MaplePrimes Questions

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Dear Friends!
I want to generate a system of the equation from the matrices! please help me in this regard! Thanks!

I was making a maple worksheet (out of functions from another), when i noticed that  execution groups and new execution groups that i was making were not showing an output.  I'm wondering why this happens, and how it can be fixed.

in more detail there are 4  [> groups of commands, the first two give the correct output when i hit enter, however, if enter is pressed on any group of commands after the second the command wont return the any output.

I believe this is a formatting issue that comes from the way i cut/paste and edited the previous worksheet down to this, but i'd like to understand so i can avoid it happening again.

3d_plot_of_Lie_derivatives_against_Dimensions_of_solutions_timer.mw

Ahh, this is a teathing problem, I had been using writeto so i could store measurements of RAM; this changes the nature of the question.

How do you implement writeto so only some things are sent there? (in the case of this worksheet the output of GTS2usage
 

Dear users!

Hope everyone fine here. I want to compare the coefficients of like power of exponential function in attachemed file. Please see and fix my problem.

Help.mw

WC51 III
Maple 2015

With straight line selected in following plot, I right-click on plot and select Line → Line Width...  In Set line width dialog I toggle line width to 1 and click OK.

I have the changes I want, but if I choose Restart → Execute Worksheet all my changes are lost.  It is often easier to use the Plot Options Toolbar instead of hunting for Maple syntax to make changes in a plot graphic.  I'm wondering if there is a way to "lock" my plot edits so that they won't disappear as I continue working on a Maple worksheet?

restart;
with(plots);
eq1 := -y^3+x^2+x*y-7; eq2 := y = (8/9)*x-2/3;
implicitplot({eq1, eq2}, x = 0 .. 4, y = 0 .. 3, axis = [gridlines = [6, color = "MidnightBlue"]], size = [1/2, 1], axesfont = ["Roman", bold, 10], thickness = 3);





 

e1 := tau-gamma*S__1-beta*S__1*S__3/(S__3*alpha__1+1)-beta*xi*S__1*S__4/(S__4*alpha__1+1)+phi*S__5 = 0;
e2 := beta*S__1*S__3/(S__3*alpha__1+1)+beta*xi*S__1*S__4/(S__4*alpha__1+1)-(gamma+eta__1+eta__2)*S__2;
e3 := eta__1*S__2-(gamma+gamma__1+delta__1+omega)*S__3;
e4 := eta__2*S__2+omega*S__3-(gamma+gamma__2+delta__2)*S__4;
e5 := delta__1*S__3+delta__2*S__4-(gamma+phi)*S__5;
solve({e1, e2, e3, e4, e5}, {S__1, S__2, S__3, S__4, S__5});
 

Hello,

      I've found that, occasionally, solve won't work if the solving variables are specified---but it will work if the variables aren't specified. For instance:

eqns:=
[x=1, -1/(exp(h)+1/exp(h))^(1/2)/(exp(h)-1/exp(h))^(1/2)*(exp(h)^2-2+1/exp(h)^2)^(1/2)*x = tanh(h)^(1/2)]:

# Works
solve(eqns);

# Doesn't work
solve(eqns, x);

I was wondering why this is, and if there is a workaround?

(I want to specify the solving variables so that solve doesn't attempt to solve for parameters---like h in this case. Also, I'm using solve as opposed to a consistency checker because, in general, I'm applying the same code to larger systems with additional variables to solve for).

Thanks!

I have a say sum(F(k) ,k=1..n)  It fails unless n is an actual integer.  But sum(F(k), k=1..a) works. Then then eval(%,a=n) completes it. That is probably correct but not necessarily a valid assumption. Details in attached. Also it is hard to understand the error message.
 

restart

``

``

Sinta := (k^2+n^2-k)*n/((k^2+n^2-2*k+1)*(k^2+n^2))

(k^2+n^2-k)*n/((k^2+n^2-2*k+1)*(k^2+n^2))

(1)

Ai := sum(Sinta, k = 1 .. n)

-((1/2)*I)*(2*n^2+I*n)*Psi(n-I*n)/(4*n^2+1)+((1/2)*I)*(2*n^2-I*n)*Psi(n+I*n)/(4*n^2+1)+((1/2)*I)*(-2*n^2+I*n)*Psi(n+1-I*n)/(4*n^2+1)-((1/2)*I)*(-2*n^2-I*n)*Psi(n+1+I*n)/(4*n^2+1)+((1/2)*I)*(2*n^2+I*n)*Psi(-I*n)/(4*n^2+1)-((1/2)*I)*(2*n^2-I*n)*Psi(I*n)/(4*n^2+1)-((1/2)*I)*(-2*n^2+I*n)*Psi(1-I*n)/(4*n^2+1)+((1/2)*I)*(-2*n^2-I*n)*Psi(1+I*n)/(4*n^2+1)

(2)

Souta := n/(k^2+n^2-k)

n/(k^2+n^2-k)

(3)

NULL

sum(Souta, k = 1 .. n)

Error, (in assuming) when calling 'Dfnt_4'. Received: 'when calling 'Dfnt_4'. Received: 'when calling 'unknown'. Received: 'invalid input: Dfnt_4 expects its 3rd argument, fpts, to be of type Or(list, piecewise), but received 0'''

 

"(->)"

Error, invalid input: evalf[10] expects 1 argument, but received 0

 

 

 

This works

sum(Souta, k = 1 .. a)

n*Psi(a+1/2-(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)-n*Psi(a+1/2+(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)-n*Psi(1/2-(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)+n*Psi(1/2+(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)

(4)

Ao := eval(%, a = n)

n*Psi(n+1/2-(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)-n*Psi(n+1/2+(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)-n*Psi(1/2-(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)+n*Psi(1/2+(1/2)*(-4*n^2+1)^(1/2))/(-4*n^2+1)^(1/2)

(5)

"(->)"

n*Psi(n+.5000000000-.5000000000*(-4.*n^2+1.)^(1/2))/(-4.*n^2+1.)^(1/2)-1.*n*Psi(n+.5000000000+.5000000000*(-4.*n^2+1.)^(1/2))/(-4.*n^2+1.)^(1/2)-1.*n*Psi(.5000000000-.5000000000*(-4.*n^2+1.)^(1/2))/(-4.*n^2+1.)^(1/2)+n*Psi(.5000000000+.5000000000*(-4.*n^2+1.)^(1/2))/(-4.*n^2+1.)^(1/2)

(6)

````

``


 

Download Summation_problem.mw

Hi all, 

Still a new visualisation problem ...
I have two random variables X and Y linked together with a dependency structure named "Gumbel copula" (just a technical stuff, if you prefer replace the bold text by "two correlated gaussian RVs X and Y").

I would like to represent on a same figure:

  • the set {(x[n], y[n]), n=1..N} of a sample of size N drawn from the joint distribution of X and Y
    (a ScatterPlot does this very well, even if I use a plot(..., style =point) in the attached file
     
  • the empirical marginal distribution of X (Histogram), put below the line "y = min({y[n], n=1..N} )" and rotated by 180 degrees
     
  • the empirical marginal distribution of Y (Histogram), put left the line "x = min({x[n], n=1..N} )" and rotated counterclockwise by 90 degrees
     

The attached file gives you two examples.

Ideally, if it's not too much to ask, I would like to have the main axes placed in more adequate locations.
For instance, the option "axes=boxed" reject the axes on the left and bottom boundary of the box which contains the 3 plots (not very astute) ; on the other side the default locations of the axes seem better but one of them "cuts" a histogram ... not very smart.
In fact the position of the axes in the "scatter plot" part is good enough for me and I would like to "add" the two histograms without changing these axes.

Is there a solution to do that (I think I read here a rather close problem where a "zom" of a plot was put within this same plot?) ?
TIA


GumbelCopula.mw

The deltoid plane curve with parameter a  is the set of all points (x, y) in the plane satisfying the equation 
     (1)   (x^2+y^2)^2 - 8ax(x^2 - 3y^2) + 18a^2(x^2+y^2)=27a^4

and the same curve may be described by the parametric equations:

     (2)    
                  x = a (2 cos(t) + cos(2 t))
,      
                  y = a (2 sin(t) - sin(2 t))


(a) Using equation (1) and the command implicitplot graph the deltoid curves with parameters a = 1, 2, and 3 on the same axes.

(b) Using equations (2) and the plot command graph the deltoid curves with parameters a = 1, 2, and 3 on the same axes.

Remark: You will need to do some experimenting with the ranges of the plots and the option numpoints in question (a) to get a decent picture. Note that you can copy and paste equations (1) and (2) and with some judicious editing can save yourself the trouble of typing them. 


      

 

Using spacecurve I've created a diagram of two curves in space, they are reflections in the plane x=y (in this diagram they are labeled k[a1]=k[a2]). How can I add the plane to the diagram?

My intuition is to use 3dplot, and then combine them with display. The problem with that is that 3d plot wants a function of the form z=f(x,y) rather than x=y.

(here is the code for the diagram)

spacecurve({[(5*10^(-4)*100)/C[T], 100*10^(-3)/C[T], C[T]], [100*10^(-3)/C[T], (5*10^(-4)*100)/C[T], C[T]]}, C[T] = 10 .. 100, labelfont = [TIMES, 32], axesfont = [TIMES, 32], titlefont = [TIMES, 32], captionfont = [TIMES, 32], labels = [conjugate(k)[a1], conjugate(k)[a2], conjugate(C)[T]], tickmarks = [[0.1e-2 = k[a1], 5*10^(-3) = 5*k[a1], 10^(-2) = 10*k[a1]], [0.1e-2 = k[a1], 5*10^(-3) = 5*k[a1], 10^(-2) = 10*k[a1]], [0 = 0, 10 = (1/10)*C[T], 50 = (1/2)*C[T], 100 = C[T]]], view = [0 .. 0.1e-1, 0 .. 0.1e-1, 0 .. 100])

I am currently trying to evaluate the performance of different methods for the same calculation and use codegen:cost to give me an overview on the rough computational effort for the results. I stumbled over the function counts not matching my own count in the optimized Matlab code generated by Maple.

Minimal example:

with(codegen):
Fcn1 := sqrt(a):
cost(Fcn1);
Fcn2 := sqrt(sin(q)):
cost(Fcn2);

The first expression gives me "2*functions+multiplications", the second one "3*functions+multiplications".

So my question: Does anyone know, why the square root is counted as two functions while the sine is counted correctly as one?

Good day everyone,

I am trying to write a finite difference Method solution for an ODE and its giving me problem solving the algebraic simplifications generated. Please, any one with useful informations. Below is the attached file

FDM1.mw
 

restart

with(ODETools)

with(student)

with(plots)

with(plottools)

xmin := 0; xmax := 6

n := 60

`σ__1` := .5

ode:=diff(f(eta),eta$3)+f(eta)*diff(f(eta),eta$2)=0

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta)) = 0

(1)

bc1:=df(xmin)=2*sigma__1;

df(0) = 1.0

(2)

bc2:=d2f(xmin)=0;

d2f(0) = 0

(3)

bc3:=d3f(xmax)=2;

d3f(6) = 2

(4)

dfde:=proc(h)(f[k+1]-f[k-1])/2/h;end proc:

dfde(h);

(1/2)*(f[k+1]-f[k-1])/h

(5)

d2fde2:=proc(h)(f[k+1]-2*f[k]+f[k-1])/h^2;end proc:

d2fde2(h);

(f[k+1]-2*f[k]+f[k-1])/h^2

(6)

d3fde3:=proc(h)(f[k+1]-3*f[k]+3*f[k-1]-f[k-2])/h^3;end proc:

d3fde3(h);

(f[k+1]-3*f[k]+3*f[k-1]-f[k-2])/h^3

(7)

d2fde2f:=proc(h)(f[k+2]-2*f[k+1]+f[k])/h^2;end proc:

d2fde2f(h);

(f[k+2]-2*f[k+1]+f[k])/h^2

(8)

d2fde2b:=proc(h)(f[k]-2*f[k-1]+f[k-2])/h^2;end proc:

d2fde2b(h);

(f[k]-2*f[k-1]+f[k-2])/h^2

(9)

 

dfdef:=proc(h)(f[k+1]-f[k])/h;end proc:

dfdef(h);

(f[k+1]-f[k])/h

(10)

h:=xmax/(n-1)

6/59

(11)

stencil:=subs(diff(f(eta),eta$3)=d3fde3(h),f(eta)=f[k],diff(f(eta),eta$2)=d2fde2,ode);

(205379/216)*f[k+1]-(205379/72)*f[k]+(205379/72)*f[k-1]-(205379/216)*f[k-2]+f[k]*(diff(diff(f[k], eta), eta)) = 0

(12)

bcEqs:=[subs(k=0,dfdef(h))=rhs(bc1),subs(k=0,d2fde2f(h))=rhs(bc2),
subs(k=n-1,d2fde2b(h))=rhs(bc3)];

[(59/6)*f[1]-(59/6)*f[0] = 1.0, (3481/36)*f[2]-(3481/18)*f[1]+(3481/36)*f[0] = 0, (3481/36)*f[59]-(3481/18)*f[58]+(3481/36)*f[57] = 2]

(13)

eqs:=Vector(n-2):
cnt:=0:

for k from 1 to n-2 do
    cnt:=cnt+1:
    eqs(cnt):=stencil;
end do:

eqs:

eqs := [op(convert(eqs, list)), op(bcEqs)]; vars := [seq(f[k], k = 0 .. n-1)]; map(print, eqs)

(205379/216)*f[2]-(205379/72)*f[1]+(205379/72)*f[0]-(205379/216)*f[-1] = 0

 

(205379/216)*f[3]-(205379/72)*f[2]+(205379/72)*f[1]-(205379/216)*f[0] = 0

 

(205379/216)*f[4]-(205379/72)*f[3]+(205379/72)*f[2]-(205379/216)*f[1] = 0

 

(205379/216)*f[5]-(205379/72)*f[4]+(205379/72)*f[3]-(205379/216)*f[2] = 0

 

(205379/216)*f[6]-(205379/72)*f[5]+(205379/72)*f[4]-(205379/216)*f[3] = 0

 

(205379/216)*f[7]-(205379/72)*f[6]+(205379/72)*f[5]-(205379/216)*f[4] = 0

 

(205379/216)*f[8]-(205379/72)*f[7]+(205379/72)*f[6]-(205379/216)*f[5] = 0

 

(205379/216)*f[9]-(205379/72)*f[8]+(205379/72)*f[7]-(205379/216)*f[6] = 0

 

(205379/216)*f[10]-(205379/72)*f[9]+(205379/72)*f[8]-(205379/216)*f[7] = 0

 

(205379/216)*f[11]-(205379/72)*f[10]+(205379/72)*f[9]-(205379/216)*f[8] = 0

 

(205379/216)*f[12]-(205379/72)*f[11]+(205379/72)*f[10]-(205379/216)*f[9] = 0

 

(205379/216)*f[13]-(205379/72)*f[12]+(205379/72)*f[11]-(205379/216)*f[10] = 0

 

(205379/216)*f[14]-(205379/72)*f[13]+(205379/72)*f[12]-(205379/216)*f[11] = 0

 

(205379/216)*f[15]-(205379/72)*f[14]+(205379/72)*f[13]-(205379/216)*f[12] = 0

 

(205379/216)*f[16]-(205379/72)*f[15]+(205379/72)*f[14]-(205379/216)*f[13] = 0

 

(205379/216)*f[17]-(205379/72)*f[16]+(205379/72)*f[15]-(205379/216)*f[14] = 0

 

(205379/216)*f[18]-(205379/72)*f[17]+(205379/72)*f[16]-(205379/216)*f[15] = 0

 

(205379/216)*f[19]-(205379/72)*f[18]+(205379/72)*f[17]-(205379/216)*f[16] = 0

 

(205379/216)*f[20]-(205379/72)*f[19]+(205379/72)*f[18]-(205379/216)*f[17] = 0

 

(205379/216)*f[21]-(205379/72)*f[20]+(205379/72)*f[19]-(205379/216)*f[18] = 0

 

(205379/216)*f[22]-(205379/72)*f[21]+(205379/72)*f[20]-(205379/216)*f[19] = 0

 

(205379/216)*f[23]-(205379/72)*f[22]+(205379/72)*f[21]-(205379/216)*f[20] = 0

 

(205379/216)*f[24]-(205379/72)*f[23]+(205379/72)*f[22]-(205379/216)*f[21] = 0

 

(205379/216)*f[25]-(205379/72)*f[24]+(205379/72)*f[23]-(205379/216)*f[22] = 0

 

(205379/216)*f[26]-(205379/72)*f[25]+(205379/72)*f[24]-(205379/216)*f[23] = 0

 

(205379/216)*f[27]-(205379/72)*f[26]+(205379/72)*f[25]-(205379/216)*f[24] = 0

 

(205379/216)*f[28]-(205379/72)*f[27]+(205379/72)*f[26]-(205379/216)*f[25] = 0

 

(205379/216)*f[29]-(205379/72)*f[28]+(205379/72)*f[27]-(205379/216)*f[26] = 0

 

(205379/216)*f[30]-(205379/72)*f[29]+(205379/72)*f[28]-(205379/216)*f[27] = 0

 

(205379/216)*f[31]-(205379/72)*f[30]+(205379/72)*f[29]-(205379/216)*f[28] = 0

 

(205379/216)*f[32]-(205379/72)*f[31]+(205379/72)*f[30]-(205379/216)*f[29] = 0

 

(205379/216)*f[33]-(205379/72)*f[32]+(205379/72)*f[31]-(205379/216)*f[30] = 0

 

(205379/216)*f[34]-(205379/72)*f[33]+(205379/72)*f[32]-(205379/216)*f[31] = 0

 

(205379/216)*f[35]-(205379/72)*f[34]+(205379/72)*f[33]-(205379/216)*f[32] = 0

 

(205379/216)*f[36]-(205379/72)*f[35]+(205379/72)*f[34]-(205379/216)*f[33] = 0

 

(205379/216)*f[37]-(205379/72)*f[36]+(205379/72)*f[35]-(205379/216)*f[34] = 0

 

(205379/216)*f[38]-(205379/72)*f[37]+(205379/72)*f[36]-(205379/216)*f[35] = 0

 

(205379/216)*f[39]-(205379/72)*f[38]+(205379/72)*f[37]-(205379/216)*f[36] = 0

 

(205379/216)*f[40]-(205379/72)*f[39]+(205379/72)*f[38]-(205379/216)*f[37] = 0

 

(205379/216)*f[41]-(205379/72)*f[40]+(205379/72)*f[39]-(205379/216)*f[38] = 0

 

(205379/216)*f[42]-(205379/72)*f[41]+(205379/72)*f[40]-(205379/216)*f[39] = 0

 

(205379/216)*f[43]-(205379/72)*f[42]+(205379/72)*f[41]-(205379/216)*f[40] = 0

 

(205379/216)*f[44]-(205379/72)*f[43]+(205379/72)*f[42]-(205379/216)*f[41] = 0

 

(205379/216)*f[45]-(205379/72)*f[44]+(205379/72)*f[43]-(205379/216)*f[42] = 0

 

(205379/216)*f[46]-(205379/72)*f[45]+(205379/72)*f[44]-(205379/216)*f[43] = 0

 

(205379/216)*f[47]-(205379/72)*f[46]+(205379/72)*f[45]-(205379/216)*f[44] = 0

 

(205379/216)*f[48]-(205379/72)*f[47]+(205379/72)*f[46]-(205379/216)*f[45] = 0

 

(205379/216)*f[49]-(205379/72)*f[48]+(205379/72)*f[47]-(205379/216)*f[46] = 0

 

(205379/216)*f[50]-(205379/72)*f[49]+(205379/72)*f[48]-(205379/216)*f[47] = 0

 

(205379/216)*f[51]-(205379/72)*f[50]+(205379/72)*f[49]-(205379/216)*f[48] = 0

 

(205379/216)*f[52]-(205379/72)*f[51]+(205379/72)*f[50]-(205379/216)*f[49] = 0

 

(205379/216)*f[53]-(205379/72)*f[52]+(205379/72)*f[51]-(205379/216)*f[50] = 0

 

(205379/216)*f[54]-(205379/72)*f[53]+(205379/72)*f[52]-(205379/216)*f[51] = 0

 

(205379/216)*f[55]-(205379/72)*f[54]+(205379/72)*f[53]-(205379/216)*f[52] = 0

 

(205379/216)*f[56]-(205379/72)*f[55]+(205379/72)*f[54]-(205379/216)*f[53] = 0

 

(205379/216)*f[57]-(205379/72)*f[56]+(205379/72)*f[55]-(205379/216)*f[54] = 0

 

(205379/216)*f[58]-(205379/72)*f[57]+(205379/72)*f[56]-(205379/216)*f[55] = 0

 

(205379/216)*f[59]-(205379/72)*f[58]+(205379/72)*f[57]-(205379/216)*f[56] = 0

 

(59/6)*f[1]-(59/6)*f[0] = 1.0

 

(3481/36)*f[2]-(3481/18)*f[1]+(3481/36)*f[0] = 0

 

(3481/36)*f[59]-(3481/18)*f[58]+(3481/36)*f[57] = 2

(14)

sol := fsolve([op(eqs)])

{f[-1] = 74076407.16, f[0] = 74076407.19, f[1] = 74076407.29, f[2] = 74076407.42, f[3] = 74076407.63, f[4] = 74076407.95, f[5] = 74076408.39, f[6] = 74076408.95, f[7] = 74076409.58, f[8] = 74076410.32, f[9] = 74076411.18, f[10] = 74076412.19, f[11] = 74076413.34, f[12] = 74076414.66, f[13] = 74076416.09, f[14] = 74076417.69, f[15] = 74076419.41, f[16] = 74076421.28, f[17] = 74076423.28, f[18] = 74076425.45, f[19] = 74076427.75, f[20] = 74076430.17, f[21] = 74076432.77, f[22] = 74076435.56, f[23] = 74076438.51, f[24] = 74076441.69, f[25] = 74076445.02, f[26] = 74076448.54, f[27] = 74076452.24, f[28] = 74076456.15, f[29] = 74076460.29, f[30] = 74076464.65, f[31] = 74076469.16, f[32] = 74076473.82, f[33] = 74076478.70, f[34] = 74076483.79, f[35] = 74076489.05, f[36] = 74076494.52, f[37] = 74076500.21, f[38] = 74076506.15, f[39] = 74076512.33, f[40] = 74076518.79, f[41] = 74076525.57, f[42] = 74076532.68, f[43] = 74076540.10, f[44] = 74076547.80, f[45] = 74076555.85, f[46] = 74076564.23, f[47] = 74076572.94, f[48] = 74076581.93, f[49] = 74076591.18, f[50] = 74076600.60, f[51] = 74076610.20, f[52] = 74076620.03, f[53] = 74076630.05, f[54] = 74076640.28, f[55] = 74076650.64, f[56] = 74076661.19, f[57] = 74076671.90, f[58] = 74076682.68, f[59] = 74076693.50}

(15)

``


 

Download FDM1.mw

 

Hello,

      I've been using frontend in conjuction with pdsolve to handle a very large system of PDEs; namely the system has many irrelevant functions that can be frozen with frontend, allowing pdsolve to solve it much quicker.

     However, I ran into a strange case, which I've included in the attached file. Namely, for this particular system of equations, the system is a PDE with respect to t, so I freeze functions of x. However, doing so produces incorrect results (labelled bad) compared to the unfrozen case (labelled good). The frontend command is working as expected (this can be verified by uncommenting the frontend/print commands); it seems that pdsolve is treating the frozen and unfrozen cases differently.

     Any idea why this behavior is occuring? (The obvious solution is to simply not use frontend here; again, I'm using this same code for much larger PDEs and frontend speeds up exceution by an order of magnitude).

Thanks!

MWE.txt

I have a system of nonlinear ODE and I solved them numerically using rkf45. the solution is in time domain and the variables are x, y and z.

I want to convert this solution from time domain to frequency domain. I know that I should do FFT, but I do not know how to do this. 

I tried to do this, but I found that I have to get the solution of x, y and z in the form of vector with respect to time. I am not sure if I understant this problem corectly.

Thanks in advance.

A sphere, radius one unit with centre at the origin of the x, y and z axes is cut along the planes x=0, y=0 and z=0 yielding 8 total pieces, none of which are cubes.

For a sphere radius 2, cut along the planes x=2,y=2,z=2, x=-2,y=-2,z=-2, you get  64 pieces of which 8 are cubes.

Derive an expression for the number of  pieces you will get with a sphere radius integer R, cut along the planes x,y,z=1, x,y,z=-1, x,y,z=2,…     x,y,z=R-1, x,y,z=1-R  

How many of those pieces will be cubes?

The question arose from slicing onions!

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