The function

110 Reputation

4 Badges

3 years, 44 days

MaplePrimes Activity


These are questions asked by The function

Dear everyone,

Ive been looking at this problem ive got, but i cant find out what would make it work. I can put numbers in a matrix, i can solve 3 unknowns with 3 equations. But now there is a variable, and the variable causes the solution to give either: 1 solution, no solution, or more than one solution. 

I can "brush the matrix" as they say it in Dutch, but im learning Maple so it can go a lot faster than manual labor. So is there a way to do it with Maple? 

Figure 4.1 gives from left to right: 1 solution, no solution, and an indefinite amount of solutions.

Questions: For what value of p, a. gives 1 solution? b. gives no solution? c. gives more than 1 solution?

Now i got Maple to tell me how much p x1, x2, and x3 was, but its not bringing me anywhere to where i can get to a solution. 

A manual method is given in this example. Manual is way way to slow for what i need to achieve, ive got some large big books with mechanics, dynamics, statics, fluid mechanics, concrete calcutions, chemistry etc. waiting for me. I can never achieve that manually in the same time as i could do it with maple. :S 

#example of how to do a solve of a matrix with 3 variables:

stelsel := {x1+x2+x3 = 4, 3*x1+x2-5*x3 = -10, 3*x1+2*x2-x3 = 1}

{x1+x2+x3 = 4, 3*x1+x2-5*x3 = -10, 3*x1+2*x2-x3 = 1}

(1)

solve(stelsel, {x1, x2, x3})

{x1 = -7+3*x3, x2 = 11-4*x3, x3 = x3}

(2)

#now the example number 15

stelsel := {(2-p)*x2+x3 = 2, -x1+2*x2+(2-p)*x3 = 0, (1-p)*x1+2*x2+2*x3 = p+3}

{(2-p)*x2+x3 = 2, -x1+2*x2+(2-p)*x3 = 0, (1-p)*x1+2*x2+2*x3 = p+3}

(3)

solve(stelsel, {x1, x2, x3})

{x1 = -(p^2-6)/(p^2-4*p+4), x2 = -(2*p-5)/(p^2-4*p+4), x3 = 1/(-2+p)}

(4)

``

Thank you!

Greetings,

The Function

Download Mapleprimes_Question_Book_2_Paragraph_4.2_Example_15.mw

Hello! 

Ive stubled into something odd. Here it is: 

Now I really wonder how they figured how i was suppose to figure out how this is suppose to go. They did not explain all that much on how to find this out. This specifically is a "Computerized Question." 

It says "find the smallest number "n", so that A^n=I"

This was the result when i tried to solve it:

"Error, (in Engine:-Dispatch)" I have no idea what is wrong. 

Any way, 

Greetings,

The Function 

#Opdracht 9

A := `<,>`(`<|>`(0, 1, 0, 0), `<|>`(0, 0, 1, 0), `<|>`(1, 0, 0, 0), `<|>`(0, 0, 0, -1))

Matrix(%id = 18446746584512210574)

(1)

solve(A^n = I, n)

Error, (in Engine:-Dispatch) badly formed input to solve: not fully algebraic

 

``

Download Mapleprimes_Question_Book_2_Paragraph_4.1_Question_9.mw

Dear all,

im having trouble with adding two matrixes when they are only described with a letter asigned to them. I really dont know why it is the case. But it has got the be done with letters only because later on i will need this a lot with vector calculations etc. 

Thank you!

A := matrix([[2, -5, 1], [0, 3, -2]]); B := matrix([[0, 2, -1], [7, 1, 3]]); matrix(A+B)

A := Matrix(2, 3, {(1, 1) = 2, (1, 2) = -5, (1, 3) = 1, (2, 1) = 0, (2, 2) = 3, (2, 3) = -2})

 

array( 1 .. 2, 1 .. 3, [( 1, 2 ) = (2), ( 2, 2 ) = (1), ( 2, 3 ) = (3), ( 1, 1 ) = (0), ( 2, 1 ) = (7), ( 1, 3 ) = (-1)  ] )

 

Error, (in matrix) invalid argument A+B

 

help("matrix")

(Matrix(2, 3, {(1, 1) = 2, (1, 2) = -5, (1, 3) = 1, (2, 1) = 0, (2, 2) = 3, (2, 3) = -2}))+(Matrix(2, 3, {(1, 1) = 0, (1, 2) = 2, (1, 3) = -1, (2, 1) = 7, (2, 2) = 1, (2, 3) = 3}))

Matrix(%id = 18446746769648450246)

(1)

A+B

A+B

(2)

evalc(%)

A+B

(3)

evalf(%)

A+B

(4)

``


Greetings,

The Function

Download Mapleprimes_Question_Book_2_Paragraph_4.1.mw

Hello everybody, im at it again. Math with maple. I should be picking up some speed again to plow through this Dutch math book that explains Maple. It creeps me out.. But hey, im learning Maple in the process, and that is what its all about!

The example shows what is done. I made question a. happen, and the answer was right. The thing is with question b. they ask to plot the phase vectors alpha of f(t), g(t), and s(t), although there is no variable t in the phase vector. So how on earth will i plot it in the complex plane?

The literal translation of quesion b is: "check graphically the answer of part a. by drawing the phase vectors f(t), g(t), and s(t) in the complex plane."

I cant get it done. 

Would anyone know the right question. There were no graphs displayed at the answers in the back of the book. 

Thank you!

Greetings,

The Function

Opdracht 2

a.

"f(t):=3*cos(2*t-Pi/(4))"

proc (t) options operator, arrow, function_assign; 3*cos(2*t-(1/4)*Pi) end proc

(1)

"g(t):=4*cos(2*t+Pi/(6))"

proc (t) options operator, arrow, function_assign; 4*cos(2*t+(1/6)*Pi) end proc

(2)

f(t)+g(t)

3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi)

(3)

smartplot(3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi))

 

3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi)

3*sin(2*t+(1/4)*Pi)+4*cos(2*t+(1/6)*Pi)

(4)

3*exp(I*(0-(1/4)*Pi))

(3/2)*2^(1/2)-((3/2)*I)*2^(1/2)

(5)

4*exp(I*((1/6)*Pi))

2*3^(1/2)+2*I

(6)

NULL

3*sqrt(2)*(1/2)-(1/2)*(3*I)*sqrt(2)+2*sqrt(3)+2*I

(3/2)*2^(1/2)-((3/2)*I)*2^(1/2)+2*3^(1/2)+2*I

(7)

evalf(%)

5.585421959-.121320343*I

(8)

arctan((2-3*sqrt(2)*(1/2))/(3*sqrt(2)*(1/2)+2*sqrt(3)))

arctan((-(3/2)*2^(1/2)+2)/((3/2)*2^(1/2)+2*3^(1/2)))

(9)

evalf(%)

-0.2171747628e-1

(10)

5.585421959*cos(2*t-0.2171747628e-1)

5.585421959*cos(2*t-0.2171747628e-1)

(11)

smartplot(5.585421959*cos(2*t-0.2171747628e-1))

 

b.

"fc(t):=3*(e)^(I*(0-Pi/(4)))"

proc (t) options operator, arrow, function_assign; 3*exp(-((1/4)*I)*Pi) end proc

(12)

"gc(t):=4*(e)^((I*Pi)/(6))"

4*exp(((1/6)*I)*Pi)

(13)

"sc(t):=(3 sqrt(2))/2-(3 &ImaginaryI; sqrt(2))/2+2 sqrt(3)+2 &ImaginaryI;"

proc (t) options operator, arrow, function_assign; (3/2)*sqrt(2)-((3/2)*I)*sqrt(2)+2*sqrt(3)+2*I end proc

(14)

``

Download Mapleprimes_Question_Book_2_Paragraph_3.9_Question_2_b.mw

It may seem like a stupid question, but why is the first answer not complete, but the second imput is the same as what the book states?

I know you dont need to be able to read Dutch to understand that A = amplitude (amplitude), Omega = Hoekfrequentie (Angular frequency), and Varphi= Fasehoek (phase angle). 

They state that fc(0)= A*e^i*varpi, but it clearly shows that the equation on the left of the equal sign gives a "2" and on the right it gives "2+ 2*i*sqrt3".

Why is this so? It does not make things more easy if you can make "mistakes" like that by just asking the answer, but maple spits out half an answer... 

He thanks a lot guys!

Greetings,

The Function 

p.s.
I scanned my math books, i like it, i can search in the book now with the "ctrl+f" function in Adobe reader. Get a scanner. Mine is a Canon Lide 400. Not regretting it.. :) 
Download Mapleprimes_Book_2_Question_3.6.mw
 

omega := 5

5

(1)

`&varphi;` := (1/3)*Pi

(1/3)*Pi

(2)

A := 4

4

(3)

"f(t):=A*cos(omega*t+`&varphi;`)""+A*I*sin(omega*t+`&varphi;`)"
 

proc (t) options operator, arrow, function_assign; A*cos(omega*t+varphi) end proc

 

(4*I)*sin((1/3)*Pi+5*t)

(4)

f(0)

2

(5)

A*exp(I*`&varphi;`)

2+(2*I)*3^(1/2)

(6)

NULL

``

Download Mapleprimes_Book_2_Question_3.6.mw

1 2 3 4 5 6 7 Page 3 of 7