Wikipedia:Reference desk/Archives/Mathematics/2014 March 7

= March 7 =

Cos(pi/2)
My own sketch of a cosine wave and Wolfram Alpha both agree that cos(pi/2) is equal to zero. My TI-83 calculator tells me that it's 0.999624... Why is there a difference? Thanks, Dismas |(talk) 16:51, 7 March 2014 (UTC)


 * You are working in degrees. Wolfram Alpha works in radians for functions of pi (as does everyone else).    D b f i r s   16:55, 7 March 2014 (UTC)


 * Ah! Yeah, I forgot about that...  Thanks,  Dismas |(talk) 17:06, 7 March 2014 (UTC)

Finding two unknowns in one equation
I had the problem

e=(Ae^(-3))+Be

My teacher tells me that just from that we can find that A=0 and B=1. To demonstrate this, she gave me the following example:

3apples = Aoranges + Bapples

Since there are no oranges on the left side of the equal sign, A must equal 0 and B must equal 3. I see that but then apples and oranges are simply variables. e is a number. Isn't it possible that somewhere along the number line with its infinite values that there is another answer that will give you something that you can multiply times e in order to get Ae^-3?

And sorry for the formatting. I deal with the math functions so rarely that I've forgotten how to use them and don't have much time right now to re-learn them for this Q.

Thanks, Dismas |(talk) 17:16, 7 March 2014 (UTC)


 * It's trivial to show that for any A value, $$B = -A\,e^{-4} + 1$$, so yes of course there are infinite solutions. Though since $$e^{-4}$$ is irrational, the only solution with both A and B as integers is the one you already gave.  Dragons flight (talk) 17:33, 7 March 2014 (UTC)


 * (WP:EC)Sure, check $$A= e^4, \; B=0 $$. That gives us right hand side = $$Ae^{-3} +Be=e^4\cdot e^{-3} + 0 \cdot e= e+0 = e $$, which is the left hand side. If you isolate ("solve for") B, then you get $$B = \frac{e^4-A}{e^4} = 1-Ae^{-4}$$, which is just the equation of a line (if you write A->x, B->y, you get something of the form y=mx+b). So there are infinitely many points (A,B) that satisfy the equation. Pick any A you want, and the last equation will tell you what B has to be. Unless there's some transcription error, or other bits of the problem, something is very wrong here... your teacher might just be showing you how to "guess and check", but that's no excuse to ignore the uncountably many solutions to the equation. Also, here's what Wolfram alpha shows . Perhaps the interest is in integer solutions, but if that's the case, proving/finding integer solutions is a much more complex subject, e.g. Diophantine equations. SemanticMantis (talk) 17:41, 7 March 2014 (UTC)

I asked another math teacher and she agrees with me that there are infinitely many solutions. I hope that I misunderstood the first teacher in that we were trying to find a solution and not necessarily the only one. Thank you for the responses! Dismas |(talk) 20:59, 7 March 2014 (UTC)
 * As hinted by Dragon's flight, the question probably meant to ask to find integer A and B. And then the only answer is indeed $$A=0,\ B=1$$, since $$e^4$$ is irrational. -- Meni Rosenfeld (talk) 19:08, 9 March 2014 (UTC)

What's the average change in cents for a purchase in US$2 or more?
What's the average change in cents for a purchase in US$2 or more? Does it differ if I say above $10 for various different items? Thanks. μηδείς (talk) 23:23, 7 March 2014 (UTC)


 * Averaged over what? —Tamfang (talk) 01:38, 8 March 2014 (UTC)
 * Erm I'd guess about 1 cent. So many things are sold at $5.99 or $5.95.--Salix alba (talk): 06:20, 8 March 2014 (UTC)
 * There is a related observation that does not suffer from the skewing of psychological factors that lead to strange pricing patterns like this: the wear of keys on a keyboard – the '1' gets used substantially more than the '9'. If you were to take the prices in a store and adjust them so that they produced a smooth distribution, I guess you'd get an expected average of more than $0.50 in change in coins from low prices, tending to exactly $0.50 as the range of the prices under consideration moves to larger values. —Quondum 06:46, 8 March 2014 (UTC)


 * Psychological pricing has a nice plot of the distribution of the final digit of prices--its 9 about 60% of the time. --Mark viking (talk) 07:05, 8 March 2014 (UTC)
 * Yeah, I do remember reading a decade ago that a little more than about 1/3 numerical expressions begin with the figure 1. This seemed to be an artifact of the expressions 1, 10-19, 100-199, 1000-1999, etc.,   That's not at all what I am asking, since change based on those amounts would be 8-99, 81-999, etc.... μηδείς (talk) 07:27, 8 March 2014 (UTC)
 * That would be Benford's Law, which would be skewed in this case because of psychological pricing. Prices are more likely to be 999.99 than 1000.00, so the first digit 9 will be overrepresented. 7 and 8 will likely still be rare. Staecker (talk) 12:37, 8 March 2014 (UTC)


 * To answer the question, you have to have a probability distribution of all prices. Perhaps this could be checked empirically in certain situations. E.g. you could save all your receipts and calculate and answer. Or do similar if you were e.g a store owner. You could assume a uniform distribution of all whole-cent prices, but that is almost certainly wrong. SemanticMantis (talk) 22:23, 8 March 2014 (UTC)


 * I don't think this is something the reference desk can answer, given that it would be mainly speculation. I'm sure such a distribution would be highly clustered, which isn't ideal for statistical methods.--Jasper Deng (talk) 09:22, 9 March 2014 (UTC)
 * I agree that it can't be answered without further specification. And even if specified, it's unclear if volunteers here would have access to the right kinds of data. However, I don't agree that clustering is a problem. Say OP was interested in "what is the average (e.g. mean or median) value of change in cents of a WalMart purchase over $2 (paid in cash), in the 2000-2010 period?" -- then, regardless of clustering in the distribution, WalMart could easily answer the question, based on their data. SemanticMantis (talk) 17:09, 9 March 2014 (UTC)
 * Most states and localities have sales tax, but the rates are quite variable -- so that's an additional complication. If you start with "x.99" or "x.95" pricing and then add variable sales taxes, the story will get very complicated very quickly. Looie496 (talk) 17:32, 9 March 2014 (UTC)
 * The reason why I say statistical methods wouldn't be desirable is that, since a census is impossible, you will have to take a (simple random) sample and make a confidence interval for the mean. But a clustered, irregular distribution without a well-behaved pdf (let alone a cdf) would make that almost impossible to do except by numerically estimating.--Jasper Deng (talk) 19:24, 9 March 2014 (UTC)
 * Ah, I see what you mean now, thanks for clarifying. I was mostly thinking of an empirical approach. And even then, you're right that tight clusters and e.g. multi-modality in the pdf will make that hard to sample effectively. SemanticMantis (talk) 21:16, 9 March 2014 (UTC)