On Picking Stocks and Mutual Funds: An Experimental Proposal

[Important disclaimer: I am not a registered investment advisor. If you choose to try the experiment described below, which is shared with everyone for free, the risk is yours alone.]

Lots of people have jumped into the stock market lately, many of them with little experience. That isn’t stopping them, though; they’re creating Robinhood accounts and trading penny stocks at a blinding rate. I’m writing this to propose an experiment individuals can use to evaluate their own skills at picking stocks and mutual funds. This experiment does require some experience investing, but not much.

First, divide your investment capital into three parts of equal size. With the first part, buy either QQQ or SPY, or equal amounts of each. These are ETFs which track well-known stock indices: the technology-heavy NASDAQ 100 and the broader S&P 500, respectively. This chart shows how the S&P 500 has performed over the last five years.

stock chart

This first third of the investment money is analogous to a control group; it is the standard by which we measure the performance of the other two-thirds.

The second third is to be invested in no-load, open-ended, actively-managed mutual funds. You should try to choose the best ones you can find. Two that I’ve bought shares of, myself, are TRBCX (a fund which invests mostly large-cap blue-chip stocks), and WAMCX (a fund investing in small-cap, rapidly-growing companies’ stocks).

The final third is made of stocks you pick and buy/hold/sell yourself, using whatever criteria you choose. If you run this experiment for six months (or longer), and this third is the largest at the end of this period, you have good evidence that your stock-picking methods work well, justifying continuing investing in this manner.

If, on the other hand, your hand-picked mutual funds do best, then the data supports the idea that you’re a good fund-picker, but should probably not invest in individual stocks.

Finally, what if SPY and/or QQQ do best? If that is the case, I wouldn’t try to beat the market with my own picks, but would simply invest in open-ended index funds, or ETFs.

I’m trying this experiment myself. So far, my hand-picked mutual funds are doing best. If the running of this experiment continues in this way, I’ll abandon the picking of individual stocks, in favor of mutual funds. Of course, it’s always possible to study and learn new things. If this running of the experiment yields the answer “funds,” there’s nothing stopping me from studying stock-picking some more, and then repeating the experiment to see if I get better results for the choosing of individual stocks. One could also abandon the experiment, but continue to invest in those particular stocks which did well.

There are some types of investment I haven’t mentioned here. Cash, money market funds, CDs, and bonds were not mentioned because interest rates are at historic lows, limiting the return one can get from such investments. As for derivatives, such as options or futures, those have not been included simply because I have no experience with derivatives. Traditional hedges against inflation, such as gold and silver, are not included because inflation is almost non-existent.

One last thing: Robinhood’s customers will have a hard time doing this experiment, for that company does not offer mutual funds. Schwab does, though; that’s the broker I use — and the one I recommend. You can find them online at http://www.schwab.com.

George Carlin, on Change

change machine

On numerous occasions, I have repeated this experiment, in keeping with the scientific method. I have obtained the same null result as Carlin obtained, each and every time.

An Asymmetrical Static Equilibrium Physics Problem Involving Pulleys and Hanging Masses

An interesting phenomenon in physics, and physics education, is the simplicity of symmetric situations, compared to the complexity of similar situations which are, instead, asymmetrical. Students generally learn the symmetrical versions first, such as this static equilibrium problem, with the hanging masses on both left and right equal. 

static equilibrium pulley setup

The problem is to find the measures of the three angles shown above, with values given for all three masses. Here is the setup, using physical objects, rather than a diagram.

100_170_100

The masses on the left and right are each 100 g, or 0.100 kg, while the central masses total 170 g, or 0.170 kg. Since all hanging masses are in static equilibrium, the forces pulling at the central point (at the common vertex of angles λ, θ, and ρ) must be balanced. Specifically, downward tension in the strings must be balanced by upward tension, and the same is true of tension forces to the left and to the right. In the diagram below (deliberately asymmetrical, since that’s coming soon), these forces are shown, along with the vertical and horizontal components of the tension forces held in the diagonal strings.

static equilibrium pulley setup force diagram

Because the horizontal forces are in balance, Tlx = Trx, so Mlgcosλ = Mrgcosρ — which is not useful now, but it will become important later. In the symmetrical situation, all that is really needed to solve the problem is the fact that the vertical forces are in balance. For this reason, Tc = Tly + Try, so Mcg = Mlgsinλ + Mrgsinρ. Since, due to symmetry, Ml = Mand λ = ρ,  Mr may be substituted for Ml, and ρ may be substituted for  λ, in the previous equation Mcg = Mlgsinλ + Mrgsinρ, yielding Mcg = Mrgsinρ + Mrgsinρ, which simplifies to Mcg =2Mrgsinρ. Cancelling “g” from each side, and substituting in the actual masses used, this becomes 0.170 kg = 2(0.100 kg)sinρ, which simplifies to 0.170 kg = (0.200 kg)sinρ, then 0.170/0.200 = sinρ. Therefore, angle ρ = sin-1(0.170/0.200) = 58°, which, by symmetry, must also equal λ. Since all three angles add up to 180º, the central angle θ = 180° – 58° – 58° = 64°. These answers can then be checked against the physical apparatus.

53_theta_53

When actually checked with a protractor, the angles on left and right are each about 53° — which is off from the predicted value of 58° by about 9%. The central angle, of course, is larger, at [180 – (2)53]° = 74°, to make up the difference in the two smaller angles. The error here could be caused by several factors, such as the mass of the string itself (neglected in the calculations above), friction in the pulleys, or possibly the fact that the pulleys did not hang straight down from the hooks which held them, but hung instead at a slight diagonal, as can be seen in the second image in this post. This is testable, of course, by using thinner, less massive string, as well as rigidly-fixed, lower-friction pulleys. However, reducing the error in a lab experiment is not my objective here — it is, rather, use of a simple change to turn a relatively easy problem into one which is much more challenging to solve. 

In this case, the simple change I am choosing is to add 50 grams to the 100 g already on the right side, while leaving the central and left sides unchanged. This causes the angles where the strings meet to change, until the situation is once more in static equilibrium, with both horizontal and vertical forces balanced. With the mass on the left remaining at 0.100 kg, the central mass at 0.170 kg, and the mass on the right now 0.150 kg, what was an easy static equilibrium problem (finding the same three angles) becomes a formidable challenge. 

100_170_150

For the same reasons as before (balancing forces), it remains true that Mlgcosλ = Mrgcosρ (force left = force right), and, this time, that equation will be needed. It also remains true that Mcg = Mlgsinλ + Mrgsinρ (downward force = sum of the two upward forces). The increased difficulty is caused by the newly-introduced asymmetry, for now Ml ≠ Mr, and λ ≠ ρ as well. It remains true, of course, that  λ + θ + ρ° = 180.

In both the vertical and horizontal equations, “g,” the acceleration due to gravity, cancels, so Mlgcosλ = Mrgcosρ becomes Mlcosλ = Mrcosρ, and Mcg = Mlgsinλ + Mrgsinρ becomes Mc = Mlsinλ + Mrsinρ. The simplified horizontal equation, Mlcosλ = Mrcosρ, becomes Ml²cos²λ = Mr²cos²ρ when both sides are squared, in order to set up a substitution based on the trigonometric identity, which works for any angle φ, which states that sin²φ + cos²φ = 1. Rearranged to solve it for cos²φ, this identity states that  cos²φ = 1 – sin²φ. Using this rearranged identity to make substitutions on both sides of the previous equation Ml²cos²λ = Mr²cos²ρ yields the new equation Ml²(1 – sin²λ) = Mr²(1 – sin²ρ). Applying the distributive property yields the equation Ml² – Ml²sin²λ = Mr² – Mr²sin²ρ. By addition, this then becomes -Ml²sin²λ = Mr² – Ml² – Mr²sin²ρ. Solving this for sin²λ turns it into sin²λ = (Mr² – Ml² – Mr²sin²ρ)/(-Ml²).

Next, Mc = Mlsinλ + Mrsinρ (the simplied version of the vertical-force-balance equation, from above), when solved for sinλ, becomes  sinλ = (Mrsinρ – Mc)/(- Ml). Squaring both sides of this equation turns it into sin²λ = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(- Ml.

There are now two equations solved for sin²λ, each shown in bold at the end of one of the previous two paragraphs. Setting the two expressions shown equal to sin²λ equal to each other yields the new equation (Mr² – Ml² – Mr²sin²ρ)/(-Ml²) = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(- Ml)², which then becomes (Mr² – Ml² – Mr²sin²ρ)/(-Ml²) = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(Ml)², and then, by multiplying both sides by -Ml², this simplifies to Mr² – Ml² – Mr²sin²ρ = – (Mr²sin²ρ – 2MrMcsinρ + Mc²), and then Mr² – Ml² – Mr²sin²ρ = – Mr²sin²ρ + 2MrMcsinρ – Mc². Since this equation has the term – Mr²sin²ρ on both sides, cancelling it simplifies this to  Mr² – Ml² = 2MrMcsinρ – Mc², which then becomes Mr² – Ml² + Mc² = 2MrMcsinρ, and then sinρ = (Mr² – Ml² + Mc²)/2MrM= [(0.150 kg)² – (0.100 kg)² + (0.170 kg)²]/[2(0.150 kg)(0.170 kg)] = (0.0225 – 0.0100 + 0.0289)/0.0510 = 0.0414/0.510 = 0.812. The inverse sine of this value gives us ρ = 54°.

Having one angle’s measure, of course, makes it far easier to find the others. Two paragraphs up, an equation in italics stated that sinλ = (Mrsinρ – Mc)/(- Ml). It follows that λ = sin-1[(Mrsinρ – Mc)/(- Ml)] = sin-1[(0.150kg)sinρ – 0.170kg)/(-0.100kg)] = 29°. These two angles sum to 83°, leaving 180° – 83° = 97° as the value of θ.

31_theta_58.png

As can be seen above, these derived values are close to demonstrated experimental values. The first angle found, ρ, measures ~58°, which differs from the theoretical value of 54° by approximately 7%. The next, λ, measures ~31°, also differing from the theoretical value, 29°, by about 7%.The experimental value for θ is (180 – 58 – 31)° = 91°, which is off from the theoretical value of 97° by ~6%. All of these errors are smaller than the 9% error found for both λ and ρ in the easier, symmetrical version of this problem, and the causes of this error should be the same as before.

Beginning the Fractiles-7 Refrigerator Experiment

To begin this experiment, I first purchased two refrigerator-sized Fractiles-7 sets (available at http://fractiles.com/), and then, early on a Sunday, quietly arranged these rhombus-shaped magnets on the refrigerator in our apartment (population: 4, which includes two math teachers and two teenagers), using a very simple pattern.

160207_0000

Here’s a close-up of the center. There are 32 each, of three types of rhombus., in this double-set, for a total of 96 rhombic magnets, all with the same edge length.

160207_0001

The number of possible arrangements of these rhombi is far greater than the population of Earth.

The next step of the experiment is simple. I wait, and see what happens.

It should be noted that there is a limit on how long I can wait before my inner mathematical drives compel me to play with these magnets more, myself — but I do not yet know the extent of that limit.

Constructing “Nightday” — An Experiment Involving Sleep

Waning-Gibbous-Moon

The last workweek having left me rather tired, I went to bed early Friday, after work, and then, having slept all I could sleep, I then woke back up quite early Saturday morning, before sunrise, and couldn’t get back to sleep. I was tired all day Saturday, but not too weary to think. What I thought was simple: as tired as I am, it sure would be nice to have a three-day weekend this weekend. Next, I thought, yeah, this would be nice, but that won’t make it happen. Finally, I realized that I actually could, perhaps, come up with some hopefully-clever and effective way to get the three-day weekend I want . . . and, having had an idea to do exactly that, I’m trying it right now.

I’ve tested the 24-hour sleep/wake cycle before, trying to find ways to lengthen that period of time. (Ever wanted more hours in the day? Well, I actually tried to make that happen, once, but the results were less than successful.) This time, however, I’m not trying to get extra hours in a day, but an extra day in the weekend — by simply using shorter “days,” and thus making “room,” temporally, to add an extra sleep-period and wake-period into the weekend. So, Friday, I fell asleep around 5:00 pm, and did so without the prescribed medication (which includes sedatives) which I usually take at bedtime, since it wasn’t that late yet . . . so I simply fell asleep because I was tired.

Without the sedatives I am used to taking, of course, I didn’t stay asleep anything like a full eight hours, and instead “popped” back awake at around 9 pm, which was less than an hour ago, as I write this. Rather than sedating and returning to sleep, however, I took the other medication I am prescribed for this time of the day (such as that needed to regulate blood pressure), and then made my “morning” coffee, which I am enjoying now . . . to begin the extra “day” I’m attempting to add to this weekend, between Saturday and Sunday. My hypothesis is that I can deliberately alter my sleep/wake cycle in such a way that I have three (shorter) sleep/wake cycles in two calendar days, thus giving myself a three-day weekend, of a sort, and enjoy the benefits of a three-day weekend as a result. If, come Monday, I feel like I’ve had a three-day weekend — in that I feel unusually well-rested — I will consider this experiment to “create” a working illusion of a three-day weekend, without any actual extra time, to be a success (subject to the opinion of my doctors, to whom I will describe all of this).

I plan to stay awake until roughly dawn on Sunday, and then go to sleep until, well, whenever I wake up. I’ll then have a shortened post-sleep Sunday wakefulness-period, go to sleep at a reasonable hour Sunday night, and get a good, full night’s sleep then, before going to work on Monday.

Right now, therefore, I’m having the middle “day” of what feels, subjectively, like a three-day weekend, and having it at night, between what seems, now, like it was yesterday (the shortened Saturday), and what I anticipate as my shortened Sunday, after I sleep again, tomorrow. Since it’s easier to talk about this extra “day” I’m having tonight if I give it a name, I’m doing so: I’m calling it “Nightday.”

Some readers may object that I’ve merely come up with an overly-convoluted way to analyze a four-hour Saturday-afternoon nap. I’ll concede that they do have a point . . . but if my calling this “Nightday,” and telling myself that I’m enjoying a three-day weekend, actually turns out to help me feel and act more rested next week, then I’ll take those benefits and run with them, regardless of what any critics tell me (unless, of course, my doctors are among those critics). If this experiment has only beneficial results, and passes medical review, then I’ll likely use more Nightdays to get additional three-day weekends in the future, whenever I need, or simply want, them.

~~~

Important disclaimer: nothing in this blog-post should be taken as any form of medical advice, for I am not medically trained. I have taken the precaution of discussing my practice of occasionally inventing and conducting experiments such as this with my own physicians, and will continue to do so. No one should attempt to replicate this experiment without first consulting their own physician(s).

[Image credit: the photo of the Moon shown above was found here — https://photographylife.com/moon-waning-gibbous. It isn’t identical in appearance to the current waning gibbous Moon, having been photographed quite some time earlier, but it is close.]

An Ethical Dilemma Involving a Polyhedral Crystal

I just ordered a crystal rhombcuboctahedron on eBay because I like its geometrical properties, despite the mystical claims in the item listing. I did so with the full knowledge and expectation that these claims are almost certainly false, because, well, they’re mystical claims.

Rhombicubocta

Here’s my ethical dilemma: would it be ethical to lab-test those claims, then post the results in the feedback I leave?

[Image created using Stella 4d, available at http://www.software3d.com/Stella.php. This isn’t a picture of the crystal on eBay; it is made of quartz, and not rainbow-colored. It is of the same shape, however.]

Free the Frozen People!

150127_0000

After seeing this sign in a local grocery store, I carefully searched the entire frozen food section, but I could find neither the frozen Mexican, nor the frozen Asian. Since they were gone, but the sign indicates they were there at one point, I concluded that the experiment was over, and hoped they had thawed out both experimental test subjects, found them still healthy after a few days in cryogenic suspended animation, and sent them home, each with a fat check to compensate them for the huge risk they just took.

However, even with compensation and signed consent forms, I still have certain ethical reservations about scientists performing this sort of experiment on actual human beings. Why not freeze, thaw, refreeze, and rethaw mice, instead? Is PETA really that scary?

Are they still doing these experiments, in my town or elsewhere? If so . . . free the frozen people!

There is one last thing about this whole thing which I just can’t figure out, though, and that’s this: why were they storing their frozen, experimental, human test subjects in the middle of a central Arkansas grocery store in the first place?

An Experiment Involving Augmentation of Octahedra with More Octahedra, Etc.

I’m going to start this experiment with a single octahedron, with faces in two colors, placed so that two faces which share an edge are always of different colors.

1

Next, I will augment the red faces — and only the red faces — with identical octahedra.

2

The regions with four blue, adjacent faces look as though they might hold icosahedra — but I checked, and they don’t quite fit. I will therefore continue the same process — augmenting only the red faces with more octahedra of the original type.

3

I’ve now decided that I definitely like this game, so I’ll keep playing it.

4

Immediately above, at the fourth of these images, some of the octahedra have started to overlap slightly, but I’m choosing to not be bothered by that — I’m continuing the now-established pattern, just in order to see where it takes me.

5

The regions of overlap are now far more obvious, but I’m continuing, anyway. Why? Because this is fun, that’s why! Right now, Stella 4d, the program I use to do these polyhedral manipulations, is chugging away on the next one. (This program is avilable at http://www.software3.com/Stella.php.) Ah, it’s ready — here it is!

6

Rather than repeat this process again, I now have another question: what would the convex hull of this figure look like? (A convex hull of a non-convex polyhedron is the smallest convex polyhedron which can contain a given non-convex polyhedron.) With Stella 4d, that’s easily answered.

Convex hull

I must admit this: that was nothing like what I expected — but such unexpected discoveries are a large part of what makes these polyhedral investigations with Stella 4d so much fun. And now, to close this particular polyhedral journey, I will have Stella 4d produce, for me, the dual of the convex hull shown above. (In case you aren’t familiar with duality regarding polyhedra, it describes the relationship between the octahedron, with which this post began, and the familar cube. Basically, with duals, faces and verticies are “flipped” over edges, although that is an extremely informal and imprecise way to describe the at the process.)

dual of Convex hull

And with that, my friends, I bid you good night!

My Aqua Regia Story

This is my twentieth year teaching, but only the first year when I have not taught at least one class in chemistry, and I miss it. One of my fondest memories of chemistry lab involves the one time I experimented with aqua regia — a mixture of acids which, unlike any single acid, can dissolve both gold and platinum, the “noble metals.” I had read a story of a scientist’s gold Nobel Prize being protected from the Nazis by dissolving it in aqua regia, and then recovering the gold from solution after World War II had ended. Having read about this, I wanted to try it myself, and also thought it would make an excellent lab for classroom use — if I could figure out how to recover the gold, and also learn what precautions would be needed to allow high school students to perform this experiment safely. For sensible and obvious reasons, I conducted a “trial run” without students present, but with another chemistry teacher nearby, since aqua regia, and the gases it produces when dissolving gold, are quite dangerous. Someone else has put a video on YouTube, showing aqua regia dissolving gold, so you can see something much like what I saw, simply by watching this video.

First, I obtained one-tenth of a troy ounce of gold, which cost about $80 at the time. I had read about the extreme malleability of gold, one of the softest metals, and wanted to see evidence for it for myself — so, before I prepared the the aqua regia, I used a hammer to try flattening the gold sample into a thin sheet. That didn’t work, but it didn’t take long for me to figure out why — I had accidentally bought gold coin-alloy, which is 10% copper, not pure gold. Since this alloy is far less malleable than pure gold, my attempt to flatten it had failed, but I also knew this would not pose a problem for my primary experiment — the one involving aqua regia. Also, I didn’t have another spare $80 handy, to purchase another 1/10 troy ounce of pure gold, so I proceeded to make, for the first time in my life, a small amount of aqua regia — Latin for “royal water.”

Unlike what is shown in the video above, I prepared the acid-mixture first, before adding the gold, using a slightly-different recipe:  the traditional 1:4 ratio, by volume, of concentrated nitric acid to concentrated hydrochloric acid. Both these acids look (superficially) like water, but the mixture instantly turned yellow, and started fuming, even before anything was added to it. Wearing full protective gear, I watched it for a few minutes — and then, using tongs held by gloved hands, lowered my hammer-bashed sample of gold into the fuming, yellow mixture of concentrated acids.

It worked. It was a fascinating reaction, and a lot of fun to watch. At approximately the same time that the last of my gold sample dissolved, something occurred to me:  I had failed to research how to recover the dissolved gold from the resulting solution! No problem, I thought — I can figure this out. (I am seldom accused of lacking self-confidence, even when I’m wrong.)

My first idea was to use a single-replacement reaction. Many times, I have had students extract pure silver from a solution of silver nitrate by adding a more-active metal, such as copper. The copper dissolves, replacing the silver in the silver nitrate solution, and silver powder forms, as a precipitate, on the surface of the copper. Thinking that a similar process could be used to precipitate out the gold from my gold / aqua regia mixure, I simply added come copper to the reaction beaker. The corrosive properties of my aqua regia sample had not yet been exhausted, though, and so the remaining aqua regia simply “ate” the copper. The result was a mess — I had only succeeded in turning an already-complicated problem into an even-more-complicated problem, by adding more chemicals to the mixture. More attempts to turn the gold ions back into solid gold dust, using other chemicals, followed, but all of them failed. Finally, I used a strong base, sodium hydroxide, to neutralize the still-acidic mixture, and then, disgusted by my failure to recover the gold, found a way to safely dispose of the mixture, and did so.

In retrospect, I think I know where I messed up — I should have neutralized the remaining acids in the mixture with sodium hydroxide first, before adding copper to cause the gold to precipitate out, in a no-longer-acidic solution of ions with much less hydronium present. That, I think, will work, and I do intend to try it sometime — after doing more research first, to increase my level of certainty, and also after waiting for the current price of gold to drop to less-expensive levels. Right now, after all, a tenth of a troy ounce of gold costs roughly $120, not a mere $80.

As for the lost $80, I’m not upset about that anymore. I definitely learned things while doing this, and now view the $80 spent as simply the cost of tuition for an educational experience.