Posts - MaplePrimes

5 tips for helping kids stay engaged with remote...

by: Karishma 885

October 14 2020

9 2

A few weeks ago a television station in Toronto asked me if I’d share some tips on how parents could help their kids stay engaged with remote learning. My initial reaction was to run for the hills – appearing on live TV is not my cup of tea. However my colleagues persuaded me to accept. You can see a clip of that segment here - I’ve included it in this post because otherwise someone on the marketing team would have ;-)

My tips are based on a wide variety of experiences. My role at Maplesoft requires me to speak with educators at all levels, and remote learning has been a hot topic of conversation lately, as you can imagine. As well, in my past life (i.e. life before kids) I was a high school math tutor, and now as a parent I’m in the thick of it helping my son navigate Kindergarten remotely.

So here are my 5 tips on how parents of elementary and high-school aged children can help their kids stay engaged with remote learning. If you have other tips, including suggestions for university students, feel free to leave them in the comments sections. And if these tips help you, please let me know. It will have made the stress of my appearance on TV worthwhile!

Tip 1: Look for the positives

These are unprecedented times for kids, parents and teachers. Over the course of the last 6-7 months, learning as we’ve grown to know it has changed radically. And while the change has been incredibility difficult for everyone, it’s helpful to look for the positives that remote learning can bring to our children:

Remote learning can help some kids focus on their work by minimizing the social pressures or distractions they may face at school.
Older kids are appreciating the flexibility that remote learning can offer with respect to when and how they complete their work.
Younger kids are loving the experience of learning in the presence of mom and dad. My 4 year old thinks it’s awesome that I now know all the lyrics to the songs that he learns in school.
As many remote learning classrooms include students from across the school board, this can provide kids with the opportunity to connect with their peers from different socio-economic backgrounds living across the city.

Tip 2: Don’t shy away from your kid’s teacher

While some kids are thriving learning from home, we know that others are struggling.

If your high school student is struggling at school, do whatever it takes to convince them to connect with their teacher. If your child is younger, make the connection yourself.

In my role, I’ve had the opportunity to work with many teachers, and rest assured, many of them would welcome this engagement. They want our kids to succeed, but without the face-to-face classroom interaction it’s becoming increasingly more difficult for them to rely on visual cues to see how your child is doing and if they are struggling with a concept.

So I encourage you to reach out to your kid’s teacher especially if you notice your child is having difficulty.

Tip 3: Get creative with learning

Another benefit of remote learning is that it presents us with a unique opportunity to get creative with learning.

Kids, especially those in middle school and high school, now have the time and opportunity to engage with a variety of different online learning resources. And when I say online learning resources, I mean more than just videos. Think interactive tools (such as Maple Learn), that help students visualize concepts from math and science, games that allow students to practice language skills, repositories of homework problems and practice questions that allow kids to practice concepts, the list goes on.

Best of all, many content providers and organizations, are offerings these resources and tools available for free or at a substantially reduced cost to help kids and parents during this time.

So if your child is having difficulty with a particular subject or if they are in need of a challenge, make sure to explore what is available online.

Tip 4: Embrace the tech

To be successful, remote learning requires children to learn a host of new digital skills, such as how to mute/unmute themselves, raise their hands electronically, turn on and off their webcam, toggle between applications to access class content and upload homework, keep track of their schedule via an electronic calendar, etc. This can be daunting for kids who are learning remotely for the first time.

As a parent you can help your child become more comfortable with remote learning by setting aside some time either before or after class to help them master these new tools. And since this is likely new to you, there are some great videos online that will show you how to use the system your school has mandated be it Microsoft Teams, Google Classroom or something else.

Tip 5: It’s a skill

Remember that remote learning is a skill like any other skill, and it takes time and practice to become proficient.

So remember to be patient with yourself, your kids, and their teachers, as we embark on this new journey of learning. Everyone is trying their best and I truly believe a new rhythm will emerge as we progress through the school year.

We will find our way.

Spline кривые формирования с учетом экстремальных...

by: Tatyana 10 Product: Maple

October 12 2020

0 0

Examples of mathematical models for the formation of spline curves based on polynomials of various orders that simulate certain trajectories are given.
Mathematical models of the formation of a spline curve, taking into account the extreme derivatives of the initial orders, are presented. The construction of the spline curves of the hermit and bezier of various orders, consisting of different segments, is considered. The presented models are systematic and universal, i.e. can be used to generate any polynomial curves with specified extreme derivatives of the vectors.

100_nodes_not_closed.mw Example_1.mw Example_2.mw Example_3.mw Example_4.mw Example_5.mw Example_6.mw Example_7.mw

«Формирования линий OC и бриджей по линии MAT для области с островами».OC_MAT_MA_bridge.mw

An example with orientable and non-orientable...

by: one man 1370 Product: Maple 17

October 08 2020

7 0

As an addition to the post.
Non-orientable surface in the sequence of orientable surfaces. In the picture we see the equations corresponding to the current surface plot.
Just entertainment.
surfaces.mw

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

Timely DataFrames Example: 2020 US Presidential...

by: Scot Gould 1059 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.

>

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

_m2211970420352

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:

>

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.

>

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

>

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:

>

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

_m2211956910784

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.)

>

_m2211960016288

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.

>

_m2211969565344

(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

>

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

_m2211957755840

Indexing by row labels:

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

>

_m2211977321984

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.

>

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

_m2211957755840

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.

>

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:

>

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:

>

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

_m2212010910496

Appending a column:

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

>

_m2212009017568

Mathematics of adding the entries of a column:

How many electoral votes are available? add them up.

>

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:

>

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?

>

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

Summing the two EV values, we obtain the total number of electoral votes.

>

Download DataFrames_Example.mw

Workaround for the problem of installing MapleCloud...

by: ecterrab 14670 Product: Maple 2020

September 24 2020

9 8

Hi,
Some people using the Windows platform have had problems installing MapleCloud packages, including the Maplesoft Physics Updates. This problem does not happen in Macintosh or Linux/Unix, also does not happen with all Windows computers but with some of them, and is not a problem of the MapleCloud packages themselves, but a problem of the installer of packages.

I understand that a solution to this problem will be presented within an upcoming Maple dot release.

Meantime, there is a solution by installing a helper library; after that, MapleCloud packages install without problems in all Windows machines. So whoever is having trouble installing MapleCloud packages in Windows and prefers not to wait until that dot release, instead wants to install this helper library, please email me at physics@maplesoft.com.

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

2D Input operator assignment syntax

by: acer 32747 Product: Maple

September 16 2020

5 11

Caution, certain kinds of earlier input can affect the results from using the 2D Input syntax for operator assignment.

>

The following now produces a remember-table assignment,
instead of assigning a procedure to name f, even though by default
Typesetting:-Settings(functionassign)
is set to true.

>

With the previous line of code commented-out the following
line assigns a procedure to name f, as expected.

If you uncomment the previous line, and re-execute the whole
worksheet using !!! from the menubar, then the following will
do a remember-table assignment instead.

>

Download operator_assignment_2dmath.mw

Visualize the behaviour of a mechanical system

by: mmcdara 7896 Product: Maple

September 15 2020

10 0

HI,

This post concerns the simulation of a physical system whose behavior is governed by ODEs.
It is likely that some people will think that all which follows is nothing but embellishments or a waste of time.
And in some sense they will be right.
I was thinking the same until I received some sharp remarks at the occasion of a few presentations of my works.
So experience has proven me that doing a presentation in front of project managers with only 2D curves often leads to smiles, not to speak about those who raise their eyes to heaven in front of the poverty of the slides.
Tired of this attitude, I decided to replace these 2D curves with a short film, which of course does not reveal more than what these 2D curves were already revealing, but which is pretty enough for the financing keeps going on.

For those of you who might regret this situation, just consider this work as a demonstration of the capabilities of Maple in 3D rendering.

PS: all the display outputs have been removed to avoid loading an unnecessary huge file.
The two last commands must be uncommented to play the animation.

Download ODE_Movie.mw

Ellipticaly polarized electromagnetic wave incident...

by: Louis Lamarche 105 Product: Maple

September 14 2020

5 0

EllipticalyPolarizedWaveRefectionByMirrorFinal.mw

Three level Newton-Raphson optimization code

by: radaar 202 Product: Maple

September 04 2020

1 3

Newton raphson method is used for optimization of functions and is based on taylor series expansion. Here is the code for a three level newton raphson method.

>

$restart; with(Student[MultivariateCalculus]); ff := proc (xx) xx^3-2*xx+2 end proc; ii := 0; XX[ii] := 2; while ii < 25 do GR := Gradient(ff(xx), [xx] = [XX[ii]]); GR1 := evalf(GR[1]); HESS := Student[VectorCalculus]:-Hessian(ff(xx), [xx] = [XX[ii]]); HESS1 := evalf(HESS[1]); YY[ii] := XX[ii]-GR1[1]/HESS1[1]; GR := Gradient(ff(xx), [xx] = [YY[ii]]); GR1 := evalf(GR[1]); HESS := Student[VectorCalculus]:-Hessian(ff(xx), [xx] = [YY[ii]]); HESS1 := evalf(HESS[1]); ZZ[ii] := YY[ii]-GR1[1]/HESS1[1]; GR := Gradient(ff(xx), [xx] = [ZZ[ii]]); GR1 := evalf(GR[1]); HESS := Student[VectorCalculus]:-Hessian(ff(xx), [xx] = [ZZ[ii]]); HESS1 := evalf(HESS[1]); ii := ii+1; XX[ii] := ZZ[ii-1]-GR1[1]/HESS1[1]; printf("%a\n", XX[ii]) end do$

.8180854533
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>

(1)

>

(2)

>

Download THREE_LEVEL_NEWTON_RAPHSON_METHOD1.mw

what are the main things you like to see improved...

by: nm 11968 Product: Maple

August 28 2020

5 43

What are the things you most like to see improved/add to next version of Maple?

This is my list for a starter:

1. Improve the debugger. Debugger is very useful but needs more work. At least be able to see code listing in larger view as one steps in for example. See Matlab debugger for inspiration.

2. Improve Latex. It still does not do fractions well. Posted about this before.

3. Eliminate hangs when using timelimit(). On long runs, random hangs happen when timelimit() do not expire as requested. Posted about this before.

Use of MapleSim to teach Vector Analysis

by: Lenin Araujo Castillo 1790 Product: MapleSim 2018

August 26 2020

3 0

In the present work we are going to demonstrate the importance of the study of vector analysis, with modeling and simulation criteria, using the MapleSim scientific software from MapleSoft. Nowadays, the majority of higher education centers direct their teaching of vector analysis in an abstract way and there are few or no teachers who carry out applications using modeling and simulation. (In spanish)

IPN_CICATA_2020.pdf

Expo_MapleSim_CICATA.zip

Maple 2020.1.1 update

by: eithne 2072 Product: Maple 2020

August 26 2020

1 11

We have just released an update to Maple, Maple 2020.1.1. This update includes the following:

Correction to a problem that occurred when printing or exporting documents to PDF. If the document included a 3-D plot, nearby text was sometimes missing from the printed/exported document.
Correction to an issue that prevented users from installing between-release updates to the Physics package

This update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2020.1.1 download page. If you are a MapleSim user, you can obtain this update from the corresponding MapleSim menu or MapleSim 2020.1.1 download page.

In particular, please note that this update fixes the problems reported on MaplePrimes in Maple 2020.1 issue with print to PDF and installing the Maplesoft Physics updates. As always, we appreciate the feedback.

The strandbeest

by: Rouben Rostamian 8596 Product: Maple 2020

August 24 2020

12 4

The strandbeest is a walking machine developed by Theo Jansen. Its cleverly designed legs consist of single-degree-of-freedom linkage mechanisms, actuated by the turning of a wind-powered crankshaft.

His working models are generally large - something of the order of the size of a bus. Look for videos on YouTube. Commercially made small toy models are also available. This one sells for under $10 and it's fun to assemble and works quite well. Beware that the kit consists of over 100 tiny pieces - so assembling it is not for the impatient type.

Here is a Maple worksheet that produces an animated strandbeest. Link lengths are taken from Theo Jansen's video (go to his site above and click on Explains) where he explains that he calculated the optimal link lengths by applying a genetic algorithm.

Here is a Maple animation of a single leg. The yellow disk represents the crankshaft.

And here are two legs working in tandem:

Here is the complete beest, running on six legs. The crankshaft turns at a constant angular velocity.

The toy model noted above runs on twelve legs for greater stability.

Download the worksheet: strandbeest.mw

The Friis Transmission Equation

by: Steve Roper 180 Product: Maple 2020

August 22 2020

9 1

This may be of interest to anyone curious about why the effective area of an isotropic antenna is λ^2/4π.

Friis Transmission Equation

	Initialise

The Hertzian Dipole antenna

The Hertzian Dipole is a conceptual antenna that carries a constant current along its length.

By laying a number of these small current elements end to end, it is possible to model a physical antenna (such as a half-wave dipole for example). But since we are only interested in obtaining an expression for the effective area of an Isotropic Antenna (in order to derive The Friis Transmission Equation) the Hertzian Dipole will be sufficient for our needs.

Maxwell's Equations

Since the purpose of a radio antenna is to either launch or to receive radio waves, we know that both the antenna, and the space surrounding the antenna, must satisfy Maxwell's Equations. We define Maxwell's Equations in terms of vector functions using spherical coordinates:

Maxwell–Faraday equation:

(3.1)

Ampère's circuital law (with Maxwell's addition):

(3.2)

Gauss' Law:

(3.3)

Gauss' Law for Magnetism:

(3.4)

Where:

E is the electric field strength [Volts/m]

H is the magnetic field strength [Amperes/m]

J is the current density (current per unit area) [Amperes/m²]

ρ is the charge density (charge per unit volume) [Coulombs/m³]

ε is Electric Permittivity

μ is Magnetic Permeability

Helmholtz decomposition

The Helmholtz Decomposition Theorem states that providing a vector field, (F) satisfies appropriate smoothness and decay conditions, it can be decomposed as the sum of components derived from a scalar field, (Φ) called the "scalar potential", and a vector field (A) called the "vector potential".

F = -VΦ + V×A

And that the scalar (Φ) and vector (A) potentials can be calculated from the field (F) as follows (image from https://en.wikipedia.org/wiki/Helmholtz_decomposition):

Where:

r is the vector from the origin to the observation point (P) at which we wish to know the scalar or vector potential.

r' is the vector from the origin to the source of the scalar or vector potential (i.e. a point on the Hertzian Dipole antenna).

V'·F(r') is the Divergence of the vector field (F) at source position r'.

V'×F(r') is the Curl of the vector field (F) at source position r'.

Calculating the Scalar Potential for the magnetic Field, H

We know that the Divergence of the magnetic field (H) is zero:

(4.1.1)

And so the magnetic field (H) must have a scalar vector potential of zero:

(4.1.2)

Calculating the Vector Potential for the magnetic Field, H

We know that the Curl of the magnetic field (H) is equal to the sum of current density (J) and the rate of change of the electric filled (E):

(4.2.1)

Since the Hertzian Dipole is a conductor, we need only concern ourselves with the current density (J) when calculating the vector potential (A). Integrating current density (J) over the volume of the antenna, is equivalent to integrating current along the length of the antenna (L).

We know that Maxwell's Equations can be solved for single frequency (monochromatic) fields, so we will excite our antenna with a single frequency current:

(4.2.2)

We can simplify the integral for the vector potential (A) by recognising that:

1.	Our observation point (P) will be a long way from the antenna and so (r) will be very large.

2.	The length of the antenna (L) will be very small and so (r') will be very small.

Since |r|>>|r'|, we can substitute |r-r'| with r.

Because we have decided that the observation point at r will be a long way from the antenna, we must allow for the fact that the observed antenna current will be delayed. The delay will be equal to the distance from the antenna to the observation point |r-r'| (which we have simplified to r), divided by the speed of light (c). The time delay will therefore be approximately equal to r/c and so the observed antenna current becomes:

(4.2.3)

Since the length, L of the antenna will be very small, we can assume that the current is in phase at all points along its length. Working in the Cartesian coordinate system, the final integral for the vector potential for the magnetic field is therefore:

A__H_ := (int(I__0*exp(I*omega*(t-r/c))*_k/r, z = -(1/2)*L .. (1/2)*L))/(4*Pi)

(1/4)*I__0*exp(I*omega*(t-r/c))*_k*L/(Pi*r)

(4.2.4)

We will now convert to the spherical coordinate system, which is more convenient when working with radio antenna radiation patterns:

The radial component of the observed current (and therefore vector potential), will be at a maximum when the observer is on the z-axis (that is when θ=0 or θ=π) and will be zero when the observer is in the x-y-plane:

(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*cos(theta)/(Pi*r)

(4.2.5)

The angular component of the observed current (and therefore vector potential), in the θ direction will be zero when the observer is on the z-axis (that is when θ=0 or θ=π) and will be at a maximum when the observer is in the x-y-plane:

-(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*sin(theta)/(Pi*r)

(4.2.6)

Since the observed current (and therefore vector potential) flows along the z-axis, there will be no variation in the ϕ direction. That is to say, that varying ϕ will have no impact on the observed vector potential.

(4.2.7)

And so the vector potential for the magnetic field (H) expressed using spherical coordinate system is:

(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*cos(theta)*_r/(Pi*r)-(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*sin(theta)*_theta/(Pi*r)

(4.2.8)

Calculating the Magnetic Field components

The Helmholtz Decomposition Theorem states that providing a vector field (F) satisfies appropriate smoothness and decay conditions, it can be decomposed as the sum of components derived from a scalar field (Φ) called "scalar potential", and a vector field (A) called the vector potential.

F = -VΦ + V×A

And so the magnetic field, H will be:

H_ = ((1/4)*I)*I__0*L*exp(I*omega*t-I*omega*r/c)*sin(theta)*omega*_phi/(c*Pi*r)+(1/4)*I__0*L*exp(I*omega*t-I*omega*r/c)*sin(theta)*_phi/(Pi*r^2)

(4.3.1)

We see that the magnetic field comprises two components, one is inversely proportional to the distance from the antenna (r) and the other falls off with r². Since we are interested in the far-field radiation pattern for the antenna, we can ignore the r² component and so the expression for the magnetic field reduces to:

H_ := I*omega*I__0*L*exp(I*omega*(t-r/c))*sin(theta)*_phi/(4*Pi*c*r)

((1/4)*I)*omega*I__0*L*exp(I*omega*(t-r/c))*sin(theta)*_phi/(Pi*c*r)

(4.3.2)

We can further simplify by substituting ω/c for 2π/λ:

H_ := (I*2)*Pi*L*I__0*exp(I*omega*t-(I*2)*Pi*r/lambda)*sin(theta)*_phi/(4*Pi*lambda*r)

((1/2)*I)*L*I__0*exp(I*omega*t-(2*I)*Pi*r/lambda)*sin(theta)*_phi/(lambda*r)

(4.3.3)

Calculating the Poynting Vector

We know that the magnitude of the Poynting Vector (S) can be calculated as the cross product of the electric field vector (E) and the magnetic field vector (H) :

S = -E x H which is analogous to a resistive circuit where power is the product of voltage and current: P=V*I.

We also know that the impedance of free space (Z) can be calculated as the ratio of the electric field (E) and magnetic field (H) vectors: Z = E /H =

This is analogous to a resistive circuit where resistance is the ratio of voltage and current: R=V/I.

This provides two more methods for calculating the Poynting Vector (S):

S = -E·E/Z which is analogous to a resistive circuit where power, P=V²/R, and:

S = -H·H*Z which is analogous to a resistive circuit where power, P=I²R.

Since we have obtained an expression for the magnetic field vector (H), we can derive an expression for the Poynting Vector (S):

S_ = (1/4)*L^2*I__0^2*(exp(I*omega*t-(2*I)*Pi*r/lambda))^2*sin(theta)^2*Z*_r/(lambda^2*r^2)

(5.1)

We can separate out the time variable part to yield:

S_ := S__0*(exp(I*omega*t-(I*2)*Pi*r/lambda))^2*_r

S__0*(exp(I*omega*t-(2*I)*Pi*r/lambda))^2*_r

(5.2)

Where:

S__0 := L^2*I__0^2*sin(theta)^2*Z/(4*lambda^2*r^2)

(1/4)*L^2*I__0^2*sin(theta)^2*Z/(lambda^2*r^2)

(5.3)

And we can visualise this radiation pattern using Maple's plotting tools:

$AntennaPattern := plot3d(sin(theta)^2, phi = 0 .. 2*Pi, theta = 0 .. Pi, coords = spherical, scaling = constrained, size = [800, 800], labels = [x, y, z], title = ["The Electromagnetic radiation pattern of a Hertzian Dipole\n\nThe blue arrow represents the axis of the antenna", font = [Times, bold, 20]])$

So the Hertzian Dipole produces a electromagnetic radiation pattern with a pleasing doughnut shape :-)

Calculating Antenna Gain

We can calculate the total power radiated by the Hertzian Dipole by integrating the power flux density over all solid angles dΩ=sin(θ) dθ dφ. Since we have expressed power flux density in terms of watts per square meter, we multiply the solid angle by r² to convert the solid angle expressed in steradians into an area expressed in m².

P__tx := int(int(S__0*r^2*sin(theta), theta = 0 .. Pi), phi = 0 .. 2*Pi)

(2/3)*L^2*I__0^2*Z*Pi/lambda^2

(6.1)

We can now use this power to calculate the power flux density that would be produced by an isotropic antenna by dividing the total transmitted power by the area of a sphere with radius r:

S__Isotropic := P__tx/(4*Pi*r^2)

(1/6)*L^2*I__0^2*Z/(lambda^2*r^2)

(6.2)

The Gain of the Hertzian Dipole is defined as the ratio between the maximum power flux density produced by the Hertzian Dipole and the maximum power flux density produced by the isotropic antenna:

G__HertzianDipole := S__0/S__Isotropic

(3/2)*sin(theta)^2

(6.3)

$Gain := polarplot(G__HertzianDipole, theta = -Pi .. Pi, axis[radial] = [color = "Blue"], angularorigin = top, angulardirection = clockwise, size = [800, 800], labels = [x, z], title = ["The Gain of a Hertzian Dipole over an isotropic antenna \n\nThe blue arrow represents the axis of the antenna", font = [Times, bold, 20]])$

Calculating Radiation Resistance

The input impedance of the Hertzian Dipole will have both a real and a reactive part. The reactive part will be associated with energy storage in the near field and will not contribute to the Poynting Vector in the far-field. For an ideal antenna (with no resistive power loss) the real part will be responsible for the radiated power:

(2/3)*L^2*I__0^2*Z*Pi/lambda^2 = I__0^2*R__rad

(7.1)

(2/3)*L^2*Z*Pi/lambda^2

(7.2)

Calculating the power received by a Hertzian Dipole

If an electromagnetic field (E) is incident on the Hertzian Dipole antenna, it will generate an Electro-Motive Force (EMF) at the antenna terminals. The EMF will be at a maximum when the transmitter is on the x-y-plane (that is when θ=π/2) and will be zero when the transmitter is on the z-axis.

For and incident E-field:

(8.1)

The z-axis component will be:

(8.2)

The z-axis component of the E-field will create an EMF at the antenna terminals that will draw charge out of the receiver to each tip of the antenna. We can calculate the work done per unit charge by integrating the z-axis component of (E) over the length of the antenna (L):

V__emf := int(E__z, z = -(1/2)*L .. (1/2)*L)

(8.3)

In order to extract the maximum possible power from the antenna, we will form a conjugate match between the impedance of the antenna and the load. This means that the load resistance must be the same as the radiation resistance of the antenna. The voltage developed across the load resistance will therefore be half of the open circuit EMF:

(1/2)*E__0*exp(I*omega*t)*sin(theta)*L

(8.4)

And so the power delivered to the load will be:

P__rx := V__pd^2/R__rad

(3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z)

(8.5)

Calculating the Effective area of an Isotropic Antenna

We can also calculate the power received by the Hertzian Dipole by multiplying the power flux density arriving at the antenna with the effective area of an isotropic antenna and the gain of the Hertzian Dipole relative to an isotropic antenna:

(3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z) = (3/2)*sin(theta)^2*A__Isotropic*S__rx

(9.1)

We can express the incident power flux density in terms of electric field strength and wave impedance:

(3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z) = (3/2)*sin(theta)^2*A__Isotropic*E__0^2*(exp(I*omega*t))^2/Z

(9.2)

Rearranging, we obtain an expression for the effective area of the isotropic antenna:

(1/4)*lambda^2/Pi

(9.3)

The Friis Transmission Equation

We can calculate the power flux density that would be produced by an isotropic antenna at a distance r from the antenna by dividing the total transmitted power P_tx by the area of a sphere with radius r:

P__tx/(4*Pi*r^2)

(10.1)

And so the power flux density that would be produced by an antenna with gain G_tx is:

(1/4)*G__tx*P__tx/(Pi*r^2)

(10.2)

We can calculate the power received by an isotropic antenna by multiplying the power flux density incident onto the antenna with the effective area of an isotropic antenna: