I have studied most posts about plot quality, still confused.

When I export 2d graphs and import the to microsoft word, the quality is horrible. How can I presents my results in a reseanably good way? 

what command should I add to my code that the quality increase notoceably? 

Any suggestion about 3d plots? 

If I export as an eps file, how can I import that in microsoft word? I did, but I guess they are not compatibble. 

thank you 

Hello, I have a somewhat math and Maple question I'm hoping some can help with.

I have this curve,

curve:=alpha^2*beta^10-alpha^2*beta^9+4*alpha*beta^7-2*alpha*beta^6-2*alpha*beta^5-alpha*beta^4+alpha*beta^3-2*beta+1;

and if I solve this system (numerically);

sol:= solve([curve=0,diff(curve,beta)=0],[alpha,beta]);

I get 6 special points (8 actually but two are critical). So I'll refer to them by subscript "i".

If I then do a coordinate transformation by;

alpha=exp(u)

beta=exp(v)

and convert the 6 coordinates and curve, I'll get everything in terms of (u,v) coordinates.

curve2:=subs([alpha=exp(u),beta=exp(v)],curve);

soluv:=map(ln,sol);  (this is just pseudo - I don't know how to do it this way)

So now the 6 points are referred to by (u_i,v_i).

Next, I want to expand this curve locally around these six points, using the following (where "z" is the local coordinate);

u -> u_i + z^2

v -> v_i + sum(a_j*z^j,j=1..n)

where n is reasonable, though around 15.

curve3(i):=subs(u=z^2 + cat(Ubp,i),curve2);

curve4(i,n):=subs(v=cat(Vbp,i)+sum(a[i,j]*z^j,j=1..n),curve3(i));

Here, I'm not sure of the pros/cons of cat() vs a[i,j]....

Anyway, I'll then have an equation in terms of only the local coordinate "z". If I then solve each coefficient of "z", at each order, I should then be able to determine the power series v(z). I reason that since the curve is initially equal to zero, that every non-zero power of z will have a coefficient/equation (in terms of unknowns a[i,j]) that should be equal to zero.

This is analagous to solving differential equations with power series...

However, I'm a little lost in implementing this,

I am currently trying, for instance,

l1:= series(curve4(1,10),z,0,9);

e1:= seq(coeff(l1,z,i),i=1..4):

s1:=seq(a[1,j],j=1..4):

sol:=solve(e1,s1);

I believe I am doing something wrong though bcause every odd power is zero.

Thank you a lot for any suggestions and/or help,

sbh

PS>

Ubp and Vbp are those 6 points - I just kept them as symbols initially because I was still getting odd-powered coefficients as zero, regardless of their actual values. Hence a little bit of the "math" side of the problem...

Hi;

To compute higher-order lie derivatives we need to pass a vector field to LieDerivative(..) function. What follows is the result of LieDerivative(..) command:

To compute 2nd-order lie derivative we should first create a vector field as follows:

L1fh := ((1/2)*(x+u)*(-2*y+2*x)/sqrt((y-x)^2+L^2))*D_x + ((1/2)*(y+v)*(2*y-2*x)/sqrt((y-x)^2+L^2))*D_y + (L^2/sqrt((y-x)^2+L^2))*D_L;

Is there a function to extraxt components of an expression which is the result of the LieDerivative(..)? For example how we can extract the first term. i.e. :

 

 

sample code:

with(DifferentialGeometry):

DGsetup([x, y, L], R3);

h := sqrt((y-x)^2 + L^2);
 
f := evalDG((x+u)*D_x + (y+v)*D_y + L*D_L);

L1fh := LieDerivative(f, h);

simplify(L1fh);
       
L2fh := LieDerivative(L1fh, h);

L1fh := ((1/2)*(x+u)*(-2*y+2*x)/sqrt((y-x)^2+L^2))*D_x + ((1/2)*(y+v)*(2*y-2*x)/sqrt((y-x)^2+L^2))*D_y + (L^2/sqrt((y-x)^2+L^2))*D_L;

L2fh := LieDerivative(L1fh, h);

simplify(L2fh);

 

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!

✐♦❝❄t✉⑥♠❧❄❞❢❤❡⑦⑧❛❋⑨⑤⑦⑧❛✯ts❝⑩❦✑❝❄❜❚qs❴❇❛ ✈❇❜❶q✉❴❷♥♦❧❄t✇q✉❤❡❧❄❞★❦sr❇⑥✤❝❄❜✫qs❴❇❛❹❸✫❤❡❛⑧⑥✜❧❺✐❇✐❻⑦⑧❛⑧q✉❧❈❼❽❜❀r❇✐♦♣✒qs❤❡❝❄✐

>teksbiasa:=`Hello!`;

teksbiasa:=Hello!

>nilaiASCII:=convert(teksbiasa,bytes);

nilaiASCII:=[72,101,108,108,111,33]

>E:=nilaiASCII+~nops(nilaiASCII);

E:=[78,107,114,114,117,39]

Hi, how i need to modify my command so the length of each word can be detected if enter a sentence instead of 1 word ?

For example, if i entered >> `Hello! Bob`, so the length of each word is [6, 3], so what i need is

>teksbiasa:=`Hello! Bob`;

teksbiasa:=Hello! Bob

>nilaiASCII:=convert(teksbiasa,bytes);

nilaiASCII:=[72, 101, 108, 108, 111, 33, 32, 66, 111, 98]

and my E will be >> E:=[72+6, 101+6, 108+6, 108+6, 111+6, 33+6, 32, 66+3, 111+3, 98+3] which is >>

E:= [78, 107, 114, 114, 117, 39, 32, 69, 114, 101]

Thanks for help~=]]
Have a nice day~=]]

Hello,

I am trying to find the interpolation of a 3D function.

I have the vector x, the vector y and a matrix M at my disposition.
I know how to do the pointplot3d , but I can not find the function: f(x,y) 

any suggestion?

Thank you for help

Hello,

I tried to rearrange below set of equations to have the equations in term of P[0], P[1], P[2], P[3], P[4], P[5] and P[6]. I used the symbol := for function definition for all Ps except one of them. Thus maple will rearrange that excepted one. However, I got error massage stating "Error, (in P[3]) too many levels of recursion" when I tried to rearranged equations for P[0].

Can I get help to rearranged them. 

P[0](s) = (P[1](s)*mu[1]+P[2](s)*mu[2]+1)/(s+3*lambda+3*sigma)

P[1](s) = (3*P[0](s)*lambda+3*P[3](s)*mu[1]+P[4](s)*mu[2])/(s+mu[1]+2*lambda+2*sigma)

P[2](s) = (3*P[0](s)*sigma+P[4](s)*mu[2]+P[3](s)*mu[1])/(s+mu[2]+2*lambda+2*sigma)

P[3](s) = 2*lambda(P[1](s)+P[2](s))/(s+2*mu[1]+lambda)

P[4](s) = 2*sigma(P[1](s)+P[2](s))/(s+2*mu[2]+sigma)

P[5](s) = lambda(P[3](s)+P[4](s))/s

P[6](s) = sigma(P[3](s)+P[4](s))/s

Thank you for your help

Is it possible to create a sumif function, like the one in excel?

I know that the following expression (a:=6*s*sqrt(9*s^2+32)+18*s^2+32)

can be rewritten as

However, none of the following maple functions is abble to give the factored result:

factor(a)

simplify(a)

combine(a)

Someone could help me to understand what is going on, please?

Let us consider the maximum value of the polynomial

x^4+c*x^2+x^3+d*x-c-1

on the interval x=-1..1 as a function g of the parameters c and d. General considerations suggest its continuity. However, a plot3d of g does not  confirm it.  Also the question arises "Is the function g(c,d) bounded from below?". Here is my try with the DirectSearch and NLPSolve:

Download bound.mw

Hello

I am trying to slve the second order differential equation with initial conditions  t0=0.dy/dt=0,y0=10000

-(diff(y, t, t))-9.81+0.563e-3*(0.1832e-2*abs(diff(y, t))+0.51702e-1*abs(diff(y, t))^(3/2)+.4*(diff(y, t))^2) = 0

using 4th order runge kutta.do i need to declare a step parameter like (D(y))(t) = u or is a command that can be applied automatically?

Thanks

 rk4.mw

>nombor1:=[4,6,2];

nombor1:=[4,6,2]

Hi, anyone know hot i need to continue my command to get 462 from [4,6,2]?

Thank you~=]]

I have an equation for r(t) that involves 3 (slidable) constants; an equation for phi(t) that involves the same three constants and is written in terms of arctan; theta is a slidable constant. How do I plot this on an x,y,z plot? I want an animation in terms of t.

>mylist:=462;

mylist:=462

>kekuncirahsia:=proc(n) local c,d,r,sum; c:=n; while (c>9) do sum:=0; d:=c; while (d>9) do r:=irem(d,10); sum:=sum+r; d:=iquo(d,10);od; c:=sum+d;print(c);od;end;

>kekuncirahsia(mylist);

12

3

Hi, anyone know how i need to write my command to get this >> kekuncirahsiafinal:=3

I just want it to print the last digit..

Thank you~=]]

>mylist:=468;

mylist:=468

Hi, how i need to continue my command to get [4,6,8] from 468?

Thank you~=]]