Comment by Double_Cast | Hacker News Reader

Comment by Double_Cast | Hacker News Reader

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> One thought that comes to mind is that I'm not convinced the bottom part of the triangle should be there. It implies a direction connection that I'm not sure exists.

The operator in question is actually quite common.

> Buying stocks. Suppose you buy $1000 worth of stocks each month, no matter the price (dollar cost averaging). You pay $25/share in Jan, $30/share in Feb, and $35/share in March. What was the average price paid? It is 3 / (1/25 + 1/30 + 1/35) = $29.43 (since you bought more at the lower price, and less at the more expensive one). And you have $3000 / 29.43 = 101.94 shares. The “workload” is a bit abstract — it’s turning dollars into shares. Some months use more dollars to buy a share than others, and in this case a high rate is bad.

https://betterexplained.com/articles/how-to-analyze-data-usi... (harmonic averages section)

The concept describes various processes which contribute towards identical workloads at different speeds (or rates).

I'm not sure how to relate what you said to my post? Wrong reply button?

> One thought that comes to mind is that I'm not convinced the bottom part of the triangle should be there. It implies a direction connection that I'm not sure exists.

I should have included this quote to begin with. I'll put it in now.

In any case. A direct connection does exist. If you follow the link, there are other examples where the (+) operator would be useful besides parallel resistance.

Now I see. Thanks.

I still think it implies a degree of symmetry that will confuse students, though, and that the notation can be further improved from what was proposed there. The operation in question is not trilaterally symmetric, and using a trilaterally symmetric symbol is probably misleading. The root symbol is arbitrary, but at least it doesn't promise nonexistent symmetries.

The root symbol (as a multi-valued function) returns the roots of unity (multiplied by a scalar). Which is like, the posterchild of imaginary symmetries. </pedantry>

More seriously. The selling point of the triangle operator is that it highlights isomorphisms between the three more traditional operators as a cyclic group. Which is a symmetry, just not the commutativity that students might naively expect. So I suppose it's a double-edged sword. I agree that it sucks as an operator. Nonetheless, the video could help a lot of students.

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