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October 5

Median range

Is there such a thing as the "median range" of a set of numbers? I don't recall encountering the term before, but the article Case Western Reserve University#Undergraduate profile says

Median SAT scores (25% - 75%) were between 1280 and 1450. The median range for ACT scores was 29 to 33.

Whittier Law School#Admission statistics says

Median Range of LSAT: 150-154

Median Range of GPA: 2.75-3.28

Kleinbrook, Texas says

the subdivision had a median price range of $95,000-$151,000.

Traces, Texas says

subdivision had a median price range of $77,000-$124,000 U.S. dollars.

Williamsville, New York says

The median age range was 45–49 years.

Is median range accepted mathematical terminology? Loraof (talk) 01:01, 5 October 2016 (UTC)[reply]

I have not come across the term before either, but I'm not a statistician. When I was in school, the statistic in your median SAT scores example was called interquartile range. --100.34.204.4 (talk) 01:47, 5 October 2016 (UTC)[reply]

As for Kleinbrook data, you may try asking the author of these edits: Special:Diff/256392404, Special:Diff/256392887.
For Williamsville ask the author of Special:Diff/525445794. Etc... --CiaPan (talk) 12:04, 5 October 2016 (UTC)[reply]

Thanks. Our article interquartile range defines it as the difference between the upper and lower quartiles--i.e., as a measure of dispersion. I guess the best way to express the interval is interquartile interval. Even if it's not a technical math term, it's accurate. Loraof (talk) 15:34, 5 October 2016 (UTC)[reply]

Geometry Origami problem, take 2

So, now that we have determined that it is theoretically possible to fold a 4-inch square into a 1.38-inch 16-gon (but with at least 10 layers), how exactly can this be done? (Yes, the steel mesh has arrived -- in fact, it arrived quite a while ago -- and it looks and feels flexible enough that even 10 layers of folds might be possible!) Keep in mind the other condition -- one of the faces of the 16-gon (or whatever-gon, as long as the number of sides is 16 or more) must be smooth, so all the folds must be on the other face. 2601:646:8E01:7E0B:240D:7200:4A16:32C2 (talk) 09:41, 5 October 2016 (UTC)[reply]

Russian roulette

A couple of questions about Russian roulette: (1) With only 1 bullet in the cylinder, how much less is the actual probability of dying than the theoretical probability of 1/6? (1.1) With this probability, if you play Russian roulette once a day with only 1 bullet, is it even remotely possible to survive for 7 years? (2) Likewise, for any non-symmetric distribution of bullets in the cylinder, how does the actual probability of dying differ from the theoretical probability for that number of bullets? (Question inspired by the movie The Deer Hunter.) 2601:646:8E01:7E0B:240D:7200:4A16:32C2 (talk) 09:55, 5 October 2016 (UTC)[reply]

The probability of surviving n days is (1 - 1/6)ⁿ = (5/6)ⁿ. That means e.g. the chance of surviving a week is (5/6)⁷, or about 28%.. The chance of surviving 30 days is (5/6)³⁰ or about 0.4%. Even that seems barely possible. Anything much longer, such as a year or seven, would not seem even remotely possible.

But that is assuming the chance is exactly 1/6. I imagine in reality the chance is somewhat different. A gun is built for reliability not randomness when spinning. Maybe even someone with experience can judge the timing to make sure they always stop the magazine at a safe position, reducing the probability to zero. The maths in this case is probably only an approximate guide.--JohnBlackburne^words_deeds 10:04, 5 October 2016 (UTC)[reply]

Well, the American Association of Neurological Surgeons seems to think that head shots are only 95% likely to kill the victim. Of course, this depends on a lot of variables. If you assume an equal likelihood for all 6 chambers, the chance of 7 years without firing a shot is roughly 5*10^-203 or 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000005 (give or take a zero - I'm prone to off-by-one errors). If you stop once you have lost, but still survive, that will dominate your outcome (i.e. you have the 5% survival chance of a head shot - the chance of no shot is negligible). But revolvers with 5 chambers are not uncommon, and the confederates used the 9+1 chamber LeMat Revolver in the US civil war. And apparently Casimir Lefaucheux had a 20 chamber revolver at one time - that would increase your chances significantly (but still not to a good level). --Stephan Schulz (talk) 10:36, 5 October 2016 (UTC)[reply]

But the whole point of my asking is, it's NOT an equal likelihood for all 6 chambers -- unless the bullets are distributed symmetrically, the chambers with the bullets will be heavier and would tend to end up on the bottom, thus (for most scenarios, and always if there's only 1 bullet) lowering the odds of shooting yourself. 2601:646:8E01:7E0B:240D:7200:4A16:32C2 (talk) 22:12, 5 October 2016 (UTC)[reply]

Just to expand on the first answer...

(1) is hard to answer without more details of the "randomization" process. But assume said probability is p, and is the same every day (plus, independent tries from day to day). Then the probability to survive n days is P(n,p)=(1 - p)n. Assume n=2500 (a bit less than 7 years) and define "remotely possible" as P=0.1%, then we must have p=1-exp(ln(1-P)/n)<0.3%.

Basically, you need to cheat, but even with a lot of magazine-spinning practice a 0.3% fail rate is probably hard to achieve.

The somewhat paradoxal lesson is that if you make many tries, an event that is unlikely at each try is virtually guaranteed to happen at some point. The Russian roulette example might seem silly, but there has been real life applications, for instance the Borislav Ivanov case where (to make it simple) an unknown guy skyrocketed in the chess ratings and "coincidentally" his moves matched computers' much more closely than any historical record of world champions. (Bad players routinely match computers' top moves, but even excellent players deviate often enough that this guy was suspicious from day one.) He was eventually excluded without the use of statistics, because it was easier (for PR and/or the legal teams), but many - myself included - hold that the statistics, properly done, are just as proof as catching him red-handed. (Long news story about the statistician hero of the case: here.) Tigraan^{Click here to contact me} 11:12, 5 October 2016 (UTC)[reply]

Similarly, see the totalitarian principle in quantum physics. --69.159.61.230 (talk) 22:47, 5 October 2016 (UTC)[reply]

Inverse cube expression and spirals

What are the mathematical proof steps of the connection between a inverse-cube central force expression and the form of the trajectory (on spirals) of a (material) mobile point subjected to the central force with the given mathematical expression?--213.233.84.13 (talk) 11:44, 5 October 2016 (UTC)[reply]

Take Newton's second law for the case at hand, express in polar coordinates, solve for r(t) for some initial condition. Anything that decreases monotonically to zero is a spiral in the plain meaning of the term, though you might have a more stringent definition. (Note: I am not sure the result is indeed a spiral; isn't a 1/r^3 force conservative for a 1/r^2 potential?) Tigraan^{Click here to contact me} 16:19, 5 October 2016 (UTC)[reply]

The article Cotes spiral says that these spirals are the solutions for the motion of a particle moving under an inverse-cube central force. The article has an external link to a Google book by N. Grossman with a short description of proof that leads to Cotes spiral for inverse cube force. Is that derivation straight forward? I want to add the proof steps into Cotes spiral article. Something may be missing from that source, it seems so. Thoughts ? (Thanks!)--213.233.84.7 (talk) 21:49, 5 October 2016 (UTC)[reply]

Binet equation has a derivation, or most of one. But you need to take Binet as a given and its derivation perhaps is not straightforward. --RDBury (talk) 02:43, 6 October 2016 (UTC)[reply]

October 6

Spherical Trigonometry

Can someone prove rigorous the spherical trigonometric identities/laws without using Linear algebra? יהודה שמחה ולדמן (talk) 12:37, 6 October 2016 (UTC)[reply]

Historically linear algebra is much later than spherical trigonometry, so the answer is yes. Bo Jacoby (talk) 16:31, 6 October 2016 (UTC).[reply]

OK, I'd like to see formal proofs for the Cosine rules and sine rules. יהודה שמחה ולדמן (talk) 17:50, 6 October 2016 (UTC)[reply]

Do you mean along the lines of Spherical law of cosines#Proof without vectors? Probably the 1886 Todhunter book has this. --Mark viking (talk) 20:57, 6 October 2016 (UTC)[reply]

I think we need a sketch review for this on that page. יהודה שמחה ולדמן (talk) 22:35, 6 October 2016 (UTC)[reply]

What is the value of these sums

$\sum _{k=0}^{\infty }{\frac {1}{k\times 10^{100}+1}}$

and

$\sum _{k=0}^{\infty }{\frac {1}{k^{k+1}\times 10^{100}+1}}$

175.45.116.99 (talk) 02:41, 7 October 2016 (UTC)[reply]

The first sum diverges, and the second is extremely close to 1 (

\lim _{x\rightarrow \infty }\sum _{k=0}^{\infty }{\frac {1}{k^{k+1}\times x+1}}=1

). 24.255.17.182 (talk) 04:06, 7 October 2016 (UTC)[reply]

To elaborate just a bit, the first one is

\sum _{k=0}^{\infty }{\frac {1}{k\times 10^{100}+1}}\approx {\frac {1}{0+1}}+\sum _{k=1}^{\infty }{\frac {1}{k\times 10^{100}}}=1+{\frac {1}{10^{100}}}\sum _{k=1}^{\infty }{\frac {1}{k}},

where the last sum is the divergent harmonic series. Loraof (talk) 16:20, 7 October 2016 (UTC)[reply]

The value of the second is approximately

1.000000000....

(99 zeroes in total)

....000013839

Dbfirs 00:03, 11 October 2016 (UTC)[reply]

October 7

Beats me

I’m not crash hot with math, and confess this claim by an oil company is doing my head in. Their promotion/exercise is based on a price of $2 per litre, and a discount of 6 cents per litre. Basically, they claim that by accumulating discounts rather than redeeming them at the time of sale, you will double your savings. Following are their examples given.

Redeemer. Buys 50 litres of gas, value $100, but pays $97 because of the 50 x 6c discount. After doing this four times the redeemer has spent $400 for a total savings of $12.

Accumulator. This is complicated because the promoter hasn’t exactly compared apples with apples. What they show is (a) someone making seven $40 purchases but accumulating the 6c litre discounts each time, and (b) then buying $100worth of gas for only $76 which they say incorporates all of the accumulated discounts and the discounts for this last sale.

The promoter claims a total spend of $380 for a total savings of $24.

Anyone? Moriori (talk) 23:03, 7 October 2016 (UTC)[reply]

The Accumulator first pays $280 for 140 gallons and a stack of coupons worth $8.40, then buys 50 gallons for $97 minus $8.40, so the total spending is $280 + $97 - $8.40 or $368.60 for 190 gallons, saving $11.40, not $24. The $24 makes sense only if the coupon is worth more than 6¢ if you save it for later. The $400 or $380 is the price before any discounts, not after. Your skepticism is well warranted. —Tamfang (talk) 00:14, 8 October 2016 (UTC)[reply]

Ahhhhh, so my logic wasn't so fuzzy after all, because that's what I had figured. The problem is, the oil company indicates that the discount is always 6 cents. A guy I know is totally sucked in, and won't listen to anyone who says otherwise. Cheers.Moriori (talk) 00:27, 8 October 2016 (UTC)[reply]

I can almost get their results, if I add the following assumptions (beyond gasoline costing $2 a litre):

1) For every purchase of gas, $40 or over, they get a 6 cent per litre discount coupon, which can either be used on that purchase or accumulated.

2) Up to 8 discount coupons may be used for each purchase, giving you up to 48 cents off per litre, on purchases up to $120 in gasoline.

3) Once a coupon is used it can not be used again.

Note that the assumptions might have also been defined in litres, so purchases of 20 litres or more get a coupon, which can be used for up to 60 litres. So, using these assumptions:

Redeemer buys 50 litres at $100 and uses discount of 6 cents per liter, for $3 off, each time. He does this 4 times, for 200 litres of gas purchased at $400, and a savings of $12.

Accumulator first buys 20 liters at $40 to get the 6 cents per litre coupon, which he saves. He does this 7 times, buying $280 worth of gas and getting 7 coupons. The 8th time he buys $120 worth of gas (60 litres), getting one more coupon, and uses all 8 coupons, which gives him 48 cents off per litre, times 60 litres, so $28.80 in savings.

Under both scenarios he buys $400 (200 litres) of gas, less the discount, but the discount is higher for the accumulator, because he used the coupons on larger purchases, so the discount applies to more litres.

Of course, in the real world, the accumulator's strategy may fail, for many reasons. More trips to the gas station to get those small amounts of gas may waste fuel and depreciate the car more than the value of the savings. The coupons may be lost. The promotion may end before they are redeemed. Etc. StuRat (talk) 00:39, 8 October 2016 (UTC)[reply]

Ummmm. Lets take a apples v apples scenario where both end up buying the same number of litres.

Redeemer buys 50 litres at $100 and uses discount of 6 cents per liter, for $3 off. He does this 4 times, for 200 litres of gas purchased at $400, and a total accumulated discount of $12.

Accumulator, on 5 occasions, buys 20 liters at $40 to get the 6 cents per litre discount. His accumulated discount is 5 x $1.20 = $6 which is correct for 100 litres at discount of 6 cents a litre. At his next purchase, he buys 100 litres of gas worth $200. If -- as advertised -- the discount is 6 cents a litre, then he has purchased a total of 200 litres of gas and should get a total accumulated discount of $12. No? Moriori (talk) 01:20, 8 October 2016 (UTC)[reply]

I think you and Tamfang are assuming that if you buy n liters, you get a coupon good for $0.06n off any purchase. StuRat is assuming that if you buy n liters, you get a coupon good for 3% off any purchase, that moreover can be stacked so that 8 coupons get you 24% off a single purchase. StuRat's assumption is consistent with the oil company's examples if you assume that the quoted total costs are before the discount, not after (i.e. the redeemer spends $388 and the accumulator spends $356). -- BenRG (talk) 01:57, 8 October 2016 (UTC)[reply]

Correct, except I would define it as a cents per litres discount rather than a percentage of dollars spent. So, eight 6 cents per litre coupons can be stacked, for up to 48 cents per litre off, total. StuRat (talk) 14:24, 8 October 2016 (UTC)[reply]

If this is a real-life promotion, is there a web site where we can consult the actual rules? --69.159.61.230 (talk) 08:29, 8 October 2016 (UTC)[reply]

Here's a url with a bit about it - https://www.aa.co.nz/aasmartfuel/how-aa-smartfuel-works/#tips Note the video at the top right which includes "...you can accumulate those savings to get a bigger discount per litre, meaning your fuel will be cheaper for sure, maybe even free". The "bigger discount per litre" is not consistent with their earlier illustration which quite clearly showed the price per litre is $2 and the discount is 6 cents a litre. Anyway, looks as if you do this often enough and you get free gas. Yeah right! Moriori (talk) 00:12, 10 October 2016 (UTC)[reply]

Looking through that website, it looks like the limit to which discounts can apply is 50 litres, but you can apply as many stacked discounts as you like, potentially getting that 50 litres free. However, they don't list the minimum purchase required to get each discount, or what each discount is, as apparently each retailer sets that differently. But, presumably, you need to buy lots of gas at full price to get that 50 litres "free". StuRat (talk) 00:35, 10 October 2016 (UTC)[reply]

Aha, there's a basic point that hasn't been covered: buying fuel isn't the only way for members to get the points. At the top of the linked page is a link "shop and swipe", pointing to this page. Obviously if you accumulate points by buying other stuff, then you can spend more of them when you do buy fuel. --69.159.61.230 (talk) 03:56, 10 October 2016 (UTC)[reply]

October 10

Name for such progressions

What is the name of a progression where there are two or more ratios which alternate every step.

E.g., two ratios 5 and 2

1,5,10,50,100,500,1000,5000,10000,...

Three ratios 5, 2, 10

1,5,10,100,500,1000,10000,...

Or 2, 5, 10

1,2,10,100,200,1000,10000,...

Or 2, 3, 5, 10

1,2,6,30,300,600,1800,9000,90000,...

--Lüboslóv Yęzýkin (talk) 06:39, 10 October 2016 (UTC)[reply]

I don't know a name for such progressions, but they are geometric progressions added together:

(1, 10, 100, 1000, ... ...) + (5, 50, 500, 5000, ... ...)

Dolphin (t) 11:25, 10 October 2016 (UTC)[reply]

Interleave sequence. -- ToE 12:17, 10 October 2016 (UTC)[reply]

Well, it's a specific kind of interleave sequence. And for the ones with three or more ratios, it's a generalization of that, if the article's definition is complete. --69.159.61.230 (talk) 18:32, 10 October 2016 (UTC)[reply]

Preferred number. Bo Jacoby (talk) 05:40, 11 October 2016 (UTC).[reply]

October 11

How to isolate a variable

X = AB

Y = AC

Z = C/B

I know the exact values of X, Y, and Z. Now I want to solve for one of the unknowns A, B, or C. Is it possible?98.20.64.142 (talk) 18:00, 11 October 2016 (UTC)[reply]

Yes it can be solved. Since you say that X, Y, and Z are knowns and not variables, then to solve for the unknowns, by a method known as the system of equations method can be employed. The basic principle is that any system of equations is soluble so long as you have at least as many equations as unknowns. This walks you through the methods. --Jayron 32 18:27, 11 October 2016 (UTC)[reply]

Except these are not polynomial equations (the variables are not joined by addition or subtraction).98.20.64.142 (talk) 19:15, 11 October 2016 (UTC)[reply]

(ce) I assume that Jayron meant a system is solvable if you have at least as many unknowns as equations, not vice versa. (Otherwise, A=1, A=2 would be solvable.) But either way, not all such systems are solvable in the sense of having an analytic solution. For the one variable, one equation case, see Abel-Ruffini theorem or quintic equation. For multiple equations and variables, see system of polynomial equations. Regarding the particular system the OP is inquiring about, try solving the first equation for A, then substituting that expression for A into the other equations, then solving the next one for B and substituting into the last equation, then solving that one for C, then using that expression for C in the previous expressions, etc. Loraof (talk) 19:34, 11 October 2016 (UTC)[reply]

That's just it though. I only know values for X, Y, and Z! I need to isolate either A, B, or C from the first three. But how?98.20.64.142 (talk) 19:49, 11 October 2016 (UTC)[reply]

XZ = (AB)(C/B) = AC = Y. Your equations are not independent. For the values of X, Y, and Z you know, do they satisfy XZ = Y? If not, then there are no solutions. If so, then there is a family of solutions. -- ToE 20:06, 11 October 2016 (UTC)[reply]

Yes, they do satisfy that condition. Can you point me in the direction of finding that family of solutions?98.20.64.142 (talk) 20:15, 11 October 2016 (UTC)[reply]

If your X, Y, and Z, do satisfy XZ = Y, then you can characterize your family of solutions with any one of A, B, or C. For instance, {(A, X/A, Y/A)}. -- ToE 20:19, 11 October 2016 (UTC)[reply]

I see what you mean. That doesn't help much for my problem though.98.20.64.142 (talk) 20:35, 11 October 2016 (UTC)[reply]