Recent content by tnich

I think the point of the change of variables is to get rid of the ##mgx## term in the Hamiltonian by completing the square. Note that there is no linear ##y## term in the resulting Hamiltonian. As to how to do the change of variables, try this: 1) Solve ##y = x+\frac{mg} {mω^2} = x+\frac g ω^2##...

Yes, you are right, and that has been the point all along. $$\begin{align}&\bar D \bar E+ \bar B \bar C \bar D + && \bar ACDE + BCD+\bar B\bar CD\bar E\nonumber\\ =&\bar D \bar E+ \bar B \bar C \bar D+\bar B \bar C \bar D \bar E + && \bar ACDE + BCD+\bar B\bar CD\bar E\nonumber\\ =&\bar D \bar...

##\bar B \bar C \bar D+\bar B \bar C \bar D\bar E=\bar B \bar C \bar D(1+\bar E)=\bar B \bar C \bar D(1)=\bar B \bar C \bar D##

Circled in blue

Also, I wonder why you are including your random walk process in the measurement equation rather than in the state propagation equation?

You must be looking at a different diagram than the one you posted. The one I see has ##1##s in the squares for ##B=C=D=E=0##. You don't seem to have any argument with the ## \bar B \bar C \bar D## term since you have it in your solution. But ## \bar B \bar C \bar D + \bar B \bar C \bar D\bar E...

It looks to me as though you are trying to add quantities with incommensurate units. ##g## has units of acceleration, and ##v## has units that are neither velocity or acceleration. That suggests that there is an error somewhere in the derivation of the ##-g+v## term. Maybe you have assumed that...

Your result is equivalent to the solution in the book since ##\bar B \bar C D\bar E + \bar B \bar C \bar D\bar E = \bar B \bar C \bar E##. I would choose the book's solution over yours since it has one less AND operation.

There are several substitutions you need to make to find the inverse LaPlace transform of this, and I think you have applied one of them incorrectly: ##sin(\omega t) \rightarrow \frac {\omega}{s^2+\omega^2}##.

That's an interesting point. So if less mass is rolling, the moment of inertia is smaller, and the can of liquid picks up speed more quickly, at least to start. I think that effect would be very brief, though.

For what it's worth, I found E&M (even with a good professor) a lot more difficult than Linear Algebra.

If you use similar cans and fill them with different materials, the moment of inertia of the can itself will be the same. Assuming it has a uniform density, the moment of inertia of the food inside is proportional to its weight. If you put the two together, a denser substance inside the can will...

Generally, the less turbulent motion within the can, the faster it will roll. So a can full of cranberry sauce should roll faster than a can full of chicken broth, and the chicken broth should move faster than a can of chicken noodle soup. Things moving inside the can do rub against each other...

The recurrence relation for Chebyshev polynomials of the first kind is ##T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x)##. So to calculate ##T_{n}(x)##, you just need a loop. Start with ##T_0=1##, and ##T_1=x## and iterate until you get ##T_n##.

Suppose you bet amount ##2^{k-1}b## each time. If you stop betting after the 15th bet, then you lose a total of ##(2^{15}-1)b = 32767b##, that is 32767 times your initial bet ##b##. On the other hand, if you win at the ##k##th bet for ##k<15##, your total winnings are exactly ##b##. Supposing...