Maple Questions and Posts

How to obtain multiple solution...

Asked by: Madhukesh J K 140

October 07 2020

1 1

How to obtain the multiple solution and graph given in the paper.

Stefan Blowing and Slip Effects on Unsteady Nanofluid Transport Past a Shrinking Sheet: Multiple Solutions

https://doi.org/10.1002/htj.21470

Can anyone help to get solutions.

Maple helps in simulating multiscale-multidomain...

by: Dr. Venkat Subramanian 431 Product: Maple

October 06 2020

1 0

We recently published a paper on multiscale-multidomain simulation of battery models.

https://iopscience.iop.org/article/10.1149/1945-7111/abb37b

Some challenges are listed at

https://twitter.com/UT_MAPLE/status/1311356957941522438

Maple and symbolic math can play a critical role in solving many challenging problems. For example, consider a seemingly-trivial problem

uxx+uyy = 0

x = 0 and x = 1, ux = 0 for all y

y = 1, u = 0, for all x

y = 0, u =1, 0<=x<=0.5

y = 0, uy = 0, 0.5<x<=1

There is a singularity at (0.5,0) and most numerical methods will have trouble there. In these equations, uxx means the second derivative of u with respect to x.

Maple can help solve this problem with conformal mapping to achieve arbitrary precision. As of today, machine precision is not possible with any numerical method even with millions of Degrees of Freedom. The Maple code is given below. A FEM code is given below as well. Models like this can benefit from Maple adding parallel sparse direct and iterative solvers.

>

restart;

>	Digits:=15;

(1)

The domain is tranformed from Z = X+IY to w. The points tranformed are

>	Zdomain:=[[0,0],[1,0],[1,1],[0,1]];

(2)

>	wdomain:=[0,1,1+a,a+2];

(3)

>	eq1:=diff(Z(w),w)=-I*K1/sqrt(w)/sqrt(w-1)/sqrt(w-a-1)/sqrt(w-a-2);

(4)

a is not known apriori. The value of a should be found to make sure [1,1] in the Z coordinate is transformed to [1+a] in the w coordinate.

>	a:=sqrt(2)-1;

(5)

Value of K1 is found using the transformation of [1,0] to 1 in the w coordinate

>	eq11:=1=int(rhs(eq1),w=0..1);

(6)

>	K1:=solve(eq11,K1);

K1 := -(1/2)*(2^(1/2)/EllipticK(2^(1/2)/2))

(7)

The height Y in the Z coordinate is found by integrating from 1 to 1+a in the w coordinate.

>	simplify(int(rhs(eq1),w=1..1+a));

(2*I)*2^(1/2)*EllipticK((2^(1/2)-1)/(2^(1/2)+1))/(EllipticK(2^(1/2)/2)*(2^(1/2)+1))

(8)

>

evalf(%);

(9)

The choice of a = sqrt(2)-1 gives the height of 1 for Z coordinate.

>	eval(simplify(int(rhs(eq1),w=0..1+a)));

(2^(1/2)*EllipticK(2^(1/2)/2)+EllipticK(2^(1/2)/2)+(2*I)*2^(1/2)*EllipticK((2^(1/2)-1)/(2^(1/2)+1)))/(EllipticK(2^(1/2)/2)*(2^(1/2)+1))

(10)

>

evalf(%);

(11)

integration from 0 to 1+a in the w coordinate gives 1,1 in the Z coordinate.

>	simplify(int(rhs(eq1),w=0..1/sqrt(2)));

EllipticF(2^(1/2)/(2+2^(1/2))^(1/2), 2^(1/2)/2)/EllipticK(2^(1/2)/2)

(12)

>

evalf(%);

(13)

Integrating w from 0 to wmid =1/sqrt(2) gives the point 0.5,0 in the Z coordinate

>	wmid:=1/sqrt(2);

(14)

Next w domain is transformed to Z2 domain Z2 = X2+IY2

>	Z2domain:=[[0,0],[1,0],[1,H],[0,H]];

(15)

>	wdomain2:=[0,1/sqrt(2),1+a,a+2];

(16)

>	eq2:=diff(Z2(w),w)=-I*K2/sqrt(w)/sqrt(w-wmid)/sqrt(w-1-a)/sqrt(w-a-2);

eq2 := diff(Z2(w), w) = -(2*I)*K2/(w^(1/2)*(4*w-2*2^(1/2))^(1/2)*(w-2^(1/2))^(1/2)*(w-2^(1/2)-1)^(1/2))

(17)

K2 is found based on the transformation of wmid to [1,0] in Z2 coordinate.

>	eq21:=1=int(rhs(eq2),w=0..wmid);

eq21 := 1 = -2*K2*EllipticK(1/(((2+2^(1/2))^(1/2))))/(2^(1/2)+1)^(1/2)

(18)

>	K2:=solve(eq21,K2);

K2 := -(1/2)*((2^(1/2)+1)^(1/2)/EllipticK(1/(((2+2^(1/2))^(1/2)))))

(19)

The corner 0,1 in the Z coordinate is mapped by integrating eq2 from 0 to 1 in the w coordinate

>	int(rhs(eq2),w=0..1);

1+EllipticF((2-2^(1/2))^(1/2), ((2^(1/2)+1)/(2+2^(1/2)))^(1/2))*I/EllipticK(1/(((2+2^(1/2))^(1/2))))

(20)

>	corner:=evalf(%);

(21)

This is the point 1,0 in the original coordinate.

The height in the Z2 coorinate is found by integrating eq2 from wmid to 1+a.

>	ytot:=int(rhs(eq2),w=wmid..1+a);

ytot := EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))*I/EllipticK(1/(((2+2^(1/2))^(1/2))))

(22)

>

evalf(%);

(23)

The magnitude in the Y direction is given by the coefficient of I, the imaginary number

>	Ytot:=coeff(ytot,I);

Ytot := EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))/EllipticK(1/(((2+2^(1/2))^(1/2))))

(24)

The analytial solution in the Z2 corordinate is a line in Y2 to satisfy simple zero flux conditions at X2 = 0 and X2 = 1.

>	phianal:=1+b*Y2;

(25)

The value of phi is zero at Y2 = Ytot (originally the cathode domain in the Z domain)

>	bc:=subs(Y2=Ytot,phianal)=0;

bc := 1+b*EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))/EllipticK(1/(((2+2^(1/2))^(1/2)))) = 0

(26)

>	b:=solve(bc,b);

b := -EllipticK(1/(((2+2^(1/2))^(1/2))))/EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))

(27)

The analytical solution is given by

>

phianal;

1-EllipticK(1/(((2+2^(1/2))^(1/2))))*Y2/EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))

(28)

>	evalf(phianal);

(29)

The potential at the corner is given by substituting the imaginary value of corner for Y2 in phinanal)

>	phicorner:=subs(Y2=Im(corner),phianal);

phicorner := 1-.559419351518322*EllipticK(1/(((2+2^(1/2))^(1/2))))/EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))

(30)

>	evalf(phicorner);

(31)

local flux/current density calculation, written in terms of w is

>	curr:=b*rhs(eq2)/rhs(eq1);

curr := -(2^(1/2)+1)^(1/2)*2^(1/2)*EllipticK(2^(1/2)/2)*(w-1)^(1/2)/(EllipticK(((2^(1/2)+1)/(2+2^(1/2)))^(1/2))*(4*w-2*2^(1/2))^(1/2))

(32)

average flux/current density calculation for the anode

>	currave:=int((curr),w=0..wmid)/wmid;

currave := -(1/2)*((2^(1/2)+1)^(1/2)*(2^(1/2)-1)*EllipticK(2^(1/2)/2)*(2^(3/4)+2^(1/4)+arctanh(2^(3/4)/2))*2^(1/2)/EllipticK(2^(3/4)/2))

(33)

>	Digits:=25:

The average current density at Y =0, local current density at X = 0,Y=0 and potential at X=1,Y=0 (Corner) can be used to study convergence of FEM and other numerical methods

>	evalf(currave),evalf(subs(w=0,curr)),evalf(phicorner);

(34)

>

Download Conformalmapping.mws

This FEM code is for solving Laplace's equation with primary current distribution considered in Model 1.
This code is based on FEM weak-form. Biquadratic Lagrange shape functions (9nodes in an element) are used.

>

restart;

>	with(LinearAlgebra):

>	Lx:=1: #length in X

>	Ly:=1: #length in Y

>	nx:=10: #number of elements in X (even numbers only)

>	ny:=10: #number of elements in Y, to be kept same as nx in this version

>	hx:=Lx/nx: #element size x

>	hy:=Ly/ny: #element size y

Procedure to perform numerical integration on shape functions to obtain local matrices (can be replaced with analytical integration for this particular problem)
-Shape functions are also used as weight functions in applying weak formulation. Numerical integration is done using Simpson's rule.
-Local cartesian coordinates x,y are converted to natural coordinates zeta and eta. This transformation is not required for this simple geomerty but useful in general. zeta and eta are obtained by scaling x and y with hx/2 and hy/2, respectively, in this code.

>	localmatrices:=proc(a1,a2,a3,b1,b2,b3,q1,q2) global Kx,Ky,Nx,Ny,zeta,eta,c; local A,dAdzeta,dAdeta,y,x,J,terms,i,j,k,l,dx,dy,fx,fy,fxy,fyy,dzeta,deta,J1,J2;

>	A:=[(1-zeta)zeta(1-eta)eta/4,-(1-(zeta)^2)(1-eta)eta/2,-(1+zeta)zeta(1-eta)eta/4,(1-(eta)^2)(1+zeta)zeta/2,(1+zeta)zeta(1+eta)eta/4,(1-(zeta)^2)(1+eta)eta/2,-(1-zeta)zeta(1+eta)eta/4,-(1-(eta)^2)(1-zeta)(zeta)/2,(1-(zeta)^2)*(1-(eta)^2)]; #bi quadratic langrange shape functions

>	dAdzeta:=diff(A,zeta);

>	dAdeta:=diff(A,eta);

>	y:=[-a1,-a2,-a3,0,a3,a2,a1,0,0];x:=[-b1,0,b1,b2,b3,0,-b3,-b2,0];

>	J:=simplify(add(dAdzeta[i]x[i],i=1..9)add(dAdeta[i]y[i],i=1..9)-add(dAdzeta[i]y[i],i=1..9)add(dAdeta[i]x[i],i=1..9));

>	Nx:=[seq((simplify(add(dAdeta[i]y[i],i=1..9))dAdzeta[j]-simplify(add(dAdzeta[i]y[i],i=1..9))dAdeta[j])/simplify(J),j=1..9)];

>	Ny:=[seq((-dAdzeta[j]simplify(add(dAdeta[i]x[i],i=1..9))+dAdeta[j]simplify(add(dAdzeta[i]x[i],i=1..9)))/simplify(J),j=1..9)];

>	Kx:=Matrix(nops(A),nops(A),datatype=float[8]):

>	Ky:=Matrix(nops(A),nops(A),datatype=float[8]): c:=Vector(nops(A),datatype=float[8]):

>	terms:=20:#number of terms for numerical integration dzeta:=2/terms: deta:=2/terms:

>

for i from 1 to nops(Nx) do #loop to obtain local matrices

for j from 1 to nops(Ny) do
Kx[i,j]:=0;
Ky[i,j]:=0;

for k from 0 to terms do #outer loop double integration, integration in zeta

if k = 0 then fx[k]:= subs(zeta=-1,Nx[i]*Nx[j]*J); fy[k]:= subs(zeta=-1,Ny[i]*Ny[j]*J);
elif k = terms then
fx[k]:= subs(zeta=-1+(k*dzeta),Nx[i]*Nx[j]*J);
fy[k]:= subs(zeta=-1+(k*dzeta),Ny[i]*Ny[j]*J);
elif irem(k,2) = 0 then
fx[k]:= 2*subs(zeta=-1+(k*dzeta),Nx[i]*Nx[j]*J);
fy[k]:=     2*subs(zeta=-1+(k*dzeta),Ny[i]*Ny[j]*J);
else fx[k]:= 4*subs(zeta=-1+(k*dzeta),Nx[i]*Nx[j]*J);
fy[k]:=     4*subs(zeta=-1+(k*dzeta),Ny[i]*Ny[j]*J); end if;

for l from 0 to terms do #inner loop double integration, integration in eta

if l = 0 then fxy[l]:= subs(eta=-1,fx[k]); fyy[l]:= subs(eta=-1,fy[k]);
elif l = terms then fxy[l]:= subs(eta=-1+(l*deta),fx[k]); fyy[l]:= subs(eta=-1+(l*deta),fy[k]);
elif irem(l,2) = 0 then fxy[l]:= 2*subs(eta=-1+(l*deta),fx[k]); fyy[l]:= 2*subs(eta=-1+(l*deta),fy[k]);
else fxy[l]:=4*subs(eta=-1+(l*deta),fx[k]); fyy[l]:=4*subs(eta=-1+(l*deta),fy[k]); end if;
Kx[i,j]:=Kx[i,j]+fxy[l];
Ky[i,j]:=Ky[i,j]+fyy[l];

end do;

end do;
Kx[i,j]:=Kx[i,j]*dzeta*deta/9;
Ky[i,j]:=Ky[i,j]*dzeta*deta/9;
end do;
end do:

>

end proc:

>	n:=nx*ny; #total number of elements

(1)

>	Nx1:=nx*2+1: #number of nodes in x in one row

>	N:=Nx1(2ny+1); # total number of nodes/equations

(2)

>	K:=Matrix(N,N,storage=sparse): # global K matrix

>	C:=Vector(N,storage=sparse): # global c matrix

>	L2G:=Matrix(n,9): #mapping matrix - each row has node numbers for each element

>	l:=1:k:=1: localmatrices(hy/2,hy/2,hy/2,hx/2,hx/2,hx/2,0,0): kx:=copy(Kx):ky:=copy(Ky):c0:=copy(c):

>

for i from 1 to n do #modifying,adding and assembling matrices to get global matrix

if i<=nx/2 then
a1:=copy(kx); a2:=copy(ky); a3:=0; a1[1..3,1..9]:=IdentityMatrix(3,9); a2[1..3,1..9]:=Matrix(3,9,shape=zero); a4:=a1+a2; c:=copy(c0); c[1..3]:=1.0;

elif i=nx/2+1 then a1:=copy(kx); a2:=copy(ky); a3:=0; a1[1,1..9]:=IdentityMatrix(1,9); a2[1,1..9]:=Matrix(1,9,shape=zero); a4:=a1+a2; c:=copy(c0); c[1]:=1.0;

elif i>nx*(ny-1) then a1:=copy(kx); a2:=copy(ky); a3:=0; a1[5..7,5..7]:=IdentityMatrix(3,3);
a1[5..7,1..4]:=ZeroMatrix(3,4);
a1[5..7,8..9]:=ZeroMatrix(3,2); a2[5..7,1..9]:=Matrix(3,9,shape=zero); a4:=a1+a2; c:=copy(c0); c[5..7]:=0;

else a1:=kx; a2:=ky; a3:=0;a4:=a1+a2; c:=c0; end if;

L2G[i,1..9]:=Matrix([l,l+1,l+2,l+2+Nx1,l+2+Nx1*2,l+1+Nx1*2,l+Nx1*2,l+Nx1,l+1+Nx1]):

k:=k+1:

if k>nx then k:=1; l:=l+Nx1+3;

else l:=l+2; end if:

indx2:=L2G[i,1..9]:
indx2:=convert(indx2,list):

C[indx2]:=C[indx2]+c[1..9];
c[1..9];

for i1 from 1 to 9 do
indx1:=L2G[i,i1]:
K[indx1,indx2]:=K[indx1,indx2]+a4[i1,1..9]:
end do:

end do:

>	phi:=LinearSolve(K,C,method=SparseLU): #linear set of equations solved using Sparse LU solver

>	phi_at(1,0):=phi[Nx1];

(3)

>	dNdy:=copy(Ny):

>	dNdy_bottom_left:=subs(eta=-1,zeta=-1,dNdy):

>	current_at(0,0):=add(dNdy_bottom_left[i]*phi[L2G[1,i]],i=1..nops(Ny));

(4)

>	if irem(nx/2,2)=0 then current_at(0,0.25):=add(dNdy_bottom_left[i]phi[L2G[nx/4+1,i]],i=1..nops(Ny)); else dNdy_bottom_center:=subs(eta=-1,zeta=0,dNdy): current_at(0,0.25):=add(dNdy_bottom_center[i]phi[L2G[(nx/2+1)/2,i]],i=1..nops(Ny)); end if;

(5)

>

Download FEM_2D.mws

Combining algebraic expression...

Asked by: sunit 105

October 06 2020

0 3

Hi,

It might be a really silly question, but I am wondering is it possible to simplify expression like this

a^2+b^2+2*a*b+c^2 . Just by looking it we know that we can write it in the form of (a+b)^2+c^2. This is the basic exmaple I come up with. I have very lengthy expressions in maple, which can be factorize like this, but factor command will not work as it will try to factorize entire expression. So I am wondering is it doable in maple or I have to do it manually by collecting terms and check whether they can be factorize or not.

Thanks in Advance.

With Regards

sunit

how to use Jordan form inside a proc?...

Asked by: nm 11353

October 06 2020

2 1

I know this has to do with name scoping issue. But I do not see how to fix it now.

Calling J,Q:=LinearAlgebra:-JordanForm(A,output=['J','Q']); works in global, but not inside a proc.

What is the correct way to use this inside a proc?

Download jordan_issue.mw

how much costs maple 9.5 licencel ?...

Asked by: carrijo 5

October 06 2020

1 1

Hi,

I would like to instal Maple 9.5 in my laptop, once I already have some few programs for his version. I don't know how much cost Maple 9.5, and how do download it.

I live in Brazil.

Thanks in advance.

Carrijo.

jose.carrijo@gmail.com

To eliminate some entries of a list that are not ...

Asked by: Magma 70

October 06 2020

0 2

Let L=[a₁,a₂,...,a_n] be a list of positive real numbers.

My Question:

How to write a procedure to find a minimal interval such as [b,c]

provided that this interval covers entries of L which are close together.

Example:

let L:=[8.1 , 2.03 , 3.5 , 0.05 , 4.1]. Then the output of the procedure is the interval [2.03 , 4.1].

In fact we want to eliminate some entries of L that are not close to other entries of L.

Thanks in advance

Why do the summation index not count as a god matr...

Asked by: Eric Ragnarsson 50

October 06 2020

1 3

Hi

I am on my way to construct a matrix for centripetal and Coriolis forces for a robot arm. However this requeers that I use a sum where the matrix index is the sumations index.

So I started testing things out before I constructed the full expresion, this is what I did.

According to the error page, this is du to that k in M[1,k] is an bad index, however in the sum this would be k=1 och k=2 whitch are vallid.

Do anybody know a way around this problem or can tell me what I did wrong.

Many thanks

Eric Ragnarsson

How to add a legend (color bar with number) on the...

Asked by: winfredgg 30

October 06 2020

5 2

plots:-contourplot(16.70196911*2^(1/2)*((x^2 + 0.1*y)/((1 - x)*(3*x^2 + 0.2*y)))^(1/2)/(4.373839156*(x^2 + 0.1*y)/((1 - x)*(3*x^2 + 0.2*y)) + 1)^(1/2), x = 0.001 .. 1, y = 0.001 .. 1, contours = 20, thickness = 0, coloring = ["blue", "yellow"], axes = "boxed", filledregions = true)

I have tried many times, such as adding legendstyle = [position=correct],legend = true command, but it always report errors.

Can anyone help me solve this problem? Thank you.

find Databases for the NASA Seven-Coefficient Poly...

Asked by: 9009134 110

October 06 2020

0 7

Hi.

in the ThermophysicalData[Chemicals] package that compute the coefficients for different species how I can find that coefficients for seven coefficients not nine of them

in other words, I am seeking to find Databases for the NASA Seven-Coefficient Polynomial Fits for Calculating Thermodynamic Properties of Individual Species.

Best

Disappointing results regarding complex expr simpl...

Asked by: zphaze 20

October 06 2020

0 4

Hi,

I'm trying to make Maple reduce an extensively simple complex equation, something so simple anybody with basic knowledge of complex numbers would take about 2 mins to solve manually.

Yet, I can't seem to get Maple to do it, which is disappointing. I can only hope the issue is on my side (I would be glad to be the one in the wrong).

example.mw

I expect at the last line a reduction to s^2 + 2*z*omega__n * s + omega__n^2

I tried different combinations of simplification functions without success. Simplify doesn't work much better.

Basically looking at something like this, but on Maple (this comes from Maxima), if possible *WITHOUT* the need for assume() on any scale (just like Maxima and Mathematica) :

Thanks!

why the Elzaki transform is not working for deriva...

Asked by: saher 35

October 06 2020

2 4

I made a customised procedure of Elzaki transform which is working for functions but it is not for derivatives.Elziki_costume.mw

does Maple have option for fraction free eigenvect...

Asked by: nm 11353

October 05 2020

0 3

I remember seeing sometime ago an option called something like "fraction free" in LinearAlgebra. But may be I was looking at something else or different package. I can't remember now. I searched the help pages now and googled and can't find it.

In Maple, when asking for eigenvectors of matrix, I'd like the vectors to come out fraction free, like with Mathematica.

It is ofcourse easy to write code to post process this and remove the fractions.

But before I do this, I thought to ask. Here is an example

restart;
A:=Matrix([[48,-30,-14,1],[65,-41,-19,0],[17,-10,-5,3],[-35,22,10,0]]);
(e,v):=LinearAlgebra:-Eigenvectors(A);

In Mathematica:

Anyone knows if such option exists somewhere?

Maple 2020.1

If cannot do cos(x)>0 and sin(x)>0...

Asked by: Thomas Dean 327

October 05 2020

0 3

Heck Example 15.5 must have worked for an old version of maple.

restart;
## plot two functions and color the region between
sine :=   plot(sin(x), x=0..4*Pi, color=black,thickness=3):
s    :=   plot(sin(x), x=0..4*Pi, color=red, filled=true):
cosine := plot(cos(x), x=0..4*Pi, color=black,thickness=3):
c      := plot(cos(x), x=0..4*Pi, color=red, filled=true):
f := x -> if cos(x)>0 and sin(x)>0 then
              min(cos(x),sin(x))
          elif cos(x)<0 and sin(x)<0 then
              max(cos(x),sin(x))
          else 0
          end if;
b := plot(f(x), x=0..4*Pi, filled=true, color=green):

display([sine, cosine, b, s, c]):

Gives the error.

Error, (in f) cannot determine if this expression is true or false: 0 < cos(x)
and 0 < sin(x)

I tried verify(cos(x),0,less_than) and verify(sin(x),0,less_than), etc., but that makes f(x) always return 0.

f := sin(x);
g := cos(x);
plottools:-transform(unapply([x,y+g],x,y))(plot(f-g,x=0 .. 4*Pi,filled=true));
Works, but, I can not remember how that works.

Is it possible to use if in maple 2020.

PDEtools:-Solve vs. solve. Error vs. no solution ...

Asked by: nm 11353

October 05 2020

1 1

In this example, PDEtools:-Solve throws an error, while solve returns empty solution.

Why the different behavior? Should PDEtools:-Solve also return empty solution like solve?

I noticed this, when I changed my code from using solve to using PDEtools:-Solve

restart;
eq:=[ eta1+2*eta2 = 0, eta1+2*eta2 = a1+a3, eta2 = a2+2*a3];
PDEtools:-Solve(eq,[a1, a2, a3]);


solve(eq,[a1, a2, a3])

interface(version)
Standard Worksheet Interface, Maple 2020.1, Windows 10, July 30 

   2020 Build ID 1482634


Physics:-Version()

The "Physics Updates" version in the MapleCloud is 832 and is 

   the same as the version installed in this computer, created 

   2020, October 3, 5:34 hours Pacific Time.

Maple 2020.1

Timely DataFrames Example: 2020 US Presidential...

by: Scot Gould 1039 Product: Maple 2020

October 05 2020

2 3

DataFrames: An example from the 2020 U.S. Presidential election

(Or why DataFrames are more powerful and readable than spreadsheets.)

In this example of working with DataFrames, the goal is to use a spreadsheet from a website, which contains polling data, to estimate the probability each of the two candidates from the major parties will win the US Presidential election in November. I first tried doing the calculations with a spreadsheet, but I discovered DataFrames was far more powerful. Warning: This worksheet uses live data. Hence the outcome at the end of the worksheet is likely to change daily. A more extensive example with even more common DataFrame operations should be available soon.

How the US Presidential election works - highly simplified version: In the US there are only two parties for which their candidate could win the election: the Democratic party and Republican party. The Republican party is often referred to as the "Grand Old Party", or GOP. Each state executes its own election. The candidate who receives the most votes wins the states "electoral votes" (EV). The number of the electoral votes for each state is essentially proportional to the population of the state. A candidate who receives a total of 270 or more EVs out of 538, is declared the president of the US for the next term, which starts January 20 of 2021.

Creating DataFrame from web based data:

First I download the data from the website. It is a CSV spreadsheet.

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restart; interface(displayprecision = 3); interface(rtablesize = [6, 8]); web_data := Import("https://www.electoral-vote.com/evp2020/Pres/pres_polls.csv")

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Each row contains information about a poll conducted in one of the states. The first poll starts on row 2, hence the number of polls are:

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Now I want to create a new DataFrame containing only the most useful information. In web_data, many are the columns are not important. However I do want to keep the column label names from those columns I wish to retain.

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web_data_cols := [1, 3, 4, 5, 6]; column_labels := convert(web_data[1, web_data_cols], list)

Because the first poll in web_data is labeled 2, I would like to relabel all the polls starting from 1

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Creating a DataFrame from a Matrix or another DataFrame: (with row labels and column labels)

Now I can build the DataFrame that I will be working with:

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poll_data := DataFrame(web_data[2 .. (), web_data_cols], 'columns' = column_labels, 'rows' = row_labels)

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What each column means

* "Day" - day of the year in 2020 when the poll within the state was halfway completed. The larger the value, the more recent the poll.

* "State" - the state in the US where the poll was conducted. The candidate that receives the most votes "wins the state".

* "EV" - the number of electoral votes given to the candidate who receives the most votes within the state.

* "Dem" - the percentage of people who said they are going to vote for the candidate from the Democratic party.

* "GOP" - the percentage of people who said they are going to vote for the candidate from the Republican party.

Sorting:

By using the sort function, using the `>` operator, I can see which polls are the more recent. (If you run the worksheet yourself, the outcome will change as more polls are added to the website spreadsheet.)

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Selecting Unique entries - by column values:

For the my simple analysis, I will use only the most recent poll, one from each state. Hence, using AreUnique, I can pull the first row that matches a state name. This new DataFrame called states.

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(Note, one of the "states" is the District of Columbia, D.C., which is why there are 51 rows.)

Removing a column: (and relabeling rows)

This next example isn't necessary, but shows some of the cool features of DataFrames.

Since there is only 1 entry per state, I'm going to remove the "State" column and relabel all the rows with the state names

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state_names := convert(states["State"], list); states := DataFrame(Remove(states, "State"), 'rows' = state_names)

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Indexing by row labels:

This allow me to to display information by individual states. What is the data for California, Maine and Alaska?

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Mathematics with multiple-columns:

My preference is to work with fractions, rather than percentages. Hence I want all the values in the "Dem" and "GOP" to be divided by 100 (or multiplied by 1/100). Treating each column like a vector, the multiplication is performed individually on each cell. This is what the tilda, "~", symbol performs.

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states[["Dem", "GOP"]] := `~`[`*`](states[["Dem", "GOP"]], 1/100.); states

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Mathematics: using a function to calculate a column

For the next action, I want to use the power of the Statistics package to create a "probability of winning the state" function.

For simplicity, I will assume the outcome of the voting in a state is purely random, but is conditional to popularity of each candidate as measured by the polls. I'll assume the likelihood of an outcome follows a normal (Gaussian) distribution with the peak being at point where the difference of the polling of the two candidates is zero. (Note, other than 2016, where there was an unusually larger percentage of undecided voters on election day, this simple model is reasonable accurate. For example, in 2012, of the states which appeared to be the "closest", the winner over-performed his polling in half of them, and under-performed in the other half with a mean difference of nearly zero.) From previous elections, the standard deviation of differences between polling values and the actual outcome is at most 0.05, however, it does increase with the fraction of undecided voters.

To mathematically model this situation, I have chosen to use the "Cumulative Density Function" CDF in the Statistics package. It will calculate the probability that a candidate polling with fraction f1 wins the election if the other candidate is polling with fraction f2. The variable u is the fraction of undecided voters. It is included in the calculation to increase the spread of the possible outcomes.

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win_prob := Statistics:-CDF(Statistics:-RandomVariable(Normal(0., 0.5e-1+(1/4)*u)), f1-f2)

1/2+(1/2)*erf((1/2)*(f1-f2)*2^(1/2)/(0.5e-1+(1/4)*u))

Converting this expression into a function using the worst named function in Maple, unapply:

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Now I can calculate a DataFrames column of the "win probability", in this case, for the candidate from the Democratic platy. By apply the function, individually, using the columns "Dem" and "GOP", I produce:

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dem_win_prob := `~`[win_prob_f](states["Dem"], states["GOP"], `~`[`-`](1, `~`[`+`](states["Dem"], states["GOP"])))

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Appending a column:

I can add this column to the end of the states with the label "DemWinProb":

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Mathematics of adding the entries of a column:

How many electoral votes are available? add them up.

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While the number of EV a candidate wins is discrete, I can use the "win probability" from each state to estimate the total number of EV each of the candidates might win. This means adding up number of EV in each state times, individually, the probability of winning that state:

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Currently, the candidate from the Democratic party is likely to win more then 300 electoral vtes.

What about for the candidate from the Republican / "GOP" party?

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gop_win_prob := `~`[win_prob_f](states["GOP"], states["Dem"], `~`[`-`](1, `~`[`+`](states["Dem"], states["GOP"]))); GOP_EV := round(add(`~`[`*`](states["EV"], gop_win_prob)))