Maple Questions and Posts

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So I have this expression

f:=(coth(x)^(1/3)-tanh(x)^(1/3))*(coth(x)^(2/3)+tanh(x)^(2/3)+1)

which Maple can not simplify?

I need to do it like this

`assuming`([expand(simplify(add(`~`[simplify]([op(combine(expand((coth(x)^(1/3)-tanh(x)^(1/3))*(coth(x)^(2/3)+tanh(x)^(2/3)+1))))]))))], [x > 1])

Is this actually true or what is happening here?

Hi

I have a first oder PDE, I use pdsolve I obtained a solution depend on function F

condition_unique_solution.mw

 

My question: The boundary condition  f(x,y) = 1 is supplied on the line y = k x, where k is a constant. For which k
does there exist a unique solution for f(x, y)?

 

Many thanks for your help

 

Hi there

I'm an old user of Maple, but I've never been able to plot functions with unit. You can see my latest attempt down below

b := 120*Unit('mm');
h := 200*Unit('mm');
V := 8*Unit('kN');

I__x := (1/12)*b*h^3

Q(x):=(1/2)*((1/4)*h^2-(100*Unit('mm')-x)^2)*b 
tau(x):=V*Q(x)/(I__x*b)

plot(Q(x(Unit('mm')), units), x = 0*Unit('mm') .. 100*Unit('mm'))

Plot_function_with_units.mw

If anyone is able to help me with this problem, I would greatly appreciate it.

Hi 

I solve the laplace equation written in polar coordinates in annular domain.
The code run without any error 

But there is no solution displayed after running the code, note that I use Maple 18

Laplace_annulardomain.mw

Many thinks for your help

I want to divide each row of Marix A by diagonal element. In for loop, when I assign dividing results to the letter , the type of A still remains matrix, but if I choose another letter (like B) results are stored in Table. Why? How can I assign to a matrix?

Also I can't figure out why maple doesn't show elements of table(see worksheet file).

I should notice that my main problem is assigning not dividing.

worksheet.mw

 

i want to design a packaging container to hold 320 sphere-shaped chocolates that each has a diameter 1.8 cm and weights about 3.2g each. i hope can get all posible shape using maple18 .

 

Good morning everyone, 

I have a problem, when I try to evaluate the definite integral below, Maple can not provide a result. What can I do so that the Maple can calculate this integral?

This is the Maple code with the result:

 

 

restart

with(VectorCalculus):

with(LinearAlgebra):

with(CodeGeneration):

N := 1:

M := 2:

``

for i to N do rpv1 || i := 0; rpv2 || i := 0; rpv3 || i := 0; for j to M do rpv1 || i := VectorCalculus:-`+`(rpv1 || i, Typesetting:-delayDotProduct(diff(VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^j, s), Phi || i || j)); rpv2 || i := VectorCalculus:-`+`(rpv2 || i, Typesetting:-delayDotProduct(diff(VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^j, s), `ϕ` || i || j)); rpv3 || i := VectorCalculus:-`+`(rpv3 || i, Typesetting:-delayDotProduct(diff(VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^j, s), gamma || i || j)) end do; rp || i := Matrix([[rpv1 || i], [rpv2 || i], [rpv3 || i]]) end do:

``

for i to N do rppv1 || i := 0; rppv2 || i := 0; rppv3 || i := 0; for j to M do rppv1 || i := VectorCalculus:-`+`(rppv1 || i, Typesetting:-delayDotProduct(diff(diff(VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^j, s), s), Phi || i || j)); rppv2 || i := VectorCalculus:-`+`(rppv2 || i, Typesetting:-delayDotProduct(diff(diff(VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^j, s), s), `ϕ` || i || j)); rppv3 || i := VectorCalculus:-`+`(rppv3 || i, Typesetting:-delayDotProduct(diff(diff(VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^j, s), s), gamma || i || j)) end do; rpp || i := Matrix([[rppv1 || i], [rppv2 || i], [rppv3 || i]]) end do:

``

``

for i to N do for j from 0 to 0 do U || i || j := 0 end do end do:

for i to N do for j from 0 to 0 do V || i || j := 0 end do end do:

for i to N do for j from 0 to 0 do W || i || j := 0 end do end do:

for i to N do for j from 0 to VectorCalculus:-`+`(M, -2) do U || i || (VectorCalculus:-`+`(j, 1)) := VectorCalculus:-`+`(U || i || j, VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^VectorCalculus:-`+`(j, 1)) end do end do:

for i to N do for j from 0 to VectorCalculus:-`+`(M, -2) do V || i || (VectorCalculus:-`+`(j, 1)) := VectorCalculus:-`+`(V || i || j, VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^VectorCalculus:-`+`(j, 1)) end do end do:

for i to N do for j from 0 to VectorCalculus:-`+`(M, -2) do W || i || (VectorCalculus:-`+`(j, 1)) := VectorCalculus:-`+`(W || i || j, VectorCalculus:-`*`(VectorCalculus:-`+`(s, VectorCalculus:-`-`(xi || i)), 1/L || i)^VectorCalculus:-`+`(j, 1)) end do end do:

for i to N do f || i := VectorCalculus:-`+`(Typesetting:-delayDotProduct(VectorCalculus:-`*`(Typesetting:-delayDotProduct(E, A), 1/mu), VectorCalculus:-`+`(VectorCalculus:-`+`(rpp || i, VectorCalculus:-`-`(VectorCalculus:-`*`(rpp || i, 1/evalc(norm(Re(rp || i), 2))))), VectorCalculus:-`*`(Typesetting:-delayDotProduct(rp || i, Typesetting:-delayDotProduct(rp || i^%T, rpp || i)), 1/evalc(norm(Re(rp || i), 2))^3))), Typesetting:-delayDotProduct(g, e3)) end do:

for i to N do for j to VectorCalculus:-`+`(M, -1) do fun || i || j := int(VectorCalculus:-`*`(U || i || j, Row(f || i, 1)), s = xi || i .. L || i) end do end do;

fun11

(int((s-xi1)*(E*A*(2*Phi12/L1^2-2*Phi12/(sqrt((gamma11/L1+2*gamma12*s/L1^2-2*gamma12*xi1/L1^2)^2+(`ϕ11`/L1+2*`ϕ12`*s/L1^2-2*`ϕ12`*xi1/L1^2)^2+(Phi11/L1+2*Phi12*s/L1^2-2*Phi12*xi1/L1^2)^2)*L1^2)+(Phi11/L1+(2*(s-xi1))*Phi12/L1^2)*((2*(Phi11/L1+(2*(s-xi1))*Phi12/L1^2))*Phi12/L1^2+(2*(`ϕ11`/L1+(2*(s-xi1))*`ϕ12`/L1^2))*`ϕ12`/L1^2+(2*(gamma11/L1+(2*(s-xi1))*gamma12/L1^2))*gamma12/L1^2)/((gamma11/L1+2*gamma12*s/L1^2-2*gamma12*xi1/L1^2)^2+(`ϕ11`/L1+2*`ϕ12`*s/L1^2-2*`ϕ12`*xi1/L1^2)^2+(Phi11/L1+2*Phi12*s/L1^2-2*Phi12*xi1/L1^2)^2)^(3/2))/mu+g*e3)/L1, s = xi1 .. L1))*e[x]

(1.1)

``

NULL

``


 

Download integral.mw

 

Thank you !

If I have checked the Editable button just below the working window, then the temperature would be very high in the next time when I start Maple 2019. I do not what is going on. But when I unchecked the Editable button, and wait for several seconds, then the temperature and the load of my laptop are on the normal state.  Is this a bug for Maple 2019? My OS is Debian Stretch, that is,

$ uname -a
Linux debian 4.9.0-9-amd64 #1 SMP Debian 4.9.168-1 (2019-04-12) x86_64 GNU/Linux

 

Hi

I try to solve the laplace equation with some special boundary conditions.

But, i get the follwoing error

Error, (in pdsolve/sys) the given system is not polynomial in the variables {f}

 

 laplace_equation.mw

Thank you for any help

 

This worksheet is a modification to Kitonum's excellent http://www.mapleprimes.com/posts/202222-Contour-Curves-With-Labels.

The mod adds the ability to display labelled contours for expressions in x and y defined parametrically.

Your comments are welcome.

Contourplot_with_labels.mw


 

with(LinearAlgebra)

A := Matrix(4, 4, {(1, 1) = 1, (1, 3) = -1, (1, 4) = 3, (2, 2) = 2, (2, 3) = 1, (3, 1) = -1, (3, 2) = 1, (3, 3) = 6, (3, 4) = -1, (4, 1) = 3, (4, 3) = -1, (4, 4) = 10}, fill = 0)

Matrix(%id = 18446746512315154430)

(1)

b := Matrix(4, 1, {(1, 1) = 0, (2, 1) = -2, (3, 1) = -1, (4, 1) = -1})

Matrix(%id = 18446746512315153574)

(2)

x := Matrix([[x1], [x2], [x3], [x4]])

Matrix(%id = 18446746512315146838)

(3)

f := proc (x) options operator, arrow; (1/2)*Transpose(x).A.x+Transpose(b).x end proc

proc (x) options operator, arrow; Typesetting:-delayDotProduct(Typesetting:-delayDotProduct((1/2)*LinearAlgebra:-Transpose(x), A), x)+Typesetting:-delayDotProduct(LinearAlgebra:-Transpose(b), x) end proc

(4)

(1/2)*Transpose(x).A.x+Transpose(b).x

Matrix(%id = 18446746512315132494)

(5)

while g(vk) < 10^(-6) do k end do

Error, cannot determine if this expression is true or false: (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -1, (1, 4) = 3, (2, 1) = 0, (2, 2) = 2, (2, 3) = 1, (2, 4) = 0, (3, 1) = -1, (3, 2) = 1, (3, 3) = 6, (3, 4) = -1, (4, 1) = 3, (4, 2) = 0, (4, 3) = -1, (4, 4) = 10})) . vk+(Matrix(4, 1, {(1, 1) = 0, (2, 1) = -2, (3, 1) = -1, (4, 1) = -1})) < 1/1000000

 

v0 := Matrix([[0], [1], [0], [0]])

Matrix(%id = 18446746512315172734)

(6)

g := proc (x) options operator, arrow; A.x+b end proc

proc (x) options operator, arrow; Typesetting:-delayDotProduct(A, x)+b end proc

(7)

alpha0 := solve(diff(f(v0-g(v0)*alpha)[1, 1], alpha) = 0)

1/10

(8)

v1 := v0-alpha0*g(v0)

Matrix(%id = 18446746512315163582)

(9)

v(k+1) := vk-`&alpha;k`*g(vk)

vk-`&alpha;k`*(Matrix(%id = 18446746512315154430).vk+Matrix(%id = 18446746512315153574))

(10)

`&alpha;k` := solve(diff(f(vk-g(vk)*alpha)[1, 1], alpha) = 0)

Hi, 
Essentially i am trying to programe an iterative loop where v(K+1) can be found fro v(K), I'm not sure how to programe a loop but I know this is not a hard thing to do so I am struggling, any help would be appreciated! Thanks. 
Edit: Also alpha K must be found at each stage by optimising f(v(k+1))
 

Download Optimisation_coursework.mw

Hello everyone!

I'm having some problem with this equation:

solve(0.1 = 23.714*(-0.93205)^2/(20.3+61.4*.884^x), x)

I'm trying to solve for x, but i keeps saying "Warning, solutions may have been lost."

Any ideas?
 

Hello,

What is the minimum period of the following equation.


 

d := evalf(expand((100+100*cos(6*t)+200*cos(12*sqrt(2)*t))^2))

40000.-2580480000.*cos(t)^2*cos(1.414213562*t)^6+604800000.*cos(t)^2*cos(1.414213562*t)^4-51840000.*cos(t)^2*cos(1.414213562*t)^2+2621440000.*cos(t)^6*cos(1.414213562*t)^12-7864320000.*cos(t)^6*cos(1.414213562*t)^10+8847360000.*cos(t)^6*cos(1.414213562*t)^8-4587520000.*cos(t)^6*cos(1.414213562*t)^6+1075200000.*cos(t)^6*cos(1.414213562*t)^4-92160000.*cos(t)^6*cos(1.414213562*t)^2-3932160000.*cos(t)^4*cos(1.414213562*t)^12+0.1179648000e11*cos(t)^4*cos(1.414213562*t)^10-0.1327104000e11*cos(t)^4*cos(1.414213562*t)^8+6881280000.*cos(t)^4*cos(1.414213562*t)^6-1612800000.*cos(t)^4*cos(1.414213562*t)^4+138240000.*cos(t)^4*cos(1.414213562*t)^2+1474560000.*cos(t)^2*cos(1.414213562*t)^12-4423680000.*cos(t)^2*cos(1.414213562*t)^10+4976640000.*cos(t)^2*cos(1.414213562*t)^8+720000.*cos(t)^2+274560000.*cos(1.414213562*t)^4-5125120000.*cos(1.414213562*t)^6+0.4942080000e11*cos(1.414213562*t)^8-0.2811494400e12*cos(1.414213562*t)^10+0.1013841920e13*cos(1.414213562*t)^12+0.1677721600e12*cos(1.414213562*t)^24-0.1006632960e13*cos(1.414213562*t)^22+0.2642411520e13*cos(1.414213562*t)^20-0.3984588800e13*cos(1.414213562*t)^18+0.3810263040e13*cos(1.414213562*t)^16-0.2406481920e13*cos(1.414213562*t)^14-5760000.*cos(1.414213562*t)^2+10240000.*cos(t)^12-30720000.*cos(t)^10+34560000.*cos(t)^8+1320000.*cos(t)^4-16000000.*cos(t)^6

(1)

``


 

Download period

 

 

When typing 

z:=exp(I*2*Pi/3);
convert(z,'sincos')

Maple evaluates the intermediate result which is cos(2*Pi/3)+I*sin(2*Pi/3) and  gives

Is there a way to tell it not to do this? I'd like to see the result as when typing

'cos(2*Pi/3)+I*sin(2*Pi/3)'

Is there an option or method to tell Maple not to immediate evaluation in the above? it can do evaluate next time the expression is used.

 

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