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Hi!

In a paper due to Borwein

http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P172.pdf

it is shown a (very beautiful) graph of the zeros of a partial sum of the Zeta-Riemann, where he indicates that the plot is "the normalized zeros of the 5th partial sum of the Zeta function". Somebody know how one can plot this with Maple?

Thank you!

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Here is my Maple 16 code:

 I expected to get outuput

a [a,b,c]

a [a,c,b]

But I get no output.

Why?

 

 

 

Every time I try to type in a procedure I get the error:  

Error, unterminated procedure

immediately after typing in the first line.  How can I type in the remaining lines of my procedure?

Note:  in maple 7 (years ago) I never had this problem.

Hi everyone. This problem is driving me nuts. I'm pretty sure it's a glitch but I'm not sure how to solve it. I'm trying to do some data analysis with Maple:

(as a side note, even if I remove the for loop but don't execute the restart command the error remains, however if I get rid of the for loop and execute the restart command it is fine.)

Any help would be greatly appreciated. As it stands this is really driving me insane.

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We are running Maple16 in a virtual lab setup using Citrix, when exporting to PDF and saving the file to the local machine the PDF comes out as a 0kb file and the PDF is corrupt.  Has anyone else had an issue like this before?

hi all

 

i have tow equations and i want plot q vs.A , that A change beetwin 0..phi. vn and a are definite and positive.

f is a function of y.

becouse i have a physical problem so infinity can be 20 , and a can be 10 ,

so whit terms , f[infinity]=f[-infinity]=1 and f'[infinity]=f'[-infinity]=0 .

also phiB[y]=q*int(f,y=-infinity..y) , but i want plot q vs phiB[infinity] when phiB[infinity] change between 0..phi

plz plz help me

 

(diff(f,y))^2-(f[y]^4)/4+q^2/(f[y]^2)+(1+2*q^2)*(f[y]^2)/2=piecewise(y<-a,1/4+2*q^2,-a<y<a,vn*(f[y]^2)-vn+1/4+2*q^2,y>a,1/4+2*q^2);

phiB[infinity]=q*int(f,y=-infinity..infinity):
A:=phiB[infinity]:

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Hello,

I've tried to find the solution to my problem, but none of my attempts was succesful.

I have a function which is one-to-one in a particular domain which I am interested in. I would like to get the inverse function of it only in this domain. Here is my function and plot for xp=0..10000:

x := xp-> (-1)*720.5668720*sinh(0.2043018094e-3*xp-0.8532729286)+84952.59423+4.014460003*10^5*arcsinh(0.1219272144e-1*sinh(0.2043018094e-3*xp-0.8532729286)-0.2032498888)

I would appreciate any help,

Iza

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Hello everyone, 

I have some problems with the "isolve" command on Maple. I am trying to solve for integer a very easy system of equations. When I type the commands

 

restart; 

n := 2;
isolve({sum(a[k], k = 1 .. n)-1 = 1}, d)


I get the expected {a[1] = 2-d, a[2] = d}. However, if I add conditions a[1],a[2] >= 0, that is the commands



restart;
n := 2;
isolve({ge(a[1], 0), ge(a[2], 0), sum(a[k], k = 1 .. n)-1 = 1}, d)


I get the warning "Warning, solutions may have been lost". What am I doing wrong? Is there a way to get Maple to give me the possible values?

 

Thank you in advance,

David

Hi everyone. Please excuse me in advance, as I am new on this website and to Maple.

So I am using Maple 16 and I want it to give me the derivative of a function that has constants in it, for example f(x)=(x-a)(x-b). I wrote the (probably stupid) commands on Maple:

> f(x):=(x-a)(x-b);

> diff(f(x), x);

but the result is not the expected (2x-a-b), but rather (D(x))(x-b)-(D(a))(x-b). What would be the right things to write to get what I need?

 

Thank you in advance,

David

Hello,

I use a map() command to get the values of function for each element of vector. Here is an example of a simple task:

restart:

A:=[1,2,8];

f1:=x->2*x;
f1_table_A:=Vector([map(f1,A)]);

I would like to do the same with piecewise function but it doesn't work. Here is an example:

restart:
with(plots):
f:=2*x:
g:=x^2:

h:=x->piecewise( 0<=x and x<= 5, f, 5<x and x <= 10, g ) ;
A:=[1,2,8]:
h_table_A:=map(h,A):

Is it possible to work with piecewise function? Maybe someone has an idea how to do it in different way?

Kind regards,

Iza

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