Wikipedia:Reference desk/Archives/Mathematics/2011 August 20

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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)