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March 25

Figuring out percentages

I'm math clueless. I'm sure this is a very simple matter for many of you but not for me. If someone could not only give me an answer, but tell me how you reached an answer, I'd appreciate it. I am trying to figure out what an item depreciated by, by percentage, adjusting for real dollars. The item, an antique, was purchased in 1995 for $500. I have calculated using Wikipedia's inflation template that 500 1995 dollars is $999.78 today. The item sold at auction this year for $575, so in real dollars it went down in value. So my question is, how much by percentage did it go down in real dollars—is there a formula I can plug this type of calculation into?--141.155.143.65 (talk) 02:03, 25 March 2011 (UTC)[reply]

The important thing for percentages it to figure out the base. That is, the thing you need to divide by. In your case, the base is "real dollars", by which you mean inflation adjusted dollars. So, your base is $720.48 (noting that you used a dynamic template, so your figure above will change). Thus, we get that your price of $575 is $575/$720.48 or 0.798 or 79.8% of the expected value. This is a reduction of 20.2% from the expected value (100% - 79.8%). StuRat (talk) 02:13, 25 March 2011 (UTC)[reply]

Hmm. Thanks much. I understand what you did, meaning that I could take your math and do the same calculation again for a similar type problem with a different set of numbers and reach a correct conclusion, but I don't understand why it works and that's the problem with math for me in general. With language, I just get it. With math, my understanding is so much on the surface. I am memorizing a formula, and memorizing what it is for, with no deeper understanding of it; it has no qualia for me. Such is life.--141.155.143.65 (talk) 02:33, 25 March 2011 (UTC)[reply]

Which part wasn't apparent ? My selection of the base ? That we divided $575 by it ? That we multiplied the result by 100 to get a percentage ? That we subtracted that percentage from 100% to get the final answer ? StuRat (talk) 03:09, 25 March 2011 (UTC)[reply]

It just does not make sense to me. You have grasp of a foreign language that I can only navigate by a phrase book. I don't think you can educate me on this because I am limited in this area. I can understand how to use the result, and I can blindly follow the formula to reach the result but having a true sense of why it works... It is an innate limitation. You mind has a capacity in this area that mine simply does not.--141.155.143.65 (talk) 03:43, 25 March 2011 (UTC)[reply]

It might help if we take it step by step. For example, the base is the "whole" or the expected value, which was $720.48, does that make sense ? StuRat (talk) 03:50, 25 March 2011 (UTC)[reply]

If I had to guess, your confusion stems in part from not really understanding percentages (a fairly common condition). A percentage is just a way of expressing a comparison between two quantities (it's a single number expressing the relationship between two numbers) with the additional twist that you pick one of the quantities to be the reference (what Stu called the "base"). Frequently, it is expressing a comparison between a part of something and the whole, but that's not always the case, and you can confuse yourself if you try to pigeonhole thinking of one of the numbers as "the whole". (I think this is the case here, so just forget about "the whole" for now.) So to calculate a percentage, you have to figure out what two things you're comparing, and which one's the reference. In this instance, you're trying to compare the price today with the price in 1995, where the price in 1995 is serving as the reference. A final caveat of percentages is that the two quantities have to be in the same units. 2 inches is not 50% of 4 kilometers, and finding what percentage 2 kg is of 3 days is just nonsensical. In your case, while the units are nominally the same (both dollars), they aren't exactly the same. Generally, any time you need to "covert" between two things, they're in different units. So before you can compare the two prices, you'll have to convert them to a common unit first. Which unit to use? Well, today's dollars are probably most relevant to you today, and probably easier since you already did the conversion. (A side note: if the conversion is just multiplication by a constant, the percentage should work the same either way. If the conversion is more complicated, like Fahrenheit to Celsius, you'll either know which direction the conversion should be made in, or calculating the percentage is a meaningless exercise to begin with.) Once you have your two quantities, have decided which is the reference/base, and have converted both of them to common units, finding the percentage is just a mechanical divide-and-multiply-by-100 (which isn't anything special, but just the way percentage is defined). A final note - as you hopefully now realize, there isn't anything magical about percentages. Generally, unless you know why you want things expressed as a percentage, calculating a percentage isn't going to make things any clearer. The percentage numbers themselves are only meaningful if you know what, exactly, are the two quantities you're comparing. I've found that expressing things as percentages just to express them as percentages engenders more confusion than it usually solves. -- 174.21.235.112 (talk) 17:33, 25 March 2011 (UTC)[reply]

An item was bought at a price which would be $720.48 today, and is sold this year for $575. You have difficulty in computing the loss measured in percentages. If a problem is hard, try first another problem which is less hard. Suppose the item was bought at a price which would be $700 today, and is sold this year for $500. This problem is a little easier because the numbers are not that confusing. You have lost $200, because you gave $700, and you got only $500 and 700−500=200. This is called subtraction. Out of your $700 you have lost $200. How much is this in percentage, that is, per $100? Your seven $100-bills has resulted in a loss of $200. So the loss must be divided 7 ways to obtain the loss per $100. So you must compute ${\frac {200}{7}}$ which is 28 dollars per 100 dollars, or 28 per cent. This is called division. If you understand this example you may be ready to solve your original problem which is ${\frac {720.48-575}{7.2048}}$ which is 20 dollars per 100 dollars, or 20 per cent. Bo Jacoby (talk) 07:36, 25 March 2011 (UTC).[reply]

I think you have a mistake there. You aren't using the $575 price at which the item was just sold. StuRat (talk) 07:54, 25 March 2011 (UTC)[reply]

Thanks. I'm correcting it. Bo Jacoby (talk) 08:24, 25 March 2011 (UTC).[reply]

Perhaps the problem is the strange concept of calculating a "loss" based on an unreal starting-point. In reality, you have actually made a 15% profit on the antique, when, if you believe the inflation calculator, "average" consumer items would have given you a profit of almost 44.1% over the intervening years. You could thus claim to have "lost" about 29% of the "expected profit". This is not necessarily a better way of looking at the problem, but I always feel uneasy about calculating a loss based on "real dollars" when they are not real at all. You have not taken into consideration the auction charges (on purchase and on sale?), nor the enjoyment of the antique over the years. What would the $500 have been used for if the antique had not been purchased. All this doesn't answer your question. It just suggests a different way of looking at the problem. Dbfirs 07:33, 27 March 2011 (UTC)[reply]

Erdős–Graham problem

For every 1<m∈ℕ, let I_m:={n∈ℤ|1<n≤m}. How large must m be to guarantee that for every 2-coloring of I_m, there is a monochromatic subset S of I_m such that

\sum _{n\in S}{\frac {1}{n}}=1?

--84.61.170.180 (talk) 16:35, 25 March 2011 (UTC)[reply]

You are asking us to figure out something that Erdős couldn't? Looie496 (talk) 17:43, 25 March 2011 (UTC)[reply]

Variance of matrix-valued random variables

Hi. If X is a matrix-valued random variable with zero mean, then there seem to be two ways to calculate the variance: $E(X^{T}X)$ or $E(XX^{T})$ . If 'X' is not square, these two aren't even the same size (although the nonzero eigenvalues agree). Can anyone give me pointers to literature that discusses this? cheers, Robinh (talk) 19:59, 25 March 2011 (UTC)[reply]

Well, the situation I am familiar with is where X is a column vector. In that case,

E(X^{T}X)

is nothing of general importance, but

E(XX^{T})

is the covariance matrix, which is extremely important. If X were a row vector, the reverse would be true. Looie496 (talk) 22:50, 25 March 2011 (UTC)[reply]

Thanks for this. Do you have any information that might be relevant to my question, which was about matrices? Robinh (talk) 01:28, 26 March 2011 (UTC)[reply]

I'm going to wax speculative, so feel free to ignore me if you want. Something to consider is that it may depend on what you want to do with the information contained in the "variance" that you compute. Referring back to the covariance (I'll just assume that everything is Gaussian), the covariance matrix contains precisely the information needed to transform the vector, by a deterministic matrix, to a standard Gaussian variable in n-dimensions. One might ask an analogous question of a matrix, specifically what information will suffice to put a random matrix into some convenient normal form. I think the answer depends on what kinds of transformations are allowed:

X\mapsto AX

,

X\mapsto XB

, or

X\mapsto AXB

where A and B are deterministic square matrices (usually corresponding to a change of basis). If only one kind of transformation seems to be natural, then that will probably suggest one or the other definition. For instance, if X is a column vector, then the relevant basis changes are

X\mapsto AX

, so the appropriate variance is

E(XX^{T})

, since this transforms well under such mappings. It may be that

E(XX^{T})

and

E(X^{T}X)

are both useful and may serve complementary roles. In this context, the singular value decomposition seems relevant. Sławomir Biały (talk) 02:44, 26 March 2011 (UTC)[reply]

(OP) thanks, Sławomir. SVD does seem to be relevant. Thinking about your comment about Gaussian variables leads me to Matrix normal distribution, which has two covariance matrices. This is pretty much what I was asking for. Cheers, Robinh (talk) 03:02, 26 March 2011 (UTC)[reply]

Power series

If it makes a difference, these questions are in relation to positive real numbers. By "power series" I mean something like a_0 + a_1*(x - c) + a_2*(x - c)^2 + a_3*(x - c)^3 + ...

1. What is the power series for the gamma function? The article says it is an analytic function, so it ought to have one, right? But I can't seem to find it documented anywhere...

2. Do functions like x^x, x^(x^x) and x^(x^(x^x)) have power series?

86.181.206.155 (talk) 21:26, 25 March 2011 (UTC)[reply]

What do you mean by "have a power series"? You need to have a point about which you expand the function as a power series. See, for example, Taylor series or Laurent series. Functions don't have a power series. They have different power series depending on which point you expand about. The Gamma function has a well defined power series expansions at each point of the positive real numbers (it has a simple pole at zero). Again, the function x^x have well defined power series expansions at each point of the positive real numbers. It might seem off that the same function can be represented by different power series, but it can be made rigorous. See for example the function germ article. — Fly by Night (talk) 00:43, 26 March 2011 (UTC)[reply]

You can work out a Taylor series for the gamma function explicitly in a neighborhood of any point using the polygamma function. Check Wolfram Alpha for the first few terms. Sławomir Biały (talk) 00:55, 26 March 2011 (UTC)[reply]

Except the non-positive integers, surly? The Gamma function is meromorphic. — Fly by Night (talk) 00:59, 26 March 2011 (UTC)[reply]

It will be a Laurent series about the poles. My point is that this can be done explicitly. Sławomir Biały (talk) 01:09, 26 March 2011 (UTC)[reply]

(edit conflict) I don't think there's a known, explicit power series expansion like a₀ + a₁(z − w) + a₂(z − w)² + …, where we have explicit expressions for the a_k, is there? I've just noticed that in-line HTML looks terrible in Firefox. I'm going back to Google Chrome. — Fly by Night (talk) 01:11, 26 March 2011 (UTC)[reply]

I guess "explicit" is somewhat ill-defined. It doesn't seem worth arguing over. Sławomir Biały (talk) 01:19, 26 March 2011 (UTC)[reply]

Ok, so there is a well-known series for the logarithm of the Gamma function

\log \Gamma (z+1)=-\log(z+1)+z(1-\gamma )+\sum _{n=2}^{\infty }(-1)^{n}(\zeta (n)-1)z^{n}/n

There are similar expansions about other points, but they involve Hurwitz zeta functions. There are also similar formulas for the reciprocal Gamma function. For the Gamma function itself, there are formulas involving summing polygamma functions over partitions of n, with explicit multinomial coefficients. Sławomir Biały (talk) 02:22, 26 March 2011 (UTC)[reply]

Note that the structure of these expansion formulae are easy to understand: The coefficient of z^n for n >= 2 in the expansion of Log(Gamma) is proportional to the summation of 1/zp^n, where zp are the poles of the Gamma function. Count Iblis (talk) 17:10, 26 March 2011 (UTC)[reply]

Monoid ring

Let R be a commutative ring and M a monoid. Is the monoid ring R[M] the free-est R algebra on M? I know it's supposed to be a ring but it has the obvious scalar multiplication on it. The wikipedia article group ring says if M is a group the R[M] is the free-est algebra on groups of units; I'm considering the left adjoint to the functor that sends an R algebra to the underlying monoid. Money is tight (talk) 22:31, 25 March 2011 (UTC)[reply]

It is the same for monoids. Sławomir Biały (talk) 00:34, 26 March 2011 (UTC)[reply]