I have resolved the roots of a series both numerically & analytically.  Let me qualify numerical  versus analytical.  Analytically I evaluate the series without substituting values for the various parameters of the series.  I then differentiate the series, then substitute in the appropriate parametric values, & then solve.  By this method I obtained 5 complex roots.

The numerical approach has values already assigned to the parameters of the series.  I then differentiate & solve.  I obtain only REAL roots in this instance.  I then restricted these results to obtain the solution I believe to be correct given by result (7).

I cannot seem to steer the solver in the analytic case to obtain the correct REAL result that I am expecting.  Can anyone help on this?

Before any website moderator thinks this is the same question as Error-in-Isinternal-Too-Many-Levels, it is not.  I have resolved that question.  This is a different question, but on the same problem!

reconcile_solns.mw

hi, i have an expression "f" and i want to substitute the expression with Square root with another namee like phi, how should i do that? i should not use solve commands since the expression is more complicated and i have just pick part of it. tnx in advance.

 

Download problem.mw

Hi,

I tried a Animation with 10 randoms HeatMaps:

restart:

with(Statistics): with(LinearAlgebra); randomize(): with(plots):

f:= rand( 1..2 ):

for n to 10 do :
RM := Matrix(10, 10, proc (i, j) options operator, arrow; (-1)^f() end proc);
G[n] := HeatMap(RM, axis = [gridlines = [10, color = blue], thickness = 2]) ;
end do
display(seq(G[n], n = 1 .. 10), insequence = true);

But this is nor work as well. How do i this?

Regards.
 

Hello,

I want numerical randoms whose out are 1 or -1 only, how do i it?

I tried this:

with(Statistics):

RM := LinearAlgebra:-RandomMatrix(10, generator = -1 .. 1)

But this is not work as well.

Regards.

Result.mw
 

Download Result.mw

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

Download tmp.mw

Minimum:
1. Maple Tour
2. Maple Quick Start
3. Quick Help
4. Quick Reference Card
5. Math Apps
6. Plotting Guide
...
https://drive.google.com/file/d/1ZAlFQ8_MbuKNsr2PDIyEHmIoogKInBmE/view?usp=sharing

Is that only for my students?

In worksheets:
https://drive.google.com/file/d/1cfA8WKPXSQQxJQR7KbyYHBY6g4OjHaSn/view?usp=sharing

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?