Hi.

I have a procedure that takes multiple parameters, in which I want to fix all but one, and then define a single variable function that maps x to this remaining unspecified parameter, so I can give it to the plot() routine. I would like to define a number of such single variable functions for different values of the parameters and plot them all on the same axes.

Ideally, I would like to find a way to make the following, instead of returning 14, return 21,...

I am a novice user in Maple with some previous Matlab experience. My question is the following: I am thinking about doing some Maple calculations. The calculations will involve (in the easiest case) about 15 (symbolic) variables , and it will require all kinds of symbolic differantation, integration etc. such as F(x)=b(t) where x is the symbolic variables and t is the differentiation parameter (all symbolic). I would like to know how many variables Maple can deal...

There have been many times when I want to see what the graph a function looks like, but don't have numbers to put in there. Sure, I could just plug in numbers, but that often affects the graph - such as slope. Is there any way for Maple to show me the shape of a graph with a symbolic function?

Why does this happen to Maple 15?

`assuming`([sum(k*p*Beta(k, p+1), k = 1 .. infinity)], [p > 1]); eval(%, p = 2)

Is it possible to show with Maple that for any real p>1 the series converges to p/(p-1), e.g.,

`assuming`([sum(k*p*Beta(k, p+1), k = 1 .. 1000)], [p > 1]): subs(p = 2, %): evalf(%)

How do I show this symbolically? Thanks.

Maple evaluates

> sum(1/z^2, z = 1 .. infinity)

to (1/6)*Pi^2. What do I do when sumation is over all even positive integers? Is there any closed-form symbolic formula for this case? Above is the minimal example. What I need is to compute

> sum(2*binomial(m-1, k-1)*binomial(n-1, k-1)/binomial(m+n, m), k = 1 .. infinity)

where k belongs to a set of only even positive integers, not any posint. The second expression evaluates to 2*m*n/((m+n-1...

I am trying to calculate inverse laplace of a 3*3 matrix (answer1 matrix). Find below the maple commands i used.

>with (linear algebra):

A:=<alpha,p11,e11|0,o11,-e11|a,m11,0>

N:=<s,0,0|0,s,0|0,0,s>

k:=N-A

Z:=MatrixInverse(k)

B:=<0,-m11,0>

C:=<1,0,0>

X0:=<isalpha0,ilalpha0,vcalpha0>

answer1:=Z.(X0+(ScalarMultiply(B,uinvalpha/s))+(ScalarMultiply(C,ealpha/s)))

I would like to generate a symbolic sum of the first "n" derivatives of a function.

For example, f = 1/(1-x^4)

sum(f^(n),n=0..3)

The response from Maple is not as expected. The first term is: -1/(4*(x-1)) and the subsequent terms are less recognizable.

If I use "seq" with the same syntax, I end up with a list that includes the function and the first 3 derivatives.

I have also tried:

sum(diff(f,x$n),n=0..3)

With f := i->`if`(i = 0, 1, alpha), I'd like to evaluate expression A below entirely symbolically, i.e. with l=1..3 being replaced with l=1..h in

sum('sum(sum(f(i)*f(i+j), i = 0 .. l-1), j = 1 .. h-l)', l = 1 .. 3); subs(alpha = .3, h = 3, %);#A

but using A with l= 1 .. h provides a different number, the one one obtains with

sum(sum(sum(f(i)*f(i+j), i = 0 .. l-1), j = 1 .. h-l), l = 1 .. h); subs(alpha = .3, h = 3, %); #BExpressions...

1.For win 8, does there exist an arm version maple in future?

2.Will it possible to use gpu acceleration for symbolic computation in maple?

Hello,

I want to use the values of the solutions of

sol := solve({x+y=a, x-y=b}, {x,y});

I would like to be able to continue some symbolic computationg, for example

dernier:= sol[1]/sol[2];

but I get an error message.

How can I take only the values of x (and not the expression x= something) to compute, for example "dernier"?

Thanks in advance,

I use SOLVE command

>solu=solve({x+y=a, x-y=b}, {x, y});

solu = {x = (1/2)*b+(1/2)*a, y = -(1/2)*b+(1/2)*a}

>solu[1];solu[1]

(which shows the solution value for x is not solu[1]).

I would like to use the values of x and y in SOLU to continue with some symbolic computation.

How do I recover -FROM SOLU- the values of X and Y (solutions)?

Jean-Jacques

Thank you

Your answer makes sense if Maple evaluates from innermost to outermost, so the substitution in T2c takes place AFTER Test is evaluated with a symbolic value of "n".

But it doesn't explain why the "assuming" examples work - or don't work. If "Test" is evaluated first, then none of the examples should work; if the assumption is passed through "simultaneously" (whatever that means), then the wording of the error message doesn't make sense...

hi ,friends

now ,i get a matrix of symbolic expression,it's difficulty to find the simplest forms using the common command as like simplify or combine

p := Matrix(4, 4, {(1, 1) = (1/2)*exp((1/2)*J[1]-(1/2)*J[2]+(1/2)*J[3])+(1/2)*exp(-(1/2)*J[1]+(1/2)*J[2]+(1/2)*J[3]), (1, 2) = 0, (1, 3) = 0, (1, 4) = (1/2)*exp((1/2)*J[1]-(1/2)*J[2]+(1/2)*J[3])-(1/2)*exp(-(1/2)*J[1]+(1/2)*J[2]+(1/2)*J[3]), (2, 1) = 0, (2, 2) = (1/2)*exp(-(1/2)*J[1...

Let a:= (2x-1)/sqrt(1-4x^{2}) , b:=sqrt(1-2x)/sqrt(1+2x)

then c:=combine(a/b,radical,symbolic) is 1, it should be -1 !

(Multiply b top and bottom by sqrt(1-2x) to see this).

Am I misunderstanding combine?

Hi everyone,

I'm quite sure that this question has already been answered but I couldn't find any satisfactory explanation (my bad).

I'm wondering why the following expression

sum(0^k, k = 0 .. 0);yields 0 as result while

0^0;returns 1

and

sum(1, k = 0 .. 0);also return 1.

I understand that 0^0;

is an indeterminate form so that its evaluation is a matter of convention but Maple's approach...

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