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June 3

How Many Triangles?

Problems like this are relatively easy to solve with a little planning and some brute force. I've been working on a similar problem, but with the medians drawn in for each of the sixteen smallest triangles. (Sorry, no image.) With so many triangles, it's much harder to develop a strategy that avoids double counting. Any thoughts about how I should approach this? Is there a general solution or strategy for problems like this? For what it's worth, I've found 482 so far. 24.1.231.240 (talk) 06:08, 3 June 2009 (UTC)[reply]

You could number the vertices v₁ to v_n and then for each vertex v_i, find all verticies v_j connected by a straight path where j > i, and then for each such pair, find any v_k which connects via a straight path to both v_i and v_j, where k > j, and where v_k are not colinear. This brute force method will give you all triangles and no duplicates. This approach is probably not practical as a manual method.

Another approach (one you may have already taken) would be to attempt to identify each uniquely sized and shapes triangle that could possibly exist in the structure, including rotations and reflections, and then for each one, pick a vertex and find all possible positions for that vertex within the structure. This more closely matches that approach typically used in the original problem and easier to perform manually. -- Tcncv (talk) 06:39, 3 June 2009 (UTC)[reply]

The first problem (for all size n of the triangle) is contained in the OEIS : [1]. Maybe your variant is there also: try introducing the number of triangles for small values of n. In any case a closed formula seems easy to obtain (but why should one do it? leave it where it is). --pma (talk) 17:58, 4 June 2009 (UTC)[reply]

Modelling "value" for used cars

I'm in the market for a used car, and I've been trying to figure out a simple score that (in a handwavey sense) captures something approximating "value". It's just a first approximation (naturally condition and history will affect the real value), but as the dealers are all rather far away (in vexingly diverse directions) this should hopefully allow me at least to reject the overly ambitious. Given the same model (and ignoring age) leaves mileage and asking price.

I've been calculating the difference between the new price of the model (it's still made) minus the asking price; call that the saving. A large saving is good, a small one bad. Then I'm simply dividing the saving by the mileage (as a big mileage is bad, a small one good). Assuming that car prices vary linearly with mileage (they don't, it's of a gentle exponential) does the logic of my calculation hold? That is, if V is (original-asking)/mileage, is it true to say that a higher V is (practical things aside) a better "value"? 87.115.17.103 (talk) 13:14, 3 June 2009 (UTC)[reply]

Mileage and price are nonnegative variables. So the approximate linear relationship may be between the logarithms of the mileage and price. Bo Jacoby (talk) 17:00, 3 June 2009 (UTC).[reply]

If the mileage is small then it doesn't matter. The OP did mention that the linear relationship is an approximation to the slightly better approximation of the value decaying exponentially with mileage. -- Meni Rosenfeld (talk) 20:14, 3 June 2009 (UTC)[reply]

I'm not sure this is valid. Let's say the original price is 10,000$ and it depreciates at 1$/mile. If you are offered a car at 9,999$ with negligible mileage, then by your method it has infinite value - even though all you are saving is 1$. On the other hand, a 9,000$, 500 mile car will have a value of only 2, even though you are saving 500$ compared to the true value.

Your calculation would work if you were buying many cars and limited by the total mileage you can buy. But you're only buying one car, and you want to gain as much as possible from this purchase. To do this you need to determine how much each mile decreases the value of the car to you, and simply maximize the difference ValueToYou - AskingPrice. -- Meni Rosenfeld (talk) 20:14, 3 June 2009 (UTC)[reply]

The minute somebody buys a new car and drives it off the dealer lot, it becomes a used car. I've always heard that the value drops about $1000 at that instant. After that, there's a slower decline in value (you could get a used car price guide and plot a curve through it to see the shape). Also, brand new cars often have manufacturing defects, which get sorted out under warranty during the car's first year. Cars more than 3 years old have usually developed some problems from normal wear and tear. So there are some folks who claim that the most problem-free cars you can buy are used cars between 1 and 3 years old, AND you get the savings resulting from the instant depreciation that the first owner incurred.

That all said, this is probably about as good a time as it gets to buy a new car (dealers are desperate to sell). After a series of used cars I bought my first new one several years ago (I'm still driving it) and although I knew it that decision wasn't optimal in pure financial terms, I found it to be a much more satisfying experience than buying a used car, due to freedom from uncertainty about undisclosed problems with the car, etc. 207.241.239.70 (talk) 03:12, 4 June 2009 (UTC)[reply]

u dv

Okay, I haven't learned how to do this. I keep getting "parse error" or something similar when I try to make it appear.

In my calculus book, the first integral on the list of integrals is (integral sign) u dv = uv - (integral sign) v du. List of integrals on Wikipedia doesn't have it, so it must be somewhere else.

I was looking to see how Wikipedia explained this, just out of curiosity. I know how my calculus book does.Vchimpanzee · talk · contributions · 21:44, 3 June 2009 (UTC)[reply]

Integration by parts, no? Angus Lepper^{(T, C)} 21:45, 3 June 2009 (UTC)[reply]

Yep, that's integration by parts all right (it's easy enough to derive, just integrate the product rule for differentiation and rearrange it a bit). To write in on Wikipedia you type this: <math>\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u</math> which displays as:

\int u\,\mathrm {d} v=uv-\int v\,\mathrm {d} u

. --Tango (talk) 21:56, 3 June 2009 (UTC)[reply]

"just integrate the product rule for differentiation and rearrange." Although slightly pedantic, it is important to understand that the functions involved in "integration by parts" must be continuously differentiable. Otherwise, the obvious proof is incorrect. However, this is just a side comment to Tango's post above. --PS T 01:00, 4 June 2009 (UTC)[reply]

Yeah, I knew where it came from, but I just forgot what it was called and how it would appear in Wikipedia. I could have opened the calculus book up and found that term. Thanks.Vchimpanzee · talk · contributions · 13:25, 4 June 2009 (UTC)[reply]

Indeed, my parenthetical asides do sometimes lack rigour! --Tango (talk) 16:37, 4 June 2009 (UTC)[reply]