Wierd, but thanks! It must have been the >assume(0<a<1); that I had in there...
K

On the same theme, I have some functions to plot;
A := exp(-100*z);
B := (100*z)^(-1);
C := (99/(1/z-1))^(-1);
E := piecewise(z < .5, (99/(1/z-1))^(-1), z >= .5, (1/99)*exp(-4*(z-.5)));

loglogplot([A, B, C, E], z = 0.01 .. 1);

But then I try and plot on the transformed axes...

plottools[tranform]((x,y)->[y,x])(loglogplot([A, B, C, E], z = 0.01 .. 1);

and I get an error message "Error, invalid input: unknown uses a 2nd argument, y, which is missing"
... which I don't really get... any ideas as to how to overcome this issue?

Thanks
k

no worries, I'm sorted... It seems to work if I just use,
plot([seq(c[hat], t = [0,0.1,0.5,1])], z = 0 .. 1);

wierd.

Thanks! Axes are successfully switched... but when I try and get a graph I just get an empty plot...

c[hat] := sum(A[n]*Phi[n]*exp(-L[n]*t), n = 1 .. 3); 

which gives...

c[hat] := (-.1658604333+.1466606263*exp(1)+3.270186914*10^(-11)*exp(-1))*exp(-z)*sin(2.028757838*z)*exp(-2.557929183*t)+(-0.9398056825e-1-0.3748789972e-1*exp(1)+1.147695299*10^(-10)*exp(-1))*exp(-z)*sin(4.913180442*z)*exp(-12.56967102*t)+(-0.6075825679e-1+0.1511194787e-1*exp(1)+3.061266146*10^(-10)*exp(-1))*exp(-z)*sin(7.978665710*z)*exp(-32.32955328*t)

times = [t = 0, .1, .5, 1];
plot([seq(c[hat], t = times)], z = 0 .. 1);

.. and I get axes with no plot or error message
... am I doing something silly?
k

*should read*
c[hat] := Sum(A[n]*Phi[n](z)*exp(-lambda[n]*t), n = 1 .. 7)

ok, new question(ish)...I've got

eq1:= tan(sqrt(lambda*alpha-(alpha/2)^2))=-2/alpha*(sqrt(lambda*alpha-(alpha/2)^2));

Then, if I sub in a value for alpha, I can use fsolve to get a particular value for the smallest non zero lambda

eq2:=subs(alpha=2,eq1);

lambda1:=fsolve(eq2,lambda,Pi/2..3*Pi/2);

I was wondering, is there anyway of plotting this lamba1 as a function of alpha?

Thanks again,

Thanks, that's certainly given me something to think about.
Basically, I have a sturm liouville type problem where lambda is the eigenvalue... I'm expecting there to be infinitley many lamdas that satisfy eq1 since:

plot([tan(x), -(2/5)*x], x=-2*Pi..2*Pi, y=-4..4, color=[red,blue],discont = true); 

where x is beta and I've subbed alpha=5
I was looking for a general lambda_n and then investigate my solution for the slowest decaying exponential for n=1...

Ok, I've realised I've made a mistake, I infact want a plot of lamba in terms of alpha... which makes it more complicated I think.
I've tried following the method above...
 

eq1:= tan(beta)= -(2/alpha)*beta;

eq2:= beta=sqrt(lambda*alpha - (alpha/2)^2);

eq3:=eval(eq2,beta = solve(eq1,beta));

eq4:=solve(eq3,lambda);

plot(eq4,alpha=-2*Pi..2*Pi);

... which seems to be ok, until I get to the plot where I just get a straight line of gradient approx 0.25... which isn't what I expected...
Have I made an error somewhere?

Thanks
k

Thanks! Looks much simpler now, I'll try it out.

Just to clarify...

tan(beta) = -(2/alpha)*beta

k