Items tagged with convert convert Tagged Items Feed

 

hi.i encountered this erroe  [Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system] with solving set of differential equation.please help me.thanks a lot  

dsys3 := {`1`*h1(theta)+`1`*(diff(h1(theta), theta, theta))+`1`*(diff(h2(theta), theta))+`1`*(diff(h2(theta), theta, theta, theta))+`1`*h3(theta)+`1`*(diff(h3(theta), theta, theta))+`1`*(diff(h1(theta), theta, theta, theta, theta)) = 0, `1`*h2(theta)+`1`*(diff(h2(theta), theta, theta, theta, theta))+`1`*(diff(h2(theta), theta, theta))+`1`*(diff(h1(theta), theta))+`1`*(diff(h1(theta), theta, theta, theta))+`1`*(diff(h3(theta), theta))+`1`*(diff(h3(theta), theta, theta, theta)) = 0, h3(theta)^5*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h3(theta), theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h3(theta), theta, theta, theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h1(theta)*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h1(theta), theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h2(theta), theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h2(theta), theta, theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h3(theta)^4*(diff(h2(theta), theta, theta, theta, theta, theta, theta))*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)-beta*h3(theta)^3*`1`-chi*ln(h3(theta))^2*`1`/kappa-chi*`1`/kappa-2*chi*ln(h3(theta))*`1`/kappa = 0, h1(0) = 0, h1(1) = 0, h2(0) = 0, h2(1) = 0, h3(0) = 1, h3(1) = 1, ((D@@1)(h1))(0) = 0, ((D@@1)(h1))(1) = 0, ((D@@1)(h2))(0) = 0, ((D@@1)(h2))(1) = 0, ((D@@1)(h3))(0) = 0, ((D@@1)(h3))(1) = 0, ((D@@2)(h3))(0) = 0, ((D@@2)(h3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 600, numeric, output = listprocedure);
%;
Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

 

you can change help of older maple version to 18 by this command:

HelpTools:-Database:-ConvertAll():

for example if you download DirectSearch optimization package it's help don't open in maple 18 because in maple 18 .hdb converted to .help and you can do this convert by HelpTools:-Database:-ConvertAll():

DirectSearch version 2 created for maple 13 and i converted it's help to 18.

after i typed this command maple 18 wrote: 

"Converting G:\\Program Files\\Maple 18\\lib\\DirectSearch.hdb to G:\\Program Files\\Maple 18\\lib\\DirectSearch.help"
Warning, .hdb help databases are deprecated, 'G:\Program Files\Maple 18\lib\DirectSearch.hdb' will not be used, see ?HelpTools,Migrate help page for more information.

and when try again it worked properly and DirectSearch help opened.

I am using Maple 15 to numerically solve a system of differential algebraic euqations (DAE) with given initial conditions, and I've tried rfk45_dae and rosenbrock_dae solver, but both solver responded in error like this

 

Error, (in dsolve/numeric) cannot numerically solve complex DAE initial value problems, the system must be converted to a real system

 

I don't understand what is a real system, and how could i convert it to a real system.

 

This is an addition to the following post:

http://www.mapleprimes.com/questions/141795-Unit-Conversion-Problem

 

But I use the clickable facilities of Maple.  Here is the problem:

>32[[degC]];

                         32[[degC]]
right-click -> Units -> Replace units -> degF

                         288           
                         --- [[degF]]
                          5            

>evalf[5]( (2) );
                       57.600[[degF]]

but if I do that:


>convert(32, 'temperature', 'degC', 'degF');

                              448
                              ---
                               5
                           
>evalf[5]( (4) );
                             89.600

Why the conversion is bad when you try to do it by the clickable way????????????

 

--------------------------------------
Mario Lemelin
Maple 18 Ubuntu 13.10 - 64 bits
Maple 18 Win 7 - 64 bits messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987

I have two 6x1 Matrices which are the results of a calculation process in Maple. One with a set of equations and the other one with a set of variables: 

A := [0, f(x6), f(x6), 0, 0, f(x6)];

b := [x1, x2, x3, x4, x5, x6];

I'd like to solve the following system:

for i from 1 to 6 do

eq[i] := A[i] = b[i]:

od;

which is

eq[1] := 0 = x1;

eq[2] := f(x6) = x2;

eq[3] := f(x6) = x3;

...

 

If I type in the eqations manually, and execute "s := solve({eq[1],..,eq[6]},{x1,..,x6})" everything solves fine.

If I use the "for i from..." - structure, and execute "s := solve({eq[1],..,eq[6]},{x1,..,x6})" I get an empty space as solution.

I've tried to convert both matrices into lists, but it doesn't work.

Could it be that Maple doesnt know that x6 has to be the x6 in the function f(x6) ?

Can anyone tell me how to solve this please?

i am trying to solve 6 ODE with boundary condition


restart

with*plots

with*plots

(1)

Eq1 := (1-theta(eta)/theta[r])*(diff(f(eta), eta, eta, eta))+(diff(f(eta), eta, eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(1-theta(eta)/theta[r])*(diff(diff(diff(f(eta), eta), eta), eta))+(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(2)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(3)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(4)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(5)

Eq5 := (1+s*theta(eta))*(diff(theta(eta), eta, eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(1+s*theta(eta))*(diff(diff(theta(eta), eta), eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(6)

Eq6 := 2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

(7)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(8)

fixedparameter := [M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1];

[M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1]

(9)

Eq7 := eval(Eq1, fixedparameter);

(1+(1/10)*theta(eta))*(diff(diff(diff(f(eta), eta), eta), eta))-(1/10)*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))+(1+(1/10)*theta(eta))^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-.5*(diff(f(eta), eta))+.5*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(10)

Eq8 := eval(Eq2, fixedparameter);

G(eta)*(diff(F(eta), eta))+F(eta)^2+.5*F(eta)-.5*(diff(f(eta), eta)) = 0

(11)

Eq9 := eval(Eq3, fixedparameter);

G(eta)*(diff(G(eta), eta))+.5*f(eta)+.5*G(eta) = 0

(12)

Eq10 := eval(Eq5, fixedparameter);

(1+.1*theta(eta))*(diff(diff(theta(eta), eta), eta))+.1*(diff(theta(eta), eta))^2+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)+.3333333333*H(eta)*(theta[p](eta)-theta(eta)) = 0

(13)

Eq11 := eval(Eq6, fixedparameter);

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+.5*theta[p](eta)-.5*theta(eta) = 0

(14)

bcs2 := F(10) = 0;

F(10) = 0

(15)

bcs3 := G(10) = -f(10);

G(10) = -f(10)

(16)

bcs4 := H(10) = n;

H(10) = n

(17)

bcs5 := theta(10) = 0;

theta(10) = 0

(18)

bcs6 := theta[p](10) = 0;

theta[p](10) = 0

(19)

L := [.2];

[.2]

(20)

for k to 1 do R := dsolve(eval({Eq10, Eq11, Eq4, Eq7, Eq8, Eq9, bcs1, bcs2, bcs3, bcs4, bcs5, bcs6}, n = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta[p](eta)], numeric, output = listprocedure); Y || k := rhs(R[5]); YP || k := rhs(R[6]); YJ || k := rhs(R[7]); YS || k := rhs(R[2]) end do

``


Download hydro.mw

restart

with*plots

with*plots

(1)

Eq1 := (1-theta(eta)/theta[r])*(diff(f(eta), eta, eta, eta))+(diff(f(eta), eta, eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(1-theta(eta)/theta[r])*(diff(diff(diff(f(eta), eta), eta), eta))+(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(2)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(3)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(4)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(5)

Eq5 := (1+s*theta(eta))*(diff(theta(eta), eta, eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(1+s*theta(eta))*(diff(diff(theta(eta), eta), eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(6)

Eq6 := 2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

(7)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(8)

fixedparameter := [M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1];

[M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1]

(9)

Eq7 := eval(Eq1, fixedparameter);

(1+(1/10)*theta(eta))*(diff(diff(diff(f(eta), eta), eta), eta))-(1/10)*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))+(1+(1/10)*theta(eta))^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-.5*(diff(f(eta), eta))+.5*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(10)

Eq8 := eval(Eq2, fixedparameter);

G(eta)*(diff(F(eta), eta))+F(eta)^2+.5*F(eta)-.5*(diff(f(eta), eta)) = 0

(11)

Eq9 := eval(Eq3, fixedparameter);

G(eta)*(diff(G(eta), eta))+.5*f(eta)+.5*G(eta) = 0

(12)

Eq10 := eval(Eq5, fixedparameter);

(1+.1*theta(eta))*(diff(diff(theta(eta), eta), eta))+.1*(diff(theta(eta), eta))^2+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)+.3333333333*H(eta)*(theta[p](eta)-theta(eta)) = 0

(13)

Eq11 := eval(Eq6, fixedparameter);

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+.5*theta[p](eta)-.5*theta(eta) = 0

(14)

bcs2 := F(10) = 0;

F(10) = 0

(15)

bcs3 := G(10) = -f(10);

G(10) = -f(10)

(16)

bcs4 := H(10) = n;

H(10) = n

(17)

bcs5 := theta(10) = 0;

theta(10) = 0

(18)

bcs6 := theta[p](10) = 0;

theta[p](10) = 0

(19)

L := [.2];

[.2]

(20)

for k to 1 do R := dsolve(eval({Eq10, Eq11, Eq4, Eq7, Eq8, Eq9, bcs1, bcs2, bcs3, bcs4, bcs5, bcs6}, n = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta[p](eta)], numeric, output = listprocedure); Y || k := rhs(R[5]); YP || k := rhs(R[6]); YJ || k := rhs(R[7]); YS || k := rhs(R[2]) end do

``


then i get this error

Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

i dont know where i need to change after view it one by one..

Download hydro.mw

Dear Mapleprimes,

I have been struggling with a problem in the last couple of days. I wish to export a Maple plot to LaTeX while ensuring font consistency. While searching for solutions online, I found the psfrag package in LaTeX. So far, however, I have been unsuccesful in making this work. As as test, I attempted to export plot(x^2) to LaTeX. I used the following code to convert to .eps which worked fine:

plotsetup(ps, plotoutput = `plot1.eps`, plotoptions = `portrait, noborder,height=5in,width=5in`);plot(x^2);

Then in LaTeX, I have:

\documentclass{article}

\usepackage{graphicx}

\usepackage{psfrag}

\begin{document}

\begin{figure}[!h]
\centering
\psfrag{x}{$ \alpha $}
\includegraphics[scale=0.5]{plot1.eps}
\end{figure}
\end{document}

However, no replacements are made. After intense Google searching I found the following post http://www.mapleprimes.com/posts/43255-Trouble-Replacing-Maple-Axes-Labels which to sum up argues that this was only possible with earlier versions of Maple.

Does anyone know if the problem has been resolved?

Does anyone know any other ways to ensure font consistency for plots imported from Maple to LaTeX?

Thank you very much in advance!

C

Hi all,

 

Say I have some list like this,

tmp:=[[0, 0, 1], [0, 1, 0], [0, 1, 1], [0, 1, 2], [1, 0, 0], [1, 0, 1], [1, 0, 2], [1, 1, 0], [1, 1, 1], [1, 1, 2], [1, 2, 0]];

 

And I have worked out some probabilities for each of them, a,b,c,d, ect.

I want to print them like this

Pr( 001 ) = 1

Pr( 010 ) = 1-phi[2]+phi[2]*(1-p[3])*(1-phi[3])

Pr( 011 ) = phi[2]*p[3]*(1-phi[3])

and so on.

I there a way to do that?

The probabilities can be extracted from a Vector. I have no problem to print them.

I dont know how to convert the 0,1,2 into the desired format as shown above.

 

This is the best I can do.

 

Also, is it possible to convert all the subscripte [] to _ when printing the output?

and get ride of all * as well.

Thanks,

 

casperyc

 

Dear all;

Please Have some one an idea to transform or convert 2nd order ODE to system of First ODE ( of course using maple).

Thanks

 

how to convert decimal to fraction without simplify

for example

convert(0.25, fraction)

expect 25/100, but not 1/4

Hi,

I have a problem with dsolve in the following code

restart;
>
n:=20;
m:=1;
cc:=-200;
zzeta:=0.1;
sefr1:=0.3;
sefr:=0.2;
MM:=0;
lambda:=0.1;
Br:=1;
nn:=3;
>
>
#u(tau):=tau;
u(tau):=421.7129935*tau-2217.587728*tau^2+8897.376593*tau^3-27612.59182*tau^4+64248.00336*tau^5-1.083977605*10^5*tau^6-10.57029600-1.080951714*10^6*tau^13+7.999517316*10^5*tau^14-4.788741005*10^5*tau^15+2.309563748*10^5*tau^16+26511.11102*tau^18-5959.001794*tau^19+1.148523882*10^5*tau^7-95.23809524*tau^21+4.545454545*tau^22-9435.563781*tau^8-2.587683745*10^5*tau^9+6.473880128*10^5*tau^10+948.0272727*tau^20-88660.41892*tau^17-1.008692404*10^6*tau^11+1.175504242*10^6*tau^12;
>
>
B := 1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2*(1-zzeta)));
eq4 := 4*B*u(tau)-(1+zzeta)*(diff(tau*(diff(theta(tau), tau)), tau))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2;


theta(tau):=sum(p^ii*theta[ii](tau),ii=0..nn);
HH:= p*((4*(1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2-(1/2)*zzeta))))*u(tau)-(1+zzeta)*(diff(theta(tau), tau)+tau*(diff(theta(tau), tau, tau)))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2)+(1-p)*(diff(theta(tau),tau$2)):
eq5:=simplify(HH):
eq6:=collect(expand(eq5),p);

eq7:=
convert(series(collect(expand(eq5), p), p, nn+1), 'polynom');


for ii to nn do
ss[ii] := (coeff(eq7, p^ii)) ;
print (ii);
end do;

ss[0]:=diff(theta[0](tau), tau, tau);

icss[0]:=theta[0](zzeta/(2*(1-zzeta)))=0, D(theta[0])(1/(2*(1-zzeta)))=1;

dsolve({ss[0], icss[0]});
theta[0](tau):= rhs(%);


for ii to nn do
ss[ii]:=evalf[5](ss[ii]);
icss[ii]:=theta[ii](zzeta/(2*(1-zzeta)))=0, D(theta[ii])(1/(2*(1-zzeta)))=0;
dsolve({ss[ii], icss[ii]});
theta[ii](tau):=rhs(%);
end do;

I would be most grateful if you help me to find this problem.

Thanks for your attention in advance

 

Dear all

I would like to convert Matlab code to Maple, is there anu idea, this is the code.

 

% Usage: [y t] = abm4(f,a,b,ya,n) or y = abm4(f,a,b,ya,n)
% Adams-Bashforth-Moulton 4-th order predictor-corrector method for initial value problems
% It uses
% Adams-Bashforth 4-step method as a precdictor,
% Adams-Moulton 3-step method as a corrector, and
% Runge-Kutta method of order 4 as a starter
%
% Input:
% f - Matlab inline function f(t,y)
% a,b - interval
% ya - initial condition
% n - number of subintervals (panels)
%
% Output:
% y - computed solution
% t - time steps
%
% Examples:
% [y t]=abm4(@myfunc,0,1,1,10);          here 'myfunc' is a user-defined function in M-file
% y=abm4(inline('sin(y*t)','t','y'),0,1,1,10);
% f=inline('sin(y(1))-cos(y(2))','t','y');
% y=abm4(f,0,1,1,10);

function [y t] = abm4(f,a,b,ya,n)
h = (b - a) / n;
h24 = h / 24;

y(1,:) = ya;
t(1) = a;

m = min(3,n);

for i = 1 : m % start-up phase, using Runge-Kutta of order 4
    t(i+1) = t(i) + h;
    s(i,:) = f(t(i), y(i,:));
    s2 = f(t(i) + h / 2, y(i,:) + s(i,:) * h /2);
    s3 = f(t(i) + h / 2, y(i,:) + s2 * h /2);
    s4 = f(t(i+1), y(i,:) + s3 * h);
    y(i+1,:) = y(i,:) + (s(i,:) + s2+s2 + s3+s3 + s4) * h / 6;
end;

for i = m + 1 : n % main phase
    s(i,:) = f(t(i), y(i,:));
    y(i+1,:) = y(i,:) + (55 * s(i,:) - 59 * s(i-1,:) + 37 * s(i-2,:) - 9 * s(i-3,:)) * h24; % predictor
    t(i+1) = t(i) + h;
    y(i+1,:) = y(i,:) + (9 * f(t(i+1), y(i+1,:)) + 19 * s(i,:) - 5 * s(i-1,:) + s(i-2,:)) * h24; % corrector
end;

number10:=`466d06ece998b7a2fb1d464fed2ced7641ddaa3cc31c9941cf110abbf409ed39598005b3399ccfafb61d0315fca0a314be138a9f32503bedac8067f03adbf3575c3b8edc9ba7f537530541ab0f9f3cd04ff50d66f1d559ba520e89a2cb2a83`:

number8:=`315c4eeaa8b5f8bffd11155ea506b56041c6a00c8a08854dd21a4bbde54ce56801d943ba708b8a3574f40c00fff9e00fa1439fd0654327a3bfc860b92f89ee04132ecb9298f5fd2d5e4b45e40ecc3b9d59e9417df7c

I first define

f:=x->convert(x, decimal, hex):

with(Bits):
str1:=convert( `Xor(f(number8), f(number10))`, bytes);

now how can I get back the alphabets, since again use of convert with bytes return the inital argument.

Moreover, I would really appreciate if someone could explain the difference between 

convert(`expr`, bytes)

convert( [expr], bytes)

 

Many regards!!

 

As the title, how to convert "sin(x)+cos(x)" and this kinds into "sqrt(2)*sin(x+pi/4)"?

Sum of two sine functions with the same cycle should can be converted into one sine function, with some amplitude gain and phase offset. 

 

> assume(a < 0);
> convert(cosh(sqrt(a)), sincos);
print(`output redirected...`); # input placeholder
/ (1/2)\
cos\(-a) /

This is what I expected.

Now

> assume(L > 0);
> assume(K > 0);
> assume(mu > 0);
> assume(mu^2 < 4*L*k);
> assume(t > 0);
> convert(cosh((1/2)*t*sqrt(mu^2-4*L*k)/L), sincos);
print(`output redirected...`); # input placeholder
/ (1/2)\
| / 2 \ |
|t \mu - 4 L k/ |
cosh|--------------------|
\ 2 L /

I wanted to obtain again the cos function. Could someone help me?
(What is the reason that convert does not work "well" in later case?)

 Thanks,  Sandor

 

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