Student "edition" found at {csi dot journalspace dot com}.

Maybe I shouldn't have started this blog now, not with everything that's been going on.

On the first day of the tenth week of classes for the first term, in my Mathematical Methods One lecture, I started on the supposedly supplementary topic in the textbook that is directly following remainder theorem and synthetic division: partial fractions.

Since I am using this even in my DIFEREQ class, I decided that these students, all of whom will be taking up Differential Calculus, will be able to use it, just like David envisioned when we both taught this same subject two terms ago.

I started out with the simplest cases, fractions (with the degree of the numerator being less than that of the denominator, of course) with the numerator of degree zero, and there only being two factors in the denominator, both of degree one and both unique.

After that was the case where the numerator is of degree one, and there were three factors in the denominator, all of degree one and all unique.

I also went into when the factors in the denominator are not unique, but I did not include that in their exercise and said we would have more examples during the next meeting.

In my class afterwards, I took up differential equations of degree one solved by inspection of frequently used integrating factors. This included d(xy) = xdy + ydx, d(y/x) = (xdy – ydx)/x^2 and d(x/y) = (ydx – xdy)/y^2.

For this one, I used what I believe is a simpler method than the one outlined in the book, where, reverse of what we had done in the previous meetings, u is substituted for xy or x/y and du for xdy + ydx instead of the other way around.

The results were the same anyway. I told them we would have our next exam two meetings later, since we had covered two more methods, and because the more exams we had, the more points they could add to their accumulation.

In my introduction to electricity and magnetism class succeeding that, I gave them a more complicated example of Kirchhoff’s Rules than we had in the lab, which only had two nodes (of three connections each), three branches (and thus three unknown currents), and three possible loops.

This circuit looked like an L made up of three squares with seven resistors and two voltage sources (more could be added of both if needed). Thus it had two nodes of three connections each and one node of four connections. There were five branches and five currents, and six possible loops.

I will continue on how I solved this next time. The final answer for session 692 will be boxed at this point. Class dismissed.