hey guys.. ive been trying to get the root of the following equation.. but watever i do i am getting only the positive value.. and i have another equation which has 5 answers.. but only 1 is being shown.. can u help me out??

 

>f0:=x->x/(sqrt(x^2 + cos(x-1))):
>f1:=D(f0):
>f2:=D(f1):
>a:=fsolve(f1(x)=0,x);
                                a:=3.639062685
 

hi, how do i perform the following operation and simplfy?

(4a- 3ab + 2b) + [ (2a- 4b) - ab ]

We are contemplating using Maple TA.  I have been playing with a trial and it looks quite good.

If you read the Maplesoft advertising,  you get the impresion that a lot of Universities out there are using TA. 

On the other hand, the (rather variable quality) mapleapps question banks resource does not really contain a huge amount of material.

Hello all,

When I tried to plot with "semilogplot" command, I've found the ticks on the x-axis and the vertical grids do not line up. This is with Maple 10. Would somebody tell me if there is a way to line them up?

with(plots):

semilogplot(log[10](x), x=1..1000, gridlines=true);

 

Thank you.

Chin Li

 

Hi there.

 

I have two vectors:

x1:= [ ( 1+sqrt(1-4*a^2) )/(2*a) , ( 1+sqrt(1-4*a^2) )/2 ];

x2:= [ ( 1-sqrt(1-4*a^2) )/(2*a) , ( 1-sqrt(1-4*a^2) )/2 ];

I have the following questions to complete:

(2)Determine when x1 and x2 are both real

(3)Determine the largest interval I=(c,d) (with d>0) such that x1 and x2 are real and distinct.

 

The answers are obvious on paper, but I have to calculate this with Maple.  How do I go about this?

 

Thanks

Probably the easiest way to explain this is to show you what I want to do. I want to draw a contourplot which shows the contour lines going through the points [0,-1] and [3/4,-7/16] (these are the expression's stationary points).  The values of f at each of these points respectively are 1 and 101/128, so I have tried to display these contours, but when the graph comes up, the contour lines don't go through the point [3/4,-7/16], although they do go through [0,-1]. The code I'm trying to use is below.

f(x,y):=x^3-3*x*y+2*y^2-3*x+4*y+3;

Dear all,

 

I'm trying to duplicate by hand the variancecovariance matrix (and stanard error vector) of a simple LinearFit of data (2nd order function of form a + bx + cx^2), using the inverse of the curvature matrix as in Chapter 15 "Modeling of Data" of the Numerical Algorithms book 3rd Edition, or the paper by Keith H. burrell "Error analysis for parameters determined in nonlinwear least-squares fits", American journal of Physics, Vol. 58, No. 2, 160-164 (1990). Unfortunately, I cannot get the same answer, even with simple unweighted examples.

Hi,

I know it is possible to find a basis from a set of vectors using the 'basis' command but is it possible to find a basis from a set of matrices?

I'm trying to make it animate slowly without having to click the 'decrease frame rate' button. Does anyone know if there's something I can put in the code to make it do this?

Thanks

I need to find an example of a function of one variable that has an antiderivative that can be expressed very simply in terms of fuctions that a 1st-year calculus student would know, but int (command name in maple) can't find an antiderivative.

Hint: Start with the antiderivative F(x), and get f(x) by differentiating it and simplifying. You might try something involving a few square roots and logarithms or exponentials or trigonometric functions.

I need to find an example of a function of one variable that has an antiderivative that can be expressed very simply in terms of fuctions that a 1st-year calculus student would know, but int (command name in maple) can't find an antiderivative.

Hint: Start with the antiderivative F(x), and get f(x) by differentiating it and simplifying. You might try something involving a few square roots and logarithms or exponentials or trigonometric functions.

     I am trying to do a "radial" polar plot.  Yes, I know Maple can do polar plots, but their axes are always in the x-y direction.  I am looking for a polar plot that is round with the angle (theta)  labeled around the circle in degrees., and the value 'R' as the length from the center of the circle.  There should also be concentric circles marked with the distance from the center point so as to provide a scale for the R value.  So the grid looks like a round spiderweb.  Can Maple do this easily?  I looked all over the docum

Hi - I'm trying to solve a PDE using pdsolve() for the Graetz problem (heat transfer in a pipe). I can solve the PDE and get a general solution, but when I try to solve the equation with the boundary conditions, Maple thinks that the boundary conditions are new functions with the same name. Here is how I am doing it now: > heat := (1-z^2)*(diff(T(y, z), y)) = (diff(z*(diff(T(y, z), z)), z))/z; > cond := T(0, z) = 0, T(y, 1) = 1, diff(T(y, 0), z) = 0; > PDE := [heat, cond]; > sol := pdsolve(PDE, T(y, z)); This throws the error:

I need to find the square root of a large number, x (about 70 digits long).  If I use x^0.5, the answer is displayed in scientific notation and  is rounded.  How can I find this number exactly?

I have 2 functions, f(x) and g(x) both continous on [a,b] (a known interval). For which value of C ε [a,b] the area under f(x) and g(x) is minimal?

I know f(x) and g(x).