Mathematics desk
< January 21	<< Dec \| January \| Feb >>	January 23 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 22

What are the exponential functions?

Let b>0. When I was a kid, my textbook defined b^n as repeated product and b^{1/n} as n-th root. Then b^r was defined for all rationals. For every real x, chose r_n->x and defined b^x=lim b^{r_n}. We had to prove that b^x is independent of the choice of r_n. Derived all properties of b^x based on b^r.

When I was an adult, another textbook developed differential and integral calculus first. Then defined log x as the integral of 1/t from 1 to x>0. Then defined b^x in terms of log-function, etc.

When I am old now, my kid asked me what is b^x. I do not know what to say because I might be very outdated, too ignorant to know any new approaches. Is there any better way other than the above ones? Thank you in advance. Twma 01:36, 22 January 2007 (UTC)[reply]

So you want to know how to describe what an exponential function is? Okay, let

b,x\in \mathbb {C} ,b\neq 0

. Then

b^{x}

is an exponential. When

x\in \mathbb {N}

, then it can be described as the number opf times b is multiplied by itself; similarly, when

x\in \mathbb {Q}

, then it is the root (to the denominator) of the aforementioned repeated multiplication (remembering that negative powers put the result under 1). Outside of that

(x\in \mathbb {C} \backslash \mathbb {Q} )

the situation is far different, and in such a case, a thorough reading of Exponentiation may be required. --JB Adder | Talk 02:39, 22 January 2007 (UTC)[reply]

I see two distinct questions here, because we have two different students: you and your kid! Is it realistic or helpful to have a definition based on complex analysis when we first encounter exponents at a young age? I think not. If we're lucky, we can lay a solid foundation that supports and encourages later, more sophisticated, superstructures, but that allows practical computation and reasoning today.

My impulse is to frame exponentiation as a homomorphism (without using the term explicitly). At first, we map positive integers,

{\begin{aligned}1&{}\mapsto b&2&{}\mapsto b\times b&3&{}\mapsto b\times b\times b&4&{}\mapsto b\times b\times b\times b\\1&{}\mapsto b&s(1)&{}\mapsto (b)\times b&s(2)&{}\mapsto (b^{2})\times b&s(3)&{}\mapsto (b^{3})\times b\end{aligned}}

I am using s here to abbreviate "successor", the primitive "add one" function of the Peano axioms. Then we gradually extend the mapping, just as we extend our number systems. We include zero, b⁰ = 1; and, crucially, we show that we have a morphism of addition,

n+m\mapsto b^{n}\times b^{m}.\,\!

This is the familiar rule that b^n+m = bⁿ×b^m, which naturally leads us to consider negative integers, and to map these to reciprocals. Then we see that the homomorphism extends to multiplication, with the rule b^n×m = (bⁿ)^m, and expand our mapping to include rational numbers, with reciprocals mapping to roots.

The expansion from rational numbers to real numbers is qualitatively more demanding in the development of number systems, and that carries over to our homomorphism. Essentially, we need some kind of continuity argument to define x ↦ b^x for real x. In teaching a youngster, I think we can stop there.

But the key idea of a homomorphism, mapping not just the numbers (including their continuity/topology), but also their algebra (addition, identity, subtraction, multiplication, reciprocal), does not stop here. We might eventually use, say, a power series to define e^z for complex z. We can even define a matrix exponential in the same way. And the theory of Lie groups and their associated Lie algebras develops this into the vital exponential map. A version of this map also works on Riemannian manifolds, with a differential structure instead of a group structure.

At all stages of development, our understanding and our computations draw on this idea of "structure maps to similar structure". So this is what I would emphasize, both to support elementary work and to provide a foundation for later growth. --KSmrq^T 15:27, 22 January 2007 (UTC)[reply]

Okay, yes, I did overdo it by going into complex analysis there, KSmrq, and I apologise to both you and Twma for doing that. --Jb-adder (from 129.180.1.224) 23:41, 22 January 2007 (UTC)[reply]

Following your suggestions, I answered my kids, one in high school and one at second year in university. Both of them, independently in two different time slots, made me a cup of tea without any comment. I have no idea whether they accept my answer. Twma 04:10, 23 January 2007 (UTC)[reply]

My experience suggests that learning mathematics depends partly on the teacher, but even more on the active participation of the student. If I stand in front of a class and give a lecture on juggling three balls in a cascade pattern, with demonstration, will any of the students walk away with newly acquired juggling skills? I don't think so. Likewise, to learn about exponents the students must do calculations, and confront challenges to their understanding. One thing that often works well is to have students teach other students. In this case, ask the university kid to explain for the high school kid, to set some exercises, and to pose some challenge questions; it'll help both of them!

Two valuable psychological benefits from teaching are, first, it instills humility (when you find out you need a deeper understanding to answer questions), and second, it instills confidence (when you finally succeed). Many younger students (and many older ones, too) find mathematics intimidating, even if they don't admit it. Heck, I find mathematics intimidating! Building courage, even just the courage to ask questions, can therefore be an important aspect of teaching. --KSmrq^T 20:11, 23 January 2007 (UTC)[reply]

I enjoy the social relationship with members of this site more than mathematics itself. Thanks. Twma 02:27, 24 January 2007 (UTC)[reply]

Linear equation solver program for TI 84+: need help

For the past month, I have been trying to make a Prgm for my TI-84+ that solves a linear equation using 1-2 variables (ex. 14x+2=7x-9)*just a unplausible example It must show work

any help would be appreciatd 71.99.94.201 02:31, 22 January 2007 (UTC)[reply]

You could have it guess solutions with a for loop. You could also use matrices to solve a system of linear equations. Or you could upgrade to a TI-89 :-). —Mets501 (talk) 02:35, 22 January 2007 (UTC)[reply]

To use matrices, input an augmented matrix, into say, [A], go to MATRIX → MATH, and press B to select the "rref(" command, short for reduced row echelon form. Then select matrix A as your argument, and press enter. If the equations make sense, to the left side will be 0s and 1s in diagonal lines, with the variables in order in the far right column. You must line up your terms, and input their coefficients. The program should be quite simple for you to do. Prompt "Number terms", ask for the coefficients, create the matrix, rref( it, and Disp. X [Mac Davis] (DESK|How's my driving?) 02:04, 23 January 2007 (UTC)[reply]

Sorry it took so long to thank you that helped. I entered the original message from my [PSP] and encountered issues when going back.

Anyone can give me an example? NEHS Class of 2009 15:43, 29 January 2007 (UTC)[reply]

Example of what specifically? [Mαc Δαvιs] X (How's my driving?) ❖ 12:14, 1 February 2007 (UTC)[reply]

cone

what is the figure formed when a right angled triangle is revolved around its hypoteneuse.04:27, 22 January 2007 (UTC)~

two cones with bases tangent? ChowderInopa 04:35, 22 January 2007 (UTC)[reply]

Something like that: two right, circular cones with coincident bases. StuRat 07:01, 22 January 2007 (UTC)[reply]

Specifically with slope angles summing to 90 degrees.86.132.239.157 12:49, 23 January 2007 (UTC)[reply]

Statistics/probability/counting question

I've been looking at a counting problem:

There are n holes and q things are distributed over those n holes so that each different way of distributing the q things is counted only once (each way (I've called them states below) is assumed equally likely)

eg 3 holes 4 things

State 1 : 4 0 0
State 2 : 3 0 1
State 3 : 3 1 0
State 4 : 2 2 0
State 5 : 2 1 1
State 6 : 2 0 2
State 7 : 1 3 0
State 8 : 1 2 1
State 9 : 1 1 2
State 10: 1 0 3
State 11: 0 4 0
State 12: 0 3 1
State 13: 0 2 2
State 14: 0 1 3
State 15: 0 0 4

So Holes with x things=
   x
   0  15
   1  12
   2  9
   3  6
   4  3

Total = 45 = No. of states times n (obviously)

I get No. of times a hole has x things =

n²(n-1)(q-x+n-2)!/(n!(q-x)!)  (equation1)

eg holes (n=3), things (q=4), holes with 2 things (x=2) = 9x2x(4-2+3-2)!/(3!(4-2)!) = 18x3!/(3!2!) = 18/2 = 9 .

I get total states =

n(q+n-1)!/(n!q!)  (equation2)

eg (3x6!)/(3!4!) =15

These equations are correct I'm convinced.

I have three (four) questions:

1. The equation doesn't work when n=1 (though the solution is trivial - there is only one possible way - that is 1 hole with all the q things). Can anyone suggest a way of getting the equation to work for n=1

2. Does this counting method have a name or an article - can anyone give more info on it - (also variations on this method such as when certain values of things in holes are forbidden). Also the equations can be rewritten to give a probability of getting x things in a given example - assuming each state is equally likely - is there a name for this probability distribution.

3. I'm stumped as to a real world system were this would apply - any simple suggestions would be appreciated, thanks.

How about if you are in charge of a small "high rollers" casino room, only admit a given number of people, and want to know the combinations possible for how many people will go to each game. For example, you might decide that if you have more than 3 people at a table, a hostess should serve that table exclusively, bringing drinks and snacks, with another hostess roaming to the smaller tables. Thus, you need to know how many hostesses might be needed for a given number of games and players. StuRat 20:30, 22 January 2007 (UTC)[reply]

Good example - I'd like an example from 'nature' if anyone can think of one..87.102.3.64 20:59, 22 January 2007 (UTC)[reply]

OK, how about prairie dogs diving into holes when a bird of prey flies past ? You can use this type of analysis to figure out how many might end up in each hole. StuRat 07:42, 24 January 2007 (UTC)[reply]

(4. I've got f(q-x+n-2)^n-2 as an approximation to the equation for no. of times a hole has x things (where f is constant for a given example) - is there a better equation?)83.100.253.194 14:24, 22 January 2007 (UTC)[reply]

To make the equation work for n = 1, keep it as it is. You will get ${\frac {1\cdot (q+1-1)!}{1!q!}}={\frac {q!}{q!}}=1$ unless I misunderstand something. By the way, it can be written more simply as ${\frac {(q+n-1)!}{(n-1)!q!}}$ or ${\binom {q+n-1}{q}}$ .
This problem is called Combination with repetition, there is a paragraph about it in the combinatorics article and you might find more by following links.
That depends on what you mean by "apply", but the situation happens when you need to choose a given number of objects from several possible kinds (and want to find out how many different possibilities you have).
The first step is to rewrite it in the more pleasent form $n{\binom {q-x+n-2}{q-x}}$ . The second is to use Stirling's approximation which should give a reasonably nice result.

-- Meni Rosenfeld (talk) 19:43, 22 January 2007 (UTC)[reply]

Reply answer 1 : ::Yes - sorry I wasn't clear - now I've labelled the equations - I meant to get equation 1 to work with n=1 - so that it returns 0 except when x=q.83.100.253.194 19:59, 22 January 2007 (UTC)[reply]

I see. This is a subtle point, but the formula will work if you use a formula like I wrote above and agree that

{\binom {-1}{0}}=1

and

{\binom {a}{a+1}}=0

for a ≥ 0, which make sense algebraically. -- Meni Rosenfeld (talk) 20:46, 22 January 2007 (UTC)[reply]

Puzzle of the missing leprechaun

I'm looking for an article on a famous puzzle. I always encountered it as the puzzle of the missing leprechaun, as described and explained on the science roadshow's site. I don't know what other name is used for this puzzle and couldn't find it in Wikipedia. Any idea? Thanks in advance. ---Sluzzelin 15:08, 22 January 2007 (UTC)[reply]

I don't know if it has a name, I've a feeling the original illustration was by Sam Loyd who drew many similar puzzles. --Salix alba (talk) 20:09, 22 January 2007 (UTC)[reply]

Thanks, Salix. That was helpful and you're right, it was created by Sam Loyd and the correct title seems to be The Disappearing Leprechaun. WP has no article on it. ---Sluzzelin 20:47, 22 January 2007 (UTC)[reply]

I first saw the puzzle in the context of Martin Gardner's Aha! Gotcha! book; in case it's useful, the closest thing I found here was the missing square puzzle that is discussed in that book immediately before the leprechauns. --Tardis 23:08, 22 January 2007 (UTC)[reply]

http://www.personalityscience.org/csh/Stories.nsf/IDList/DRHH-6BM36D moves the pieces in a similar drawing. I first saw a similar illusion (don't remember where) with figures drawn across the boundary between two concentric circles which could rotate. 1/13 rotation made a figure disappear. PrimeHunter 20:22, 23 January 2007 (UTC)[reply]

It was a version of Get Off the Earth. PrimeHunter 20:27, 23 January 2007 (UTC)[reply]

Thanks for the links and additional info, PrimeHunter and Tardis. I hadn't ever seen the Get Off the Earth puzzle before. ---Sluzzelin 23:58, 24 January 2007 (UTC)[reply]

simple statistics question

I have a list of people, with their birth dates, and, if they are dead, their death dates, if they are not, then today's date. How do I control for still being alive to see if date of birth influences length of life? Thank you! —The preceding unsigned comment was added by 207.225.65.76 (talk) 16:37, 22 January 2007 (UTC).[reply]

For dead people you could plot average lifespan vs day or month or year of birth and see if any particular value seems to be related to a longer life span - if it does you then need to check to see if you have sufficient examples for the result to be statistically relevant eg if two people born on the 12th of a month live long it doesn't mean much, but if you have say at least 100 people per day of the month and the 12th shows a considerable increase in longlevity then there may be a correlation. In general you need to compare measured values and see if they lie outside expected deviances (due to random factors). Hope that helps83.100.253.194 18:56, 22 January 2007 (UTC)[reply]

I don't think that method will work. If they only look at dead people, then of the dead people born in 1980, the average life span will be maybe 17 years of age. That implies that people born in 1980 have very short life spans, when it's really only people born in 1980 who have died by now who had short life spans. StuRat 19:09, 22 January 2007 (UTC)[reply]

(good point - but still relevant for day and month??)(excluding/ignoring any skew introduced)83.100.253.194 19:20, 22 January 2007 (UTC)[reply]

If an equal proportion of people were born in any given month for every year, then you would be right. However, if it works out that there has been a move to different birth months in recent years, then that would make it look like people born in those months tend to live longer, when it's just that most people born in those months haven't lived long enough to die yet. StuRat 07:33, 24 January 2007 (UTC)[reply]

I'd say you need to ignore recent data. For example, stats on births in the 1950's are quite useless in determining the percentage of people who will live to be 100. For those, you would need to exclusively use births from 1906 and before. On the other hand, if you want to look at how birth year effects your chances of dying by age 50, then births from 1956 and before should be considered. If your goal is to look at average length of life, then you need to use the longest possible life span, which is somewhere in the 130 year range. I'd go with 140 to be safe, meaning you should only use births from 1866 and before. I suggest doing a series of stats like this (percentages were made up):

Birth              Chances of living to age...
Year    10    20    30    40    50    60    70    80    90   100   110   120   130   140
----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- 
1866  71.0% 68.7% 63.8% 57.6% 40.8% 34.9% 14.1% 7.92% 0.91% 0.09% 0.02% 0.01% 0.00% 0.00%
1867  72.2% 69.1% 65.2% 53.4% 43.7% 28.2% 14.3% 8.25% 0.81% 0.12% 0.01% 0.00% 0.00%   -   
1868  72.8% 67.2% 63.4% 56.2% 41.7% 36.8% 13.6% 8.43% 1.34% 0.11% 0.03% 0.01% 0.00%   -
1869  73.1% 68.9% 65.7% 55.9% 43.2% 37.4% 14.7% 7.76% 0.88% 0.08% 0.01% 0.00% 0.00%   -
1870  73.1% 64.7% 64.7% 54.2% 42.0% 35.0% 13.8% 8.43% 0.97% 0.15% 0.02% 0.00% 0.00%   -
1871  72.9% 68.2% 66.3% 58.9% 44.9% 36.5% 15.1% 8.87% 1.40% 0.14% 0.03% 0.02% 0.01%   -
1872  73.5% 66.0% 64.4% 56.3% 43.2% 36.6% 14.8% 9.22% 1.01% 0.16% 0.01% 0.00% 0.00%   -
1873  74.0% 68.1% 64.0% 57.6% 42.1% 38.7% 15.3% 6.45% 1.10% 0.13% 0.01% 0.01% 0.00%   -
1874  74.3% 69.6% 65.2% 59.7% 44.1% 36.7% 15.0% 8.83% 1.08% 0.19% 0.03% 0.01% 0.00%   -
1875  74.8% 68.3% 64.7% 62.2% 43.5% 39.4% 15.6% 9.46% 0.97% 0.18% 0.01% 0.00% 0.00%   -
1876  75.3% 70.3% 65.1% 58.9% 43.7% 38.5% 15.4% 7.85% 1.53% 0.22% 0.04% 0.00% 0.00%   -
1877  75.2% 70.8% 68.3% 63.4% 45.7% 37.3% 16.2% 9.32% 1.62% 0.23% 0.02% 0.00%   -     -
  .
  .
  .

StuRat 19:01, 22 January 2007 (UTC)[reply]

Hmmm.... thanks. Let me mull that over - appreciated. 207.225.65.76 19:06, 22 January 2007 (UTC)[reply]

You're welcome. I suspect that you will see "blips" from the Spanish Flu epidemic of 1918/WW1, and WW2, a bit of random variation, and a gradual increase with birth year. StuRat 19:14, 22 January 2007 (UTC)[reply]

Simply ignoring recent data seems potentially wasteful, and could also introduce some bias. If a large proportion of your subjects are still alive, you might benefit from more sophisticated methods such as survival analysis. You may also be interested in the results on the effect of birth month on female longevity presented in this paper (Table 1). -- Avenue 07:47, 23 January 2007 (UTC)[reply]

Thanks - that's more like what I'm looking for - I want to be able to say something about people born more recently, controlling for the fact that they have not had a chance to die of old age. Perhaps comparing rates of death at each age for each cohort? I'll take a look at survival analysis - I'm interested in year of birth, not day or month. Thanks! —The preceding unsigned comment was added by 207.225.65.76 (talk) 16:50, 23 January 2007 (UTC).[reply]

Yes, what you're describing is a cohort life table [1] [2]. Here are a couple of reports describing such results for the Canadian and New Zealand populations. -- Avenue 22:34, 24 January 2007 (UTC)[reply]

Sequence

I'm mixed up on this one kinda.

If a₁=12, how do I derive the sequence a_n+1=a_n/(n+1)?

I'm thinking of this like terms in a summation to add up, but I am not distinguishing the different variables.

I'm barely remembering how to write and input summation notation into my calculator. If this can be put into this form, is this right?

\sum _{i=12}^{n}a_{n}/(n+1)

My state in understanding this is all torn up! Will a kind soul please help me? X [Mac Davis] (DESK|How's my driving?) 22:13, 22 January 2007 (UTC)[reply]

Abstractions are the soul of mathematics, but we discover them from concrete examples. Let's generate a few terms.

{\begin{aligned}a_{1}&{}=12&a_{2}&{}={\tfrac {12}{2}}=6&a_{3}&{}={\tfrac {6}{3}}=2&a_{4}&{}={\tfrac {2}{4}}={\tfrac {1}{2}}\end{aligned}}

You already have a formula that will generate any term eventually. It is also clear that the terms are always positive and tend to zero as n increases. It might be helpful to have a closed-form for a term, as follows:

{\begin{aligned}a_{1}&{}=12={\frac {12}{1}}&a_{2}&{}={\frac {12}{1\times 2}}&a_{3}&{}={\frac {12}{1\times 2\times 3}}&a_{4}&{}={\frac {12}{1\times 2\times 3\times 4}}\end{aligned}}

In other words,

a_{n}={\frac {12}{n!}}.\,\!

The next thing we might want to do is to sum these terms; we can write that as

\sum _{k=1}^{n}a_{k}=12\sum _{k=1}^{n}{\frac {1}{k!}}.

We may recognize the sum (without the factor of 12) as that for the exponential, e^x, with x = 1, except that the first term, 1, is missing. So we have

12\sum _{k=1}^{\infty }{\frac {1}{k!}}=12(e-1).

Hope this answers your question. --KSmrq^T 22:51, 22 January 2007 (UTC)[reply]

Excellent! Why is e there in the end? X [Mac Davis] (DESK|How's my driving?) 23:25, 22 January 2007 (UTC)[reply]

The e comes in when you're taking an infinite sum. There is a thing called a Taylor series expansion for functions that satisfy various conditions (being infinitely differentiable here will suffice). These expansions let you write certain functions as an infinite series. The Taylor series expansion of the function e^x centered at the point 0 (hence, a Maclaurin series) is

e^{x}=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}},

so when x = 1, the left hand side is just e, and the right hand side is what KSmrq posted above, but the first term of the sum (when k = 0) is missing; so we subtract 1. –King Bee ^{(T • C)} 00:27, 23 January 2007 (UTC)[reply]

(Fixed King Bee's series to be a power series.) :-) --KSmrq^T 01:52, 23 January 2007 (UTC)[reply]

Uhh, yeah. That k is um, kind of important. *gulp* =) –King Bee ^{(T • C)} 14:15, 23 January 2007 (UTC)[reply]