedly should be just a comment. Going back, have you tried this link? In addition, in an attempt to find an alternative solution to your problem, have you tried MANOPT, a toolbox of MATLAB? It's actually open source just like R. I encountered MANOPT on a paper whose problem boils down to solving a system of ODEs involving purely matrices.

Simple Orthographic Structure from Motion using R -- Determining Metric Constraints

I would like to build a simple structure from motion program according to Tomasi and Kanade [1992]. The article can be found below:
https://people.eecs.berkeley.edu/~yang/courses/cs294-6/papers/TomasiC_Shape%20and%20motion%20from%20image%20streams%20under%20orthography.pdf
This method seems elegant and simple, however, I am having trouble calculating the metric constraints outlined in equation 16 of the above reference.
I am using R and have outlined my work thus far below:
Given a set of images
I want to track the corners of the three cabinet doors and the one picture (black points on images). First we read in the points as a matrix w where
Ultimately, we want to factorize w into a rotation matrix R and shape matrix S that describe the 3 dimensional points. I will spare as many details as I can but a complete description of the maths can be gleaned from the Tomasi and Kanade [1992] paper.
I supply w below:
w.vector=c(0.2076,0.1369,0.1918,0.1862,0.1741,0.1434,0.176,0.1723,0.2047,0.233,0.3593,0.3668,0.3744,0.3593,0.3876,0.3574,0.3639,0.3062,0.3295,0.3267,0.3128,0.2811,0.2979,0.2876,0.2782,0.2876,0.3838,0.3819,0.3819,0.3649,0.3913,0.3555,0.3593,0.2997,0.3202,0.3137,0.31,0.2718,0.2895,0.2867,0.825,0.7703,0.742,0.7251,0.7232,0.7138,0.7345,0.6911,0.1937,0.1248,0.1723,0.1741,0.1657,0.1313,0.162,0.1657,0.8834,0.8118,0.7552,0.727,0.7364,0.7232,0.7288,0.6892,0.4309,0.3798,0.4021,0.3965,0.3844,0.3546,0.3695,0.3583,0.314,0.3065,0.3989,0.3876,0.3857,0.3781,0.3989,0.3593,0.5184,0.4849,0.5147,0.5193,0.5109,0.4812,0.4979,0.4849,0.3536,0.3517,0.4121,0.3951,0.3951,0.3781,0.397,0.348,0.5175,0.484,0.5091,0.5147,0.5128,0.4784,0.4905,0.4821,0.7722,0.7326,0.7326,0.7232,0.7232,0.7119,0.7402,0.7006,0.4281,0.3779,0.3918,0.3863,0.3825,0.3472,0.3611,0.3537,0.8043,0.7628,0.7458,0.7288,0.727,0.7213,0.7364,0.6949,0.5789,0.5491,0.5761,0.5817,0.5733,0.5444,0.5537,0.5379,0.3649,0.3536,0.4177,0.3951,0.3857,0.3819,0.397,0.3461,0.697,0.671,0.6821,0.6821,0.6719,0.6412,0.6468,0.6235,0.3744,0.3649,0.4159,0.3819,0.3781,0.3612,0.3763,0.314,0.7008,0.6691,0.6794,0.6812,0.6747,0.6393,0.6412,0.6235,0.7571,0.7345,0.7439,0.7496,0.7402,0.742,0.7647,0.7213,0.5817,0.5463,0.5696,0.5779,0.5761,0.5398,0.551,0.5398,0.7665,0.7326,0.7439,0.7345,0.7288,0.727,0.7515,0.7062,0.8301,0.818,0.8571,0.8878,0.8766,0.8561,0.858,0.8394,0.4121,0.3876,0.4347,0.397,0.38,0.3631,0.3668,0.2971,0.912,0.8962,0.9185,0.939,0.9259,0.898,0.8887,0.8571,0.3989,0.3781,0.4215,0.3725,0.3612,0.3461,0.3423,0.2782,0.9092,0.8952,0.9176,0.9399,0.925,0.8971,0.8887,0.8571,0.4743,0.4536,0.4894,0.4517,0.446,0.4328,0.4385,0.3706,0.8273,0.8171,0.8571,0.8878,0.8766,0.8543,0.8561,0.8394,0.4743,0.4554,0.4969,0.4668,0.4536,0.4404,0.4536,0.3857)
w=matrix(w.vector,ncol=16,nrow=16,byrow=FALSE)
Then create registered measurement matrix wm according to equation 2 as
by
wm = w - rowMeans(w)
We can decompose wm into a '2FxP' matrix o1 a diagonal 'PxP' matrix e and 'PxP' matrix o2 by using a singular value decomposition.
svdwm <- svd(wm)
o1 <- svdwm$u
e <- diag(svdwm$d)
o2 <- t(svdwm$v) ## dont forget the transpose!
However, because of noise, we only pay attention to the first 3 columns of o1, first 3 values of e and the first 3 rows of o2 by:
o1p <- svdwm$u[,1:3]
ep <- diag(svdwm$d[1:3])
o2p <- t(svdwm$v)[1:3,] ## dont forget the transpose!
Now we can solve for our rhat and shat in equation (14)
by
rhat <- o1p%*%ep^(1/2)
shat <- ep^(1/2) %*% o2p
However, these results are not unique and we still need to solve for R and S by equation (15)
by using the metric constraints of equation (16)
Now I need to find Q. I believe there are two potential methods but am unclear how to employ either.
Method 1 involves solving for B where B=Q%*%solve(Q) then using Cholesky decomposition to find Q. Method 1 appears to be the common choice in literature, however, little detail is given as to how to actually solve the linear system. It is apparent that B is a '3x3' symmetric matrix of 6 unknowns. However, given the metric constraints (equations 16), I don't know how to solve for 6 unknowns given 3 equations. Am I forgetting a property of symmetric matrices?
Method II involves using non-linear methods to estimate Q and is less commonly used in structure from motion literature.
Can anyone offer some advice as to how to go about solving this problem? Thanks in advance and let me know if I need to be more clear in my question.

can be written as .
can be written as .
can be written as .
so our equations are:
So the first equation can be written as:
which is equivalent to
To keep it short we define now:
(I know the spacings are terrably small, but yes, this is a Vector...)
So for all equations in all different Frames f, we can write one big equation:
(sorry for the ugly formulas...)
Now you just need to solve the -Matrix using Cholesky decomposition or whatever...

What is an algorithm for solving quadratic with linear equation simultaneously?

Need an algorithm to solve equation like these:
8*p^2+8*q^2+8*p-16=0
p^2+q^2+12*p+20=0
If anyone can point me to the name of algorithm also it'll be enough. I could have followed any matrix related algorithm but the quadratic inside these linear are causing a problem.

The Nelder-Mead Algorithm is a very easy-to-program non-linear solver that converges quickly and accurately in low dimensional problems like this one. The pseudo-code is easy to follow on Wikipedia, but you can find some coded implementations online as well.
Your cost function in this case would be the sum of the left hand sides' squares to ensure that lower costs are closer to the correct solutions.

In the case you wrote about, I'd suggest first subtracting 8 times the second equation from the first.
0 = (8*p^2+8*q^2+8*p-16) - 8*(p^2+q^2+12*p+20) = -88*p-176 = 0
p = -2
Then you are left with a simple quadratic equation in q, which you can solve using the common methods for solving quadratic equations in a single variable.

It's a non-linear problem. You'll need an iterative solution, something like Newton Raphson or BFGS.
Here's a simple approach.
Start by linearizing it:
16*p*dp + 16*q*dq + 8 = 0
2*p*dp + 2*q*dq + 12 = 0
This becomes a linear algebra problem:
| 16*p 16*q |[ dp ] [ -8 ]
| 2*p 2*q |[ dq ] = [ -12 ]
You'll start with an initial guess for (p, q):
(p, q) = (p0, q0)
You'll solve for dp and dq, using the usual linear algebra techniques.
You'll update your guesses for p and q:
(p1, q1) = (p0 + dp, q0 + dq)
You'll iterate again until your solution either diverges or converges.
There's no guarantee that it'll converge, depending on your initial guess.

VBA How to solve two equations with two unknowns

I am trying to calculate point on a line.
I got the points of the edges and one distance between one edge to the point I want to find (which is B).
A(2,4)
B(x,y)
C(4,32)
The distance between A to B is 5.
How can I calculate Bx and By? using the following equations:
d = Math.Sqr((Bx-Ax)^2 + (By-Ay)^2)
d = Math.Sqr((Cx-Bx)^2 + (Cy-By)^2)
and than compare the equations above.
Here is the equations with the points placed:
5 = Math.Sqr((Bx-2)^2 + (By-4)^2)
23.0713366 = Math.Sqr((4-Bx)^2 + (32-By)^2)
or
Math.Sqr((Bx-2)^2 + (By-4)^2) - 5 = Math.Sqr((4-Bx)^2 + (32-By)^2) - 23.0713377
How can I solve this using VBA?
Thank you!

I won't solve your equations above because they are an unnecessarily complex way to state the problem (and the existence of a solution is questionable in the presence of rounding), but all the points on the line A=(Ax,Ay) to C=(Cx,Cy) can be described as B=(Ax,Ay) + t*(Cx-Ax,Cy-Ay) with t between 0 and 1.
The distance between B and A is then given by d=t*Sqrt((Cx-Ax)^2+(Cy-Ay)^2), which you can invert to get the proper t for a given d - t=d/Sqrt((Cx-Ax)^2+(Cy-Ay)^2)
In your case, B(t) = (2,4) + t*(2,28), t=5/Sqrt(2^2+28^2) ~ 0.178 -> B ~ (2,4) + 0.178 * (2,28) ~ (2.356, 8.987).

VBA has no Symbolic Language capability. To solve this problem, there are different approach :
Transform the equations to isolate one of the unknowns, most likely to use substitution, and compute it (I recommend this for your problem.)
Transform your functions and derive them to use Newton's methods (don't do this, it's overkill.)
Use a "brute force" convergence methods : Fix a min/max for each variable and use bisection methods to find what you want (I don't recommend this because you'll most likely "fall" into a local minimum/maximum in your case.)
So basically, I'd say you go with the first way. It requires 15mins of tinkering with mathematical equations, then you're set to go.

Second order nonlinear differential equations using MATLAB

I'm trying to input a second order differential equation to solve into matlab over x = 0 to x =1. I can't figure out how. Here's the equation:
y'' = 1 + 0.1 \sqrt{1+(y')^2}
with initial conditions at zero.

Normally you solve higher-order equations by converting to a system of first order equations. Here, you would define:
y' = v
v' = 1 + 0.1 \sqrt{1 + v^2}
Define a function computing the right-hand side, and use ode45.
Note that this equation is solvable without much trouble in closed form, too, so should be a good test for how to do it.

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