Hello dear all,

I use maple 16, x64. When I run this code:

Student[Calculus1][ApproximateInt](cos(y/(-1+y)), y = 0 .. 1, method = lower, iterations = 1)

it results in "0.0667344650", exactly equal with the result of executing

Student[Calculus1][ApproximateInt](cos(y/(-1+y)), y = 0 .. 1, method = lower, iterations = 10000000000)

Any idea?

Best regards

i understand quotient group is

G/(normal subgroup)

= G composite with inverse permutation group of normal subrgoup

is this understanding correct?

If do not have subgroup or normal subgroup, how to do quotient group?

if i shift second row many times in order to find Subgroup satisfy G*Subgroup = Subgroup*G

after find Subgroup then test

inverse(g)*Subgroup*g = Subgroup

how to test whether inverse(g)*Subgroup*g belong to Subgroup?

or just use equal in inverse(g)*Subgroup*g = Subgroup?

a*b - b*a    where a , b are permutation group

how to minus this?

if a + b , then how to plus permutation group

Hello, 

I have to show cubicspline(...) and some kind of known function on the same axis, but it seems I can only show cubicspline(..) using Draw, but it cannot plot the function, so I can do only one at a time. Does anyone know how to do it?

im trying to input a number between 0-100 and have the operation return the grade a,b,c,d,f. etc though long i though this might work.

Grades:=proc(x)
local a,b,c,d,f;
a:=(100..89.5);
b:=(89.4..79.5);
c:=(79.4..69.5);
d:=(69.4..59.5);
f:=(59.4..0);
if x=(100..89.5) then
display(a);
else
if x=(89.4..79.5)then
display(b);
else
if x=(79.4..69.5) then
display(c);
else
if x=(69.4..59.5) then
display(d);
else
if x=(59.4..0) then
display(f)
end;
end;
end;
end;
end;
end;

count the number of primes less than using an if-then statement.  Implement your code where j goes from 2 to 15. 

im at a loss i need a little nudge in the right direction.

I'm trying to run two statement sequences, one after the other, numerous times. I have the statement sequences:

>for j from 1 to N do

>S[j]:=V[j]+t;

>S[j]:=S[j]+3;

>end do:

>for j from 1 to N do

>if S[j]>99 then S[j]:=0

>end if:

>end do:

I can manage to run one of them multiple times, but when I try to encompass both of them within my 

>for counter from initial to final do statementsequence

end do:

it doesn't seem to work.

Thanks in advance

assume the word equation is

a_i *a_j - a_j *a_i = 0

how to find which permutation group is a_i and a_j

my understanding is to try all rotations

a book use underscript i and j

can i see them as upper script for i rotations which is shift i times to left for second row

and try all combination and composite them in two for loop? 

Whassup homies?

http://www.mathsisfun.com/puzzles/who-lives-in-the-city--solution.html

tried to solve this using C.Loves program, but didn't quite get their solution...

Who_Lives_in_the_Cit.mw

Vars:= [PN,Name, TV, Dest,Ages,Hair,Lives]:
PN:=[$1..5]:
Name:= [Bob, Keeley, Rachael, Eilish, Amy]:
TV:=[Simpsons, Coronation, Eastenders, Desperate, Neighbours]:
Dest:= [Fra, Aus, Eng, Afr,Ita]:
Ages:= [14, 21, 46, 52, 81]:
Hair:=[afro, long, straight, curly , bald]:
Lives:= [town, city, village, farm, youth]:
Con1:= Desperate=3: Con2:= Bob=1: Con3:= NextTo(Simpsons,youth,PN): Con4:= Succ(Afr,Rachael,PN): Con5:= village=52: Con6:= Aus=straight: Con7:= Afr=Desperate: Con8:= 14=5: Con9:= Amy=Eastenders: Con10:= Ita=long: Con11:= Keeley=village: Con12:= bald=46: Con13:= Eng=4: Con14:= NextTo(Desperate,Neighbours,PN): Con15:= NextTo(Coronation,afro,PN): Con16:= NextTo(Rachael,afro,PN): Con17:= 21=youth: Con18:= Coronation=long: Con19:= 81=farm: Con20:= Fra=town: Con21:= Eilish<>straight:

read "LogicProblem.mpl"; City:= LogicProblem(Vars): with(City);

we use modern computer algebra books

i) computer the GSO of (22,11,5),(13,6,3),(-5,-2,-1) belong to R^3.

ii)trace algorithm 16.10 on computer a reduced basis of the lattice in Z^3 spanned by the vectors form(i).

trace also the values of the d_i and of D, and compare the number of exchange steps to the theoretical upper bound from section 16.3

I have computed the infinite multiplication using Maple, $\Pi_{k=3}^{\infty} ( \cos (\frac{\pi}{k} ))$, as follows, but it resulted in 0! I wonder why this happened, although maple was using exact arithmetic.

    P := Product(cos(Pi/k), k = 3 .. infinity)   

    value(P)

Note that if I use floating-point instead, it gives me the right answer, 0.1149420449.

    evalf(P)
By the way, I did not expect such a situation! Exact arithmetic should be exact! I am multiplying non-zero numbers to each other, starting from $1/2$ to $1$ as $n \rightarrow \infty$. So it should not be zero! Why such a thing happened?

Dear all;

Special thanks for all the member who help me in Maple.

My last question is:

Write a maple procedure that solves for y(1) in the initial value problem y'(x)=f(y), y(0)=1

using a Numerical stencil based on the n^{th] order taylor series expansion of y.

The procedure arguments include an arbitrary function f, an integrer n, representing the accuracy of the taylor series expansion, and N representing the number of steps between x=0 and x=1.

Given a 2x2 matrix I am struggling to write a function that would return a list (a,b, a1, a2) of 2 complex numbers followed by 2 vectors such that the set of the 2 vectors is a basis for CxC and also Ab1=ab1, Ab2=Bb2 if these exist

Any ideas would be greatly appreciated

Hi,

Please help me in solving system consist the three differential equations with three unknowns. I did already a few attempts, but I can not finish. Once in the final result was got RootOf and do not know what to do. I tried also numerically. I very very ask for some suggestions: ( 

with the boundary condition

parameters A,B,C are constans

1)

qa1 := A1*(diff(Tg(x), x, x))+A2*(diff(Tg(x), x))+(A3+A4)*Tg+A3*Tz+A4*Tw = 0;

eqa2 := B1*(diff(Tw(x), x, x))+B2*(diff(Tw(x), x))+(B3+B4)*Tw+B3*Tz+B4*Tg = 0;

eqa3 := C1*(diff(Tz(x), x, x))+(C3+C4)*Tz+C3*Tg+C4*Tw = 0

2)

On paper, the system of three equations with three unknowns I changed to system of two equations with two unknowns but still nothing. 

A[1] := 2, 

eqa1 := A[1]*C[1]*(diff(z(x), x, x, x, x))/C[3]+A[2]*C[1]*(diff(z(x), x, x, x))/C[3]+(A[1]*C[3]+A[1]*C[4]+A[3]*C[1]+A[4]*C[1])*(diff(z(x), x, x))/C[3]+(A[1]*C[3]+A[1]*C[4])*(diff(z(x), x))/C[3]+(A[3]*C[3]+A[3]*C[4]+A[4]*C[3]+A[4]*C[4]+A[3]*C[3])*z(x)/C[3]+A[1]*C[4]*(diff(y(x), x, x))/C[3]+A[2]*C[4]*(diff(y(x), x))/C[3]+(A[3]*C[4]+A[4]*C[4]+A[4]*C[3])*y(x)/C[3] = 0;

eqa2 := B[4]*C[1]*(diff(z(x), x, x))/C[3]+(B[4]*C[3]+B[4]*C[4]+B[3]*C[3])*z(x)/C[3]+B[1]*(diff(y(x), x, x))+B[2]*(diff(y(x), x))+(B[4]*C[3]+B[4]*C[4]+B[3]*C[3])*y(x)/C[3] = 0;

row := eqa1, eqa2;

sol := dsolve({row}, {y(x), z(x)});

Thank you very much for your help.

Ewa.

How to find the equation of the tangent line to f(x)=6x/squareroot(x^2+12) at the point (2,3) ?