Wikipedia:Reference desk/Archives/Mathematics/2008 May 29

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May 29

Re: a function problem

Help!

x= 1  2  3  4  5  6
y= 2  1  0 -1 -2 -3

What is the formula that works for all?

y=x ? —Preceding unsigned comment added by 76.91.173.241 (talk) 00:36, 29 May 2008 (UTC)

Try y = ax + b, and try to come up with values of a and b that fit. (Hint: b should be the difference between consecutive values of y. Also, when x = 1, y = a + b.) —Ilmari Karonen (talk) 01:23, 29 May 2008 (UTC)

I have no idea how to write this as a function, but its the difference between x and 3. 3-1=2, 3-2=1, 3-3=0, 3-4=-1. well, i guess -x+3=y? --Xtothe3rd (talk) 01:58, 29 May 2008 (UTC)

Very good. That is correct. $y=-x+3$ is the function. In order to check this, if you have a TI calculator, hit the "y=" button and type in the corresponding x part of the equation. Then hit the 2nd key and then graph. This should display a table with the corresponding y and x values. By the way, this is a linear function. Ζρς ι'β' ^¡hábleme! 07:26, 29 May 2008 (UTC)

This looks like homework, so the only appropriate answer is really the first. The rest of you shouldn't go this far. 130.95.106.128 (talk) 11:10, 29 May 2008 (UTC)

Oops. It did look like homework to me as well. However, I gave my answer at 3:00 am local time, so I overlooked the fact that the OP and Xtothe3rd are not the same person. That was my bad. Ζρς ι'β' ^¡hábleme! 17:08, 29 May 2008 (UTC)

Different differences

How many different differences can be obtained by taking only 2 numbers at a time from 3,5,2,10 and 15? One might thing the maximum number of differences would be 2 out of 5 (5*4), and that the number of different ones must be less than this. (At least I would). Yet the answer is 1440. Anyone? 83.167.112.140 (talk) 15:32, 29 May 2008 (UTC)john

Where did you hear it should be 1440? Your reasoning is correct for what I understand the problem is. Either the problem should be interpreted in some unusual manner, you haven't stated the problem correctly, or your source is dead wrong. -- Meni Rosenfeld (talk) 16:08, 29 May 2008 (UTC)

This was an on-line gre practise question. I recited the question verabatim, so I guess it must be a mistake. Thanks. 83.167.112.140 (talk) 16:27, 29 May 2008 (UTC)

Perhaps the differences need not come from the original set. For instance, take 10-2, and then repeat the process on the numbers 3, 5, 2, 10, 15, 7? That’s not how I would interpret it, but could be what the author of the question meant. GromXXVII (talk) 17:02, 29 May 2008 (UTC)

10-2=8 That interpretation either gives infinitely many possibilities (if differences can be negative) or 15 (if they cannot). I can't see any way to get 1440. In any case I would be very unhappy with an exam that included questions this loosely worded. Algebraist 18:12, 29 May 2008 (UTC)

Another stupid question; graphs of functions

Hello, nice refdesk people. Another stupid question I've been puzzling over:

For each of these functions, decide whether the function is always increasing, always decreasing, or sometimes increasing and decreasing:

y=x³ - 3x + 1

y=x³ + 3x + 1

y=1- 3x - x³

I don't actually understand the question: By the 'function is always increasing', do they mean it always has a positive gradient? Or what? :/ *confused once again* naerii - talk 18:01, 29 May 2008 (UTC)

A function y=f(x) is increasing if f(a)<f(b) whenever a<b. That's not quite the same as having positive gradient, but certainly a function with positive gradient everywhere is increasing. Algebraist 18:08, 29 May 2008 (UTC)

Thanks for making it more clear Algebraist! :) naerii - talk 21:09, 29 May 2008 (UTC)

I forgot to mention: we have an article. Algebraist 21:15, 29 May 2008 (UTC)

The word "always" in the question apparently means "everywhere", that is, everywhere on the domain of the function.  --Lambiam 04:41, 30 May 2008 (UTC)

• You need to find the gradient of each equation, as these will give you the values:
  • $y=x^{3}-3x+1,{\tfrac {dy}{dx}}=3x^{2}-3=3(x+1)(x-1)$. So increasing when x > 1, decreasing when x < -1.
  • $y=x^{3}+3x+1,{\tfrac {dy}{dx}}=3x^{2}+3=3(x^{2}+1)$. Since it contains a squared term, which cannot be negative, the function is always increasing.
  • $y=1-3x-x^{3},{\tfrac {dy}{dx}}=-3(x^{2}+1)$. Since it contains a squared term, which cannot be negative, then multiplied by a negative term, the resulting solution will always be negative, ie always decreasing. 86.153.37.241 (talk) 23:28, 1 June 2008 (UTC)

No, the first example is increasing in both $(-\infty ,-1]$ and $[1,\infty )$ and decreasing in $[-1,1]$. Algebraist 23:49, 1 June 2008 (UTC)

But when differentiated the equation is a parabola, and it can be easily seen when the equation is increasing and when it is decreasing. 86.153.37.241 (talk) 00:18, 2 June 2008 (UTC)

Yes indeed, and it's easily seen to be what I said and not what you said. Algebraist 00:20, 2 June 2008 (UTC)

That's a matter of personal opinion. 86.153.37.241 (talk) 00:22, 2 June 2008 (UTC)

How so? --Tango (talk) 00:30, 2 June 2008 (UTC)

It's a matter of personal opinion because it depends entirely on what people prefer to see as the given solution and how to get the correct one. 86.153.37.241 (talk) 00:35, 2 June 2008 (UTC)

People usually prefer to be given the correct solution... It doesn't really matter how you get to that solution, but any method that results in an incorrect solution is, by definition, incorrect. --Tango (talk) 00:58, 2 June 2008 (UTC)

(ec) Yes, it can, and it's what Algebraist said. It's a parabola that just dips below the x-axis between -1 and 1, and it above it everywhere else. --Tango (talk) 00:30, 2 June 2008 (UTC)

Probability

First, yes, this is a school question. A question on a test I took today said, if you're dealt a 5 card hand from a deck of 52 cards, what is the probability of getting exactly one King? I worked it, and I came up with ${\frac {4}{52}}*{\frac {48}{51}}*{\frac {47}{50}}*{\frac {46}{49}}*{\frac {45}{48}}=.0599$

But, according to whoever made up this test, the actual answer is $.299$. I'm stumped, and so is my teacher. So, can you please enlighten me on what the right answer is, and how to get it? Digger3000 (talk) 19:34, 29 May 2008 (UTC)

This is a hypergeometric distribution. Define a "trial" as drawing a card. Define a "success" as the drawn card being a king. In the deck, there are 52 possible trials and 4 possible successes. You are going to run 5 trials and you seek 1 success. Plug in the values to the hypergeometric distribution and you get 0.299. $0.299={{{4 \choose 1}{{52-4} \choose {5-1}}} \over {52 \choose 5}}.$ Wikiant (talk) 19:47, 29 May 2008 (UTC)

Thanks so much. Digger3000 (talk) 20:04, 29 May 2008 (UTC)

So, what you calculated originally, was, in fact, probability that the first card would be the only king. But you have to calculate the probability that any one of the cards is a king, meaning that you have to multiply the result by 5 - the probability that the second (third, fourth, fifth) card will be the only king will be the same 0.0599. And that gives you 0.299. --Martynas Patasius (talk) 20:10, 29 May 2008 (UTC)

Velocity

U=23,t=5.5 find force at m=8 —Preceding unsigned comment added by 72.14.204.136 (talk) 20:59, 29 May 2008 (UTC)

Is this a homework question? ;) Is 'm' here the mass? Is it moving at constant velocity? naerii - talk 21:08, 29 May 2008 (UTC)

Linear algebra without vector spaces

Hi. I'm working through Algebra by Michael Artin, and I'm stuck on one of the miscellaneous problems at the end of chapter 1. The problem (which has a star by it) is:

Consider a general system AX=B of m linear equations in n unknowns. If the coefficient matrix A has a left inverse A', a matrix such that A'A=I, then we may try to solve the system as follows:

AX = B

A'AX = A'B

X = A'B

But when we try to check our work by running the solution backward, we get into trouble:

X = A'B

AX = AA'B

AX = B ???

We seem to want A' to be a right inverse: AA'=I, which isn't what was given. Explain. (Hint: Work out some examples.)

Ok, so I know what's going on in this problem, and it comes down to this. If A is m x n with m<n, then it can't have a left inverse, because it represents a transformation from Rⁿ → R^m, and no transformation back from R^m → Rⁿ can be surjective.

On the other hand, if m>n, then A' can have a left inverse (depending on the rank of A), and if it does, then AA', while not being an identity matrix, is a square matrix for which B is an eigenvector with a corresponding eigenvalue of 1; in other words: A'AB = B.

Here's my problem. I'm in chapter 1 of this book, and there's nothing about vector spaces, or surjective maps, or eigenvectors, or any of that yet. All the book has presented so far are matrices as systems of equations, square matrices as products of elementary matrices, a simple version of the invertible matrix theorem, the definition of determinants, and Cramer's rule. I don't know how to use only that to establish that the coefficient matrix of an underdetermined system has no left inverse.

So... does this question make any sense, and does anyone have any idea how to help me? -GTBacchus^(talk) 21:57, 29 May 2008 (UTC)

Why should B be an eigenvector of AA' with eval 1? Suppose we have two equations in 1 unknown, x=a and 0=b. Then our 2x1 matrix A has a left inverse A', and AA' is the 2x2 matrix with rows 1,0 and 0,0. Thus B is only an evec if b=0, and indeed, if b is not zero, then there's no solution to our system of equations, even though A had a left inverse. We conclude that while A having a left inverse guarantees uniqueness of solutions, it's the right inverse that guarantees existence. Algebraist 22:08, 29 May 2008 (UTC)

Put differently: if $AX=B$, then $A'AX=A'B$, but not the other way around. -- Meni Rosenfeld (talk) 22:33, 29 May 2008 (UTC)

Applying a function (e.g., a linear transformation) to two equal quantities (an equation) is guaranteed to preserve any difference between them (create no new solutions) if and only if the function is injective. (It is guaranteed not to destroy solutions iff its domain includes the range of each quantity.) For multiplying on the left by a matrix, this means that the matrix has full column rank. If $m\times n$ A has only a left inverse, then it is "tall" ($m>n$) and has full column rank. Its left inverse, however, is "wide" and cannot have full column rank. The problem is that A is not surjective, so there are bs for which there is no solution. The equation $A\mathbf {x} =AA_{\text{left}}^{-1}\mathbf {b}$ is interesting nonetheless: the matrix $AA_{\text{left}}^{-1}$ is the projection into A's column space and arranges for a solution to exist. Projection matrices have eigenvalues of 0 and 1; it is precisely when b is in the "1" eigenspace that it is in A's column space, so it is an eigenvector as you describe. --Tardis (talk) 22:50, 29 May 2008 (UTC)

You guys rock. Thank you all for those very helpful explanations. I had to chew on it for a couple of days, but I think I've got it now. Let me try to summarize:

Since the matrix A has a left inverse, then the transformation it represents ($T\colon x\mapsto Ax$) in injective - injectiveness is a necessary condition for having a left inverse in the category of sets and mappings, so we don't need to appeal to vector spaces for that. The mapping T is injective, but we don't know that it's surjective, so B (or B's columns - we can deal with them one at a time) might not be in the range of T.

If B is in the range of T, then A', being a left inverse for A, will map B back to its preimage, namely A'B. In that case, we're all good, and even though A' may not be a right inverse for A, we know that AA' will be a square matrix that maps B to B.

On the other hand, if B is not in the range of T, then there is no solution. In that case, A' is certainly not injective, so applying its transformation to the equation AX=B creates an extraneous solution. While A'B is a solution to the equation A'AX=A'B, it is not a solution to AX=B. The point A'B in the domain of A is mapped to AA'B, which is not the same as B, although A' takes them both to A'B.

Is that all right? -GTBacchus^(talk) 20:31, 3 June 2008 (UTC)

You've about got it, but I would refine a couple of points: first, $A'=A_{\text{left}}^{-1}$ is never injective when it but not $A_{\text{right}}^{-1}$ exists — it obviously doesn't depend on B being in T's range. Also, in the same case, it should be pointed out that there are infinitely many such left inverses: we can add to any row of a $A_{\text{left}}^{-1}$ any vector that is orthogonal to all of A's columns (which is possible because it has fewer columns than the dimension of its columns) and not affect the product $A_{\text{left}}^{-1}A$. --Tardis (talk) 13:46, 4 June 2008 (UTC)