more git-fu: reset –hard after force push

April 10, 2025 Linux No comments , , ,

When I have a branch that I’m sharing only between myself, it’s occasionally helpful to force push and then reset all the other repo versions of the branch to the new one.  Yes, I know that’s inherently dangerous, and I’ve screwed up doing this a couple times.

Here’s one way to do it:

# alias gitbranch='git branch --show-current'
git fetch
git reset origin/`gitbranch` --hard

When I do this, I always first check that the branch on the repo in question is pushed to the upstream branch (i.e., I can consider the current state discardable), and sometimes I’ll also do a sanity check diff against the commit hash that came with the fetch to see what I’m going to get with the reset.

Assuming that I still want to be stupid and dangerous, is there an easier way to do it?

I asked Grok if there was an short cut for this action, and as expected, there’s a nice one:

git fetch
git reset @{u} --hard

Here the @{u} is a shorthand for the current upstream branch.

Grok suggested a one liner using && for the two commands, but when I’m being dangerous and (possibly) stupid I’d rather be more deliberate and do these separately, so that I can think between the two steps.

git diff against previous

March 30, 2025 Linux , , , ,

I’ve been using a simple home grown script (gitdiffc) to generate git diffs against the previous version for all files in a given commit.  It would run something like:

# git diff \
$(git log --pretty=format:%h -2 --reverse d86cc4b38db | tr "\n" " ")

The git log part, just gives two commit IDs. Example:

# git log --pretty=format:%h -2 --reverse d86cc4b38db
38aeb009b21
d86cc4b38db

I have a few files that I have to manually merge, as there are two many cherry-pick conflicts, so I was collecting information for that merge.  That included extracting the version that I would have wanted to cherry-pick, since :

git show d86cc4b38db:foo.cc > foo.cc.show

I also wanted a diff, of just that file, in the branch I am trying to pick from, but wasn’t sure how to get that easily without looking up the log, like I did in my ‘gitdiffc’ script that ran the log and diff command above.

I figured there had to be an easier way. Shout out to Grok, which pointed out that I can just use:

git diff d86cc4b38db^ d86cc4b38db -- foo.cc

This also means that I can simplify my ‘gitdiffc’ script so that it just runs:

# git diff d86cc4b38db^..d86cc4b38db

eliminating the nested git log command. That is simple enough that I don’t even need a script to remember how to do it. This should have been obvious, since I’ve used HEAD^ just like that so many times (i.e.: git reset HEAD^, to break up the last commit.)

Weighted geometric series

March 23, 2025 math and physics play , , ,

[Click here for a PDF version of this post]

Karl needed to evaluate the sum:

\begin{equation}\label{eqn:weightedGeometric:20}
S = \sum_{k = 0}^9 \frac{a + b k}{\lr{ 1 + i }^k}
\end{equation}

He ended up using a spreadsheet, which was a quick and effective way to deal with the problem. I was curious about this sum, since he asked me how to sum it symbolically, and I didn’t know.

Mathematica doesn’t have any problem with it, as seen in fig. 1.

fig. 1. A funky sum.

How can we figure this out?

Let’s write \( r = 1/(1+i) \) to start with, and break up the sum into constituent parts
\begin{equation}\label{eqn:weightedGeometric:40}
\begin{aligned}
S_n
&= \sum_{k = 0}^n \frac{a + b k}{\lr{ 1 + i }^k} \\
&= a \sum_{k = 0}^n r^k + b \sum_{k = 0}^n k r^k.
\end{aligned}
\end{equation}
We can evaluate the geometric part of this easily using the usual trick. Let
\begin{equation}\label{eqn:weightedGeometric:60}
T_n = \sum_{k = 0}^n r^k,
\end{equation}
then
\begin{equation}\label{eqn:weightedGeometric:80}
r T_n – T_n = r^{n+1} – 1,
\end{equation}
so
\begin{equation}\label{eqn:weightedGeometric:100}
T_n = \frac{r^{n+1} – 1}{r – 1}.
\end{equation}
Now we just have to figure out how to sum
\begin{equation}\label{eqn:weightedGeometric:120}
G_n = \sum_{k = 0}^n k r^k = \sum_{k = 1}^n k r^k.
\end{equation}
This looks suspiciously like the derivative of a geometric series. Let’s evaluate such a derivative, as a function of r:
\begin{equation}\label{eqn:weightedGeometric:140}
\begin{aligned}
\frac{d T_n(r)}{dr}
&= \sum_{k = 0}^n \frac{d}{dr} r^k \\
&= \sum_{k = 0}^n k r^{k-1} \\
&= \sum_{k = 1}^n k r^{k-1} \\
&= \inv{r} \sum_{k = 1}^n k r^k.
\end{aligned}
\end{equation}
Having summed the geometric series, we may also take the derivative of that summed result
\begin{equation}\label{eqn:weightedGeometric:160}
\begin{aligned}
\frac{d T_n(r)}{dr}
&= \frac{d}{dr} \lr{ \frac{r^{n+1} – 1}{r – 1} } \\
&= \frac{(n+1)r^n}{r – 1} – \frac{r^{n+1} – 1}{\lr{r -1}^2} \\
&= \frac{(n+1)r^n(r-1) – \lr{r^{n+1} – 1}}{\lr{r -1}^2}.
\end{aligned}
\end{equation}
Putting the pieces together, we have
\begin{equation}\label{eqn:weightedGeometric:180}
T_n = \frac{r}{\lr{r -1}^2} \lr{ (n+1)r^n(r-1) – \lr{r^{n+1} – 1} }.
\end{equation}

This means that our sum is
\begin{equation}\label{eqn:weightedGeometric:200}
S_n = a \frac{r^{n+1} – 1}{r – 1} + b \frac{r}{\lr{r -1}^2} \lr{ (n+1)r^n(r-1) – \lr{r^{n+1} – 1} }.
\end{equation}
Putting back \( r = 1/(1+i) \), and subsequent simplification, gives the Mathematica result. It’s not pretty, but at least we can do it if we want to.

Shout out to Grok that pointed out the derivative trick for the second series.  I’d forgotten that one.

Wonderful Life=Giving Radium

March 22, 2025 Incoherent ramblings

From “The Wide World” Magazine, June 1905 edition

“This remarkable substance is absolute and quick death to the germs of cancer, tumor, consumption, malaria, blood poison, ulcers and all forms of existing disease.  When it enters the system every vestige of disease is driven out, as no germ can live in its presence.”

The amusing thing is that the idea of this advertisement has a close analogue in modern medicine, as chemotherapy is essentially the same idea.  Chemotherapy is also a gross system wide attack on the body, a desperate hope that it can be used to kill off the cancer cells faster than the rest of the cellular structure of the body.  It’s the kind of treatment that future Bones, in Star Trek IV, will eventually be describing as “What is this, the dark ages?”

I have three of these “Wide World” magazines that I got in a “unknown” box of pulp fiction magazines from a second hand store when I was a kid (Argosy, Blue Boy, …).  The ads from these magazines are the best part.  I think I’ve posted a few more of them on my old blog, but I should systematically go through them for archival purposes, and share them all.

Impedance refresher.

March 21, 2025 math and physics play , , , , , , , , , , , ,

[Click here for a PDF version of this post]

Karl is taking his circuits course right now, which means that I get a chance to field some questions. I don’t get an excuse to think about this stuff any more. It’s fun material, since most of the ideas are all really simple, and you can figure out everything from first principles.

Karl just started sinusoidal circuits, which I think is a bit exciting. They are such a nice special case, as complex calculations are all effectively reduced to \( V = I R \) style computations.

Solving RLC circuits for general time dependent sources.

To contrast the simple case of sinusoidal sources, let’s consider what we have to do in order to solve a general case RLC circuit. The simple basic RC circuit sketched in fig. 1 provides a good illustrative example, even though it does not include any inductance.

With \( v \) as the voltage at the capacitor, the equations that describe the circuit are
\begin{equation}\label{eqn:impedance:20}
\begin{aligned}
v_s – v &= i R \\
i &= C \frac{dv}{dt}.
\end{aligned}
\end{equation}

We can combine these into one equation for \( v \). Letting \( \tau = RC \), that is
\begin{equation}\label{eqn:impedance:40}
v + \tau \frac{dv}{dt} = v_s.
\end{equation}
Here \( v_s = v_s(t) \) can be an arbitrary function of time. This is a simple enough differential equation, and can probably be solved in various ways (integrating factors, Fourier transforms, Laplace transforms, …)

For illustration purposes, let’s tackle this little equation with Fourier transforms, a method logically equivalent to the computation of the Green’s function for the system.

Let’s use a symmetric representation of the Fourier transform
\begin{equation}\label{eqn:impedance:60}
\begin{aligned}
F(\omega) &= \inv{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-j\omega t} f(t) dt \\
f(t) &= \inv{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{j\omega t} F(\omega) d\omega \\
\end{aligned}
\end{equation}
Recall that the Fourier transform of the derivative is just a \( j \omega \) scaled frequency domain function, which we show with integration by parts
\begin{equation}\label{eqn:impedance:80}
\begin{aligned}
\inv{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-j\omega t} \frac{df}{dt} dt
&= \inv{\sqrt{2 \pi}} \int_{-\infty}^\infty \lr{ \frac{d}{dt} \lr{ f(t) e^{-j \omega t} } – f(t) \frac{d}{dt}\lr{ e^{-j\omega t}} } dt \\
&= j \omega F(\omega).
\end{aligned}
\end{equation}
That means that the frequency domain equivalent of our system is
\begin{equation}\label{eqn:impedance:100}
V + j \omega \tau V = V_s,
\end{equation}
or
\begin{equation}\label{eqn:impedance:120}
V(\omega) = \frac{V_s(\omega)}{1 + j \omega \tau}.
\end{equation}
Inverse transformation yields
\begin{equation}\label{eqn:impedance:140}
\begin{aligned}
v(t)
&= \inv{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{j\omega t} \frac{V_s(\omega)}{1 + j \omega \tau} d\omega \\
&= \inv{2 \pi} \iint_{-\infty}^\infty e^{j\omega t} \frac{1}{1 + j \omega \tau} d\omega e^{-j\omega t’} v_s(t’) dt’ \\
&= \int_{-\infty}^\infty dt’ v_s(t’) \inv{2\pi} \int_{-\infty}^\infty \frac{e^{j\omega(t-t’)}}{1 + j \omega \tau} d\omega,
\end{aligned}
\end{equation}
or with
\begin{equation}\label{eqn:impedance:160}
G(u) = \inv{2\pi} \int_{-\infty}^\infty \frac{e^{j\omega u}}{1 + j \omega \tau} d\omega,
\end{equation}
\begin{equation}\label{eqn:impedance:180}
v(t) = \int_{-\infty}^\infty v_s(t’) G(t – t’) dt’.
\end{equation}
We just need to evaluate the Green’s function \( G(u) \) to proceed, which we can do with standard contour integration, first writing:

\begin{equation}\label{eqn:impedance:200}
\begin{aligned}
G(u)
&= \inv{2\pi j \tau} \int_{-\infty}^\infty \frac{e^{j\omega u}}{\inv{j\tau} + \omega} d\omega \\
&= \inv{2\pi j \tau} \oint \frac{e^{j z u}}{\inv{j\tau} + z} dz.
\end{aligned}
\end{equation}
This has a single pole at \( z = j/\tau \). We need an infinite semicircular contour in the lower half plane for \( u < 0 \), and can use the upper half plane infinite semicircular contour (surrounding the pole) for \( u > 0 \). That gives
\begin{equation}\label{eqn:impedance:220}
\begin{aligned}
G(u)
&= \Theta(u) \frac{2 \pi j}{2\pi j \tau} \evalbar{e^{j z u}}{z = j/\tau} \\
&= \frac{\Theta(u)}{\tau} e^{- u/\tau}.
\end{aligned}
\end{equation}

The solution to the problem, for any Fourier integrable source \( v_s(t) \), is
\begin{equation}\label{eqn:impedance:240}
\boxed{
v(t) = \int_{-\infty}^t v_s(t’) \frac{e^{- \lr{t – t’}/\tau}}{\tau} dt’.
}
\end{equation}

As a check, let’s evaluate this convolution integral for a step source \( v_s(t) = V \Theta(t) \), to find
\begin{equation}\label{eqn:impedance:260}
\begin{aligned}
v(t)
&= \frac{V e^{-t/tau}}{\tau} \int_0^t e^{t’/\tau} dt’ \\
&= V e^{-t/tau} \evalrange{e^{t’/\tau} }{0}{t} \\
&= V e^{-t/tau} \lr{ e^{t/\tau} – 1 } \\
&= V \lr{ 1 – e^{-t/\tau} }.
\end{aligned}
\end{equation}
This is the damped time domain response that we remember for an RC circuit. In Karl’s first year engineering notes, this was presented as a given (without the step factor), and he had to verify that it worked by differentiation (for \( t > 0 \).)

Solving this exactly, even for arbitrary sources, as we’ve done above, is not strictly hard, if you have all the required tools. But the first year engineering student doesn’t have all those tools to start with. This is where the beauty of the phasor techniques for sinusoidal sources comes in. Let’s now see how that works.

Phasor approach.

Let’s consider the three simplest RLC circuits, each with just a single element, and a variable voltage source. I’ll depict that element with a box as in fig. 2.

Resistor case.

If the element is a resistor with value \( R \), our equations are simple
\begin{equation}\label{eqn:impedance:280}
v_s = i R.
\end{equation}
Clearly \( i \) is directly proportional to the source voltage. In particular, if \( v_s(t) \) has a sinusoidal character, such as
\begin{equation}\label{eqn:impedance:300}
v_s(t) = V \cos(\omega t),
\end{equation}
then
\begin{equation}\label{eqn:impedance:320}
i(t) = \frac{V}{R} \cos\lr{ \omega t }.
\end{equation}
In particular, if we let \( i(t) = I \cos\lr{ \omega t } \), then we have
\begin{equation}\label{eqn:impedance:340}
I = \frac{V}{R},
\end{equation}
or \( V = I R \).

Capacitor case.

If the load element is a capacitor with capacitance \( C \), then the equation for the system is
\begin{equation}\label{eqn:impedance:360}
i = C \frac{dv_s(t)}{dt}.
\end{equation}
If we just plug in \( v_s(t) = V \cos(\omega t) \), as before, we get a bit of a mess
\begin{equation}\label{eqn:impedance:380}
i = -C \omega V \sin\lr{ \omega t }.
\end{equation}
We no longer have a nice simple proportionality relationship between the current and the voltage source, as the capacitor has introduced a phase shift into the mix.
We can figure out that phase factor by solving the equation
\begin{equation}\label{eqn:impedance:400}
-\sin x = \cos\lr{ x + \phi }.
\end{equation}
The easiest way to solve this is to express the sinusoids in complex exponential form
\begin{equation}\label{eqn:impedance:420}
\textrm{Re} \lr{ e^{j + \phi} } = \Real \lr{ j e^{j x} } = \Real \lr{ e^{j \pi/2} e^{j x} }.
\end{equation}
We see that the phase factor is a \( \pi/2 \) shift. However, even better, we have a strong hint that working with complex exponentials may be a better approach to formulating the problem.

Let’s write
\begin{equation}\label{eqn:impedance:440}
v_s(t) = V \cos\lr{ \omega t } = \textrm{Re} \lr{ V e^{j \omega t} }.
\end{equation}
Then we have
\begin{equation}\label{eqn:impedance:460}
i(t) = C V \textrm{Re} \lr{ \frac{d}{dt} e^{j \omega t} }.
\end{equation}
If we also assume that we can write
\begin{equation}\label{eqn:impedance:480}
i(t) = \textrm{Re} \lr{ I e^{j \omega t} },
\end{equation}
then if the real parts are equal, we must also have
\begin{equation}\label{eqn:impedance:500}
I e^{j \omega t} = j \omega C V e^{j \omega t},
\end{equation}
or
\begin{equation}\label{eqn:impedance:520}
I = j \omega C V.
\end{equation}
We have a \( V = I R \) relationship, which we write as
\begin{equation}\label{eqn:impedance:540}
V = I Z,
\end{equation}
where
\begin{equation}\label{eqn:impedance:560}
Z = \inv{j \omega C}.
\end{equation}
This is the phasor description of the circuit.

Inductive case.

If the circuit has an inductive load, then the system equation is
\begin{equation}\label{eqn:impedance:580}
v_s(t) = L \frac{di}{dt}.
\end{equation}
Again, we can write \( v_s(t) = \textrm{Re} \lr{ V e^{j\omega t} } \), and assume that \( i = \Real \lr{ I e^{j\omega t} } \). We then require
\begin{equation}\label{eqn:impedance:600}
V e^{j \omega t} = L \frac{d}{dt} \lr{ I e^{j \omega t} },
\end{equation}
or
\begin{equation}\label{eqn:impedance:620}
V = j \omega L I.
\end{equation}
We write
\begin{equation}\label{eqn:impedance:640}
Z = j \omega L,
\end{equation}
so once again \( V = I Z \).

Solving a more complex RLC configuration.

An example of a more complicated RLC circuit is sketched in fig. 3.

Here we have two impedances in parallel
\begin{equation}\label{eqn:impedance:660}
\begin{aligned}
Z_C &= \inv{j \omega C} \\
Z_L &= j \omega L.
\end{aligned}
\end{equation}
The parallel impedance through that reactive load is
\begin{equation}\label{eqn:impedance:680}
\begin{aligned}
Z
&= \lr{ \inv{Z_C} + \inv{Z_L} }^{-1} \\
&= \lr{ j \omega C + \inv{ j \omega L } }^{-1}.
\end{aligned}
\end{equation}
We can compute the current through \( R \) now
\begin{equation}\label{eqn:impedance:700}
I = \frac{V_s}{ R + \lr{ j \omega C + \inv{ j \omega L } }^{-1} }.
\end{equation}
We also have
\begin{equation}\label{eqn:impedance:720}
\frac{V_s – V}{R} = I,
\end{equation}
or
\begin{equation}\label{eqn:impedance:740}
\begin{aligned}
V
&= V_s – I R \\
&= V_s \lr{ 1 – \frac{R}{ R + \lr{ j \omega C + \inv{ j \omega L } }^{-1} } } \\
&= V_s \frac{\lr{ j \omega C + \inv{ j \omega L } }^{-1} }{ R + \lr{ j \omega C + \inv{ j \omega L } }^{-1} } \\
&= \frac{V_s}{ R \lr{ j \omega C + \inv{ j \omega L } } + 1 }.
\end{aligned}
\end{equation}
The complicated time response for this system is reduced to a trivial voltage divider calculation. We see that it was kind of pointless to run an inductor and capacitor in parallel, as they are both purely reactive (imaginary). That’s a detail that I didn’t remember, since it’s been decades since I did any practical circuits applications. However, the point is, by using a complex exponential source representation, these types of systems are reduced from systems of coupled differential equations to simple linear systems. Imagine how messy it would be to try to solve this system using the Green’s function methods that we used above!