if anyone could please help me out with this code i would really appreciate it

> cuberootkj := proc (m, n)

local e, tol, x0, x1, x2, maxsteps, counter;

tol := 1/200000;

e := 1;  x1 := m;  x0 := n;

while tol < e do

x2 := (2/3)*x1+(1/3)*x0/x1^2;

e := x2-surd(x0, 3)

end do;

print("root"); x1

end proc;

 

i get this error message: Error, (in cuberootkj) cannot determine if this expression is true or false: 0 < 9399973/5400000-5^(1/3)

Hello,

I want to try to find the Laurent series centered about z=1 for the function f(z)=1/[(z+1)(z+2)] in the annular region |z|>3.  When I use the laurent() function, Maple returns the laurent series that is valid for 0<|z|<2.  Is there some way to restrict the domain of the function to make Maple return a Laurent series for a different domain?

Thanks

I need to solve equation: log[2*sqrt(2+sqrt(3))](y)=log[2+sqrt(3)](y-1) I know (i did it with my hands) that y=8+4*sqrt(3) is a solution of this eqution. But i can not find this solution using Maple. How i can obtain it? P/s/ I solved with maple numerically. It is not enough for me. I need solution namely in form 8+4*sqrt(3). Because it is explicit solution. I am looking forward for answer. Thanks.

hi

I'm looking at Fourier coefficients:

Dn:= (2/L)* int(f*sin(n*Pi*x/L), x=0..L); where L = Pi

and the series

u:= Sum (Dn * sin(n*Pi*x/L)*cos(n*Pi*t/L), n=1..infinity);

When I try to sum the series:

u:= sum (Dn * sin(n*Pi*x/L)*cos(n*Pi*t/L), n=1..10);

I get the error

Error, (in NumericRange) summand is singular in the interval of summation

Now I can see the 0 / 0 thing, but the first term should be 1 and the other terms zero. The first term is 1 if you take limit as n->1

So I have to explicitly state that:

D1:= (2/L)* int(f*sin(Pi*x/L), x=0..L);

hye, how to plot the pde using pdsolve if the error that we get is unable to handle elliptic PDEs. please look at the attachment. thanks for your help. regards, zarith

we want to find the sum:

sum((6*n+5)*(6*n+3)/((12*n*n+12*n+1)*(6*n*n+6*n+1)), n = 1 .. infinity)

is is equal to:

-15-(1/2)*sqrt(3)*Pi*tan((1/6)*Pi*sqrt(3))+(1/2)*sqrt(6)*Pi*tan((1/6)*Pi*sqrt(6))-2*Psi(1/2+(1/6)*sqrt(6))+Pi*tan((1/6)*Pi*sqrt(6))+2*Psi(1/2+(1/6)*sqrt(3))-Pi*tan((1/6)*Pi*sqrt(3))

can we get rid of  the function "\Psi"?  Can we bound this function \Psi with other known constant such as zeta function?

 

 

For certain points I have determined values: z depending on x and y. I look for a method to get extrapolated values for any x-y combination in this area.

To fit a 3D curve is not feasible because the fitted curve would not deliver the exact z values for a given data set (=essential condition). For a 2D data set I would do something like a linear regression between two data points but I have no approach to do something similar with data points depending on two variables. I hope somebody can help me to solve this problem. Thanks!

 

Hi, The following worksheet shows you how to use Maple to obtain the QR Decomposition of a matrix. The worksheet demonstrates how to use the Context Menu to obtain the decomposition as well as a stepwise solution for finding the decomposition using the Gram-Schmidt process. The worksheet also contains an example of how QR Decomposition can be applied to find the least-squares solution of an overdetermined system of equations. Watch the