Ciro Santilli @cirosantilli  40

where:

Remember that $\psi$ is a 4-vetor, gamma matrices are 4x4 matrices, so the whole thing comes down to a dot product of two 4-vectors, with a modified $\psi$ by matrix multiplication/derivatives, and the result is a scalar, as expected for a Lagrangian.

Like any other Lagrangian, you can then recover the Dirac equation, which is the corresponding equations of motion, by applying the Euler-Lagrange equation to the Lagrangian.

How genes form bodies.

Cody'sLab had a nice 5 video series on making it at home! But the United States Government asked him to take it down as suggested at Video "What's Been Going On With Cody'sLab? by Cody'sLab (2019)" at youtu.be/x1mv0vwb08Y?t=84.

Here's a copy online as of 2020: www.youtube.com/watch?v=bCXB6BdMh9Y

This is the order in which you would want to transverse to read the chapters of a book.

Like breadth-first search, this also has the property of visiting parents before any children.

This is the order in which a binary search tree should be traversed for ordered output, i.e.:

This ordering makes sense for binary trees and not k-ary trees in general because if there are more than two nodes it is not clear what the top node should go in the middle of.

This is unlike pre-order depth-first search and post-order depth-first search which generalize obviously to general trees.

Has the property of visiting all descendants before the parent.

When viewed as matrices, it is the group of all matrices that preserve the dot product, i.e.:This implies that it also preserves important geometric notions such as norm (intuitively: distance between two points) and angles.

This is perhaps the best "default definition".

Let's show that this definition is equivalent to the orthogonal group is the group of all matrices that preserve the dot product.

Note that:and for that to be true for all possible $x$ and $y$ then we must have:i.e. the matrix inverse is equal to the transpose.

Conversely, if:is true, then

These matricese are called the orthogonal matrices.

TODO is there any more intuitive way to think about this?

Or equivalently, the set of rows is orthonormal, and so is the set of columns. TODO proof that it is equivalent to the orthogonal group is the group of all matrices that preserve the dot product.

The Taylor series expansion is the most direct definition of the expontial as it obviously satisfies the exponential function differential equation:

The basic intuition for this is to start from the origin and make small changes to the function based on its known derivative at the origin.

More precisely, we know that for any base b, exponentiation satisfies:And we also know that for $b=e$ in particular that we satisfy the exponential function differential equation and so:One interesting fact is that the only thing we use from the exponential function differential equation is the value around $x=0$, which is quite little information! This idea is basically what is behind the importance of the ralationship between Lie group-Lie algebra correspondence via the exponential map. In the more general settings of groups and manifolds, restricting ourselves to be near the origin is a huge advantage.

Now suppose that we want to calculate $e^1$. The idea is to start from $e^0$ and then then to use the first order of the Taylor series to extend the known value of $e^0$ to $e^1$.

E.g., if we split into 2 parts, we know that:or in three parts:so we can just use arbitrarily many parts $e^{1/n}$ that are arbitrarily close to $x=0$:and more generally for any $x$ we have:

Let's see what happens with the Taylor series. We have near $y=0$ in little-o notation:Therefore, for $y=x/n$, which is near $y=0$ for any fixed $x$:and therefore:which is basically the formula tha we wanted. We just have to convince ourselves that at ${\lim }_{n\to \infty }$, the $o(1/n)$ disappears, i.e.:

To do that, let's multiply $e^y$ by itself once:and multiplying a third time:TODO conclude.

The dot product is a positive definite matrix, and so we see that those will have an important link to familiar geometry.

The canonical undecidable problem.

One of the Holiest age old debugging techniques!

Git has some helpers to help you achieve bisection Nirvana: stackoverflow.com/questions/4713088/how-to-use-git-bisect/22592593#22592593

Obviously not restricted to software engineering alone, and used in all areas of engineering, e.g. Video "Air-tight vs. Vacuum-tight by AlphaPhoenix (2020)" uses it in vacuum engineering.

The cool thing about bisection is that it is a brainless process: unlike when using a debugger, you don't have to understand anything about the system, and it incredibly narrows down the problem cause for you. Not having to think is great!