Hello, I found an error in the following plotting. An error is with discont=true and the range <-5,5>x<-5,5> In other cases it is OK.
plot(tan, -5 .. 5, -5 .. 5, discont = true); But when I plot tan as an expression no as a function it works correctly.
plot(tan(x), x = -5 .. 5, -5 .. 5, discont = true); The whole document can be visited via View 551_error_tan_plot.mw on MapleNet or Download 551_error_tan_plot.mw
View file details
Ok, another of my "how do you do this" questions:
In solving a an equations such as:
y''-4*y'+5*y = 0
I would like to show the roots; is there a function that will just pull the roots from the equation as written or do I have to write the equation like:
m^2 + 4*m + 5 = 0
How do I get Maple to convert the output of the polar() to degrees with convert(arg, degrees)???
Does the polar() store the magnitude and angle values that is spits out into some variable somewhere that isn't documented? I can't figure this out right now; short of making a copy of the angle part of the polar()'s output to then be fed back into the convert(angle, degrees) statement.
I'm hoping someone can help me. I'm trying to find the equation of a circle using three points (2 outside points and the midpoint). I tried typing this from the Maple software help menu, but keep getting an error:
> with(geometry); _EnvHorizontalName := m; _EnvVerticalName := n;
> circle(c1, [point(A, 0, 0), point(B, 2, 0), point(C, 1, 2)],
> 'centername' = O1);
> center(c1), coordinates(center(c1));
Error, (in geometry:-circle) the coordinate names must be a list of two names
Error, (in isassign) the first argument is expected of type name
My three points are actually (-4,-9), (6.7), and the center is (1,-1). I've worked on this for days, but can't seem to get anywhere. If anyone could help I'd greatly appreciate it!
Is there a way to redefine the way constants are displayed? I would like to have ODE results displayed with C1, C2... with 1,2 as a subscript?
Thanks for any help you can offer.
Would anyone in forum community know of a legitimate online tutoring service that could provide help for me to learn how to use maple?
If so, would you please send me the hyperlink to this service? I have specific questions that the tutorals already provided through maplesoft do not answer.
Okay, here goes I am have some questions. I am solving the Mathieu differential equation whit a damping term, so the equation I am solving is:
Where I have a=0 and b=1,9, and whit the initial conditions u(0)=0 and u'(x)=1, this gives me the following: u(x) = exp(-9/10*x)*MathieuS(-81/100, 1/2*q, x)
I would like to see for which values of q, that the function u(x) is in the interval [-20;20]. I can solve the problem by drawing some plots, but I still haven’t been able to solve it without. I know this is a long short, since you guys have so little info on the problem at hand, but any help will be much appreciates.
So I have a vector X that contains my least squares coefficients. How do I incorporate these coefficients back into a polynomial function automatically using loops or the like?
Right now I have to define the funciton manually
f:=x-> X*x^3 + X*x^2.....
but I want to be able to quickly vary the degree of the polynomial without having to edit the function definition by hand each time. Any geniuses out there that can help me?
If f(x)=-4x^(2) and g(x)=2/x, find [g*f](x).
Implement a computer programme which, for a given function f(x), interval [a,b] and real numbers epsilon1>epsilon2>0, finds an approximation x~ from the inerval (a,b) with an error less than epsilon1 by bisection and, with x~ as the stating value, finds an approximation of this root by the second modification of tj=he Newton method with an error less than epsilon2.
Download 285_quantmechprob1.14.mwsView file details
I'm trying to perform an integration on a function involving a couple constants in order to normalize (setting result of integral = 1) and solve for the constant, A.
The closest I'm able to reach with respect to completing the integral is a result that involves the error function, erf.
How do I go about getting a neater closed form solution to the this integral ?
Am I illegal integration limits ?
In a worksheet I have the following
The following doesn't work, but I was wondering if there was any wild card character that would enable me to do something like this
I have been trying to come up w/ a fast way to get identity Matrices which can later accept non-zero non-diagonal entries. If I use the shape=identity parameter then the Matrix can't be manipulated.
This is what I've come up with
KronDelt :=proc(i,j) => if i=j then 1 else 0 end if end proc:
Matrix(1..4,1..4,(i,j) -> KronDelt(i,j))
Was it necessary to seperately define KronDelt ? I tried putting a mapping directly into the Matrix constructor but always got a syntax error, e. g.,
Matrix(1..4,1..4,(i,j)-> if i=j then 1 else 0 end if )
This doesn't work. Is there any way to d
I am having the hardest time plotting the qualitative behavior of the solutions of these differntial equations. I keep getting Error, (in plots/animate) no non-zero vectors found. If someone could walk me through plotting these equations, it would be greatly appreciated.
Equation1: dx/dt=x^2, x(0)=1, 0≤t<>
My goal is to obtain a formula F:= x -> fa(x)*c1_1(x)+ fb(x)*c1_2(x)+fc(x)*c2_1(x)+... where every ci_j(x) is a function of x. fa, fb, fc are known functions of x. I do not need to have it printed, I just need it to return a numerical value for every value of x I throw in.
I have obtained the solutions of ci_j's in the form
Sol:=[c1_1=f11(x), c1_2=f12(x), c1_3=f13(x), ...], N1 terms
Sol:=[c2_1=f21(x,c1_j’), c2_2=f22(x,c1_j’), c2_3=f23(x,c1_j’), ...], N2 terms
Sol:=[c3_1=f31(x,c1_j’,c2_j’), c3_2=f32(x,c1_j’,c2_j’), c3_3=f33(x,c1_j’,c2_j’), ...], N3 terms
Sol:=[c4_1=f41(x,c1_j’,c2_j’,c3_j’), c4_2=f42(x,c1_j’,c2_j’,c3_j’), c4_3=f43(x,c1_j’,c2_j’,c3_j’), ...], N4 terms