Result.mw
 

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Anybody know why the graph doesnt appear and what is the error in my coding?

Download tmp.mw

restart; N := 4; de1 := A*(diff(f(eta), eta, eta, eta))+n*(-(diff(f(eta), eta, eta)))^(n-1)*(diff(f(eta), eta, eta, eta))-S*(diff(f(eta), eta))+(2-n)*eta*(diff(f(eta), eta, eta))/(1+n)+2*n*f(eta)*(diff(f(eta), eta, eta))/(1+n)-(diff(f(eta), eta))^2-g(eta)*(diff(f(eta), eta, eta))+(M*M)*(diff(f(eta), eta)) = 0, A*(diff(g(eta), eta, eta, eta))+(-(diff(f(eta), eta, eta)))^(n-1)*(diff(g(eta), eta, eta, eta))-(n-1)*(diff(g(eta), eta, eta))*(diff(f(eta), eta, eta, eta))*(-(diff(f(eta), eta, eta)))^(n-2)-S*(diff(g(eta), eta))+(2-n)*eta*(diff(g(eta), eta, eta))/(1+n)+2*n*f(eta)*(diff(g(eta), eta, eta))/(1+n)-(diff(g(eta), eta))^2+g(eta)*(diff(g(eta), eta, eta))-(M*M)*(diff(g(eta), eta)) = 0, (1+E*j(eta))*(diff(j(eta), eta, eta))+E*(diff(j(eta), eta))^2+2*Pr*n*f(eta)*g(eta)*(diff(j(eta), eta))/(1+n)-Pr*S*(2-n)*eta*(diff(j(eta), eta))/(1+n)+Pr*(Nb*(diff(j(eta), eta))*(diff(h(eta), eta))+Nt*(diff(j(eta), eta))^2)+Pr*lambda*j(eta) = 0, diff(h(eta), eta, eta)+2*Le*Pr*n*f(eta)*g(eta)*(diff(h(eta), eta))/(1+n)-Le*Pr*S*(2-n)*eta*(diff(h(eta), eta))/(1+n)+Nt*(diff(j(eta), eta, eta))/Nb = 0, f(0) = 0, (D(f))(0) = 1, g(0) = 0, (D(g))(0) = alpha, (D(j))(0) = -b*(1-j(0))/(1+E*j(0)), (D(h))(0) = -d*(1-h(0)), (D(f))(N) = 0, (D(g))(N) = 0, j(N) = 0, h(N) = 0; d1 := subs(alpha = .2, M = .4, A = 1, S = .1, n = .5, Pr = 4, E = 1.5, Nb = .5, Nt = .2, Le = 1, lambda = .2, b = 1.2, d = .5, [de1]); da1 := dsolve(d1, numeric, output = operator, maxmesh = 2048, method = bvp[midrich], abserr = 10); with(plots); restart; N := 4; de2 := A*(diff(f(eta), eta, eta, eta))+n*(-(diff(f(eta), eta, eta)))^(n-1)*(diff(f(eta), eta, eta, eta))-S*(diff(f(eta), eta))+(2-n)*eta*(diff(f(eta), eta, eta))/(1+n)+2*n*f(eta)*(diff(f(eta), eta, eta))/(1+n)-(diff(f(eta), eta))^2-g(eta)*(diff(f(eta), eta, eta))+(M*M)*(diff(f(eta), eta)) = 0, A*(diff(g(eta), eta, eta, eta))+(-(diff(f(eta), eta, eta)))^(n-1)*(diff(g(eta), eta, eta, eta))-(n-1)*(diff(g(eta), eta, eta))*(diff(f(eta), eta, eta, eta))*(-(diff(f(eta), eta, eta)))^(n-2)-S*(diff(g(eta), eta))+(2-n)*eta*(diff(g(eta), eta, eta))/(1+n)+2*n*f(eta)*(diff(g(eta), eta, eta))/(1+n)-(diff(g(eta), eta))^2+g(eta)*(diff(g(eta), eta, eta))-(M*M)*(diff(g(eta), eta)) = 0, (1+E*j(eta))*(diff(j(eta), eta, eta))+E*(diff(j(eta), eta))^2+2*Pr*n*f(eta)*g(eta)*(diff(j(eta), eta))/(1+n)-Pr*S*(2-n)*eta*(diff(j(eta), eta))/(1+n)+Pr*(Nb*(diff(j(eta), eta))*(diff(h(eta), eta))+Nt*(diff(j(eta), eta))^2)+Pr*lambda*j(eta) = 0, diff(h(eta), eta, eta)+2*Le*Pr*n*f(eta)*g(eta)*(diff(h(eta), eta))/(1+n)-Le*Pr*S*(2-n)*eta*(diff(h(eta), eta))/(1+n)+Nt*(diff(j(eta), eta, eta))/Nb = 0, f(0) = 0, (D(f))(0) = 1, g(0) = 0, (D(g))(0) = alpha, (D(j))(0) = -b*(1-j(0))/(1+E*j(0)), (D(h))(0) = -d*(1-h(0)), (D(f))(N) = 0, (D(g))(N) = 0, j(N) = 0, h(N) = 0; d2 := subs(alpha = .2, M = .4, A = 1, S = .1, n = .5, Pr = 5, E = 1.5, Nb = .5, Nt = .2, Le = 1, lambda = .2, b = 1.2, d = .5, [de2]); da2 := dsolve(d2, numeric, output = operator, maxmesh = 2048, method = bvp[midrich], abserr = 10); with(plots); p1 := odeplot([da1, da2], [eta, diff(f(eta), eta), linestyle = 1, color = "Red", thickness = 2], labels = ["η", "f' g' "], labeldirections = [HORIZONTAL, VERTICAL]); p4 := odeplot(da2, [[eta, f(eta)], [eta, g(eta)], [eta, h(eta)], [eta, j(eta)]], color = [red, green, blue, black]); plots[display]({p1})

Dear users!

Hope everyone should be fine. I have some query regarding to set title of the following problem in 2D

beta := 1; alpha[1] := .5

f := beta+alpha[1]+sin(x*beta);

plot(f, x = 0 .. Pi, axes = boxed, size = [300, 270], labels = [x, "f ' (x)"], labelfont = ["Arial", 10, Bold], labeldirections = [horizontal, vertical], title = "β = 1, α[1] = 0.5\n", titlefont = ["Arial", 10, Bold]);

I got the following graph

but I need the following one

Want to shift title on right. Want to write alpha[1] and f like in figure 2. 

I am waiting your response. Thanks

I want to find the i-th root of this equation s := BesselJ(1, alpha)+a1*BesselY(1, alpha)+b1*BeselJ(1, alpha)+c1*BesselK(1, alpha)

When I use the Nextzero it doesnt work

Thanks

Hi,

I couldn't find a command or tool for searching a value in a multi-dimensional list/array/matrix. Does it exist?. 

I wrote this:

where

pSearch: proc for searching

A: list, array, vector or matrix

Examples with a list:

> ss := [[1, 2, 3], [2, 3, 5], [2, 4, 5]]
> j:=1:
># search first index i such that ss[i, j]=2

> nPos := 

# search next index i such that ss[i, j]=2

> nPos := 

But I am not sure if this procedure is efficient for large lists or arrays and I would suggest (if this tool does not exist yet) a more efficient one.

César Lozada

ok i think they are called "nested" sequences, i will normally do something like: 

seq(seq(seq(F(i,j,k),i=1..Ri),j=1..Rj),k=1..Rk)

but in this circumstance the number of variables and the ranges for each are specified by the user in combo boxes and text windows etc, so i need to use foldl. The below doesn't work, so i was looking for a hint on what i am doing wrong:

Thanks in advance

This guy thinks the Cayley-Dickson Construction cycles back at 1024D? I don't actually use Maple, is he just encountering a precision error?

https://www.mapleprimes.com/posts/124913-Visualization-Of-The-CayleyDickson

quote in question

"

I found no new mathematics after 1024 because higher hypercomplex numbers greater than 1024 are cyclic (they repeat all over again).  I do not offer dimensions higher than 256D to the publc Maple Application Center because the mathematics is very slow and time cosuming past 64D.  However, I did keep the code up to 256D in the public Maple Application Center.

"

I received a program for constructing Cayley-Dickson tables for my own use, but it's actually written in bc (arbitrary precision). I'm assuming Maple supports this to some extent but maybe he's encountering a precision error that looks like it's repeating? I can't verify yet because he's right about the one thing, even a 256 table took DAYS to compute. The snapshots are intriguing too but I'm assuming even a second-long video at low resolution would take forever as well.

But I still have the bc code and constructing Cayley-Dickson tables is only a few lines of codes/conditions, it's one of those 'easy for a computer, impossible for a human' kind of deals. I don't see enough complexity in the code where it would suddenly start cycling. It's kind of important to me because the implications of the Cayley-Dickson Construction going on forever are more exciting in my opinion.

By the way I'm actually personally using the bc generated tables for making music sequences/MIDI. It's not the usual multiplication and such though because that even when normalized would "stick" so to speak? It's a bit of a secret, sorry

I know how to use numapprox to fit polynomial and rational polynomial approximations to trig functions.  I would like to fit a rational expression using numerators and denominators like this:

a*x^(1/3) + b*x^(1/2) + c*x^(2/3) + d*x + e*x^(3/2) + f*x^(2) + g*x^(3)

# About twenty years ago Joe Riel posted 

#printing_just_for_fun.mws   by Joe Riel 30-6-‘98

#As part of this he had:

 d_string:="D*e*v*o*u*r*e*r":

#eta:=symb(217):

print(cat(`Does `,d_string,` = `));

eta;
#Is it possible to put the eta symbol on the same line?

#   Yes,  …but without the comma??  And…
 printf("Does %s = %a\n",d_string, eta);

#  does not give the Greek letter eta, but just eta,  while

eta;

print(eta);
#...both give the Greek letter

#  Is it possible to use printf (or similar?) to obtain output 

#  of:  Does D*e*v*o*u*r*e*r = h  

#  Oops!  In Word the h was a Greek letter eta!

#  ...and lastly, can it be done in Maple 7?

Pretty much the title.

At first I thought something's wrong with my vectors, but even something like <1,1,1>.<2,2,2> produces this error.

I have restart in the very beginning of my worksheet and with(VectorCalculus): right after.

I'm using the 64 bit version of Maple 18 on Windows 10, if this matters.

Say I have an expression like

diff(f(x),x)*cos(f(x));

and I want to evaluate it where the function f(x)=0 (without a priori knowing f(x)). Since cos(0)=1, I expect the answer to be simply diff(f(x),x); in general this won't be identically zero. However, if I try

eval(%,f(x)=0);

it replaces f(x) with zero everywhere, including in the derivative (as expected from the documentation) and returns 0.

So, my question: is there a way to evaluate an expression at a known value of a function (f(x)=0) without knowing the function?

It seems using convert/D manages to solve it, but I imagine there's a better way to do this. I've tried using RootOf(f(x)), but can't seem to make that work.

I'm trying to make a comparison between simulation data from CST and calculations done in MAPLE - the best way to do so is have both plots on excel so the formatting matches. I can export the data from CST as a text file, but am stuck when it comes to getting the data from MAPLE (specifically trying to record kz, dispf, and dispbeamf all together in one text file).

My implicitplot code works (the code for it shown below). Both dispf and dispbeamf are functions of kz.
implicitplot([dispf, dispbeamf], kz = -150 .. 150, f3 = 0.18e11 .. 0.25e11, color = [red, blue], numpoints = 5000)

Any advice would be greatly appreciated.

Hello,

I want to adjust values in a matrix. In the example matrix you will find some measurement errors (values around -4000 or 1). But these are definitely measurement errors. Is there any possibility to adjust these values to the surrounding values (~1500)...like in an interpolation. Can someone help me how I can do this with maple?

Ps.: the matrix is exported from maple into .xls

Matrix:
O233.xlsx

Thank you very much!

Dear users! 

Hope everyone fine with everything. I want to chage the format of x-axis in the attached file. I need the x-axis in degree like (-180^o, -120^o, -60^o, 0 , 60^o, 120^o, 180^o). Also I want the lable on y-axis as:

I am waiting your positive respons. 

Help_new.mw