Have anyone toldl me how to find the family of curve at Maple 14

ax^2 + B*x*y+cy^2= D

with A=1, B=1, C =1 and D =2010

Can someone send me a clear step by step instruction to install Maple 13 for 32 bit Linux?

My OS is Ubuntu 10.04 LTS netbook edition.

I don't want to install it in home folder,instead I want to create a folder(say,MAPLE) in a partition with a lot of space and then install Maple13 inside this folder.

I have a worksheet that contains various pieces of a robot's dynamic and kinematic equations.  I want to perform different design tasks, taking the base model and working off of it.  I'd like to have only 1 model of the robot and not duplicate it across all of my different design worksheets because things change in the robot model as I switch up tooling, find better means of performing a system identification, etc.

Is there any functionality that would allow...

 Here is a example of a cesaro sum used when a series fails to converge in the usual sense of a sum.

sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity) has a cesaro sum of 1/2*(a+2MRB constant-1).

Proof:

We are given that                            S=  sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity) .

Expanding the infinite series we get         S=(a-1)+(2^(1/2)-a)+(a-3^(1/3))+(4^(1/4)-a)+... .

Collecting the a's and the surds we see that S=a-a+a-a+...+(-1+2^(1/2)-3^(1/3)+4^(1/4)...) .

By Grandi's series  we know that             S=1/2*a+(-1+2^(1/2)-3^(1/3)+4^(1/4)...) .

Collecting the infinite series we get        S=1/2*a+ sum((-1)^n*(n^(1/n)), n = 1 .. infinity).

Which can be shown to be                     S=1/2*a+ sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)-1/2 .

Thus, by factoring out the 1/2, we get           S=1/2*(a+2 sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)-1) .

Therefore,                                   S=1/2*(a+2MRB constant-1) .

In order to give symbolic results for that familly of sums Maple shouldmake this identify an integral part of maple in future versions!

Marvin Ray Burns

Original investigator of the MRB constant.

Dear Maple lovers,

As a classical worksheet user from the past years, now I'd like to be a "modern" Maple user, using the 2D Input in the modern worksheet mode (I will call it mwm from now on). But, alas, I cannot even solve a simple equation!

To describe my situation let me state the following:

1. I use 64 bit Maple 14 Student Edition (single user) under 64 bit Windows 7.

2....

how can i take any linear ode and make a plot of all its solutions, if its first order and second order?

thanks in advance

I All,

I just startup on Maple and i can't get the result of adding to  matrix. I used a worksheet and the math mode. The input is in 2-D Math Notation. Could Someone help me. 

> a := matrix([[93, 43], [19, 37]]); b := matrix([[48, 20], [19, 37]]);

> a+b;                             a + b

Thank You

I want to exactly (well roughly speaking) overlay two lines from different graphs. 

So suppose I had two datasets, 

a:=[1.4,2.1,4.6,3.7,3,2.1,2,1,1.5]:
b:=[78,75,97,98,105,95,88,75,67]:

with(plots):

aa:=listplot(a):
bb:=listplot(b):

I want to overlay the two lines to compare them.  I transformed them into 3d then rotated to overlay but maple keeps scaling the axes to fit. 

Any ideas?

I download a maple document from 'Recent Questions',"to find the constants of a equation with data x and y are given", then I open it, and click 'run all' comand in the toolbar. It works very well.

But when I put cursor in one line, and pree 'enter' key, it ruturns a different form of the result, as the picture above shows.

I want to plot a piecewise function, F(x,y), so I look for piecewise command in help page, there are a lot of examples for one-variable piecewise functions, but lack of examples for two-variable piecewise functions. I don't know how to input a two-variable piecewise function and how to plot it, can you help me? Thank you very much.

Hi,

I am new to maple and this is my first question in this forum.

Please help me in finding an analytical solution to the 2nd order ode.

ode:=diff(y(x),x$2) -( a y(x) - y(x)^3 + y(x)^5 ) = 0

bcs:= D(y)(0) = 0, y(2) = 0, y(0) = 1.31

where a is a constant.

I am bit confused about the boundary conditions. I have three boundary conditions instead of two.

The first two boundary conditions yield trivial y(x) = 0 solution.so added the third bc.

Download Maple_primes_ques_27.mw