zenterix

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2 years, 363 days

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These are questions asked by zenterix

Consider the following. 

We have a matrix A with five parameters in it.

If we tell Maple to solve Ax=0, it gives us the trivial solution.

Then, if I make a matrix B where I choose specific values for those five parameters and I tell Maple to solve Bx=0 I get a non-trivial solution.

Why doesn't Maple give me a more informative result in the Ax=0 case? 

A := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = a__1, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 0, (2, 5) = b__1, (3, 1) = -216, (3, 2) = -18, (3, 3) = 0, (3, 4) = 0, (3, 5) = c__1, (4, 1) = 0, (4, 2) = 0, (4, 3) = 1, (4, 4) = 0, (4, 5) = d__1, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 1, (5, 5) = e__1}) = Matrix(%id = 36893488151959194068)NULL

with(LinearAlgebra)

GaussianElimination(A)

Matrix(%id = 36893488151959184316)

(1)

LinearSolve(A, `<,>`(0, 0, 0, 0, 0))

Vector[column](%id = 36893488151959183100)

(2)

B := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = -216, (3, 2) = -18, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 1, (4, 4) = 0, (4, 5) = 1, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 1, (5, 5) = 4}) = Matrix([[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [-216, -18, 0, 0, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 4]])NULL

GaussianElimination(B) = Matrix(%id = 36893488151959174788)NULL

LinearSolve(B, `<,>`(0, 0, 0, 0, 0))

Vector[column](%id = 36893488151959164916)

(3)

``

NULL

An even easier way to show a non-trivial solution is to just make the entire last column zero.

Download polyn_matrix.mw

Consider the following worksheet (perhaps it is better to download the worksheet and execute since the contents below aren't showing the commands used to plot the last two plots).

NULL

T = log(R)/(a+b*log(R))^2

plot(log(R)/(-1.16+.675*log(R))^2, R = 1000 .. 30000)

 

NULL

10^log[10](T) = log(R)/(a+b*log(R))^2

NULL

log[10](T) = log[10](log(R)/(a+b*log(R))^2)

NULL

w = log[10](z/(b*z+a)^2)

w = ln(z/(b*z+a)^2)/ln(10)

(1)

NULL

plot(log[10](z/(-1.16+.675*z)^2), z = log(1000) .. log(30000))

 

NULL

``

NULL

plots:-loglogplot(log(R)/(-1.16+.675*log(R))^2, R = 1000 .. 30000)

 

NULL

 

My question is about making a loglog plot of the equation

for R between 1000 and 30000.

 

The second to last plot is of w as a function of z, as in the last equation below

and the plot command is (where I have subbed in a=-1.16 and b=0.675)

In the second to last plot, of course we have negative values of w. If we were to consider the underlying values of T, they would never be negative of course.

The last plot is the command

I think this last plot is what I want (though I am not sure because I am not totally sure what plots:-loglogplot is doing).

 

My question is how to obtain this loglog plot manually. That is, I want the axes to show values of R on the x axis and T on the y axis (just like a usual loglog plot shows).

In other words, how to go from the second to last plot to the same plot but showing the corresponding R and T values instead of z and w.

Download loglog.mw


Here is a small worksheet to illustrate my question

Consider the expression

 

solve(sqrt(log[10](R)/T) = a+b*log[10](R), T) = ln(R)*ln(10)/(ln(10)^2*a^2+2*ln(10)*ln(R)*a*b+ln(R)^2*b^2)NULL

NULL

Can we somehow tell Maple to keep the logarithms in base 10?

 

NULL


I'd like for the final expression to have only base 10 logarithms.

 

When I do it with pen and paper, I get

 

It is this last expression that I would like Maple to output.

Download log10.mw

I am creating a Maple document mode worksheet in which I use the Units package. I was doing a calculation and I noticed a discrepancy when I repeated the calculation slightly more manually (but still expecting the same result). 

Here is a link to the worksheet: Units.mw

(Unfortunately, it is hit or miss for me when I try to use the option to show the contents of the worksheet here directly)

Here is a screenshot of the issue

All I am doing in the second calculation is doing some of the unit conversions myself. 

I came across this while solving a chemistry problem, and I know the answer in the book agrees with the second calculation. 

So the question is why doesn't the first calculation, which uses more of Maple's library to do the calculation, agree with the second calculation?

Consider the task of finding the eigenvalues and eigenvectors of a simple 2x2 matrix.

Usually I can insert the contents of a Maple worksheet here, but for some reason the following worksheet cannot be inserted: Eigenvectors.mw

In that worksheet I try to use LinearAlgebra:-Eigenvectors(A). The eigenvalues contain a complex term, even though they are real for the given matrix. It's not clear what criteria are used in selecting a specific eigenvector for each eigenvalue.

I then show a more manual calculation.

I was expecting to obtain a simpler solution to this problem using LinearAlgebra:-Eigenvectors(A). Is this expectation unjustified? I am asking primarily from the perspective of a user of the software. 

It would be interesting to know the answer from the perspective of someone who knows the ins and outs of the implementation of the software as well, but as a user my initial expectation is a more digestible result that doesn't rely on knowing such implementation details.

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