I've spent way too much time today trying to see how many other problems I can get to run under the measurable limit.
September 13, 2025 at 6:49 PM
I've spent way too much time today trying to see how many other problems I can get to run under the measurable limit.
I suspect you already have TPM 1.2 support. From what I understand TPM 2.0 really only helps with not needing passwords. While not supported by MS, I imagine there are ways to run Win11 without TPM 2.0. My issue is that everything really wants SSE 4.2 now and my old hardware only had SSE4.1.
June 13, 2025 at 6:48 PM
I suspect you already have TPM 1.2 support. From what I understand TPM 2.0 really only helps with not needing passwords. While not supported by MS, I imagine there are ways to run Win11 without TPM 2.0. My issue is that everything really wants SSE 4.2 now and my old hardware only had SSE4.1.
No, it was his way of saying that the school you went to only matters in getting your first job.
May 27, 2025 at 10:12 PM
No, it was his way of saying that the school you went to only matters in getting your first job.
It seems to be a trend now.
February 19, 2025 at 2:21 PM
It seems to be a trend now.
After that it's just counting how many iterations of z(n+1) = z(n)^p+c, z(0) = 0 you have to do before |z| > 2 like in the normal Mandelbrot calculation.
I imagine all of this will work on the Julia set, as well.
I imagine all of this will work on the Julia set, as well.
February 12, 2025 at 2:01 PM
After that it's just counting how many iterations of z(n+1) = z(n)^p+c, z(0) = 0 you have to do before |z| > 2 like in the normal Mandelbrot calculation.
I imagine all of this will work on the Julia set, as well.
I imagine all of this will work on the Julia set, as well.
To get arbitrary exponents I rely on two identities
z = r e ^ t = r(cos t + i sin t)
and
a^b = e^(b ln a)
then
z0 = a + bi
r0 = | z0 |
t0 = atan(b/a)
r1 = e ^ (exponent * ln( r0 ) )
t1 = t0 * exponent
z1 = r1( cos(t1) + i sin(t1) ) + c
z = r e ^ t = r(cos t + i sin t)
and
a^b = e^(b ln a)
then
z0 = a + bi
r0 = | z0 |
t0 = atan(b/a)
r1 = e ^ (exponent * ln( r0 ) )
t1 = t0 * exponent
z1 = r1( cos(t1) + i sin(t1) ) + c
February 12, 2025 at 1:55 PM
To get arbitrary exponents I rely on two identities
z = r e ^ t = r(cos t + i sin t)
and
a^b = e^(b ln a)
then
z0 = a + bi
r0 = | z0 |
t0 = atan(b/a)
r1 = e ^ (exponent * ln( r0 ) )
t1 = t0 * exponent
z1 = r1( cos(t1) + i sin(t1) ) + c
z = r e ^ t = r(cos t + i sin t)
and
a^b = e^(b ln a)
then
z0 = a + bi
r0 = | z0 |
t0 = atan(b/a)
r1 = e ^ (exponent * ln( r0 ) )
t1 = t0 * exponent
z1 = r1( cos(t1) + i sin(t1) ) + c
Is it fair if I worked on the same game four different times?
November 20, 2024 at 12:48 AM
Is it fair if I worked on the same game four different times?