Wikipedia:Reference desk/Archives/Mathematics/2007 November 3

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November 3

Limit problem with e

Hello everyone, here is an identity that is well known but I can't seem to find out why. The question is how do we prove that

$\lim _{n\rightarrow \infty }\left(1-{\frac {1}{n}}\right)^{n}=e$

Can someone shed some light on this assuming that we already know that

$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}=e$

A Real Kaiser 03:51, 3 November 2007 (UTC)

The difficulty might be that the first equality is not correct, $\lim _{n\rightarrow \infty }\left(1-{\frac {1}{n}}\right)^{n}=e^{-1}$

You can check this is reasonable since $e>1$ just like $(1+{\tfrac {1}{n}})^{n}>1$, while $e^{-1}<1$ just like $(1-{\tfrac {1}{n}})^{n}<1$.

While I think there is a clever substitution to turn the first limit into the reciprocal of the second, I would tend to define $a_{n}=(1-{\tfrac {1}{n}})^{n}$, then consider $\log(a_{n})=\log(1-{\tfrac {1}{n}})/{\tfrac {1}{n}}$ whose limit is an indeterminate form of ${\tfrac {0}{0}}$ and apply L'Hospital's rule. This results in some messy fractions that eventually simplify to ${\tfrac {n}{1-n}}\to -1$, so $\log(a_{n})\to -1$ and $a_{n}\to \exp(-1)=e^{-1}$. JackSchmidt 04:49, 3 November 2007 (UTC)

The way to do it is a lot easier than substitution. See what happens when you raise both sides of your equation to the (-1/n) power... SamuelRiv 06:43, 3 November 2007 (UTC)

Jack, you are absolutely right. I just made a stupid typo while I was writing the question. You guessed the question I intended. My question was assuming that

$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}=e$

how can we show that

$\lim _{n\rightarrow \infty }\left(1-{\frac {1}{n}}\right)^{n}={\frac {1}{e}}$

which is equivalent to showing that

$\lim _{n\rightarrow -\infty }\left(1+{\frac {1}{n}}\right)^{n}=e$

And I had thought about taking the natural log and seeing what happened but I hadn't done it on paper and it does work as you have shown. I guess my question was is there a simpler proof (that wouldn't involve L'Hospital's rule).

And Samuel, how can one raise both sides to the power of -1/n because the right side must be independent of n and then on the right we are taking limit? If we do it, then the right side is dependent on n.
A Real Kaiser 06:59, 3 November 2007 (UTC)

An approach that does not use L'Hospital's rule - given that

$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}=e$

then we know that the sequence $\left(1+{\frac {1}{n}}\right)^{n}$ converges to a non-zero limit, so we can conclude that

$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{-n}=e^{-1}$

Now we have

$\left(1+{\frac {1}{n}}\right)^{-n}=\left({\frac {n+1}{n}}\right)^{-n}=\left({\frac {n}{n+1}}\right)^{n}\Rightarrow \lim _{n\rightarrow \infty }\left({\frac {n}{n+1}}\right)^{n}=e^{-1}$

and also

$\lim _{n\rightarrow \infty }\left({\frac {n}{n+1}}\right)=1$

so

$\lim _{n\rightarrow \infty }\left(1-{\frac {1}{n}}\right)^{n}$

$=\lim _{n\rightarrow \infty }\left({\frac {n-1}{n}}\right)^{n}$

$=\lim _{m\rightarrow \infty }\left({\frac {m}{m+1}}\right)^{m+1}$ [ replacing n by m+1]

$=\lim _{m\rightarrow \infty }\left({\frac {m}{m+1}}\right)^{m}\lim _{m\rightarrow \infty }\left({\frac {m}{m+1}}\right)$ [note that we have already shown that both of these limits exist]

$=e^{-1}.1=e^{-1}$ Gandalf61 11:10, 3 November 2007 (UTC)

Hi Gandalf61. The article on exponentiation needs your expertise. The subsection on powers of e is not good enough. There has been some discussion on the talk page. Bo Jacoby 15:59, 3 November 2007 (UTC).

Wow Gandalf, thanks! This is exactly what I was looking for.
A Real Kaiser 21:59, 3 November 2007 (UTC)

there is also another proof:

$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}=e$

$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{na}=e^{a}$

and by replacing n by m/a :

$\lim _{{\frac {m}{a}}\rightarrow \infty }\left(1+{\frac {a}{m}}\right)^{m}=e^{a}$,

$\lim _{m\rightarrow \infty }\left(1+{\frac {a}{m}}\right)^{m}=e^{a}$

now, if a=-1 you get

$\lim _{m\rightarrow \infty }\left(1-{\frac {1}{m}}\right)^{m}=e^{-1}$

--George 04:44, 4 November 2007 (UTC)

I'm not sure this is entirely valid. You have assumed that as $m\to +\infty$, also ${\tfrac {m}{a}}\to +\infty$, but this does not hold if a is negative. -- Meni Rosenfeld (talk) 13:24, 4 November 2007 (UTC)

Metrics

What is the difference between a fluid ounce and an ounce? —Preceding unsigned comment added by 69.138.248.241 (talk) 14:11, 3 November 2007 (UTC)

See fluid ounce (volume) and ounce (mass). PrimeHunter 14:51, 3 November 2007 (UTC)

If you have water or something of the same density as water, and the gravity is one g, then one fluid ounce will also have a weight of approximately one avoirdupois ounce. StuRat 17:17, 3 November 2007 (UTC)

Calculating odds of a stock market close

Hi everyone, I am curious to know how to calculate the probability that the Dow will close up or down by a certain number of points. Is this even possible? --Yoyoceramic 19:34, 3 November 2007 (UTC)

It's probably possible but so horribly complicated that it is probably impractical to even try. A math-wiki 21:54, 3 November 2007 (UTC)

If you have a mathematical model for the Dow index that models it as a stochastic process, it may be possible to calculate such a probability. In that case, whether that is a meaningful thing to do depends very much on the question whether the model is descriptively adequate.  --Lambiam 22:00, 3 November 2007 (UTC)

You can always model probability. The question is how accurate you want it. You can calculate the probability that two random numbers are coprime exactly. If you try something similar for the Dow, starting with the basic elements and their behavior and working your way up, you will never get that kind of exact accuracy. There are many influences on the Dow that are very complex themselves. Someone at Apple deciding not to bother with putting tri-band functionality on the iPhone because he had a fight with his girlfriend that morning. That can be an influence on the Dow. But more important than this insane complexity is the chaotic nature of the stock market. It feeds back into itself, causing small changes to be magnified. Just one investor buys one share less, everything else staying the same, may radically change what the Dow will do tomorrow. So let's say you want to base your probability model on as good a measurement of the state of the world in general and the stock market specifically as you can get. If you're off by just a couple of dollars here or there, your model will diverge wildly from reality after just a couple of days.

Of course, you don't need a perfect model to get a probability. If I go by the actual knowledge that I have of how the stock market works (ie. none whatsoever), I can get a pretty decent probability estimate: .5 probability for ending up, .5 probability for down. Given two options, that what any probability will be with no additional information. You can expand this a little by looking at the last ten days, and dividing the number of days the Dow ended up by ten to get you estimate for the it ending up tomorrow.

Which probability model you prefer depends rather on what you want to use it for. One of the most accurate ways of predicting the weather, by average precision, is just predicting yesterday's weather. That may get you a decent score over a whole year, but if you're commanding an oil tanker, you want a weather prediction that focuses on predicting sudden changes. I guess it's just the same with models for the stock market. risk 22:40, 3 November 2007 (UTC)

Linear Separability

Is there an algorithm I could apply to find all linearly separable functions $f:\{0,1\}\times \{0,1\}\mapsto \{0,1\}$? Of course I can set up a system of inequalities for each of these functions and solve it or derive a contradiction, but I think there must be a more elegant way to do so. Do you know one? Falk Lieder 21:28, 3 November 2007 (UTC)

Well, for this simple case it's easy. For fourteen of the sixteen cases, it's easy to exhibit a line that separates the 0s from the 1s. For the other two (which are really the same, by symmetry), it's not difficult to prove that no line can do it. Are you looking for a way to do it for more general domain and codomain sets, or something? —Keenan Pepper 01:05, 4 November 2007 (UTC)

I also found that among the Boolean functions with two arguments all except for the XOR and the NOT XOR function are linearly separable. Proving that these two are not linearly separable was not a problem. However finding the separating line for each and every of the remaining fourteen functions seems tedious. Assume I wanted to find all pairs of linearly separable subsets in $\mathbb {R} ^{2}$. In this case there are infinitely many pairs of subsets. How would you solve this problem? Falk Lieder 19:54, 6 November 2007 (UTC)

For every line there are pairs of subsets of R² that are separated exclusively by that line. It is not quite clear what you mean by "finding" all pairs. There are awfully many of those for any given line.  --Lambiam 23:34, 6 November 2007 (UTC)

You are right. My question was not well thought out. I'm sorry. Falk 20:28, 7 November 2007 (UTC)

For larger numbers of variables, some of the references here might be helpful: the titles seem to imply that they listed all such functions for seven and eight variables. —David Eppstein 05:20, 4 November 2007 (UTC)

(after edit conflict) A table of the number of linearly separable hypercubes is given in Linearly separable. Apparently this was the topic of an M.Sc. Thesis. They appear to be the same as the so-called threshold functions on n variables (sequence A000609 in the OEIS). There is a treatment in Section 7.1.1 of Volume 4A(one) of TAOCP, which can be found on-line as a "pre-fascicle" for alpha-testing: Pre-Fascicle 0b (Boolean basics), p29ff.  --Lambiam 05:51, 4 November 2007 (UTC)

Triple Integrals in Cylindrical Coordinates

Use cylindrical coordinates to find the indicated quantity

Volume of the solid bounded above by the sphere x² + y² + z²= 9, below by the plane z=0 and laterally by the cylinder x²+y²=4.

I tried it with the z limits of integration from 0 to r² r limits of integration from 0 to 2cos(θ) θ limits of integration from 0 to л/2

with the function f(dz,dr,dθ) as r dzdrdθ

Then for an answer i got something like 6.02 or very very close to 2л.

I am curious if this volume makes sense for an answer?? —Preceding unsigned comment added by 64.113.82.26 (talk) 22:58, 3 November 2007 (UTC)

Subtracting the cylinder equation from that of the sphere immediately tells us that z² = 5, where z is the height of the intersection circle. Archimedes told us that the area of a sphere is the same as that of a circumscribed cylinder. (He was so proud of this discovery he asked that it be illustrated on his tombstone.) So we can treat this as a big cylinder (radius 3) stacked on top of a small cylinder (radius 2). Then we need neither integrals nor cylindrical coordinates, we just multiply base times height, twice, and add. That should give an easy way to check the answer you get by doing it the required way. --KSmrq^T 23:53, 3 November 2007 (UTC)

That is for area, but isn't the question for volume? Somewhat symbolically, if you have ∫∫∫rdzdrdθ = ∫∫∫rdrdθdz, and your function is radially symmetric (independent of θ), you can simplify ∫∫rdrdθ = (∫rdr)×(∫dθ) = ¹/₂r²×2п = πr² (surprise – but not really), leaving the volume to be determined as π∫r²dz. Directly using the formulas for the volumes of a cylinder and a spherical cap gives me a result that is numerically close to 34.9 33.1. The cylinder alone already has a volume that is much larger than 2п. —Preceding unsigned comment added by Lambiam (talk • contribs) 06:23, 4 November 2007 (UTC)

Oops, good catch; I was mixing dimensions. The volume of the cylindrical part is still base times height, and substantially exceeds the proposed number. The volume of the spherical cap is less than that of its enclosing cylinder, and does require a different treatment — though still a simple one. (Archimedes proved as well that the volume of a sphere is ²⁄₃ that of the enclosing cylinder; but unlike the area result this doesn't apply to caps.)

Treating the cap as a cone would give an underestimate volume of ⁴⁄₃π(3−√5), while treating it as a cylinder would give an overestimate volume three times larger; that should suffice for a sanity check. To get an exact result while not providing the homework calculation, let's treat the cap as stacked cylinders with changing radius. At height 3 the radius is 0, at height √5 the radius is 2, and between at height 3−t the radius is (3²−(3−t)²)^1/2. The cap volume should then be

$\int _{0}^{3-{\sqrt {5}}}6\pi (t-t^{2})\,dt\approx 5.\,\!$

By my calculations, the total volume appears to be approximately 33. Back to you, Lambiam! :-) --KSmrq^T 16:03, 4 November 2007 (UTC)

Right, now that I recompute it, I also get (a bit more than) 33.  --Lambiam 19:37, 4 November 2007 (UTC)

Another way (which is easier if you happen to have access to a CAS) is to calculate $\int _{0}^{2}2\pi r{\sqrt {9-r^{2}}}\ dr$ (the justification is left as an exercise to the reader), which indeed is slightly over 33. -- Meni Rosenfeld (talk) 17:42, 4 November 2007 (UTC)