I have a program to find principal stresses. In it, is a cubic equation that I would like to solve and display 3 results. Here are my questions:
1. is solve() function correct one to use in ti-basic to solve the cubic equation? if not, which is correct?
2. When executing the program in home screen how to display three values of cubic equation?
Here is my program:
princstr(xx,yy,zz,xy,xz,yz)
Prgm
a = xx+yy+zz
b = xx*yy+yy*zz+zz*xx-xy^2-yz^2+xz^2
c = xx*yy*zz+xy*yz*xz+xz*xy*yz-xz*yy*xz-xz*yy*xz-yz*yz*xx-zz*xy*xy
solve(s^3-a*s^2+b*s-c=0,s)
Return s
EndPrgm
In the home screen it just returs
Done
Thank you in advance!
P.s. Edits are welcome.
You are not displaying S. Return does not work like return in other languages. In TI-Basic, Return does not return a value; it return control to the calling program. If you want to display it, you should do
:Disp S
:Return
Related
I am Trying to solve a problem in hacker rank, Which title is breaking best and worst. but the function I write didn't give proper output for any of the test case but in my visual code ide it is giving the correct result.
The problem is find out the sequence of increasing number and decreasing. finally return the result in two space separated integers. I've write the function to the point and my IDE is giving the accurate result but in hacker rank it's giving error
here is the link of problem
my solution approach
function breakingRecords(scores) {
let increase = 0;
let decrease =0;
{code that properly increase the value of
increase and decrease.}
}
const result = (increase,decrase);
return result;
}
What I made mistake in my code is I return my value normally in two variable but those value I have to return in array like
const result = [increase,decrase];
return result;
That's it!!
I am relatively new to Modelica (Dymola-environment) and I am getting very desperate/upset that I cannot solve such a simple problem as a random number generation in Modelica and I hope that you can help me out.
The simple function random produces a random number between 0 and 1 with an input seed seedIn[3] and produces the output seed seedOut[3] for the next time step or event. The call
(z,seedOut) = random(seedIn);
works perfectly fine.
The problem is that I cannot find a way in Modelica to compute this assignment over time by using the seedOut[3] as the next seedIn[3], which is very frustrating.
My simple program looks like this:
*model Randomgenerator
Real z;
Integer seedIn[3]( start={1,23,131},fixed=true), seedOut[3];
equation
(z,seedOut) = random(seedIn);
algorithm
seedIn := seedOut;
end Randomgenerator;*
I have tried nearly all possibilities with algorithm assignments, initial conditions and equations but none of them works. I just simply want to use seedOut in the next time step. One problem seems to be that when entering into the algorithm section, neither the initial conditions nor the values from the equation section are used.
Using the 'sample' and 'reinit' functions the code below will calculate a new random number at the frequency specified in 'sample'. Note the way of defining the "start value" of seedIn.
model Randomgenerator
Real seedIn[3] = {1,23,131};
Real z;
Real[3] seedOut;
equation
(z,seedOut) = random(seedIn);
when sample(1,1) then
reinit(seedIn,pre(seedOut));
end when;
end Randomgenerator;
The 'pre' function allows the use of the previous value of the variable. If this was not used, the output 'z' would have returned a constant value. Two things regarding the 'reinint' function, it requires use of 'when' and requires 'Real' variables/expressions hence seedIn and seedOut are now defined as 'Real'.
The simple "random" generator I used was:
function random
input Real[3] seedIn;
output Real z;
output Real[3] seedOut;
algorithm
seedOut[1] :=seedIn[1] + 1;
seedOut[2] :=seedIn[2] + 5;
seedOut[3] :=seedIn[3] + 10;
z :=(0.1*seedIn[1] + 0.2*seedIn[2] + 0.3*seedIn[3])/(0.5*sum(seedIn));
end random;
Surely there are other ways depending on the application to perform this operation. At least this will give you something to start with. Hope it helps.
When I run the following code, the two output matrices (diffInDiffOne & diffInDiffTwo) are the same. My guess is that coeffs is not being replaced after each loop but I have no idea why . I think that the coefficients matrix is being overwritten but I have no idea how. I tried changing the for loop order but this surprisingly didn't solve my issue either:
local treatments treat_one treat_two
matrix diffInDiffOne = J(1,9,.)
matrix diffInDiffTwo = J(1,9,.)
foreach treatment in `treatments' {
reg science inSchool#`treatment'#male
matrix coeffs=e(b)
if treat_one==`treatment'{
matrix diffInDiffOne = diffInDiffOne\coeffs
}
if treat_two==`treatment'{
matrix diffInDiffTwo = diffInDiffTwo\coeffs
}
}
matrix list diffInDiffOne
matrix list diffInDiffTwo
When I list the matrix they are both the same, depsite the fact that two regressions give different answers. Any help with this issue is much appreciated. Thanks
This code appears at first sight to reduce to
reg science inSchool#treat_one#male
matrix li e(b)
reg science inSchool#treat_two#male
matrix li e(b)
apart from the detail of adding nine missing values to the matrix.
However, that is not your code, so what is biting you? I guess at something much more subtle.
You should need to be very careful with the if command. Variables evaluated in if commands are evaluated in their first observation. So, the first time round the loop
the conditions are
if treat_one[1] == treat_one[1]
if treat_two[1] == treat_one[1]
The second time, it is
if treat_one[1] == treat_two[1]
if treat_two[1] == treat_two[1]
If it is true in your data that treat_one[1] == treat_two[1] the effect will not be as you may imagine.
If you want to test for equality of strings, do something like
if "`treatment'" == "treat_one"
You may have in mind something more like
foreach treatment in treat_one treat_two {
reg science inSchool#`treatment'#male
matrix `treatment' = e(b)
matrix list `treatment`
}
You seem to be wanting to write very complicated code for rather simple problems. A while back, I recommended thinking in terms of do-files rather than programs. That may be advice to reconsider.
I have a simple SAS (version 9.2) program as follows,
proc model;
cdf('normal',log(V/100)+1)=0.5;
bounds V>0;
solve V/solveprint;
run;
It throws exception that says jacobian matrix to be singular,
The Newton method Jacobian matrix of partial derivatives of the
equations with respect to the variables to be solved is singular.
What is the possible cause of this error?
Update: I have simplified the problem a bit. When modified to "cdf('normal', X)=0.5", it works without exception.
Update2: bounds is updated to V>0; but exception still there
What input data set are you passing to proc model? For example, this code works consistently:
data a;
v=100;
run;
proc model data=a;
cdf('normal',log(V/100)+1) = 0.5;
bounds V>0;
solve V / solveprint;
run;
quit;
And gives a solution of V=36.78794
But changing the input data somewhat (see below) will consistently give a singular Jacobian matrix error.
data a;
v=0.00001;
run;
proc model data=a;
cdf('normal',log(V/100)+1) = 0.5;
bounds V>0;
solve V / solveprint;
run;
quit;
You are asking SAS to solve a function that has no solution. You are asking for the value of V>1000 that makes this equation true. But there are no such values because log(1000/100+1) is about 3.3, and the CDF of a Normal random variable with mean 0 and standard deviation 1 evaluated at 3.3 is 0.9995. Any larger value of V will just move the function closer to 1, not toward 0.5, so there is no answer to your question.
By telling you that the matrix of partial derivatives is singular, SAS is just using fancy math speak for "your function doesn't have a solution". (Really what it's saying is, "I've turned your question into an equivalent maximization problem, and that problem doesn't have a maximum, so I can't help you.")
I am using VBScript in a Manifold GIS Database to verify the distance between two longitude and latitude points using Trig function. The script for finding the distance run without any problems but the script for verifying whether the O-D is valid had syntax error. I really hope any of you could help me with this problem.
There were 2 new active columns that I created : Distance and VerifyDistance. Distance column was created to find the distance using Trig function. It ran successfuly but may need improvements in its messy structure. The VerifyDistance used the IF conditions and checked the value in another column named "Valid O/D". If the value is "OK", it will return the Distance value; otherwise, it will return a text value saying "O-D points are invalid" in this column.
As you can see, I am a newb in programming language. The VBScript can be seen below:
Function Distance
Distance = sqr((111.21*Record.Data("Work Y-coord") - 111.21*Record.Data("Home Y-coord"))^2 + (85.30*Record.Data("Work X-coord") - 85.30*Record.Data("Home X-coord"))^2)
End Function
Function VerifyDistance
If Record.Data("Valid O/D") = "OK"
VerifyDistance = Record.Data("Distance")
'document.write("Invalid O-D Points")
Else
VerifyDistance = "O-D Points are invalid."
End If
End Function
You are missing Then here.
If Record.Data("Valid O/D") = "OK" Then