Recent content by phyzguy

I used these books to build a 6" telescope when I was in high school. Your brother and I should sit down and have a beer!

Well, O-rings should work. Again, a picture would help a lot. You say the old solution, "leaked air and did not work". Maybe the O-rings just need to be replaced and re-greased with vacuum grease. As for passing in low voltage, what you want is a feedthrough. You can buy one, or you can...

As @DaveE said, Ampere's law is always valid and you can draw any loop you choose. Sometimes the symmetry of the problem is such that you can draw the loop in a way that makes it easy to solve the problem. But in most cases you draw the loop and you don't know how the magnetic field varies...

Since there is no evidence magnetic monopoles exist, there is no defined unit for their magnetic charge.

How low pressure are you talking about? O-rings work OK down to about 10^-7 Torr. Lower than that you want metal-metal seals. A picture or drawing of the holes would help us give better advice.

You may have to do it numerically, or use some approximations. Are the values of alpha, beta, mu and sigma completely arbitrary, or do you have some prior knowledge of what they are? For example, for alpha = 5, beta = 0.5, mu=4.0, sigma=1.0, the second term (the Gaussian) is much narrower than...

Elaborating a little more on my example, and how u substitution is basically the inverse of the chain rule. If you have:$$ \frac{d}{dx} \frac{sin^2(x)}{2}$$ Applying the chain rule, you get: $$ \frac{d}{dx} \frac{sin^2(x)}{2} = \frac{d}{d(sin(x))}\frac{sin^2(x)}{2} \frac{d}{dx} sin(x) = sin(x)...

Another example is: $$ I = \int \sin{x} cos{x} dx $$ It's not obvious how to do this. But if I set $$ u = \sin{x}, du = \cos{x} dx $$, then we have: $$ I = \int u du = 1/2 u^2 = 1/2 \sin^2{x} $$

I loved those books when I was young!! I remember "Tom Swift and his Flying Lab" quite well.

Yes, but many people have said that all the LLM's are doing is parroting back what they have been trained. The fact that have solved a problem which no one had solved before and which couldn't have been in their training data proves that they are doing more than that.

Using the Leavitt's law formula in this Wikipedia article, I get an absolute magnitude of -5.14.

There are cases where LLMs have decoded ancient text that no human had ever decoded before. If they are just parroting back what they were trained on, that couldn't happen. What they are doing when training is building a model of the world inside their neural networks, just like the model of...

One thing to think about. This Hubble release says that "It is the spiral galaxy NGC 4603, the most distant galaxy in which Cepheid variables have been found" NGC 4603 is at a distance of 33 Mpc. Yet you have points on your graph going out to 140 Mpc. So it looks like you have done something...

Define what intelligence is and we can answer your question. The classical measure of intelligence has been an IQ test. Based on that, there is no doubt that LLMs are intelligent.

So to understand where the problem is, you need to show us what you did. To show your work, take one of the points, let's say the one farthest to the right. (1) Which galaxy is that? (2) What was the recession velocity? With units! (3) How did you calculate the distance? Show us what data...