ablmf's profile - activity

I have tried

s = sum(1/(2^x-1), x, 1, oo)
s.n()

But I got

cannot evaluate symbolic expression numerically

The sum does not have a simple form, but it is finite. So is there a way to evaluate it numerically in sage?

Can you draw a picture to show the solution is wrong? Also can you try different algorithms? Also you can directly use scipy to solve the problem.

You probably should use a SMT solver like z3.

Actually, I found that you can write a Cython program with .spyx extension and run it with sage. sage will compile it first and then run the compiled program. This is not efficient if we have to run the script many times. But it is fine for me since I am running a very long simulation and the compilation time of the program is negligible.

To use Cython in sage, according to the document, you can either write Cython code in a sage notebook, load a .spyx file from command line, or create a .pyx file and add it to the sage library.

My question is, can we write a standalone cython script and run it with sage? Just like a normal sage standalone script.

Let's say I am taking derivatives of an expression involving unknow function.

var('x,a,b');

f=(x^(a+b)).function(x,a,b);

h=function('h',nargs=1)(x);

g=h(f(x,a,b));

dg=diff(g,x);

dg

This gives the output

(a + b)*x^(a + b - 1)*D[0](h)(x^(a + b))

How do I replace D[0](h)(x^(a + b)) with something like direvative_of_h(x^(a + b))? In this simple case, I can just do this manually using .operands, but if I have a rather complicated equation, how I can I do it?

Let's say I am taking derivatives of an expression involving unknow function.

var('x,a,b');

f=(x^(a+b)).function(x,a,b);

h=function('h',nargs=1)(x);

g=h(f(x,a,b));

dg=diff(g,x);

dg

This gives the output

(a + b)*x^(a + b - 1)*D[0](h)(x^(a + b))

How do I replace D[0](h)(x^(a + b)) with something like direvative_of_h(x^(a + b))? In this simple case, I can just do this manually using .operands, but if I have a rather complicated equation, how I can I do it?

I have the following expression $$ \frac{\left(-1\right)^{n} - 2 n - 1}{4 {\left(2 n + 1\right)}} $$ It clearly equals $$ \frac{\left(-1\right)^{n}}{4 {\left(2 n + 1\right)}} -\frac 1 4. $$

Is there anyway to make sage show this?

This has two solutions in $\mathbb C$, $\pm \sqrt{2\sqrt{3}-3}/2$.

I am trying to solve this equation in sage $$ \sqrt{-4 \, z^{2} + 2 \, \sqrt{-4 \, z^{2} + 1} - 1} = 0. $$ But when I try the code

var('z')
eq = sqrt(-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1) == 0
solve(eq,z)

I get

[z == -1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1), z == 1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1)]

Is there any way to actually solve it in sage?

I have tried

sage: assume(0<u<pi/2)

But I still get

sage: simplify(1-cos(u)^2)
-cos(u)^2 + 1

I am trying to make the following code work.

t = var('t')
f = zeta(1/2+i*t).abs()
ff = fast_callable(f, vars=[t], domain=CDF)
print find_root(ff, 0, 40)

There are actually 6 roots between 0 and 40. But find_root could not find any of them. Is there any walkaround?

Let $$h(t) = \frac{\sinh(t)}{t}.$$ Let $$ f_i(t) = h\left(\frac{t}{2^i} \right)^{2^i}, $$ where $i\ge 0$ is an integer.

Is there anyway to get a series expansion of $f_i(t)$ without replacing $i$ with a fixed integer?