How do I create a pointplot with different values of a parameter?

I tried

pointplot([seq([k, a(k, a0)], k = 0 .. 10, a0 = [0, 50, 100, 150, 200])]);
Error, invalid input: seq expects its 3rd argument, step, to be of type numeric, but received a0 = [0, 50, 100, 150, 200]
NULL;

ImportGraph reads one or more graphs from a file. For a file containing multiple graphs, the supported formats are digraph6, graph6, and sparse6. I notice that one option is "output = graph or list, or iterator". However, in my programming with Maple, I was reminded several times that list is only good for a small amount of data.  

Although I can convert the list to an array using the convert function, the time spent in the conversion process needs to be considered.

restart: 
with(GraphTheory):
L:=ImportGraph("C:/Users/eul10c.g6", graph6, output=list):#31026 connected eulerian graphs on 10 vertices graphs and Change to your path here.
convert(L,Array) 

So does it make more sense to provide an "output=Array" as an output option?

eul10c.txt

I am trying to decide O of times fo a computer process. I thought it was n^2*log(N) or something like that.

With 12  points, it may be hard to determine if O(N^4), etc.

See big-o.mw

When resizing a textplot the text is not scaled. 

I am trying to use "solve" to solve a set of 2 equations defined by using summation function "sum". But I got an error saying "Error, (in sum) summation variable previously assigned, second argument evaluates to 3 = 1 .. 2". 

The Maple code is listed here

restart;
N := 2:
var:= seq(alpha[n], n = 1 .. N):
v(x):=sum(alpha[n]*phi[n](x), n = 1 .. N):
R(x):=diff(v(x), x $ 4) - p/EI:
eqs := seq('int'('R(x)'*phi[n](x), x = 0 .. L) = 0, n = 1 .. N):
for n to N do
    phi[n](x):= sin((2*n - 1)*Pi*x/L):
end do:       
solve({eqs}, {var});
Error, (in sum) summation variable previously assigned, second argument evaluates to 3 = 1 .. 2

Hi,

While playing with this, I got:

> sum(Beta(k,1/2)/(2*k+1)^2,k=1..infinity)
Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation

I don't see how this can be singular, as Beta(k,1/2) is trivially bounded for k>=1, and sum(1/(2*k+1)^2) is absolutely convergent.

Maple is still able to evaluate numerically the inert sum (I get 0.3361376233). If I replace Beta with the expression with GAMMA, I get a result with MeijerG, that doesn't seem to help much for numerical evaluation, as it takes a long time and gives the wrong result -1.966769953*10^10 - (6.886827095*10^8)*I.

I have no reason to expect a closed form for this sum, but the error is surprising. And I don't know MeijerG well enough to explain the numerical error either. It fails with Maple 2020, 2021 and 2022.

Hi,

I try to animate the tangent in a moving point on the curve

ideas ?

Thanks

QuestionAnim.mw

Hey, I've got simmilar problem. I will show it on the same example, but here my expressions are defined as functions and both graphs should be on the same plot. With my version I get:

Warning, expecting only range variable x in expression a*x^2 to be plotted but found name a

My version is:

;

Version below works fine, but I wanted to have this plot in a worksheet not in popup window:

Thanks in advance,

Iza

I try to simplify the expression by putting exponential function into equation but the maple shows error which I can't fix it.simplification.mw

Hi,

I am trying to use Maple to solve a PDE involving composite functions.  How do I program this?  Here is an attached file describing what I am trying to do.

Partial_Differential_Equations_of_Composite_Functions.mw

Hi I was trying to difine a constant c that exists in range(0,d). The constant c has two values and is depends on its x location.  The first value "a"  has a duration "e" , the second value "b" has a duiration "f". Start from x =0, the constant c has value "a", then "b", then "a"......

 How should I define this constant in "piecewise"?

I have the following example.

> A:= Vector(3,symbol=v)

A:= 

>v[1]:= 5

v[1]:=5

> A

Is there any way such that v[1] is actually referring to the first element in the vector? In other words, changing v[1] actually modifies the vector A.

If the question does not make sense, then here is the actual scenario that I faced.

I need to solve a system of equations in the form of 

where a[i], c[i] refers to the entries in the vector.

solve returns the following:

I then did an assign(%). However, the entries in A and C are not changed. I want the answer returned by solve to be properly applied to the entries of the vectors.

I have been trying to integrate the following an expression using Maple's command PathInt:

PathInt(4*x*y*(y^4+2*x*y-2)/sqrt((1-(2*x*y)^2+(-3*y^2+1)^2)*(1+(y^2-1)^2)), [x, y] = Path(`<,>`(-(1+sqrt(2))*cos(t)-(sqrt(2)-1)*sin(t), cos(t)-sin(t)), t = 0 .. 2*Pi)).

However, Maple does not return a result, but rather an integral. Is there an alternative way to solve the integral above?

Thanks in advance.

A user wondered why an example of integration by parts from the Calculus Study Guide was immediately showing the final answer instead of the parts steps shown in the Guide. 

We suggest users pay special attention to the "Initialize" rows of the Guide example(s) where converting the integral to inert form is discussed. 

Using an inert form of the integral ensures that Maple does not evaluate the integral unexpectedly. 

Download CSGG-6-1-4.mw