Wikipedia:Reference desk/Archives/Mathematics/2011 August 21

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August 21

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Arithmetic mean of two numbers > Geometric mean of two numbers.

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Show that   for positive x and y and  . Widener (talk) 04:53, 21 August 2011 (UTC)[reply]

Yes, this is the inequality of arithmetic and geometric means. What have you got so far? —Bkell (talk) 04:56, 21 August 2011 (UTC)[reply]
Oh there's actually a wikipedia page on this. Thanks. Widener (talk) 05:26, 21 August 2011 (UTC)[reply]

Circle geometry

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Hello. There's a circle geometry problem which has been frustrating me. Take a triangle ABC and let the bisector of angle CAB meet BC in D. Produce CB to K so that BK = AC. Construct the circle around A, K, and D and produce AB so that it cuts this circle at P. Prove that BP = DC. Thanks for any help with the proof. —Anonymous DissidentTalk 08:55, 21 August 2011 (UTC)[reply]

BK x BD = AB x BP Circle#Chord
therefore AC x BD = AB x BP
BD/CD = AB/AC Angle bisector theorem
therefore AB x CD = AC x BD = AB x BP
and so CD = BP Thincat (talk) 13:31, 21 August 2011 (UTC)[reply]
Thanks for that. I've never seen or used the angle bisector theorem, so you've been doubly helpful. —Anonymous DissidentTalk 13:52, 21 August 2011 (UTC)[reply]

Triangles

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can you tell me about centroid, circumcentre, orthocentre & incentre in isosceles and equilateral triangle separately, I mean just tell the relation existing between them i.e., they coincide or collinear etc, in both isosceles and equilateral triangles.--Krishnashyam94 (talk) 09:14, 21 August 2011 (UTC)[reply]

For any triangle the centroid, circumcentre and orthocentre lie on the Euler line. The incentre of an isosceles triangle also lies on this line. For an equilateral triangle they are all the same point. The Euler line article covers all this (and much more beside). Thincat (talk) 11:53, 21 August 2011 (UTC)[reply]

range of a transformation on the set of polynomials

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Given integers   Let   be distinct points in   and   denote the set of polynomials of degree at most  . Define a transformation   by   where  . What is the dimension of the range of T?

Experimentation leads me to believe that the answer is  , the reason for this being that a polynomial of degree   can fit   points with coordinates (x,f(x)) exactly, so this is the dimension of the range unless   in which case the dimension of the range is the dimension of   (since   is the range of T) which is  . Can someone provide a formal proof of this? Widener (talk) 09:33, 21 August 2011 (UTC)[reply]

Assuming this is correct of course. Widener (talk) 09:42, 21 August 2011 (UTC)[reply]
It's correct. You've already given the argument for why it's dimension at least  . Now suppose towards contradiction that the range contained   independent vectors. Then by appropriate linear combinations, it contains a vector   with at least   zeros which is not the zero vector. So let  . Then   is a polynomial of degree at most   with at least   zeros, which is impossible.--Antendren (talk) 10:14, 21 August 2011 (UTC)[reply]
Thanks, actually, I don't know why a polynomial of degree   can fit   points, I just know that it can, could you prove that too? Widener (talk) 10:51, 21 August 2011 (UTC)[reply]
Oh and by the way, what do you mean by "m + 1 zeros" ? Widener (talk) 10:57, 21 August 2011 (UTC)[reply]
For the first, suppose you want   for  . Then let  . You can easily check that this is a polynomial of degree   behaving as desired.
"m+1 zeros" meant there are m+1 different values of   you can plug in to the polynomial to get zero. But the fundamental theorem of algebra says that any polynomial can be re-written as  , where the   are the zeros and   is some constant. So the degree of a polynomial must be at least as large as the number of zeros it has (actually it must be exactly the same, if you count right).--Antendren (talk) 11:17, 21 August 2011 (UTC)[reply]
To nitpick a little, a polynomial can't necessarily be factored into linear terms like that unless the field it's over is algebraically closed, which   is not. Still the degree of a polynomial does provide an upper bound for the number of roots. Rckrone (talk)

Number Triangle

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Is anything known about the number triangle where each element is the sum of the three above it? What I am looking for is some kind of generating function in terms of the coordinates of the number, (similar to the relationship of binomial coefficients to pascals triangle). Here is the triangle I mean

1
1    1
2    2    1
4    5    3    1
9    12   9    4    1
21   30   25   14   5    1
51   76   69   44   20   6   1
....

Thank you. — Preceding unsigned comment added by 192.76.7.207 (talk) 14:25, 21 August 2011 (UTC)[reply]

http://oeis.org/A064189 Algebraist 16:03, 21 August 2011 (UTC)[reply]

math (pressing homework question)

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a press will produce 3000 parts per hour with a two person crew. if 60000 parts must be produced, how many hours will a four person crew be required to work,using two presses, in order to complete the work order? — Preceding unsigned comment added by 71.56.97.86 (talk) 16:13, 21 August 2011 (UTC)[reply]

Show us your work so far and we will point out any errors. StuRat (talk) 20:26, 21 August 2011 (UTC)[reply]
Also don't you think the title 'math' is a little unspecific? It is a bit like putting a label on each item in a clothes store just saying 'clothes'. Dmcq (talk) 21:05, 21 August 2011 (UTC)[reply]
I added to it. StuRat (talk) 21:49, 21 August 2011 (UTC)[reply]
Good one, I like that ;-) Dmcq (talk) 06:46, 22 August 2011 (UTC)[reply]
The OP may also wonder "If a hen and a half lays an egg and a half in a day and a half, how many and a half that lay better by half will lay half a score and a half in a week and a half?" Wikipedia lacks an article on Hen and a half. -- 110.49.233.66 (talk) 13:26, 22 August 2011 (UTC)[reply]
I'd create the article, but it would only be a half-hearted attempt inevitably producing half-assed results. StuRat (talk) 21:44, 22 August 2011 (UTC) [reply]
There are 1500 parts produced per press person hour, with a linear increase if the number of presses, persons or hours increases, which I hope is enough for you to answer the problem.→86.155.185.195 (talk) 21:54, 22 August 2011 (UTC)[reply]