Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

The other night, first time using Maple in more than 20 years, I asked for the solution to this, looking for u1 in terms of u2:

equ := u2 - u1/(-u1^2 + 1)

solve(equ, [u1])

==> u1 = (-1 + sqrt(4*u2^2 + 1))/(2*u2)

But this is sheer nonsense. The correct solution should be:

u1 = (-1 + sqrt(u2^2 + 1))/u2   ;; note the lack of 4x scaling inside the Sqrt, nor the 2x in the denominator.

What the heck??

I want collect 1/G(xi) & G'(xi)/G(xi) also 1/G(xi)*G(xi)/G(xi)  when they have power ,and give me what i looking for, i can do by hand but it take time any one can do this by maple code? like this picture below, and if possible find some arbitrary parameter

collect.mw

After more than 25 years of leading research in areas such as differential equations, special functions, and computational physics, Edgardo's role with Maplesoft will shift at the end of 2024 as he returns to academic research. At Maplesoft, he will transition into the position of Research Fellow Emeritus. In this role, Edgardo will remain engaged with many of his cherished projects, though he will not have as much time to maintain the intense level of activity that characterized his work for so many years.

Many of you know Edgardo personally or have interacted with him here or on the Maple Beta Forum. I hope you'll join me in wishing him the very best as he begins this new chapter.

how this integro-differential equation can be solved?
any assumption or suggestion is appreciated. tnx in advance

restart

eq:=diff(y(x),x)=y(x)+exp(x)*exp(-3*x)/2+int(exp(x+t)*y(t),t=0..x);

IC:=y(0)=1;

diff(y(x), x) = y(x)+(1/2)*exp(x)*exp(-3*x)+int(exp(x+t)*y(t), t = 0 .. x)

 

y(0) = 1

(1)

eq2:=g(x)=int(exp(x+t)*y(t),t=0..x);

g(x) = int(exp(x+t)*y(t), t = 0 .. x)

(2)

IC_2:=eval(eq2,[x=0,y=1])

g(0) = 0

(3)

sys:={diff(eq2,x),subs(rhs(eq2)=lhs(eq2),eq)}

{diff(g(x), x) = int(exp(x+t)*y(t), t = 0 .. x)+exp(2*x)*y(x), diff(y(x), x) = y(x)+(1/2)*exp(x)*exp(-3*x)+g(x)}

(4)

dsolve(sys union {IC,IC_2},numeric)

Error, (in dsolve/numeric/process_input) input system must be an ODE system, got independent variables {t, x}

 

Download integro-diffrential_problem.mw

Hey all Maple experts.I could really use some help with  diff,D,Diff

restart

interface(version)

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

(1)

with(Physics[Vectors])

NULL

CompactDisplay(A_(x, y, z, t), `ϕ`(x, y, z, t), v_(x, y, z, t), F_(x, y, z, t), v__x(x, y, z, t), v__y(x, y, z, t), v__z(x, y, z, t), A__x(x, y, z, t), A__y(x, y, z, t), A__z(x, y, z, t), quiet)

macro(Av = A_(x, y, z, t), `ϑ` = `ϕ`(x, y, z, t), Vv = v_(x, y, z, t), Fv = F_(x, y, z, t))

show, ON, OFF, kd_, ep_, Av, vartheta, Vv, Fv

(2)

Fv = q*('-VectorCalculus[Nabla](`ϑ`)'-(diff(Av, t))+`&x`(Vv, `&x`(VectorCalculus[Nabla], Av)))

F_(x, y, z, t) = q*(-Physics:-Vectors:-Nabla(varphi(x, y, z, t))-(diff(A_(x, y, z, t), t))+Physics:-Vectors:-`&x`(v_(x, y, z, t), Physics:-Vectors:-Curl(A_(x, y, z, t))))

(3)

Av = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, Vv = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k

A_(x, y, z, t) = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, v_(x, y, z, t) = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k

(4)

subs[eval](A_(x, y, z, t) = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, v_(x, y, z, t) = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k, F_(x, y, z, t) = q*(-Physics[Vectors][Nabla](varphi(x, y, z, t))-(diff(A_(x, y, z, t), t))+Physics[Vectors][`&x`](v_(x, y, z, t), Physics[Vectors][Curl](A_(x, y, z, t)))))

F__x*_i+F__y*_j+F__z*_k = q*(-(diff(varphi(x, y, z, t), x))*_i-(diff(varphi(x, y, z, t), y))*_j-(diff(varphi(x, y, z, t), z))*_k-(diff(A__x(x, y, z, t), t))*_i-(diff(A__y(x, y, z, t), t))*_j-(diff(A__z(x, y, z, t), t))*_k+(-v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_i+(-v__z(x, y, z, t)*(diff(A__y(x, y, z, t), z))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), y))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__y(x, y, z, t), x)))*_j+(v__y(x, y, z, t)*(diff(A__y(x, y, z, t), z))-v__y(x, y, z, t)*(diff(A__z(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__z(x, y, z, t), x))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_k)

(5)

map(Component, F__x*_i+F__y*_j+F__z*_k = q*(-(diff(varphi(x, y, z, t), x))*_i-(diff(varphi(x, y, z, t), y))*_j-(diff(varphi(x, y, z, t), z))*_k-(diff(A__x(x, y, z, t), t))*_i-(diff(A__y(x, y, z, t), t))*_j-(diff(A__z(x, y, z, t), t))*_k+(-v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_i+(-v__z(x, y, z, t)*(diff(A__y(x, y, z, t), z))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), y))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__y(x, y, z, t), x)))*_j+(v__y(x, y, z, t)*(diff(A__y(x, y, z, t), z))-v__y(x, y, z, t)*(diff(A__z(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__z(x, y, z, t), x))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_k), 1)

F__x = -v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))*q+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))*q+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))*q-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z))*q-(diff(varphi(x, y, z, t), x))*q-(diff(A__x(x, y, z, t), t))*q

(6)

collect(F__x = -v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))*q+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))*q+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))*q-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z))*q-(diff(varphi(x, y, z, t), x))*q-(diff(A__x(x, y, z, t), t))*q, [q, v__x(x, y, z, t), v__y(x, y, z, t), v__z(x, y, z, t)])

F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q

(7)

convert(F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q, Diff)

F__x = (v__y(x, y, z, t)*(Diff(A__y(x, y, z, t), x)-(Diff(A__x(x, y, z, t), y)))+(Diff(A__z(x, y, z, t), x)-(Diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(Diff(varphi(x, y, z, t), x))-(Diff(A__x(x, y, z, t), t)))*q

(8)

convert(F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q, D)

F__x = (v__y(x, y, z, t)*((D[1](A__y))(x, y, z, t)-(D[2](A__x))(x, y, z, t))+((D[1](A__z))(x, y, z, t)-(D[3](A__x))(x, y, z, t))*v__z(x, y, z, t)-(D[1](varphi))(x, y, z, t)-(D[4](A__x))(x, y, z, t))*q

(9)

 
Hello everyone, in the result of this command execution process, it appears that the symbols for partial derivatives and derivatives in equation (8) are displayed incorrectly. What should I do?

Download error_display.mw

I am trying to compute partial derivatives of some complicated expression which include summations. First, I noticed that sum behaves differently if I use 1D vs. 2D math. Why?

Questions:
  1. Partial derivative of a summation: why is it not just 2*X[i]?
  2. Partial derivative of a double summation: how to define the nested structure of a double summation with j<>i?
  3. System of n+1 equations: how to define and solve for it?

For 3., each i equation is the partial derivative of my complicated expression with summations with respect to X[i], where i ranges from 1 to n. The last equation is the partial derivative with respect to X_r (a fixed variable).

Thanks.

restart

A := sum(X[i]^2, i = 1 .. n); eq[1] := diff(A, X[i]) = 0

sum(X[i]^2, i = 1 .. n)

 

sum(2*X[i], i = 1 .. n) = 0

(1)

B__wrong := sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n); B__correct := 'sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n)'

n^2*X__r^2*w^2+sum(sum(X__r*w*X[j]+X[i]*X[j], j = 1 .. n)+n*X__r*w*X[i], i = 1 .. n)

 

sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n)

(2)

eqs := seq(eq[i], i = 1 .. n); vars := seq(X[i], i = 1 .. n)

Error, range bounds in seq must be numeric or character

 

Error, range bounds in seq must be numeric or character

 
 

NULL

Download equations_with_summations.mw

On some occasions I have seen output like this

where the user name is replaced by a placeholder.
How to achieve this?

Let P(u,v) = -u⁴+88u³v-146u²v²+88uv³-v⁴+2u²+40uv+2v²-1, with P(u,v)=0,

and u0 = sqrt(2)/6 i,  v0 = -sqrt(2)/6 i. We see that P(u0,v0)=0.

Direct substitution of implicitdiff(P(u,v),v,u) at u0, v0 leads to 0/0.

For obtaining the true value of dv/du at (u0,v0) we have written in Maple:

solve(limit(subs(v=v0+k*(u-u0), implicitdiff(P(u,v),v,u)), u=u0)-k, k);

There are two solutions: 49/113 - 72*sqrt(2)/113*I, and 49/113 + 72*sqrt(2)/113*I.

However we were unable to determine the true value of d²v/du² at (u0,v0).

QUESTION:  How to determine it?

Hi. I'm comparing two 7x3-man teams lists. I can see there are at least 2 teams that have common members, new[1] and past[4], new[4] and past[3]. There may be others but the chat gpt code below doesn't definitively find any of them.

common.mw

The company library with all the good stuff has passed a 100 Mb size, and I wonder if it is possible to find out, which parts of the library that use most of the storage space?

After exertion with ordinary differential equations now relaxation:

Determine the formation law, limit and sum limit for
u_n+3=(13/12)*u_n+2 - (3/8)*u_n+1 + (1/24)*u_n .
Starting values ​​u_1=0, u_2=1, u_3=1.

We’re thrilled to announce the launch of our new Student Success Platform! Over the past several months, our academic team has dedicated itself to understanding how we can better support institutions in addressing their concerns around student retention rates. The numbers tell a concerning story: In the U.S., nearly 25% of first-year undergraduates don’t complete their studies, and in STEM fields, the numbers are even higher. In both STEM programs and non-STEM programs with math gateway courses, struggles with math are often a key reason students do not, or cannot, continue their studies. This has a profound impact on both the students’ futures and the institution’s revenue and funding.

From what we’re hearing from institutions and instructors, one of the most pressing issues is the lack of readiness among first-year students, particularly in math courses. With larger class sizes and students arriving with varying levels of preparedness, instructors face challenges in providing the personalized support that is essential. Additionally, many students don’t fully utilize existing resources, such as office hours or TA sessions, which increases their risk of falling behind and ultimately dropping out.

Our new Student Success Platform is designed to tackle these issues head-on. It combines all of our existing tools with exciting new features to help students succeed on their own terms—without adding to instructors' already busy workloads. The early feedback has been fantastic, and we can’t wait for you to see the impact it can make.

You can read more about the Student Success Platform here: https://www.maplesoft.com/student-success-platform/

 

If we calculating it take to much time but if we make a procedure it will be more effectable for such example, i want the exact and approximat and error

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(inttrans)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

eq := diff(y(x, t), `$`(t, 2))+(1+x)*(diff(y(x, t), x))-y(x, t) = 2*y(x, t)^3

diff(diff(y(x, t), t), t)+(1+x)*(diff(y(x, t), x))-y(x, t) = 2*y(x, t)^3

(2)

eq1 := laplace(eq, t, s)

s^2*laplace(y(x, t), t, s)-s*y(x, 0)+laplace(diff(y(x, t), x), t, s)*x-laplace(y(x, t), t, s)-(D[2](y))(x, 0)+laplace(diff(y(x, t), x), t, s) = 2*laplace(y(x, t)^3, t, s)

(3)

eq2 := subs({y(x, 0) = 1, (D[2](y))(x, 0) = 1}, eq1)

s^2*laplace(y(x, t), t, s)-s+laplace(diff(y(x, t), x), t, s)*x-laplace(y(x, t), t, s)-1+laplace(diff(y(x, t), x), t, s) = 2*laplace(y(x, t)^3, t, s)

(4)

eq3 := s^2*laplace(y(x, t), t, s) = s-laplace(diff(y(x, t), x), t, s)*x+1+laplace(diff(y(x, t), x), t, s)+2*laplace(y(x, t)^3, t, s)+laplace(y(x, t), t, s)

s^2*laplace(y(x, t), t, s) = s-laplace(diff(y(x, t), x), t, s)*x+1+laplace(diff(y(x, t), x), t, s)+2*laplace(y(x, t)^3, t, s)+laplace(y(x, t), t, s)

(5)

eq4 := expand(eq3/s^2)

laplace(y(x, t), t, s) = 1/s-laplace(diff(y(x, t), x), t, s)*x/s^2+1/s^2+laplace(diff(y(x, t), x), t, s)/s^2+2*laplace(y(x, t)^3, t, s)/s^2+laplace(y(x, t), t, s)/s^2

(6)

NULL

"u[0](x):=invlaplace(1/s+1/(s^2),s,x)"

proc (x) options operator, arrow, function_assign; invlaplace(1/s+1/s^2, s, x) end proc

(7)

u[0](x)

1+x

(8)

n := N

N

(9)

k := K

K

(10)

f := proc (u) options operator, arrow; u^3 end proc

proc (u) options operator, arrow; u^3 end proc

(11)

for j from 0 to 3 do A[j] := subs(lambda = 0, (diff(f(seq(sum(lambda^i*u[i](x), i = 0 .. 20), m = 1 .. 2)), [`$`(lambda, j)]))/factorial(j)) end do

(1+x)^3

 

3*(1+x)^2*u[1](x)

 

3*(1+x)*u[1](x)^2+3*(1+x)^2*u[2](x)

 

u[1](x)^3+6*(1+x)*u[1](x)*u[2](x)+3*(1+x)^2*u[3](x)

(12)

A[0]

(1+x)^3

(13)

y[0] := 1+x

1+x

(14)

y[1] := invlaplace(2*laplace(A[0], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[0], x), x, s)/s^2, s, x)+invlaplace(laplace(y[0], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[0], x), x, s)/s^2, s, x)

(1/10)*x^2*(x^3+5*x^2+10*x+10)-(1/2)*x^3+(1/6)*x^2*(x+3)-(1/2)*x^2

(15)

y[1] := expand((1/10)*x^2*(x^3+5*x^2+10*x+10)-(1/2)*x^3+(1/6)*x^2*(x+3)-(1/2)*x^2)

(1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2

(16)

"u[1](x) :=y[1]  "

proc (x) options operator, arrow, function_assign; y[1] end proc

(17)

NULL

A[1]

3*(1+x)^2*((1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2)

(18)

y[2] := invlaplace(2*laplace(A[1], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[1], x), x, s)/s^2, s, x)+invlaplace(laplace(y[1], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[1], x), x, s)/s^2, s, x)

(1/840)*x^4*(7*x^5+63*x^4+212*x^3+476*x^2+672*x+420)-(1/60)*x^4*(x^3+6*x^2+10*x+20)+(1/420)*x^4*(x^3+7*x^2+14*x+35)-(1/60)*x^3*(x^3+6*x^2+10*x+20)

(19)

y[2] := expand(%)

(1/120)*x^9+(3/40)*x^8+(5/21)*x^7+(7/15)*x^6+(17/30)*x^5+(1/12)*x^4-(1/3)*x^3

(20)

" u[2](x):=y[2]"

proc (x) options operator, arrow, function_assign; y[2] end proc

(21)

A[2]

3*(1+x)*((1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2)^2+3*(1+x)^2*((1/120)*x^9+(3/40)*x^8+(5/21)*x^7+(7/15)*x^6+(17/30)*x^5+(1/12)*x^4-(1/3)*x^3)

(22)

y[3] := invlaplace(2*laplace(A[2], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[2], x), x, s)/s^2, s, x)+invlaplace(laplace(y[2], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[2], x), x, s)/s^2, s, x)

(1/10810800)*x^5*(7623*x^8+99099*x^7+518778*x^6+1634490*x^5+3647930*x^4+5167305*x^3+4221360*x^2+900900*x-1081080)-(1/25200)*x^5*(21*x^6+210*x^5+750*x^4+1680*x^3+2380*x^2+420*x-2100)+(1/831600)*x^5*(63*x^6+693*x^5+2750*x^4+6930*x^3+11220*x^2+2310*x-13860)-(1/25200)*x^4*(21*x^6+210*x^5+750*x^4+1680*x^3+2380*x^2+420*x-2100)

(23)

y[3] := expand(y[3])

(286/945)*x^9+(131/336)*x^8+(1/7)*x^10+(17/70)*x^7-(1/40)*x^6-(1/20)*x^5+(1/12)*x^4+(1091/23100)*x^11+(11/15600)*x^13+(11/1200)*x^12

(24)

NULL

addingterm := y[0]+y[1]+y[2]+y[3]

1+x+(37/60)*x^5+(2/3)*x^4+(1/3)*x^3+x^2+(2351/7560)*x^9+(781/1680)*x^8+(101/210)*x^7+(53/120)*x^6+(1/7)*x^10+(1091/23100)*x^11+(11/15600)*x^13+(11/1200)*x^12

(25)


 

Download aproximate_and_exact_solution.mw

a table like that

 

there is four formula for calculate them which i know them by name of author the first one is adomian second one is (BiazarShafiofAdomian) which one member of mableprimes write code for me,but i don't know how use for all kind function maybe in future i upload this program for fix this issue, the third one is by zhao which is i think i easy for calculate just  i need someone one to wite the program and do some test for some example i  upload some picture in case for getting algorithm to writting and have some example for testing  so  lets see who can do this algorithm is very usfule when we solve ODE or PDE by LDM, also last method is by taking integral have a good method, in this question this algorithm is zhao which is usfull one

Hi!

I am using a proceure to conpute de integral of a function by he Simpson's rule. My function is defined from a function and a procedure, but I am getting the error  "Error, (in w) invalid input: hfun2 expects its 1st argument, t, to be of type numeric, but received (1/10)*i+1/20"

As you can see in the attaxhed file, I have tried several ways to compute the integral but always returns the above error. Please, can yo help me?

Thanks

forum.mw

First 70 71 72 73 74 75 76 Last Page 72 of 2228