MaplePrimes Questions


 

Download example.mw

I tried to solve some linear differential equations,

but the result is incomplete.

The complete solutions have been derived by hand.

How can I get the right result using maple?

The example is from determining symmetres of differential equations.

Thank you!

 

How to solve this problem? I want to display plot of differential equation system

this download link my problem https://drive.google.com/file/d/0B-qKE-5zgVbLeWVMd0xkMFY1Y00/view?usp=sharing

Thank you :)

  1. I have a set of equations described on different lines and I want them to be executed together and their answers to be displayed in seperate lines after that. How do I do that?
  2. This is particulalry important because, at times I will have to document multiple equations sequentially and later on have their answers reffere to. This way its more structured and I can group equations falling into certain category.

Please find the attached Minimum working example. Any advise is appreciated. File link: [MWE.mw]

 

Thanks

 

 

Hello,

i look for an Maple animation about similar triangles? Thank you

HI MaplePrimes,

The input -

rsolve(f(n)=f(n-1)+10*f(n-2),f(k))

returns a large expression.

My had calculations reduce this to

f(k) = [(41-19*sqrt(41))/820]*[((1-sqrt(41))/2)^k+((1+sqrt(41))/2)^k)].

There may be an error.

We let f(1)=1 and f(2)=2.

The sequence, starting with 1 should read -

1,2,12,32,152,472,...

What is the correct expression for f(k)?

 

Regards,

Matt

Consider a standard initial/boundary value problem for the heat equation on the interval x ∈ [0,1]:

restart;
pde := diff(u(x,t),t) = diff(u(x,t),x,x);
ic := u(x,0) = f(x);
bc := u(0,t)=0,  u(1,t)=0;

Then
pdsolve({pde, ic, bc});
produces the expected Fourier series solution.

However, if we change the interval to x ∈ [-1,1], as in:
bc := u(-1,t)=0,  u(1,t)=0;
pdsolve({pde, ic, bc});

then Maple fails to return a solution.  Why?

Dear all

I need a help how can i solve for example the following PDEs with Initial condition and boundary condition given at x=-1, and x=1.

pde:=diff(u(t,x),t)=diff(u(t,x),x$2);

ics:=u(0,x)=sin(x);

Bcs:=diff(u(t,-1),x)=0;

Bcs:=diff(u(t,1),x)=0;

Many thanks

 

 

 

In using Isolate in RootFinding to compute roots of a real polynomial, the output contains, say, z= some number.  How to get rid of the "z =" so that I can declare that "some number" to be some variable?

I am writing a matrix to file where each matrix element is placed on a new line. An example of such code:

restart:
HH:=Matrix([[x**2,x**z,z**12],[2*x,5**y,6**1]]):
filenameHH:=fopen("test.txt",WRITE,TEXT):

nRow,nCol :=LinearAlgebra[Dimension](HH);
for i from 1 to nRow do
   for j from 1 to nCol do
      fprintf(filenameHH,"%a\n",HH[i,j]):
   od:
od:
fclose(filenameHH):

When this is printed to file, it will print as the power sign ^ regardless of the input it is given. This particular output file will be read in by another language, and due to this the power symbol "^" is not desired, but instead " ** ". By converting the matrix into a string format and applying string operations, this can be done; but is there a simpler way to do it?

-Yeti

Hello,

I am solving a simple system of algebraic equations in a "for" loop. I would like to write all the values of solutions to a matrix, but I only get the symbols. Here is my code:

A:=1:E:=0.1:L0:=<5,5,5,5>:m0g:=0.05:u:=<-5/4,-1/4,-3/4,-1/4>:v:=<-1/2,-3/2,1/2,1/2>:w:=<0,1,0,-1>:
Force:=Matrix(m,2);
for i from 1 to m do
 r1||i:=H||i*L0(i)/(A*E)+H||i/m0g*(arcsinh(V_A||i/H||i)+arcsinh((m0g*L0(i)-V_A||i)/H||i))-sqrt(u(i)^2+v(i)^2);
 r2||i:=V_A||i*L0(i)/(A*E)*(m0g*L0(i)/(2*V_A||i)-1)+H||i/m0g*(sqrt(1+((m0g*L0(i)-V_A||i)/H||i)^2)-sqrt(1+(V_A||i/H||i)^2))-abs(w(i));
 rozw||i:=fsolve({r1||i, r2||i}, {H||i,V_A||i},H||i=0..1);Force(i,1):=H||i;Force(i,2):=V_A||i;subs(rozw||i,Force(i,..));
end do:
evalf(Force);

Thanks in advance,

Iza

I need to create randomly several {1,2,3}-sequences of length 15. I execute:

> with(RandomTools):

> w:=Generate(list(posint(range = 3), 15)):
 

The problem is that when I close the program and then open and run it again, it always produces the same sequences. 

What should I do in order to make it work really randomly? That is, to start at a "random" sequence as well?

Thanks!

Can any one help me correct my doc wherein I want to record 5 minutes with microphone and play it for the same duration through my speaker with a click of button component. I find problem in file path specification.This document in my PC is  in E:\A_PhD2017\MAPLEPhDFiles\RequestCorrection_Microphone_Speaker_from_primetime.mw

Components added are only three: microphone0, Speaker0 and Button0

``

``

``

Edit*Data*Available*Action*contains*the*following*commands

rate := DocumentTools:-GetProperty("Microphone0", samplerate); 1; R := DocumentTools:-GetProperty("Microphone0", value)

Array(%id = 18446745571883247742)

(1)

Edit Click Action for Button contains the following commands.

R := DocumentTools:-GetProperty("Microphone0", value); 1; srate := DocumentTools:-GetProperty("Microphone0", samplerate); 1; DocumentTools:-SetProperty("Speaker0", samplerate, srate); 1; DocumentTools:-SetProperty("Speaker0", value, R)

16000

(2)

NULL

No errors displayed for both the above doc tools use``

Also nonstop continuous evaluation takes place! Where is the loop formation if at all there is any?

Thanks for answering

Ramakrishnan V


 

Download RequestCorrection_Microphone_Speaker_from_primetime.mwRequestCorrection_Microphone_Speaker_from_primetime.mw

PhDFiles\....mw

Hello,

I'm trying to write and simplify expressions involving partial derivatives of an arbitrary function, say f(x,y).

Specifically, I would like diff(f(t,y),t) to evaluate to D[1](f)(t,y), instead of 
diff(f(t,y),t)

And eval(diff(f(x,y),x),x=t) gives the same result. 

Interestingly, diff(f(2*t,y),t) does evaluate to 2*D[1](f)(2*t,y), as expected.

I could get some results by using a custom differentiation function
 

`diff/f` := proc(x,y,v) 
    if v = x then return D[1](f)(v,y):
    elif v = y then return D[2](f)(x,v):
    else return 0: # is there a fallback I can use without risking infinite recursion?
    fi:
end proc:

and it works for the simple case, but now diff(f(2*t,y),t) does not work.

All this makes me feel that there must be a way to get what I want by default, but I can't figure it out.

Is there? Thank you in advance!

Hi Mapleprimes people and robots,


My question is regarding a recursive sequence.  It can be defined non-recursively as - 


a(r) :=  0.8*3^r + 0.2*(-2)^r.

The first few terms are - 

1,2,8,20,68,188, and so on.

Here is my Maple Worksheet.
recursive_sequence_A133467.mw      recursive_sequence_A133467.pdf

I want some Maple code that will produce 30 terms of this sequence.  It is defined as

s[1]:=1:
s[2]:=2:

for n>2 we let s[n] = s[n-1] + 6*s[n-2].

Let me know if my question does not make sense.

Regards,
Matt

 

Hello,
I have a second problem which is related to converting horizontal axis to degrees ( or in terms of Pi :  0 - Pi/2 ).

Thank you for your helps

Bengu


restart;
s1 := (1/3*(4*(1.66+10^(-1))^2-x^2))^(3/2)/(x^2*(1+.66+10^(-1)))-sin(2*a);
s2 := (1/3*(4*(1.66+10^(-2))^2-x^2))^(3/2)/(x^2*(1+.66+10^(-2)))-sin(2*a);
s3 := (1/3*(4*(1.66+10^(-3))^2-x^2))^(3/2)/(x^2*(1+.66+10^(-3)))-sin(2*a);
k1 := series(s1, x);
                    (-2)                                 
       4.769067180 x     + (-0.5773502688 - 1. sin(2. a))

                           2    / 4\
          + 0.01164914507 x  + O\x /
k2 := series(s2, x);
                    (-2)                                 
       4.293792441 x     + (-0.5773502692 - 1. sin(2. a))

                           2    / 4\
          + 0.01293857500 x  + O\x /
k3 := series(s3, x);
                    (-2)                                 
       4.247636750 x     + (-0.5773502693 - 1. sin(2. a))

                           2    / 4\
          + 0.01307916821 x  + O\x /
p1 := convert(k1, polynom);
  4.769067180                                                2
  ----------- - 0.5773502688 - 1. sin(2. a) + 0.01164914507 x 
       2                                                      
      x                                                       
p2 := convert(k2, polynom);
  4.293792441                                                2
  ----------- - 0.5773502692 - 1. sin(2. a) + 0.01293857500 x 
       2                                                      
      x                                                       
p3 := convert(k3, polynom);
  4.247636750                                                2
  ----------- - 0.5773502693 - 1. sin(2. a) + 0.01307916821 x 
       2                                                      
      x                                                       
with(plots, implicitplot);
implicitplot([p1, p2, p3], a = 0 .. (1/2)*Pi, x = 0 .. 3, style = point, symbol = [box, cross, circle], labeldirections = [horizontal, vertical], labels = ["&theta; (radians)", "Switching  field (T)"], color = [blue, red, black], symbolsize = 13);

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