Hi,

I am trying to complete this assignment but I am having trouble on this question. It is asking me to use a graphical approach to estimate a 𝛿 > 0 such that for all x...

0 < | x - c | < 𝛿 implies that | f(x) - L | < ε

It says I can do this by plotting my f(x), y₁, and y₂ over the interval [c - 𝛿, c + 𝛿] with a y-range of [L - ε, L + ε]. y₁, and y₂ are banding lines defined by ( L - ε ) and ( L + ε ), respectively. The value of ε is shown below. I dont know where to start but this is what I have so far...

f(x) := ( x ( 1 - cos(x) ) ) / ( x - sin(x) );

The limit ( L ) of this function as x approaches 0 ( c = 0 ) is...

L := ( limit ( f(x), x = 0) );

The limit is 3 ( L = 3 ).

 ε = r = Student Id / 5000000   = 0621748 / 5000000   = 155337 / 1250000

 ε := 155337 / 1250000;

y₁ := ( L - ε );

y₂ := ( L + ε );

Originally, it had asked me to graph the function with the banding lines (y₁, y₂) together within the interval [-0.2, 0.2], which I have done. I just do not know how to find a value for 𝛿 in this case. Please help.

Thanks.

This doesn't work directly:

is(RootOf(_Z^3-_Z-1, index = 1)+RootOf(_Z^3-_Z-1, index = 2) = 0);
                                      FAIL

But Maple can compute the bounding interval for a polynomial RootOf (convert(..., RootOf, form=interval)), and evalrC can perform operations on complex intervals:

evalrC(RootOf(_Z^3-_Z-1, index = 1)+RootOf(_Z^3-_Z-1, index = 2));
       INTERVAL(0.6623589782, 0.5622795119, 0.6623589795, 0.5622795122)

evalrC(Re(%));
Error, (in evalrC/INTERVAL) not implemented yet: 4

Re and Im are obviously positive though. Why doesn't Maple try to do that automatically, and also, how do I check if the point is inside the interval? (Without writing my own function to handle real intervals, unions of real intervals, and complex intervals.)

the infint  summation  maple cant evaluated it  when my variables  are more than one decimal number 

;

In the link below is my code which generates numerical data of a Fourier series of an impulse function (not to be confused with the DIRAC DELTA IMPULSE).  The issue is as k increases the harmonic frequencies become faster than the time step which causes resolution issues at higher k values.  Is there a way to iteratively change the time step for each iteration of k value?  However, the range for t would have to correspondingly decrease or the file size would balloon by the corresponding change in k value.

So lets say my time step for k = 1 is 0.001 and my time range is 0..1, now for k = 1000 my time step would be 0.000001, but my time range would shrink from 0..1 to 0..0.001.  I hope this is not too confusing.  Maybe experimenting with my code will aid in understanding my request.

Thanks for any assistance

MATRIX_loop.mw

Maple will multiply two matrices for me perfectly.  

However, when I change the value in one of the cells and try to re-evaluate the product, Maple starts giving me the sum of something in all of my cells.   

Are you simply not allowed to ever change the initial values you set for a matrix?  This doesn't seem to make sense to me.  

I have attached a screen shot to show an example with a random matrix.  

is(k::OrProp(even, odd)) assuming k::integer;
                             false

is(k::Or(posint, nonposint)) assuming k::integer; # a synonym for OrProp
                             false

is(`or`(k::posint, k::negint, k = 0)) assuming k::integer;
                             false

Not including the infix or because that would fail for k=0 for different reasons.

1.

eliminate({-a+x+y+z, -b+x*y+y*z+x*z, -c+x*y*z, -P+x^4+y^4+z^4}, {x, y, z});
    [{x = (sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2+a*b*c-c^2)/(2*b*c),
      y = -(sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2-a*b*c+c^2)/(2*b*c),
      z = c/b}, {-a^4+4*a^2*b-4*a*c-2*b^2+P}],
    [{x = -(sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2-a*b*c+c^2)/(2*b*c),
      y = (sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2+a*b*c-c^2)/(2*b*c),
      z = c/b}, {-a^4+4*a^2*b-4*a*c-2*b^2+P}]

The expression for P is fine, but how are x and y square roots and z=c/b??

2.

eliminate({y*z, x+y-z}, z);
                       [{z = 0}, {x + y}]

Is this supposed to happen? I was expecting y*(x+y):

Groebner:-Basis({y*z, x+y-z}, lexdeg([z], [x, y]));
                     [x*y + y^2, z - x - y]

Although a very general question... does the rope break or not?

A 100 kg weight is hung from the center of a 1/4" polypropelene rope making an angle of 10 degrees at both ends with the horizontal.  According to some characteristics, the breaking strength of the rope is 1125 lbs or 5kN.  The safe load (safety factor of 12) drops the weight to  93.8 lbs or 0.417kN. Find the tension in the rope?  Is it strong enough?

Fairly easy to solve in Maple - just two equations two unknowns making t1 tension on one side and t2 tension on the other side

eq1 := t1*cos(80*Pi*(1/180))+t2*cos(80*Pi*(1/180)) = 100*9.8 # ΣFy=mg

eq2 := t1*sin(80*Pi*(1/180))-t2*sin(80*Pi*(1/180)) = 0 #ΣFx=0 hence basically t1=t2

solve([eq1,eq2])

                 {t1 = 2821.797537, t2 = 2821.797537}

Therefore the tension on the rope t1=t2 is 2821 N  (looks like way beyond the safety factor) so this means the rope will not break, correct?  or is it that the total tension becomes t1+t2=5642 N putting us over the 5kN and the rope breaks?

I am new to maple. how do i segregate different term in long PDE in maple

In the unrelated discussion https://mapleprimes.com/questions/222768-Interrupting-An-Evaluation-Leaves-Mserver-Running
I asked Carl Love if he had an estimate of the number of builtin procedures in Maple.
Carl answered with the code you will find in the link.
His code came up with 316 and 320 in Maple 2016.2 and 2017.2, respectively.
He mentioned that the last one alphabetically was zip.
I happened to look up the help page for type/builtin and found the statement:

"It may be used with anames to list all builtin procedures. For example anames(builtin)."

So I tried:

AB:={anames(builtin)};
nops(AB); # 234 in Maple 2017.2
member(zip,AB); #false
op(3,eval(zip)); # builtin = 589 in Maple 2017.2
showstat(zip); # Works despite being builtin.

In Maple 2016.2 the corresponding numbers were 232 and 585.
So I'm left somewhat confused.
Apparently some procedures are less builtin than others? Does this just mean that the most difficult zipping is builtin, while simple stuff like the following is done by line 16 in the available code for zip?

showstat(zip,16..17);
zip(`=`,[a,b,c], [1,2,3]);

I have saved 3 Maple workbook files on my computer, but only one of them can re-open.
when i try to open the other two a pop-up says

"A problem was ecountered while opening the workbook.

database is not opened" 

and then i get the option to press "ok". What should i do?

If I enter the following I do not get the residue... how do I make the residue command work?

The documentation for the option AllSolutions for int says that the results are always valid for all real parameter values (in the endpoints). That seems like a pretty major claim. Each of these three is already wrong for a=-1/2, b=1/2:

int(1/ln(t), t = a .. b, AllSolutions);
    piecewise(ln(a) < ln(b), piecewise(And(1 < b, a < 1), undefined, piecewise(a = 1, infinity,
    Ei(1, -ln(a)))+piecewise(b = 1, -infinity, -Ei(1, -ln(b)))), ln(b) = ln(a), 0, ln(b) < ln(a),
    -piecewise(And(1 < a, b < 1), undefined, piecewise(b = 1, infinity, Ei(1, -ln(b)))+
    piecewise(a = 1, -infinity, -Ei(1, -ln(a)))))

int(sqrt(t^2-1+I*t), t = a .. b, AllSolutions);
    piecewise(a < b, (1/2)*sqrt(b^2-1+I*b)*b+I*sqrt(b^2-1+I*b)*(1/4)-3*ln(-2*signum(0, -b, 1)^2*
    b^2+2*sqrt(b^4-b^2+1)*signum(0, -b, 1)^2+4*b*sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+2*
    signum(0, -b, 1)^2-2*signum(0, -b, 1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)+6*b^2+2*
    sqrt(b^4-b^2+1)-1)*(1/16)-3*ln((-I*(signum(0, -b, 1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)-1+I*
    sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+(2*I)*b))*(1/sqrt(-2*signum(0, -b, 1)^2*b^2+2*sqrt(b^4-b^2+1)*
    signum(0, -b, 1)^2+4*b*sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+2*signum(0, -b, 1)^2-2*signum(0, -b, 1)*
    sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)+6*b^2+2*sqrt(b^4-b^2+1)-1)))*(1/8)-(1/2)*sqrt(a^2-1+I*a)*a-I*
    sqrt(a^2-1+I*a)*(1/4)+3*ln(-2*signum(0, -a, -1)^2*a^2+2*sqrt(a^4-a^2+1)*signum(0, -a, -1)^2+
    4*a*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)+2*signum(0, -a, -1)^2-2*signum(0, -a, -1)*sqrt(2*
    sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+2*sqrt(a^4-a^2+1)-1)*(1/16)+3*ln((-I*(signum(0, -a, -1)*sqrt(2*
    sqrt(a^4-a^2+1)-2*a^2+2)+I*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)-1+(2*I)*a))*(1/sqrt(-2*
    signum(0, -a, -1)^2*a^2+2*sqrt(a^4-a^2+1)*signum(0, -a, -1)^2+4*a*sqrt(2*sqrt(a^4-a^2+1)+2*
    a^2-2)+2*signum(0, -a, -1)^2-2*signum(0, -a, -1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+2*
    sqrt(a^4-a^2+1)-1)))*(1/8), b = a, 0, b < a, -(1/2)*sqrt(a^2-1+I*a)*a-I*sqrt(a^2-1+I*a)*(1/4)+
    3*ln(-2*signum(0, -a, 1)^2*a^2+2*signum(0, -a, 1)^2*sqrt(a^4-a^2+1)+4*a*sqrt(2*sqrt(a^4-a^2+1)+
    2*a^2-2)+2*signum(0, -a, 1)^2-2*signum(0, -a, 1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+
    2*sqrt(a^4-a^2+1)-1)*(1/16)+3*ln((-I*(signum(0, -a, 1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+
    I*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)-1+(2*I)*a))*(1/sqrt(-2*signum(0, -a, 1)^2*a^2+
    2*signum(0, -a, 1)^2*sqrt(a^4-a^2+1)+4*a*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)+2*signum(0, -a, 1)^2-
    2*signum(0, -a, 1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+2*sqrt(a^4-a^2+1)-1)))*(1/8)+(1/2)*
    sqrt(b^2-1+I*b)*b+I*sqrt(b^2-1+I*b)*(1/4)-3*ln(-2*signum(0, -b, -1)^2*b^2+2*sqrt(b^4-b^2+1)*
    signum(0, -b, -1)^2+4*b*sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+2*signum(0, -b, -1)^2-
    2*signum(0, -b, -1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)+6*b^2+2*sqrt(b^4-b^2+1)-1)*(1/16)-
    3*ln(-(I*signum(0, -b, -1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)-I-sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)-
    2*b)/sqrt(-2*signum(0, -b, -1)^2*b^2+2*sqrt(b^4-b^2+1)*signum(0, -b, -1)^2+4*b*sqrt(2*
    sqrt(b^4-b^2+1)+2*b^2-2)+2*signum(0, -b, -1)^2-2*signum(0, -b, -1)*sqrt(2*sqrt(b^4-b^2+1)-
    2*b^2+2)+6*b^2+2*sqrt(b^4-b^2+1)-1))*(1/8))

int(arctan(t+2*I), t = a .. b, AllSolutions);
   piecewise(a < b, piecewise(a < 0, I*arctan(4*a/(a^2-3))*(1/2)+(1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-
   (2*I)*arctan(2*I+a)-arctan(2*I+a)*a+I*Pi*(1/2), a = 0, -I*Pi+3*ln(3)*(1/2), 0 < a, I*arctan(4*a/
   (a^2-3))*(1/2)+(1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-(2*I)*arctan(2*I+a)-arctan(2*I+a)*a-I*Pi*(1/2))+
   piecewise(b < 0, -I*arctan(4*b/(b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+(2*I)*
   arctan(2*I+b)+arctan(2*I+b)*b-I*Pi*(1/2), b = 0, -I*Pi-3*ln(3)*(1/2), 0 < b, -I*arctan(4*b/
   (b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+(2*I)*arctan(2*I+b)+arctan(2*I+b)*b+I*Pi*(1/2))+
   piecewise(And(0 < b, a < 0), -(2*I)*Pi, 0), b = a, 0, b < a, piecewise(b < 0, -I*arctan(4*b/
   (b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+(2*I)*arctan(2*I+b)+arctan(2*I+b)*b-I*Pi*(1/2),
   b = 0, I*Pi-3*ln(3)*(1/2), 0 < b, -I*arctan(4*b/(b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+
   (2*I)*arctan(2*I+b)+arctan(2*I+b)*b+I*Pi*(1/2))+piecewise(a < 0, I*arctan(4*a/(a^2-3))*(1/2)+
   (1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-(2*I)*arctan(2*I+a)-arctan(2*I+a)*a+I*Pi*(1/2), a = 0,
   I*Pi+3*ln(3)*(1/2), 0 < a, I*arctan(4*a/(a^2-3))*(1/2)+(1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-(2*I)*
   arctan(2*I+a)-arctan(2*I+a)*a-I*Pi*(1/2))-piecewise(And(0 < a, b < 0), -(2*I)*Pi, 0))

The first one probably has the correct answer inside, but it has conditions like ln(a)<ln(b), so that case never gets selected when the values are complex.

f := (z, t) -> ln(t)^2/((t^2+1)*(t-z));

int(f(z, t), t = 0 .. infinity) assuming Im(z) > 0;
       int(ln(t)^2/((t^2+1)*(t-z)), t = 0 .. infinity)

int(f(a + I*b, t), t = 0 .. infinity) assuming a::real, b > 0; # 0*infinity
       -sqrt(a^2+b^2-2*b+1)*signum(I*arctan(b, a)-I*arctan(-b, -a)-I*Pi)*
       infinity/((I*b-I+a)*(I*b+I+a))

int(f(z, t), t = 0 .. infinity) assuming Re(z) > 0, Im(z) > 0;
       int(ln(t)^2/((t^2+1)*(t-z)), t = 0 .. infinity)

int(f(a + I*b, t), t = 0 .. infinity) assuming a > 0, b > 0;
      -((3*I)*Pi^3*b+3*Pi^3*a-(16*I)*Pi^2*arctan(b/a)-(6*I)*Pi*ln(a^2+b^2)^2+(24*I)*
      Pi*arctan(b/a)^2+(6*I)*ln(a^2+b^2)^2*arctan(b/a)-(8*I)*arctan(b/a)^3-8*Pi^2*
      ln(a^2+b^2)+24*Pi*ln(a^2+b^2)*arctan(b/a)+ln(a^2+b^2)^3-12*ln(a^2+b^2)*
      arctan(b/a)^2)/(24*(-b^2+(2*I)*a*b+a^2+1))

So it looks like the first three can be made to work as well (and the result in terms of z will be much neater).

I have noticed that I don't receive e-mails anymore when contributions are submitted to my subscriptions.
I used to.
Has this happened to anyone else?

It is embarrasing to have asked somebody a question and gotten a reply you are not made aware of.

What to do about it?