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For some reason it always bugs me that Maple pulls a negative sign out in the solution.

eq:=c=a+b*exp(-k)

solve(eq,k)

      

The solution is correct, but one would expect it to be written like ..

        

.. after all that is a more conventional solution, isn't it? 

If I type the last equation into Maple it evaluates to

    

..again slightly unconventional but still correct.  Only if back quotes around the `c-a` are used can we achieve what we would write out on paper.  It would be nice if Maple would output the answer in a conventional manner ... at least in my opinion. 

Can we manipulate Maple to evaluate that solution into the conventional answer? 

 

Afterall, Maple is advertising it's "typsetting appears like it would appear in a textbook" and uses standard Math notation.  All I am trying to say is that if Maple advertises as a math standard could we also present solutions in the same manner? 

But not to get sidetracked from the question I posed, is there a way we can manipulate that solution into a "handwritten" convention?

Hi all

I'm having a hard time, making Maple plot a pretty huge expression in my project.

I have solved a differential equation with initial conditions with method=laplace. The differential equation contains a fourier serie equation, so the more accurate i want the equation to be, the larger the differential equation will be.

Maple solves the equation just fine, and i can plot the solution with 2-4 fourier parts, but when i go higher as i need, the graph ends up empty?

with 20 parts i get the following equation: 

0.*sin(52.88*t)+0.*cos(74.03*t)-0.*sin(74.03*t)-0.*cos(52.88*t)+0.*cos(200.95*t)-0.*sin(200.95*t)+0.*cos(158.65*t)-5.55*10^(-8)*sin(105.76*t)-0.*sin(116.34*t)+0.*cos(31.73*t)-.45*sin(10.58*t)+1.02*cos(10.58*t)+0.*sin(95.19*t)+0.*cos(116.34*t)+0.*sin(179.80*t)-0.*cos(179.80*t)+0.*sin(137.49*t)-0.*sin(31.73*t)-0.*cos(95.19*t)+5.53*10^(-993)*(-1.13*10^992*cos(10.61*t)+8.14*10^991*sin(10.61*t))*exp(-0.7e-1*t)+4.23*10^(-7)*cos(211.53*t)-6.69*10^(-7)*cos(63.46*t)-6.11*10^(-7)*cos(105.76*t)+5.79*10^(-7)*cos(126.92*t)+6.67*10^(-8)*sin(42.31*t)-5.88*10^(-8)*sin(148.07*t)+5.88*10^(-8)*sin(211.53*t)+7.09*10^(-7)*cos(42.31*t)+5.45*10^(-8)*sin(84.61*t)+6.40*10^(-7)*cos(84.61*t)+5.72*10^(-8)*sin(126.92*t)-9.01*10^(-7)*cos(21.15*t)+5.97*10^(-8)*sin(169.22*t)+5.06*10^(-7)*cos(169.22*t)-5.98*10^(-8)*sin(190.38*t)-4.65*10^(-7)*cos(190.38*t)-5.44*10^(-7)*cos(148.07*t)-1.33*10^(-7)*sin(21.15*t)-5.61*10^(-8)*sin(63.46*t)-0.*cos(137.49*t)-0.*sin(158.65*t)

if i plot that expression, the graph ends up empty?

I did also try to solve the equation numerical to plot it with odeplot, but when i try to solve it without the laplace method i get this error message:
"Error, (in dsolve) found the following equations not depending on the unknows of the input system:"

The differential equation is:

ode:=diff(Theta(t), t, t)+2*Zeta*omega[balanceue]*(diff(Theta(t), t))+omega[balanceue]^2*Theta(t) = M[p]/m[balanceue]

and the initial conditions:

ICS := Theta(0) = (1/8)*Pi, (D(Theta))(0) = 0;

when i do:

dsolve({ICS, ode}, Theta(t), method = laplace) it solves just fine.

 

but when i try with:

dsolve({ICS, ode}, Theta(t))

or

dsolve({ICS, ode}, Theta(t),numeric)

I get the message: 

Error, (in dsolve) found the following equations not depending on the unknowns of the input system: {Theta(0) = (1/8)*Pi, (D(Theta))(0) = 0}

It doesnt seem logical at all, is it a bug? Or can anybody help me with this problem?


Regards

Nicolai

Determine the exact solution to the initial value problem

 

y'(x)=   y(x)(20-y(x)) , y(0)=1

                 80

 

Compute a polynomial approximation to y(x). Plot this polynomial approximation together with y(x) on the same axes for x∈[0,20]. Choose different colours and linestyles for each curve.

 

Investigate whether or not it is possible to choose Order to be large enough to ensure that the plots of the polynomial approximation and y(x) are indistinguishable over the [0,20] interval? If this is possible, determine the minimum value of Order required. If you think that it is not possible, explain why not.

I have denotation like A[0], A[1], A[2], A[3]... But one package doesn't allow to use indexed variables.

I'd like to change denotation. For example, to A0, A1, A2, A3, but I don't know how to do it automatically...

I am trying to perform the following manipulation (This is a minimum working example).

 

a < b  < c;
(1)*2;

Error, invalid terms in product: a < b and b < c

 Can anyone tell if it is possible to manipulate inequalities exactly as it is the case with equations?

 

Simple Physics Problem...

Yesterday at 6:06 PM Keith Dow 0

I have two Reissner Nordstrom black holes that are near extreme. How do I show they move? 

export expression to word ...

Yesterday at 4:19 PM aTec 10

When i copy expression and past it in word, i can change the size of the picture whitout loosing the detials.

How can i export the expression to a file, such that when i will open it in word i could change the size without loosing details? much thnks :)

Dear Maple users,

My problem is as follows:

I have a factor base [2,3,5,7,11,33,34,35,36,37,38,39,40]

The numbers from 2 till 11 are primes, the rest is not. 

Then I have to factor (H+c1)(H+c2) in numbers of the factor base , where c1 and c2 go from 1 to some pre-defined limit. H=32 in my case.
And then I have to put the powers of the numbers of the factor base in a matrix. For example: (H+1)(H+1)=33² but also (H+1)(H+1)=3²*11².

That will become in matrix form [0 , 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0 ] but also (!) [0 , 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0 ].

This is not what I want! I want no double representations....

What I want is that (H+c1)(H+c2) should be represented in primes in the matrix if possible and else just represented as the other numbers.

 

hope you guys can help me!

I want to print 2+3= in the input and get exactly the same output.

And how can i do it in a program?

Hello people in mapleprimes,

I want to solve the next system of equation for B/A and C/A.

eq1:=A+B=F+G;
eq2:=k*(A-B)=kappa*(F-G);
eq3:=F*exp(I*kappa*a)+G*exp(-I*kappa*a)=C*exp(I*k*a);
eq4:=kappa*F*exp(I*kappa*a)-kappa*G*exp(-I*kappa*a)=k*C*exp(I*k*a);


But, though it is well-known, solve({eq1,eq2,eq3,eq4},{B/A,C/A})
does not work well, as the values I want to solve it for are
expressions: B/A and C/A not variables.

Then, you might thing the next works well.
eq:=subs({B=A/t,C=A/u},{eq1,eq2,eq3,eq4}):
solve(eq,{t,u});

But, this doesn't work well, with the answer was
only the ratio of t and u expressed as the following:

t = t, u = exp(I*k*a)*(exp(-I*kappa*a)*k^2-exp(I*kappa*a)*k^2-exp(-I*kappa*a)*kappa^2+exp(I*kappa*a)*kappa^2)*t/(4*kappa*k*exp(I*kappa*a)*exp(-I*kappa*a))

Isn't there nice way to solve the above system of equation, except that
sol1:=solve({eq3,eq4},{F,G});assign(sol1);
sol2:=solve({eq1,eq2},{A,B});assign(sol2);

Best wishes
taro

test.mw

restart; with(LinearAlgebra)

``

dF := -.525*exp(-7*t)+2.625*exp(-3*t)+.8*exp(-4*t);

-.525*exp(-7*t)+2.625*exp(-3*t)+.8*exp(-4*t)

(1)

``

e3 := `<,>`(1, 1, 1); E := proc (m) options operator, arrow; IdentityMatrix(m) end proc; beta := `<|>`(.1, .6, .3); S := `<|>`(`<,>`(-3, 1, 1), `<,>`(1, -5, 2), `<,>`(0, 2, -4)); S0 := -S.e3

beta := Vector[row](3, {(1) = .1, (2) = .6, (3) = .3})

 

S := Matrix(3, 3, {(1, 1) = -3, (1, 2) = 1, (1, 3) = 0, (2, 1) = 1, (2, 2) = -5, (2, 3) = 2, (3, 1) = 1, (3, 2) = 2, (3, 3) = -4})

 

S0 := Vector(3, {(1) = 2, (2) = 2, (3) = 1})

(2)

Z := `<|>`(x, y, z)

Z := Vector[row](3, {(1) = x, (2) = y, (3) = z})

(3)

ME := MatrixExponential(S+Typesetting:-delayDotProduct(S0, Z), t);

`[Length of output exceeds limit of 1000000]`

(4)

MEint := map(int, ME.dF, t = 0 .. infinity)

Error, (in int) wrong number (or type) of arguments: wrong type of integrand passed to definite integration.

 

`&beta;plus&Assign;solve`(Z = beta.MEint, Z)

"(RTABLE(18446744074195006390,VECTOR([x, y, z]),Vector[row])=RTABLE(18446744074193876574,VECTOR([.1, .6, .3]),Vector[row]).MEint) betaplus:=solve (RTABLE(18446744074195006390,VECTOR([x, y, z]),Vector[row]))"

(5)

``

1step- I want to integrate the (ME*dF) from t=0 to ∞ .

2step- Evaluate Z=<x,y,z> by solving Z=β*MEint.

Download test.mw

After installing the 18.02 update to Maple 18, the inverse Laplace transform no longer works!

I have a long expression with different order derrivatives, that is written in form like that:

-(D[1](f))(x, y)

I'd like to transform it into standard maple form like:

diff(f(x,y),x)

Is there any special procedure to achieve this goal?

Hi experts

 

In a procedure with no declared parameteres I would like to return (print) the passed arguments (expecting Matrix structures) in a modified form along with the name of the symbol holding the structure passed as arguments.

 

That is, the procedure just iterates through the _passed arguments in a for-loop to display the name of the the passed argument (the symbol), a colon and then the modified matrix structure.

My problem is that when looping from i to _npassed arguments, refering to _passed[i] gives me the evaluated form.Tat is what I need to modify the structure but not to list the symbol name.

 

Say I wanted to print a transposed version of my passed matrices. Then I would call

M:=<<1,2>|<3,4>>;

myProc(M);

And the result I want would be

M: <<1,3>|<2,4>>

But I don't know if _passed holds the symbol names or just the evaluated versions of the passed arguments??

And theoretically the passed argument (assuming a matrix) could be the matrix structure put directly in the procedure call, in which case there is no symbol to refer to.

 

I hope you get my question and can help me out.

 

Thanks

Simon

 

 
Hello
Please help solve this system

restart; 
B:=1: 
q:=1*10^3: 
l:=1: 
n:=4.7: 
M_F:=z->2*q*l*(z-l)-q*z^2/2: 
M_1:=z->piecewise((z<l), 2*q*l*(z-l)+l-z-q*z^2/2, (z>l), 2*q*l*(z-l)+l-q*z^2/2-l): 
M_2:=z->2*q*l*(z-l)+2*l-z-q*z^2/2: 
one_int:=z->int(B*(M_F(z)+X_1*M_1(z)+X_2*M_2(z))^n*M_1(z),z=0..2*l); 
two_int:=z->int(B*(M_F(z)+X_1*M_1(z)+X_2*M_2(z))^n*M_2(z),z=0..2*l);
value(one_int(z)); 
value(two_int(z)); 
eqs1:={value(one_int(z))=0,value(two_int(z))=0}; 
fsolve(eqs1);

 

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