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August 20

Recursive sequence

Define a sequence $(x_{n})$ such that $x_{1}=1$ and $x_{n}={\sqrt {1+x_{n-1}}}$ for $n\geq 2$. Prove that $1\leq x_{n}<x_{n+1}$ for all $n\geq 1$.
The most obvious approach is to use induction. The statement trivially holds for $n=1$.
Assuming $1\leq x_{n}<x_{n+1}$
$\Rightarrow 1\leq x_{n+1}$
All that's left is to show that $x_{n+1}<x_{n+2}$ but this is where I come unstuck. It is not obviously true that $x_{n+1}<{\sqrt {1+x_{n+1}}}$. Could someone show me what to do next? Widener (talk) 01:38, 20 August 2011 (UTC)

You pretty much have it. The last inequality is obviously not true for all reals, but it is true in a certain range, so it suffices to prove that x never leaves that range. A quick calculation will tell you what that range is (mainly the maximum value). Induction suffices to show that x never exceeds this maximum value. Invrnc (talk) 02:17, 20 August 2011 (UTC)

Yes, the upper bound is the golden ratio interestingly. But how does the induction show that x never exceeds this value? Widener (talk) 02:24, 20 August 2011 (UTC)

Prove that inductively. Suppose $x_{n}<\Phi$ and prove that $x_{n+1}<\Phi$. —Bkell (talk) 02:49, 20 August 2011 (UTC)

Thank you. Widener (talk) 03:51, 20 August 2011 (UTC)

Subsets

Let $I_{n}=[a_{n},b_{n}]$ be a closed interval in $\mathbb {R}$ such that $I_{n}\subseteq I_{n-1}$ for $n\geq 2$. Show that $\bigcap _{k=1}^{\infty }I_{k}$ is nonempty.
I would like to know if this is correct reasoning:
$\bigcap _{k=1}^{\infty }I_{k}$
$=\lim _{n\rightarrow \infty }\bigcap _{k=1}^{n}I_{k}$

$=\lim _{n\rightarrow \infty }I_{n}$ which contains $\lim _{n\rightarrow \infty }{a_{n}}$.
I'm not sure if taking the limit of the intersection of sets is really allowed because it's not really compatible with the formal definition of a limit. I feel that I need to be more precise and formal in my proof. Widener (talk) 04:34, 20 August 2011 (UTC)

You're right that taking the limit of the intersections is rather sketchy. Remember each intersection is a set; what's the limit of a sequence of sets? (There are ways to define it, but they would basically just come back to intersection.) However, you've also correctly identified the point you want: $a=\lim _{n\rightarrow \infty }{a_{n}}$. Prove that $a$ is an element of every $I_{n}$ (prove that it's greater than every $a_{n}$ and less than every $b_{n}$), and conclude that it's in the intersection.--Antendren (talk) 07:51, 20 August 2011 (UTC)

Statistical hypothesis testing of Poisson processes

I'm performing an experiment in which there are two Poisson processes that have been producing events for the same amount of time. So the data gathered for the experiment just consists of two numbers, n₁ and n₂, the number of events that have occurred so far for each of the two processes. The null hypothesis of the experiment is that the rate parameters of the two Poison processes are the same. How can I tell if I can reject that null hypothesis at, say, a 95% confidence level? I'd know how to test the hypothesis if I knew one of the two rate parameters exactly, but I don't have any other information available other than n₁ and n₂. n₁ and n₂ are both at least 179, in case that helps determine whether some approximation would be valid. Red Act (talk) 04:48, 20 August 2011 (UTC)

I think a chi-square test should work for you. Looie496 (talk) 06:25, 20 August 2011 (UTC)

A Bayesian approach is this. Knowing only that the observed $n_{1}$ and $n_{2}$ had poisson distributions with unknown parameters $\lambda _{1}$ and $\lambda _{2}$, the parameters have gamma distributions with mean values $n_{1}+1$ and $n_{2}+1$ and standard deviations ${\sqrt {n_{1}+1}}$ and ${\sqrt {n_{2}+1}}$. The difference $\lambda _{2}-\lambda _{1}$ has mean value $(n_{2}+1)-(n_{1}+1)$ and standard deviation ${\sqrt {(n_{2}+1)+(n_{1}+1)}}$. Now the problem is reduced to decide if you believe that zero is a sample from a distribution having mean value ${\frac {n_{2}-n_{1}}{\sqrt {n_{2}+n_{1}+2}}}$ and standard deviation one. This is the same as the chi-square test except for the '+2' in the denominator. Bo Jacoby (talk) 07:34, 20 August 2011 (UTC).

Thanks! Red Act (talk) 10:26, 20 August 2011 (UTC)

Actually the probability (or confidence) that $\lambda _{1}<\lambda _{2}$ can be computed exactly.

$P(\lambda _{1}<\lambda _{2}|n_{1},n_{2})=\int _{\lambda _{1}=0}^{\infty }{\frac {\lambda _{1}^{n_{1}}d\lambda _{1}}{e^{\lambda _{1}}n_{1}!}}\int _{\lambda _{2}=\lambda _{1}}^{\infty }{\frac {\lambda _{2}^{n_{2}}d\lambda _{2}}{e^{\lambda _{2}}n_{2}!}}$

If this number is >0.95 you may be pretty confident that $\lambda _{1}<\lambda _{2}$, and if it is <0.05 you may be pretty confident that $\lambda _{2}<\lambda _{1}$.

The expression is simplified by means of the formula

$\int _{x=a}^{\infty }{\frac {x^{n}dx}{e^{x}n!}}=\sum _{i=0}^{n}{\frac {a^{i}}{e^{a}i!}}$

and the final result is

$P(\lambda _{1}<\lambda _{2}|n_{1},n_{2})=\sum _{i=n_{1}}^{n_{1}+n_{2}}{\frac {\binom {i}{n_{1}}}{2^{i+1}}}$.

For example

$P(\lambda _{1}<\lambda _{2}|n_{1}=3,n_{2}=2)=\sum _{i=3}^{5}{\frac {\binom {i}{3}}{2^{i+1}}}$

$={\frac {\binom {3}{3}}{2^{4}}}+{\frac {\binom {4}{3}}{2^{5}}}+{\frac {\binom {5}{3}}{2^{6}}}={\frac {1}{16}}+{\frac {4}{32}}+{\frac {10}{64}}={\frac {11}{32}}$.

Bo Jacoby (talk) 15:39, 21 August 2011 (UTC).

Probabilities $P(\lambda _{1}<\lambda _{2}|n_{1},n_{2})$ are computed by the J program p=. [+/@(!%2&^&>:@])(+i.&>:) like this

  3 p 2
0.34375
  2 p 3
0.65625
  1 p 6
0.964844
  10 p 19
0.950631

Bo Jacoby (talk) 08:32, 22 August 2011 (UTC).

(1+x)^n ≥ 1+nx

Show that $(1+x)^{n}\geq 1+nx$ for natural n and $x\geq -1$.

$(1+x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}=1+nx+\sum _{k=2}^{n}{\binom {n}{k}}x^{k}$ (for n≥2. The cases n=0 and n=1 are trivial)

So now we need to show that $\sum _{k=2}^{n}{\binom {n}{k}}x^{k}\geq 0$ for $x\geq -1$. How? Widener (talk) 10:12, 20 August 2011 (UTC)

Of course it's obvious for $x\geq 0$. Widener (talk) 10:16, 20 August 2011 (UTC)

Minimize $(1+x)^{n}-(1+nx)$ on [−1,0]. Sławomir Biały (talk) 12:40, 20 August 2011 (UTC)

This is Bernoulli's inequality. You can do it inductively without much fuss. —Anonymous Dissident^Talk 00:19, 21 August 2011 (UTC)

Ratio from percent

Five oil containing trucks loaded fully with oil travel from City P to City Q. However, because of bad driving conditions, seepage occurs and so the actual quantity of oil delivered to the destination stations reduces.

The filling is done at the starting station and the containers are emptied at the destination station.

The table below gives the details about amount of oil filled at the starting station, seepage that occurred in each truck and the actual quantity of oil that was delivered to the destination (all in terms of total percentage).

Truck	Original Qty (Gallons, %)	Seepage (Gallons, %)	Delivered Qty (Gallons, %)
A	13.33	8	14.4
B	26.66	32	25.6
C	33.33	34	33.2
D	10	6	10.8
E	16.67	20	16
Total	100%	100%	100%

What is the ratio of seepage in truck B to the actual quantity of oil delivered by truck D in City Q?

I have been trying hard for the solution, but in vain. please help out - Samantha 12:49, 20 August 2011 (UTC)

Suppose there are originally X gallons of oil altogether, Y gallons lost due to seepage and Z gallons delivered. Then truck B loses 32Y/100 gallons due to seepage and truck D delivers 10.8Z/100 gallons, so the ratio of these quantities is 32Y/10.8Z. If you knew tha ratio Y/Z then you would be home and dry.

But we know that truck A starts out with 13.33X/100 gallons, loses 8Y/100 gallons and delivers 14.4Z/100 gallons. So

13.33X - 8Y = 14.4Z

and so

13.33(X/Z) - 8(Y/Z) = 14.4 (Equation 1)

Similarly, by considering the quantities for truck B, we have

26.66(X/Z) - 32(Y/Z) = 25.6 (Equation 2)

Eliminate X/Z from Equations 1 and 2 and you can find the value of Y/Z. Multiply this by 32/10.8 and you have the ratio of the seepage in truck B to the quantity delivered by truck D. Gandalf61 (talk) 13:27, 20 August 2011 (UTC)

WoW. Thanks. y/z = 1/5 and therefore Ans = 16:27. Samantha 13:36, 20 August 2011 (UTC) — Preceding unsigned comment added by 117.201.252.65 (talk)

One part of oil seeps for every five delivered? I think I'd have rephrased the question in terms of five fat men who were asked to deliver bowls of chips to some tables! ;-) Dmcq (talk) 14:33, 20 August 2011 (UTC)

derivative

I was trying to prove that the derivative of e^x is itself.I used the fact that f(a+b)=f(a)f(b) and since I don't know how to write math here, :D, I summarize it this way: I concluded that f'(x)=f'(0)f(x). now all I need to do is to prove that f'(0)=1... help!--Irrational number (talk) 17:11, 20 August 2011 (UTC)

What definition of $e^{x}$ are you using? A common definition of $e^{x}$ includes the fact that the derivative at $x=0$ is 1, as part of the definition. In other words, the number $e$ is defined to be the unique real number such that the graph of the exponential function $e^{x}$ has a tangent line of slope 1 at $x=0$. If you aren't using this definition, what is the definition of $e$ that you are using? —Bkell (talk) 17:20, 20 August 2011 (UTC)

do you mean that the f'(0)=1 is part of the function's definition? if that's the case my proof is complete...--Irrational number (talk) 17:35, 20 August 2011 (UTC)

Yes, sometimes. But you might not be using that definition. There are other ways to define the number $e$. For example, $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ is another common one. If you're using that definition, then the definition does not include the fact that the derivative of $e^{x}$ at $x=0$ is 1. So you need to go back and find out exactly what definition of $e$, and what definition of $e^{x}$, you are working with. —Bkell (talk) 17:38, 20 August 2011 (UTC)

actually, I knew from Wikipedia that the derivative for e^x is itself, using the fact that it's an exponential function (and therefore f(a+b)=f(a)f(b)) I proved that f'(x)=f'(0)f(x) which is apparently true for all exponential functions.then I wondered that if I can prove that for e^x f'(0)=1, then my proof is complete.is it like "e is the number for which f'(0) (in my equation is) is 1" and therefore my proof is somehow wrong?(sorry for my stupidity :D!)--Irrational number (talk) 17:49, 20 August 2011 (UTC)

It depends on what you have to start with. Every proof in mathematics starts with definitions, postulates, and previously proved theorems, and uses these to prove something new. So it is important to know what definitions you are using. What do you already know about $e^{x}$? Here are two common ways to define the number $e$:

1. $e$ is the unique real number such that the graph of the exponential function $e^{x}$ has a tangent line of slope 1 at $x=0$.
2. $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$.

Some textbooks use the first definition, and some textbooks use the second definition. (There are other definitions too—these are not the only ones.) As it turns out, both of these definitions define the same real number, 2.718281828…, so either definition is fine; but this fact requires proof. If you use definition 1, you know that the derivative of $e^{x}$ at $x=0$ is 1, because that's part of the definition—but you do not know immediately from that definition that $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$; that requires proof. Likewise, if you use definition 2, you know from the definition that $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$, but you do not know immediately that the derivative of $e^{x}$ at $x=0$ is 1; that requires proof.

In your proof, you are starting with some definition of $e$, along with some previously proved theorems about the function $e^{x}$ (for example, you are using the previously proved theorem that $e^{a+b}=e^{a}e^{b}$). If the definition of $e$ that you are starting with is definition 1 above, then you do not need to prove that the derivative of $e^{x}$ at $x=0$ is 1, because that fact is part of the definition. However, if the definition of $e$ that you are starting with is definition 2 above, or some other definition, then you do not know from that definition that the derivative of $e^{x}$ at $x=0$ is 1, so you will need to prove that fact from the definition you have. —Bkell (talk) 18:02, 20 August 2011 (UTC)

so what I understand is whether I need to prove that f'(0)=1 or not depends on which definition I'm using...well I prefer the easy one :D.but what about others like 2^x? how is f'(0) found for that?(which in this case is ln2)?--Irrational number (talk) 18:10, 20 August 2011 (UTC)

You say that you prefer the easy one. Right—so does everyone. :-) But if you are proving this statement for a class, you probably do not get to choose what definition to use—your instructor or your textbook has probably given a definition for $e$, and that is the definition you must use.

For $2^{x}$, use the properties of exponents and logarithms to write $2^{x}=(e^{\ln 2})^{x}=e^{(\ln 2)x}$, and then use the things you know about derivatives and about the derivative of $e^{x}$. —Bkell (talk) 18:14, 20 August 2011 (UTC)

cool! I get it now... thanks!--Irrational number (talk) 18:17, 20 August 2011 (UTC)

? (what is infinity times zero?)

Hey, I've been trying to figure this out: what's infinity times zero? 64.166.145.51 (talk) 23:51, 20 August 2011 (UTC)

• See Indeterminate form. --Kinu ^t/_c 00:00, 21 August 2011 (UTC)

Uhh... 64.166.145.51 (talk) 00:11, 21 August 2011 (UTC)

The answer really depends. What do you mean by "infinity"? What do you mean by "zero"? What do you mean by "times"? In what context is this question being asked? —Bkell (talk) 00:14, 21 August 2011 (UTC)

What I meant by my comment above was that "infinity" means lots of different things in mathematics, depending on context. For example, maybe you are thinking of calculus. Maybe you have a function $f(x)$ with $\lim _{x\to \infty }f(x)=\infty$ (that's one kind of "infinity"), and you have another function $g(x)$ with $\lim _{x\to \infty }g(x)=0$, and you are asking about the value of $\lim _{x\to \infty }f(x)g(x)$. This is one way to interpret the question, "What is infinity times zero?" In this case, the answer is indeterminate, as Kinu answered above. The answer could be anything, depending on exactly what functions $f(x)$ and $g(x)$ are:

• If $f(x)=x^{2}$ and $g(x)=1/x^{3}$, then $\lim _{x\to \infty }f(x)g(x)=0$.
• If $f(x)=ax^{2}$ and $g(x)=1/x^{2}$, then $\lim _{x\to \infty }f(x)g(x)=a$. Here $a$ can be any positive real number. If you want the answer to be a negative real number, just use $g(x)=-1/x^{2}$ instead.
• If $f(x)=x^{2}$ and $g(x)=1/x$, then $\lim _{x\to \infty }f(x)g(x)=\infty$. If you want an answer of $-\infty$, use $g(x)=-1/x$.
• If $f(x)=x^{2}(2+\sin x)$ and $g(x)=1/x^{2}$, then $\lim _{x\to \infty }f(x)g(x)$ does not exist.

So in this interpretation of your question, there is no single answer. That's why, in calculus, "$\infty \times 0$" is an "indeterminate form."

But maybe you don't mean "infinity" to be interpreted that way. Maybe what you mean by "infinity" is the cardinality of an infinite set (but even then there are infinitely many "infinities"). If that's what you mean, then by "zero" you probably mean the cardinality of the empty set, and by "times" you probably mean the cardinality of two sets' Cartesian product. Under this interpretation of the question "What is infinity times zero?", the answer is zero, because the Cartesian product of an empty set with any other set, even an infinite set, is empty.

Or maybe you have a different meaning in mind. Maybe you meant "infinity" as in the extended real number line. In this case, the expression "$\infty \times 0$" either is left undefined, so there is no answer to your question, or is considered to be 0, as is the common convention in probability and measure theory. So with this interpretation the answer might be 0, or there might be no answer, depending on the convention in use.

Or maybe you are in non-standard analysis, and by "infinity" you mean an infinite hyperreal number. In this case, as long as "zero" really means 0, and not an infinitesimal hyperreal number, then the answer to your question is 0; the product of 0 and an infinite hyperreal number is 0. But that's not true if "zero" can mean an infinitesimal hyperreal number—if you allow that interpretation (admittedly out of place in non-standard analysis), then the product could be any hyperreal number at all.

Maybe you don't mean any of these things. Maybe you have a different meaning in mind when you say "infinity times zero." My point is that your question is meaningless on its own. "Infinity" is not a single thing—it is a broad label for a wide collection of quite different concepts. —Bkell (talk) 03:46, 21 August 2011 (UTC)

TRo give a concrete example; the chances of picking a particular random place between home and town depend on the size of the place. A point has no size and the chance of picking a particular random point - and infinitely long decimal fraction of the distance, is zero. However for the whole line with its infinity of points somehow that infinity of zero probabilities has summed up to a certainty. Dmcq (talk) 10:04, 21 August 2011 (UTC)

August 21

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

Show that ${\frac {x+y}{2}}>{\sqrt {xy}}$ for positive x and y and $x\neq y$. Widener (talk) 04:53, 21 August 2011 (UTC)

Yes, this is the inequality of arithmetic and geometric means. What have you got so far? —Bkell (talk) 04:56, 21 August 2011 (UTC)

Oh there's actually a wikipedia page on this. Thanks. Widener (talk) 05:26, 21 August 2011 (UTC)

Circle geometry

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 Dissident^Talk 08:55, 21 August 2011 (UTC)

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)

Thanks for that. I've never seen or used the angle bisector theorem, so you've been doubly helpful. —Anonymous Dissident^Talk 13:52, 21 August 2011 (UTC)

Triangles

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)

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)

range of a transformation on the set of polynomials

Given integers $n,m\geq 1$ Let $x_{1},...,x_{n}$ be distinct points in $\mathbb {R}$ and $\mathbb {P} _{m}$ denote the set of polynomials of degree at most $m$. Define a transformation $T:\mathbb {P} _{m}\rightarrow \mathbb {R} ^{n}$ by $T(f)={\begin{bmatrix}f(x_{1})\\\vdots \\f(x_{n})\end{bmatrix}}$ where $f\in \mathbb {P} _{m}$. What is the dimension of the range of T?

Experimentation leads me to believe that the answer is $\min(n,m+1)$, the reason for this being that a polynomial of degree $m$ can fit $m+1$ points with coordinates (x,f(x)) exactly, so this is the dimension of the range unless $m+1>n$ in which case the dimension of the range is the dimension of $\mathbb {R} ^{n}$ (since $\mathbb {R} ^{n}$ is the range of T) which is $n$. Can someone provide a formal proof of this? Widener (talk) 09:33, 21 August 2011 (UTC)

Assuming this is correct of course. Widener (talk) 09:42, 21 August 2011 (UTC)

It's correct. You've already given the argument for why it's dimension at least $\min(n,m+1)$. Now suppose towards contradiction that the range contained $m+2$ independent vectors. Then by appropriate linear combinations, it contains a vector $v$ with at least $m+1$ zeros which is not the zero vector. So let $v=T(f)$. Then $f$ is a polynomial of degree at most $m$ with at least $m+1$ zeros, which is impossible.--Antendren (talk) 10:14, 21 August 2011 (UTC)

Thanks, actually, I don't know why a polynomial of degree $m$ can fit $m+1$ points, I just know that it can, could you prove that too? Widener (talk) 10:51, 21 August 2011 (UTC)

Oh and by the way, what do you mean by "m + 1 zeros" ? Widener (talk) 10:57, 21 August 2011 (UTC)

For the first, suppose you want $f(a_{i})=b_{i}$ for $0\leq i\leq n$. Then let $f(x)=\sum _{i=0}^{n}b_{i}\prod _{j\neq i}(x-a_{j})/(a_{i}-a_{j})$. You can easily check that this is a polynomial of degree $n$ behaving as desired.

"m+1 zeros" meant there are m+1 different values of $x$ 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 $c\prod (x-a_{i})$, where the $a_{i}$ are the zeros and $c$ 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)

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 $\mathbb {R}$ is not. Still the degree of a polynomial does provide an upper bound for the number of roots. Rckrone (talk)

Number Triangle

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)

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

math (pressing homework question)

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)

• "The reference desk will not do your homework for you." --Kinu ^t/_c 17:02, 21 August 2011 (UTC)

Show us your work so far and we will point out any errors. StuRat (talk) 20:26, 21 August 2011 (UTC)

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)

I added to it. StuRat (talk) 21:49, 21 August 2011 (UTC)

Good one, I like that ;-) Dmcq (talk) 06:46, 22 August 2011 (UTC)

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)

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)

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)

August 22

Simplify a sum involving binomial coefficients

Perhaps the sum $\sum _{i=n}^{m}{\frac {\binom {i}{n}}{2^{i}}}$ (where 0≤n≤m) can be simplified. Please tell me! The result would be very useful. (See here). Bo Jacoby (talk) 09:05, 22 August 2011 (UTC).

It's a special value of the hypergeometric function. I get

$2-{\binom {m+1}{n}}2^{-1-m}{}_{2}F_{1}(1,m+2;m-n+2;1/2).$

There are various identities one can try to play with. This is also equal to

$2-{\binom {m+1}{n}}2^{-m}{}_{2}F_{1}(1,-n;m-n+2;-1).$

I can't find an easy closed form, though. Sławomir Biały (talk) 14:18, 22 August 2011 (UTC)

Thank you very much Sławomir ! This will eventually help me speeding up my program for large values of n and m. Bo Jacoby (talk) 08:16, 23 August 2011 (UTC).

I can see it's very similar to the Binomial Theorem (bottom formula) with {{{1}}}2^-1}}, except that the binomial coefficient is upside-down. That might be a dead-end, but one could consider wandering through the proofs? SamuelRiv (talk) 21:08, 25 August 2011 (UTC)

Trajectories, gravity, acceleration, straightforward calculus

It's been too long since school when I last did any calculus. I was reading this article about the dangers of firing guns into the air earlier and it quotes a ballistics expert as saying bullets "go a long way up when they're fired". The question is, how high? If a bullet is accelerated by its firing to a speed of somewhere between 120 to 1700 ms⁻¹ (according to the muzzle velocity article) and gravity acts to decelerate it at 9.81 ms⁻², at what height does it stop going up? I believe this is a straightforward calculus problem, but I can't remember how to frame it! I can picture a graph of height against time as an inverse parabola intersecting the origin and a point on the x-axis corresponding to the time when it hits the ground (neglecting air resistance, because the bullet will reach terminal velocity on its way back down), and I know that I need the y value at the point of inflection, where the derivative (speed) is zero, but this is about as far as I get. Beorhtwulf (talk) 18:54, 22 August 2011 (UTC)

If I recall correctly, tehn we can think about it like this. If it starts at 120 to 1700m/s, and comes to instantaneous rest at the top, its average speed will be 60 to 850m/s. The time it spends doing this is the time it spends decelerating, or 120 to 1700 divided by 9.81 (assuming g is constant in both cases) which is between 12 and 173. So the total height reached is one value times the other, or 720 to 147,000m, which is quite a range if you did mean 120 to 1700m/s, I have no idea of muzzle velocity myself. Clearly this overly simplistic - 147km up is way more than the atmosphere, so clearly there's a problem modelling it like that. Grandiose (me, talk, contribs) 20:49, 22 August 2011 (UTC)

Basic approach: figure out the time at which the vertical velocity reaches zero using $v_{0}=gt$, then plug that time into $h=v_{0}t-1/2gt^{2}$. Note that for a muzzle velocity of 1700 m/s (= Mach 5), air resistance will be huge and the formula will be miles off the mark. Looie496 (talk) 21:33, 22 August 2011 (UTC)

Energy conservation tells you that hg=v²/2. Bo Jacoby (talk) 08:12, 23 August 2011 (UTC).

That would agree with my figures (thankfully) - however absurd they are. Grandiose (me, talk, contribs) 13:30, 23 August 2011 (UTC)

Unfortunately, I don't think you can neglect air resistance at any point in the bullet's motion - I think drag will be a significant force on the bullet at all points in its trajectory. There is some discussion of this in our external ballistics article; the bottom line seems to be that mathematical modelling of drag on bullets is complicated. Gandalf61 (talk) 13:47, 23 August 2011 (UTC)

Getting a Formula Given Points: Step-Like Function

I've got a series of points: for x of 1-4, y is 0. For x of 5-13 (9 points), y is 1. For x of 14-23 (10 points), y is 2. For x of 24-32 (9 points), y is 3. For x of 33-41 (9 points), y is 4. For x of 42-50 (9 points), y is 5. All further steps are 9 points wide. How would I go about making a formula that matches these points? ΣΑΠΦ (Sapph)^Talk 20:38, 22 August 2011 (UTC)

Could you clarify where this problem comes from? If it is a homework problem we won't do it, although we may be able to offer suggestions for how to approach it. Looie496 (talk) 21:25, 22 August 2011 (UTC)

It comes from a diet program. Carbs are worth points, in the amounts listed above. Instead of using a lookup table, I want to have an automated widget on my desktop/homepage that can give me results. And programming the lookup table directly is . . . inelegant. ΣΑΠΦ (Sapph)^Talk 11:26, 23 August 2011 (UTC)

See lagrange interpolation and trigonometric interpolation. Bo Jacoby (talk) 08:07, 23 August 2011 (UTC).

A "formula" is not the right tool for describing such a function (though you could get a decent fit with enough trigonometric terms). What's wrong with using a piecewise definition? -- Meni Rosenfeld (talk) 08:52, 23 August 2011 (UTC)

I should have been more specific - I'm not necessarily looking for a formula like y=5x. I'm looking for something I might plug into a spreadsheet or webpage or program to automate the calculation. So something like round(x/5+int(x/10)) or whatever would be just fine. ΣΑΠΦ (Sapph)^Talk 11:26, 23 August 2011 (UTC)

Using Iverson brackets: f(x)=[5≤x]+[14≤x]+[24≤x]+[33≤x]+[42≤x]. Bo Jacoby (talk) 13:20, 23 August 2011 (UTC).

The points you've described are almost described by simply rounding x/9 to the nearest integer, except that x-values of 23 and above should have 1 subtracted from them first, before the division (because of that one interval of width 10). The formula round(x/9) will give you the right value most of the time. To accommodate that one strange interval, either put in a line of code beforehand that subtracts 1 from x if x is greater than or equal to 23, or try something like round((x-(x>=23))/9), which will work if the >= operator yields 1 for true and 0 for false. In Excel the formula =ROUND((x-(x>=23))/9,0) will work. —Bkell (talk) 15:14, 23 August 2011 (UTC)

Bkell, that worked great. As it turned out, I found another source that stated the ratio as 19/175 - which is pretty close to 1/9, but accounts for the discrepancy at 23. Thanks again. ΣΑΠΦ (Sapph)^Talk 22:38, 23 August 2011 (UTC)

Ah, I see. So if the ratio is really 19/175, then it's not quite true that "all further steps are 9 points wide"—there is another width-10 interval somewhere down the road, and another one after that eventually, and so on. —Bkell (talk) 22:47, 23 August 2011 (UTC)

Derivation of Poiseuille's Law

Hello. I read the derivation of Poiseuille's law here. I would like to know the steps immediately prior to rearranging ${\frac {1}{\eta }}{\frac {\Delta P}{\Delta x}}={\frac {d^{2}v}{dr^{2}}}+{\frac {1}{r}}{\frac {dv}{dr}}$ by using the chain rule to ${\frac {1}{\eta }}{\frac {\Delta P}{\Delta x}}={\frac {1}{r}}{\frac {d}{dr}}r{\frac {dv}{dr}}$. Links to the calculus techniques involved would be greatly appreciated. Thanks in advance. --Mayfare (talk) 23:55, 22 August 2011 (UTC)

From the product rule we have

${\frac {1}{r}}{\frac {d}{dr}}\left[r{\frac {dv}{dr}}\right]={\frac {1}{r}}\left[r\left({\frac {dv}{dr}}\right)'+(r)'{\frac {dv}{dr}}\right]={\frac {1}{r}}\left[r{\frac {d^{2}v}{dr^{2}}}+{\frac {dv}{dr}}\right]={\frac {d^{2}v}{dr^{2}}}+{\frac {1}{r}}{\frac {dv}{dr}}.$

Not sure where the chain rule comes in. I'm also assuming the position of the square brackets on the left-hand side; the notation in the article is a little ambiguous in my eyes. —Anonymous Dissident^Talk 07:46, 23 August 2011 (UTC)

How does the product rule come to mind without knowing the next step? --Mayfare (talk) 19:27, 23 August 2011 (UTC)

Because I noticed we're considering the derivative of a product of two functions f and g, where f(r) = r and g(r) = dv/dr. —Anonymous Dissident^Talk 21:33, 23 August 2011 (UTC)

August 23

Multiplying a Factorial by Another Factorial

When it comes to dividing a factorial by another factorial, you can "cancel" them out. But when multiplying a factorial by another factorial, is there a shortcut there too, or is it just figure out what factorials are, then multiply them? --Thebackofmymind (talk) 10:10, 23 August 2011 (UTC)

Your question is like the following: When it comes to dividing an X by another X, you can "cancel" them out. But when multiplying an X by another X, is there a shortcut there too, or is it just figure out what X-s are, then multiply them?

Hope this helps.

HOOTmag (talk) 10:25, 23 August 2011 (UTC)

The only identity I can think of, for m>n, is $n!m!=(n!)^{2}(m\cdot (m-1)\cdot ...\cdot (n+1))$, that may or may not make the mental calculation easier. Widener (talk) 11:20, 23 August 2011 (UTC)

So, for example, 9! × 11! = (9!)² × 10 × 11. StuRat (talk) 09:19, 24 August 2011 (UTC)

Factorization

Help me factorize X squared + 4x + 3 + mx + 3m — Preceding unsigned comment added by Agdesi (talk • contribs) 15:02, 23 August 2011 (UTC)

Here is a hint: gather terms in the same power of x together and rewrite the expression as

$x^{2}+(m+4)x+3(m+1)$

Gandalf61 (talk) 15:05, 23 August 2011 (UTC)

Also, try to group the "unknowns" into a similar format, as in $x^{2}+((m+1)+3)x+3(m+1)$. Now, "m+1" is in both terms containing m. -- kainaw™ 16:15, 23 August 2011 (UTC)

Why the long tail is just a 2 variables graph and not a 3 variables graph??

You have the amount or % of the different products being sold. The amount of % of population. And the amount or % of products sold. So you have 3 variables, but use a 2d graph to explain long tail.201.78.194.18 (talk) 20:54, 23 August 2011 (UTC)

Have you looked at the long tail article? It's all explained there. — Fly by Night (talk) 21:51, 23 August 2011 (UTC)

August 24

Real cube roots of the identity matrix

Does a solution for $\mathbf {A} ^{3}=\mathbf {I}$ exist, where $\mathbf {I} \neq \mathbf {A} \in \mathbb {R} ^{3\times 3}$? I have been trying to find a solution for hours. Thanks, 85.250.176.130 (talk) 08:18, 24 August 2011 (UTC)

$\mathbf {A} ={\begin{pmatrix}-{\frac {1}{2}}&-{\frac {\sqrt {3}}{2}}&0\\{\frac {\sqrt {3}}{2}}&-{\frac {1}{2}}&0\\0&0&1\end{pmatrix}}$

Bo Jacoby (talk) 08:36, 24 August 2011 (UTC).

This is a perfect example of where it's easier to think of composition of linear transformations, rather than multiplication of matrices. What linear transformation composed with itself 3 times will return the identity matrix? Well, clearly, if you rotate around an arbitrary axis with angle 120°, then you get such a function, and then you only need to confirm that the function is, in fact, a linear transformation. --COVIZAPIBETEFOKY (talk) 13:09, 24 August 2011 (UTC)

A simpler solution is

$\mathbf {A} ={\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}}$

Bo Jacoby (talk) 16:46, 24 August 2011 (UTC).

... which is a rotation about the line x = y = z. Gandalf61 (talk) 17:04, 24 August 2011 (UTC)

Problem from Spivak's Calculus

Let f be two-times differentiable with f(0) = 0, f(1) = 1, and f'(0) = f'(1) = 0. Prove that |f''(a)| ≥ 4 for some a in (0,1). Spivak hints that we should try to prove that either f''(a) ≥ 4 for some a in (0,1/2), or f''(a) ≤ −4 for some a in (1/2, 1), but I'm not sure how to do that. I managed to prove the claim for the case when f has a maximum point over (0,1) in (0, 1/2), but I'm not convinced this is the approach he's suggesting. Thanks for the help. —Anonymous Dissident^Talk 21:44, 24 August 2011 (UTC)

Try a proof by contradiction. Suppose $f''<4$ on the interval $(0,1/2)$. Using $f'(0)=0$ and $f(0)=0$, find a bound for $f(1/2)$. Repeat using $f''>-4$ on $(1/2,1)$, $f'(1)=0$, $f(1)=1$ to get a different bound. Conclude a contradiction.--Antendren (talk) 22:04, 24 August 2011 (UTC)

Success! Cheers. Maybe I should stop considering contradiction as a last resort... —Anonymous Dissident^Talk 08:11, 25 August 2011 (UTC)

Actually no. I thought I had it but now things went wrong. I began with the false premises. Then I used the mean value theorem on the interval (0, x) to show that f'(x) < 4x for all x in (0, 1/2], and then the MVT again to show f(x) < 4x², whence f(1/2) < 1. Similar methods on [1/2, 1) lead merely to f(1/2) > 0, which isn't helpful. What have I done wrong? —Anonymous Dissident^Talk 12:00, 25 August 2011 (UTC)

The bound $f(x)<4x^{2}$ is too weak. Hint: $(2x^{2})'=4x$. -- Meni Rosenfeld (talk) 13:38, 25 August 2011 (UTC)

Let x=f(t) be the position of a car standing still on the position x=0, then speeding up at t=0 at maximum acceleration x ' '=4 until t=1/2, when it brakes at maximum deceleration x ' '=−4, until t=1. Then x=1 and x '=0. This is the only motion with |x ' '|≤4 that satifies the boundary conditions. But f(t) is not two times differentiable at the time t=1/2 when speeding up is suddenly changed to braking. A two times differentiable motion satisfies the sharp inequality f ' '(a)>4 for some a. Bo Jacoby (talk) 13:52, 25 August 2011 (UTC).

Given that the problem as stated is even easier, calling only for proof that |f''(a)| ≥ 4 somewhere (not strictly greater), how far can the differentiability condition be relaxed? Would it be sufficient that f be a differentiable function, twice differentiable almost everywhere, or would f need to be twice differentiable at all by finitely many locations? -- 110.49.248.74 (talk) 00:11, 26 August 2011 (UTC)

yes. If f is unbounded, we're done. Otherwise f is in L^1, so the FTC holds. Sławomir Biały (talk) 00:30, 26 August 2011 (UTC)

August 25

Logarithmic Spiral

What's the reason why the logarithmic spiral is so common in nature? The article mentions self-similarity, but could someone be more specific? 74.15.137.168 (talk) 02:49, 25 August 2011 (UTC)

Only God knows the true reason why, but man can guess. Symmetries may reuse code in the DNA. These symmetries are often broken in later stages of development or evolution. There is a high level of symmetry between arm and leg on a human, and even higher symmetry between right arm and left arm. There is also a kind of symmetry between a small boy and a big boy. The organs grow. The change of teeth is an exception. The logarithmic spiral has a continuous symmetry group, and so it may grow using a short piece of DNA code. Bo Jacoby (talk) 08:00, 25 August 2011 (UTC).

I think the fundamental underlying reason is that exponential growth and exponential decay are common in nature, and when something changes exponentially while circling around a fixed center, you naturally get a logarithmic spiral. Looie496 (talk) 16:03, 25 August 2011 (UTC)

Oh okay, so because ${\frac {dr}{d\theta }}={\frac {dr}{dt}}{\frac {dt}{d\theta }}={\frac {v}{\omega }}\propto r$, logarithmic spirals are common? 74.15.137.168 (talk) 17:18, 25 August 2011 (UTC)

That's a reasonable way of looking at it, but it also ignores all the processes (developmental, evolutionary, geo-chemical, etc) that go into forming such spirals in nature (which others allude to above). You may be interested in reading up on phyllotaxis and links therein (this is the start of the whole 'pineapple bracts follow the fibonacci sequence' business). You might also get better answers on the science desk, this isn't really a math question :) SemanticMantis (talk) 20:29, 25 August 2011 (UTC)

Proof too easy?

Good morning. I was to show rigourously that f(x)=x³-4x+6 is not one to one (injective, I think it's also called?). This is my proof (contradiction), but something doesn't sit right with me about it; it seems too easy. Have a look:
Suppose that f(m)=f(n) if and only if m=n
$m^{3}-4m+6=n^{3}-4n+6$ (m=n)
$(m^{3}-n^{3})-(4m-4n)=0$ (6s cancel, bring both to one side)
$(m-n)(m^{2}+mn+n^{2})-(m-n)4=0$ (factoring)
$(m-n)(m^{2}+mn+n^{2}-4)=0$ (grouping)
$m^{2}+mn+n^{2}-4=0$ (The expression is zero when either the first or the second factor is zero; I'm not interested in the first because it is the same as the hypothesis)
By the quadratic formula:
$(*)\qquad m={\frac {-n\pm {\sqrt {n^{2}-4(n^{2}-4)}}}{2}}$
$n={\frac {-n\pm {\sqrt {n^{2}-4(n^{2}-4)}}}{2}}$ (m=n)
$\rightarrow n={\frac {2{\sqrt {3}}}{3}}$ However, putting n back into $(*)$, I get $m={\frac {2{\sqrt {3}}}{3}},{\frac {-4{\sqrt {3}}}{3}}$, contradiction Therefore, f(m)=f(n) for at least one pair (m,n): m≠n 203.117.33.23 (talk) 23:24, 25 August 2011 (UTC) Is this proof legit? Thanks so much, and sorry for wasting so much of your time. PS: What happened to all the other pairs, because in the graph there is obviously more than one. THanks again.

Didn't look too closely at your proof above, but you're probably meant to just notice that f(2) = f(0). Staecker (talk) 00:00, 26 August 2011 (UTC)

OK now I looked at your proof. It's definitely suspicious. Your assumption, in order to derive a contradiction, is: "f(n)=f(m) iff n=m". By your argument (which is fine) this is equivalent to "$(m-n)(m^{2}+mn+n^{2}-4)=0$ iff m=n". Your use of the quadratic formula is fine, but why then substitute n for m? You're not assuming that m=n, you're assuming the more complicated "iff" statement. Staecker (talk) 00:06, 26 August 2011 (UTC)

The first derivative has two distinct real roots, so there are portions with positive and negative slope, which can't be true for a one to one function. I don't know what your standard of rigour has to be, but something could be worked up round this fact.←86.155.185.195 (talk) 12:03, 26 August 2011 (UTC)

August 26

Limit to infinity

If $\lim _{x\rightarrow \infty }f(x)$ exists, is it always true that $\lim _{x\rightarrow \infty }f(x)=\lim _{x\rightarrow \infty }f(x+1)$? Intuitively I would think so, but I'm not sure how I would prove it.Widener (talk) 06:43, 26 August 2011 (UTC)

Yes. This is easy to show using the definition of limit. -- Meni Rosenfeld (talk) 07:16, 26 August 2011 (UTC)

Same N ? Widener (talk) 08:12, 26 August 2011 (UTC)

That depends on what exactly you mean by N (that is, which limit it applies to). You can certainly use N₂ = N₁-1, but x>N ⇒ x>N-1. -- 110.49.248.138 (talk) 20:02, 26 August 2011 (UTC)

FLAW in Cauchy's proof for the inequality of arithmetic and geometric means?!

The article inequality of arithmetic and geometric means gives the following proof by Cauchy, for the subcase when n = 2.
${\begin{aligned}x_{1}&\neq x_{2}\\[3pt]x_{1}-x_{2}&\neq 0\\[3pt]\left(x_{1}-x_{2}\right)^{2}&>0\\[3pt]x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}&>0\\[3pt]x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}&>4x_{1}x_{2}\\[3pt]\left(x_{1}+x_{2}\right)^{2}&>4x_{1}x_{2}\\[3pt]{\Bigl (}{\frac {x_{1}+x_{2}}{2}}{\Bigr )}^{2}&>x_{1}x_{2}\\[3pt]{\frac {x_{1}+x_{2}}{2}}&>{\sqrt {x_{1}x_{2}}}\end{aligned}}$
How is the last step in this proof valid? Widener (talk) 08:18, 26 August 2011 (UTC)

By assumption, $x_{1}$ and $x_{2}$ are non-negative, so $x_{1}x_{2}$ and ${\frac {x_{1}+x_{2}}{2}}$ are non-negative. Square-root is an increasing function on the non-negative reals, so it preserves order.--121.74.96.240 (talk) 08:26, 26 August 2011 (UTC)

Okay, so generally, what is the proof that $x^{2}>y\Rightarrow x>{\sqrt {y}}$ for $x,y\geq 0$ ? Widener (talk) 08:38, 26 August 2011 (UTC)

Instead of showing that, I'll show $x^{2}>y^{2}\Rightarrow x>y$. Replace $y$ with ${\sqrt {y}}$ for what you want.

Now, suppose not. Then $x\leq y$. So $x^{2}\leq xy\leq y^{2}$. Contradiction.--Antendren (talk) 08:55, 26 August 2011 (UTC)

Cumulative position probabilities as a result of tossing dice

Let's say you roll a die or dice and move a token forward by that number of positions. What's the probability of hitting each position, if multiple throws are allowed ? Let me give you an example with one die:

Chance of reaching position 1 = 1/6 (the chance of rolling a 1)

Chance of reaching position 2 = 1/6 (the chance of rolling a 2) + 1/36 (the chance of rolling a 1, twice in a row)

So, the chances of hitting each spot should go up with increasing distance from the starting spot, approaching a limit. Note that the chances, when added up, will be more than 100%, since the token will land on multiple spots. So, do we have a chart of these probabilities, for one die and for when rolling a pair of dice ? Also, what is the theoretical limit probability for the cases of one die or two dice ? StuRat (talk) 18:26, 26 August 2011 (UTC)

With one standard six-sided die, your token advances an average of 3.5 positions per roll, so the limit of its probability of landing on any one position far enough along is 2/7. -- 110.49.248.138 (talk) 20:20, 26 August 2011 (UTC)

Also note (again for one die), that the probability of the token landing on any one position is 1/6 times the sum of the probabilities of landing on the previous six positions (which, as you pointed out, will sum to greater than unity). This yields an easy to calculate algorithm. For my HP-48G, I wrote the program << 6 DUPN + + + + + 6 / >>, seeded the stack with 0 0 0 0 0 1, and ran the program repeatedly, yielding:

 1: 0.166667
 2: 0.194444
 3: 0.226852
 4: 0.264660
 5: 0.308771 HI
 6: 0.360232
 7: 0.253604 LO
 8: 0.268094
 9: 0.280369
10: 0.289288
11: 0.293393 HI
12: 0.290830
...
38: 0.28571632
39: 0.28571480
40: 0.28571365
...

This agrees with your probabilities for positions 1 and 2, and approaches the limit of 2/7 ≈ 0.28571429. Note that 6 is the most likely position to land on. For two dice, you will have to weight your average appropriately, but it is easy to see that the limit is 1/7. -- 110.49.227.131 (talk) 20:59, 26 August 2011 (UTC)

Interesting. I was wondering if it would behave like this, first rising above the limit, then dropping below, and continuing to bounce back and forth at ever reducing magnitudes as it approaches the limit. If you'd be willing to do the same for 2 dice, I'd like to see those results, too. StuRat (talk) 21:30, 26 August 2011 (UTC)

Sure. For two dice the odds of throwing a 1, 2, ..., 12 are, in thirty-sixths, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1. So my quick and dirty HP-48G program becomes << 12 DUPN 0 * SWAP 1 * + SWAP 2 * + SWAP 3 * + SWAP 4 * + SWAP 5 * + SWAP 6 * + SWAP 5 * + SWAP 4 * + SWAP 3 * + SWAP 2 * + SWAP 1 * + 36 / >> (I suppose that I really should be using a loop here), and the stack is initialized to 0 0 0 0 0 0 0 0 0 0 0 1.

  1: 0
  2: 0.027778
  3: 0.055556
  4: 0.084105
  5: 0.114198
  6: 0.146626
  7: 0.182227 HI
  8: 0.166346
  9: 0.155526
 10: 0.147784
 11: 0.141275
 12: 0.134199
 13: 0.124704 LO
 14: 0.138567
 15: 0.145771
 16: 0.148355 HI
 17: 0.147890
 18: 0.145672
 19: 0.142886
 20: 0.140764
 21: 0.140745 LO
 22: 0.141558
 ...
 48: 0.142850
 49: 0.142855
 50: 0.142859
 ...

peaking on Lucky Number 7 and with the limit of 1/7 = 0.142857... . I am curious about the period of the oscillations about the limit -- whether and how it changes as the values approach the limit. My sense is that the period remains fairly steady at between 8 and 9. I'm sure that this type of hunting behavior is well studied in numerical analysis. -- 110.49.235.126 (talk) —Preceding undated comment added 02:35, 27 August 2011 (UTC).

Thanks. I hope you don't mind, but I went ahead and labelled the HI and LO points above. The distances between them seem rather chaotic. I wonder if they smooth out more later on. StuRat (talk) 05:50, 27 August 2011 (UTC)

(Edit Conflict) I don't mind at all. I've added a couple of extra points to the lists so that they end one point after an extremum, moving both of your "questioned" labels, but they are still too short to gather more than an impression on initial behavior, and they are probably approaching the point of being too long to post on this ref desk. My statement about the 8-9 period was based on casual observation of positions out in the range of 40 - 60, but I really need to run this on a computer to get better analysis -- and to avoid the hassle of copying figures over from a calculator. I'll do a bit more work on this tonight, and contact you via your talk page. -- 110.49.225.10 (talk) 07:12, 27 August 2011 (UTC)

Generating functions can often be used to investigate such questions. In the first case, if the denominator is taken to be 6ⁿ, then a generating function for the numerator is given by 1/(1-x-6x²-6²x³-6³x⁴-6⁴x⁵-6⁵x⁶. The denominator this function has a factor of 1-6x. The limit of probability can be found using partial fractions: the generating function can be written (2/7)/(1-6x)+(another term) and as long as the roots (real and complex) of the denominator of the other term have absolute value greater than 1/6, the 1-6x term dominates. I don't see an obvious way of proving this, though it's apparently the case given the table above. Note that the anon's proof above that the limit is 2/7 has a subtle flaw in that it assumes the limit exists in the first place. The analysis of two die case can be done similarly but is more complex.--RDBury (talk) 07:01, 27 August 2011 (UTC)

Wouldn't another approach be to consider the numbers in batches of 6 (or 12 for the two dice case), and construct a matrix A that calculates the probabilities of the next 6 given the probabilities of the current 6 (this matrix would be fairly complex, since you'd have to consider up to 6 tosses of the die)? Then hope it's diagonalizable, and if so, compute the limit.--Antendren (talk) 08:32, 27 August 2011 (UTC)

That's a great idea! We can even simplify the matrix by instead calculating the 6 position probabilities counting back from the very next position, given the 6 counting back from the current position. Thus, your A is my M⁶. For the one die case:

    [ 1/6 1/6 1/6 1/6 1/6 1/6 ]      [ 1 ]
    [  1   0   0   0   0   0  ]      [ 0 ]
    [  0   1   0   0   0   0  ]      [ 0 ]
M = [  0   0   1   0   0   0  ], S = [ 0 ].
    [  0   0   0   1   0   0  ]      [ 0 ]
    [  0   0   0   0   1   0  ]      [ 0 ]

And for the two dice case:

    [  0   1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 ]      [ 1 ]
    [  1    0    0    0    0    0    0    0    0    0    0    0   ]      [ 0 ]
    [  0    1    0    0    0    0    0    0    0    0    0    0   ]      [ 0 ]
    [  0    0    1    0    0    0    0    0    0    0    0    0   ]      [ 0 ]
    [  0    0    0    1    0    0    0    0    0    0    0    0   ]      [ 0 ]
    [  0    0    0    0    1    0    0    0    0    0    0    0   ]      [ 0 ]
M = [  0    0    0    0    0    1    0    0    0    0    0    0   ], S = [ 0 ].
    [  0    0    0    0    0    0    1    0    0    0    0    0   ]      [ 0 ]
    [  0    0    0    0    0    0    0    1    0    0    0    0   ]      [ 0 ]
    [  0    0    0    0    0    0    0    0    1    0    0    0   ]      [ 0 ]
    [  0    0    0    0    0    0    0    0    0    1    0    0   ]      [ 0 ]
    [  0    0    0    0    0    0    0    0    0    0    1    0   ]      [ 0 ]

Then the probability of landing on position n is the top value of Mⁿ S, which, given S, is the top left value of Mⁿ. Note that the column vectors:

[ 2/7;  2/7;  2/7;  2/7;  2/7;  2/7]

and

[ 1/7;  1/7;  1/7;  1/7;  1/7;  1/7; 1/7;  1/7;  1/7;  1/7;  1/7;  1/7]

are fixed points under their respective M -- that is, they are eigenvectors with eigenvalue 1. Does that help? (That would seem to be a necessary condition for them to be limits, but not necessarily sufficient.) -- 110.49.242.138 1(talk) 12:17, 27 August 2011 (UTC)

The eigenvectors above are overspecific, given that any scalar multiple of an eigenvector is also an eigenvector. So what I should have pointed out is that the "one-vector" (sometimes called the "summing vector")

[1; 1; ...1]

is an eigenvector with eigenvalue 1 for any such matrix with a finitely supported probability distribution, given in block form:

p_1 ... p_n-1 | p_n
--------------+----
              |
      I       |  0
              |

where sum(p_i) = 1. So the point should have been that fixed points exist, not that the specific limit alone is a fixed point. -- 110.49.241.90 (talk) 23:44, 28 August 2011 (UTC)

That the roots of the denominator of the second term have absolute value at least 1/6 can be proved with winding numbers. If z traces the circle |z|=1/6, given by z=1/6(cos t + i sin t), then the curve traced by the denominator is given by w=1 + 5/6 cos t + 4/6 cos 2t + 3/6 cos 3t + 2/6 cos 4t + 1/6 cos 5t + i(5/6 sin t + 4/6 sin 2t + 3/6 sin 3t + 2/6 sin 4t + 1/6 sin 5t). I claim that the real part of w is always at least 1/2, in fact 1/2 + 5/6 cos t + 4/6 cos 2t + 3/6 cos 3t + 2/6 cos 4t + 1/6 cos 5t = 1/6(cos 5/2t + cos 3/2t + cos 1/2)² as can be shown using various trigonometric identities. Then the curve traced by w stays to the right of the imaginary axis and therefore its winding number around 0 is 0, implying that there are no roots of the polynomial inside the original circle.--RDBury (talk) 08:25, 27 August 2011 (UTC)

OK, thanks for the feedback, everyone. 110.49.225.10 placed 100 lines of output for the 1 die case (and 150 lines of output for the 2 dice case) on my talk page. StuRat (talk) 07:21, 28 August 2011 (UTC)

I have a question about fractals.

My question is does anyone know if fractals has found any applications in psychology? I was asking this because I took a random system of thought (Theology, actually); and it seems to be based on a fractal. Please let me know. Thanks! Lighthead ^þ 22:49, 26 August 2011 (UTC)

By the way I realize that Theology is not exactly quantifiable. Roughly speaking. Lighthead ^þ 22:51, 26 August 2011 (UTC)

Let me explain so that I don't look like a complete idiot. Let's take Judeo-Christian Philosophy. God (1). Angels (.5) Humans (.25) Animals, perhaps (.125) And so on. Lighthead ^þ 23:00, 26 August 2011 (UTC)

Keep explaining- I still don't see anything like fractals in what you're describing. (And I don't know if fractals have applications in psychology- I doubt it.) Staecker (talk) 23:21, 26 August 2011 (UTC)

Well maybe I don't understand fractals then. I'm not mathematical. And thanks for doubting, I'm sure that if people would've doubted in the first place the idea of fractals would've gotten off the ground. Mandelbrot was very popular. Especially in his views. Lighthead ^þ 23:52, 26 August 2011 (UTC)

The concept of a fractal is that a part of the system has the same structure as the whole system, but at a smaller scale. For your example to represent a fractal, humans would have to be parts of God and also have the same structure as God -- in Christian theology humans may be considered to have the same form of God, but they are not considered parts of God. So that isn't really a fractal in any sense that I understand. I'm not aware of any applications of fractals to psychology. (There are claims to have found fractal structure in brain electrical activity, if that can be counted as psychology.) Looie496 (talk) 00:16, 27 August 2011 (UTC)

Well, as for religions, perhaps the "repeating the same pattern at different scales" part may apply to some time scales in the Hindu and native Central, and South American religions, in particular. I believe they have the "cycles within cycles" concept as central parts of their religions. StuRat (talk) 00:39, 27 August 2011 (UTC)

As to the people that thought that I was saying nonsense; I meant Theology from the socio-psychological standpoint. That's actually why I gave the disclaimer that I realize that God is not quantifiable. I don't believe that God is a construct, but I was attacking it from the standpoint that those from the scientific standpoint at least objectively see it as a construct. I think that StuRat came close to understanding what I'm talking about. And by the way it's not too far fetched to see psychological processes as mathematical. How do you explain game theory, and how they've found the actions of terrorist cells and dictatorial governments to follow mathematical principles? People like Staecker should avoid the knee-jerk reaction of treating everything that people say as nonsense. I might not be very mathematical, but I do have a capacity to think theoretically. Lighthead ^þ 03:00, 27 August 2011 (UTC)

I actually don't want to get too far into Theology because that was just an example; but even under the Judeo-Christian perspective, God is said to be a part of everything and therefore a part of us (from what Looie496 was saying). And also to what Looie496 was saying, yeah, electrical brain activity would count from what I'm saying. Electrical brain activity has been researched as perhaps an indicator of what people are thinking. That seems to answer most if not all of what I'm saying. Lighthead ^þ 04:12, 27 August 2011 (UTC)

August 27

Show that the set of bounded functions is infinite dimensional

Define the set of bounded functions as the set of functions for which there exists $c\in \mathbb {R}$ such that $|f(x)|<c$ for all real x. Anyway I guess the approach is to show that, given any arbitrary finite linearly independent subset of this set of functions, show that their exists another bounded function which is not in the span of them. How do you do that, or is there a better approach? Widener (talk) 03:38, 27 August 2011 (UTC)

Hint: Think about the characteristic functions of singletons — functions that are zero everywhere except at one point, where they take the value 1. Given a finite set of such functions, can you see a way to find a bounded function not in their span? --Trovatore (talk) 03:46, 27 August 2011 (UTC)

(or more to the point, a function that is in the span of all characteristic functions of singletons, but not in the span of the given subset) --Trovatore (talk) 03:49, 27 August 2011 (UTC)

Obviously any other characteristic function of singletons will do. But that only shows that the set of characteristic functions of singletons is infinite dimensional, it does not show that all bounded functions are infinite dimensional. Widener (talk) 03:55, 27 August 2011 (UTC)

The span of the set of characteristic functions of singletons is a subspace of the space of all bounded functions. --Trovatore (talk) 04:02, 27 August 2011 (UTC)

I'm guessing that if A is a subspace of B and A is infinite dimensional, that implies B is infinite dimensional. Can you prove that? Widener (talk) 04:08, 27 August 2011 (UTC)

Try it the other direction. Take a subspace of a finite-dimensional space. Can you show it's finite-dimensional? --Trovatore (talk) 04:10, 27 August 2011 (UTC)

Okay thanks. Widener (talk) 04:23, 27 August 2011 (UTC)

Previous problem with generalized dice

Suppose in the problem above, instead of a fair die you use a weighted die, so the probability of getting i is p_i (i>0). We don't need to assume the number of sides is 6 or even finite. The argument given above shows that if limit of the probability (call this q_n) that position n will be reached exists, then that limit is 1/E(i), provided of course that E(i) exists. This won't always be the case though, for example if p₂=1 and all other p_i are 0, then the probability of reaching n is 0 if n is odd and 1 if n is even, so there is no limit. There are also distributions with no expected value. But if you add conditions to allow for these possibilities then is it true that the limit always exists? Specifically, prove or disprove: If gcf{i: p_i>0}=1 then lim(n→∞) q_n = 1/E(i) if E(i) exists and lim(n→∞) q_n = 0 if E(i) does not exist.--RDBury (talk) 09:07, 27 August 2011 (UTC)

Can you solve this?

What is 12/20 - 15/20 — Preceding unsigned comment added by Lightylight (talk • contribs) 20:28, 27 August 2011 (UTC)

Yes I can solve it :) Did you try it out, what is 12-15 dollars for instance? How about if the dollars are replaced by nickels (5 cents or a twentieth of a dollar) Dmcq (talk) 20:34, 27 August 2011 (UTC)

Holy shit, I am blown away!

You have presented the problem whose solution will solve... ALL OF MATHEMATICS. --COVIZAPIBETEFOKY (talk) 21:06, 27 August 2011 (UTC)

Is the sarcasm really necessary? It's a perfectly valid question, although it does seem a little strange age-group wise. Grandiose (me, talk, contribs) 21:08, 27 August 2011 (UTC)

Yes. --COVIZAPIBETEFOKY (talk) 00:51, 28 August 2011 (UTC)

The only possible source of confusion on this problem is the order of operations. The standard order of operations specifies that the division be done before the subtraction:

(12/20) - (15/20) = (12-15)/20

You could get different answers, though, if you used a different order of operations:

12/(20-15)
    20

But, assuming you meant to use the standard order of operations, just complete the first one, noting that you will get a negative fraction. StuRat (talk) 21:58, 27 August 2011 (UTC)

Its a bit deeper than that. Problems like this are one of the biggest stumbling blocks in mathematical development, I guess about half the population, many adults included, would not be able to do it. Conceptually it represents the first generalisation of number after the integers, it involves a whole new techniques for manipulating numbers which follow counter intuitive rules. Why do you only subtract the tops not the bottoms as well? Why are the add/subtract rules so different to multiply rules? How come two different fractions are equal to each other. And to make matters worse the sums going to have a negative result as well, what the heck does minus half an apple mean?

So to try and give some meaning, take a sheet of paper, fold it in four one way, fold it in five the other way. Cut along the fold so you get 20 bits. All the bits together make one whole, each bit is 1/20th. Try the easier sum 15/20 - 12/20. Count out 15 bits take away 12, how many is left, whats that as a fraction?. Play with 1/2, 1/4, 1/5 all of which can be made from the bits, add and subtract them and see what happens. --Salix (talk): 23:22, 27 August 2011 (UTC)

Commas in non-decimal numbers

Is it okay to use commas to separate digits in numbers that are not in base ten? For example, instead of writing 10001000100011100011001011011010000100111011101100110000000011100010110010010₂, could one write 10,001,000,100,011,100,011,001,011,011,010,000,100,111,011,101,100,110,000,000,011,100,010,110,010,010₂? -Metroman (talk) 23:54, 27 August 2011 (UTC)

While the comma is a commonly used digit group separator, there are many options, and your choice should be based on context. The comma is certainly a permissible choice for your own use, but a space or an underscore ("_") is more commonly used with binary numbers. The space or thin space is more likely to be seen in general texts, and the underscore in computer science. The underscore is a permissible digit group separator symbol for constants (of any base, including decimal) in many computer languages because the comma is reserved to separate values.

Also note that you need not group only by threes as you are used to in decimal (although several cultures use different groupings of two or four in decimal numbers, and may even vary the group size within a single number as with Indians writing 7,00,00,00,000 where you may be used to 7,000,000,000 -- see lakh). Grouping a binary number in three suggests that while you are writing it in binary, you are representing it in support of octal. These days it is more common to group in fours, suggesting support of hexadecimal, or eights, marking byte (or octet) boundaries. -- 110.49.242.14 (talk) 01:16, 28 August 2011 (UTC)

It's problematic. See Digit group separator. Groups of 3 hints at decimal numbers. I wouldn't use binary groups of 3 except in special circumstances where octal numbers are very common. PrimeHunter (talk) 01:18, 28 August 2011 (UTC)

August 28

Kernel of a transformation on the set of polynomials

Given integers $n,m\geq 1$ Let $x_{1},...,x_{n}$ be distinct points in $\mathbb {R}$ and $\mathbb {P} _{m}$ denote the set of polynomials of degree at most $m$. Define a transformation $T:\mathbb {P} _{m}\rightarrow \mathbb {R} ^{n}$ by $T(f)={\begin{bmatrix}f(x_{1})\\\vdots \\f(x_{n})\end{bmatrix}}$ where $f\in \mathbb {P} _{m}$. What is the dimension of the kernel of T?

The case m<n is trivial. Otherwise any such polynomial in the kernel of T can be written as $a\left(\prod _{i=1}^{n}(x-x_{i})\right)\left(\prod _{i=n+1}^{m}(x-c_{i})\right)$ where $c_{i}\in \mathbb {C} ,a\in \mathbb {R}$ are arbitrary (free?) variables. Is it sufficient to claim that the answer is $m-n+1$ because that is the number of free variables, or do I need to show something more (like find a basis for the kernel) ? Widener (talk) 11:57, 28 August 2011 (UTC)

You know the dimension of the image from the question on Aug. 21. Use dim(ker)+dim(im)=dim(domain).--RDBury (talk) 15:27, 28 August 2011 (UTC)

Yes, I know that, but I would like to know if it can be done without knowing dim(im) beforehand. Widener (talk) 17:38, 28 August 2011 (UTC)

You've more or less already answered your own question: any polynomial in the kernel can be written (uniquely) as $\left(\prod _{i=1}^{n}(x-x_{i})\right)p(x)$ for some $p\in P_{m-n}$. Sławomir Biały (talk) 17:49, 28 August 2011 (UTC)

Partial Derivative of Volume of Cone

Hello. Why is the partial derivative of the volume of a cone not its surface area unlike the ball (solid sphere)? Thanks in advance. --Mayfare (talk) 15:24, 28 August 2011 (UTC)

The reason it works for a ball is that a small change in radius radius results in an shell with uniform thickness; it has approximately the same volume as area times the thickness. That reasoning fails for a cone. It also depends on the how you measure the size of the sphere. For example the formulas for area and volume in terms of diameter are V=1/6 π d³ and A = π d²; the derivative is off by a factor of 2 in that case. There are reasons that the derivative of volume formula is the area formula for a sphere, but it's kind of a one-off.--RDBury (talk) 15:39, 28 August 2011 (UTC)

Cone terminology

There are four different things I think of as a "cone":

1) What most people call a "cone", that is, the portion trimmed between a plane and a vertex point.

2) A "semi-infinite cone", trimmed only to the vertex point.

3) An "infinite cone", with two lobes, unbounded at both ends.

4) A "conic frustrum", with a single lobe trimmed at both ends by planes.

So, am I using the proper terminology to describe these ? Is there a way to specify the first one that makes it clear I'm not talking about the others ? StuRat (talk) 17:49, 28 August 2011 (UTC)

See the disambiguation page cone. Those aren't too different and are only a small subset of its meanings. Dmcq (talk) 18:13, 28 August 2011 (UTC)

Romanian math

I noticed that in several international math journals, a disproportionally large number of problems were submitted by Romanians. Why is this? Are Romanians naturally better at math or something? --76.211.88.37 (talk) 19:41, 28 August 2011 (UTC)

No they're terrible at maths and can't solve any problems so they're trying to get other people to help them ;-) They've done a lot of fine maths over the years. Also mathematics has a good public reputation there, the original maths olympiad was held by Romania for instance. Dmcq (talk) 21:20, 28 August 2011 (UTC)

You find that a lot of the old Eastern Block countries are the same. They weren't touched by commercialism like the West was. Things like a good education still come before being popular and famous. In the West, being clever isn't cool. — Fly by Night (talk) 22:55, 28 August 2011 (UTC)

A question about $\zeta (0)$

I know that when evaluated this produces $-{\frac {1}{2}}$

I'm not quite sure why this is the answer, but I was wondering if there was any significance of this being -.5 and whether or not there are any other values z such that $\zeta (z)=-{\frac {1}{2}}$ And if there were, what the significance of THAT would be. As you can probably tell, I'm new to this function :-) Aacehm (talk) 21:14, 28 August 2011 (UTC)