Wikipedia:Reference desk/Mathematics

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January 8

long math problems

1 × 2 × 3 × … × 98 × 99 × 100 ≈ 9.33262154 × 10^157

Is there any easier way to solve this problem than to write the entire thing into a calculator (any formulas?)?

Is there a name for this kind of math problem? --172.130.196.207 03:39, 8 January 2006 (UTC)

Yes, that's 100 factorial, or 100! in math notation. A scientific calc will allow you to enter 100 then hit the "!" key to get the answer. StuRat 03:41, 8 January 2006 (UTC)

A very good approximation can be made very fast using Stirling's approximation:

$100!\sim {\sqrt {2\pi 100}}\left({\frac {100}{e}}\right)^{100}$ deeptrivia (talk) 04:19, 8 January 2006 (UTC)

Also, you wrote the answer wrong. It's not 9.33 times 10157, it's 9.33 times 10 to the power of 157, which is much bigger. You can write it like this: 9.33 × 10^157 or like this: $9.33\times 10^{157}$ —Keenan Pepper 05:51, 8 January 2006 (UTC)

3is the answer provided by bignums. Don't believe there is any trick as there is for the equivalent summation (sum of all numbers from 1 to 100). Sdedeo (tips) 23:40, 12 January 2006 (UTC)

There are many tricks to calculate factorials faster with a computer, based on splitting the subproducts into smaller pieces, but they don't help if you're entering commands into a calculator by hand. There is indeed no closed formula like the one for sums. Fredrik Johansson - talk - contribs 13:45, 13 January 2006 (UTC)

Largest number

Infinity is the largest number possible, but it is not a number. So would infinity minus 1 to the power of negative infinity be the largest number that is a number?

No. There is no largest (natural/rational/real/complex) number. Proving that rigorously, of course, requires first rigorously defining what a number is, which can get difficult. For natural numbers, a common definition uses the Peano axioms, which explictly state as given that each number has a successor, which is greater than it.

Questions like this don't make sense unless you explicitly specify the set of objects you're working with. The terms "integer", "rational number", "complex number" and so on all have rigorous mathematical definitions, but the word "number" by itself does not. Are quaternions numbers? You can call them numbers if you want, it doesn't really matter. Same with infinity. You can call it a number if you want, but it is not an integer, a complex number, or even a quaternion, so if you want to talk about things like "infinity minus 1 to the power of negative infinity", first you have to define a number system that includes infinity and define a way to raise something to a power in it. —Keenan Pepper 05:45, 8 January 2006 (UTC)

For examples of number systems that do include infinity, see extended real number line, point at infinity, cardinal number, and ordinal number. —Keenan Pepper 05:56, 8 January 2006 (UTC)

However, note that neither the cardinals nor the ordinals have a largest number. The extended real numbers do have a largest number — positive infinity — but not a second-largest one. And the projective line is not totally ordered. —Ilmari Karonen (talk) 06:44, 8 January 2006 (UTC)

If you have some spare time, I can recommend you this page. All you ever wanted to know about big numbers. Some wikipedia articles: Ackermann function, Tetration --Joris Gillis 13:19, 8 January 2006 (UTC)

An article on Busy beaver claims "Both of these functions are noncomputable, because they grow faster than any computable function." However, I do not see the proof of this statement. Yes, the sequences are noncomputable by Turing Machines, but it does not say anything about the growth rate, does it? After all, the problem may be noncomputable, yet the answer could be 42.(Igny 01:49, 9 January 2006 (UTC))

Nevermind, if such Turing-computable function f(n)>S(n) existed, it would solve the halting problem. (Igny 03:21, 9 January 2006 (UTC))

Infinity is a number of sorts (depending on your definition of a number), however it isn't representable by digits, because of its immense (dare I say, infinate) length. The same can be said for the smallest (in magnitude) known real number, which, as far as I know (feel free to say if I'm speculating here), is representable only by $1-0.{\dot {9}}$ (I call it epsilon; came about because the smallest floating-point number representable on a computer is called machine epsilon): because of its length, it is unrepresentable in a purely digital form (using solely digits--and the decimal point, of course--and no mathematical symbols).

The smallest real number in magnitude is zero. Paul August ☎ 15:46, 11 January 2006 (UTC)

The largest singularly-named (that is, uses only one word for its name) number (in terms of magnitude) is the googolplex, which is represented as $10^{10^{100}}$. --JB Adder | Talk 22:26, 10 January 2006 (UTC)

Thankyou, Paul, for picking out my error there; I completely forgot to put non-zero in. --JB Adder | Talk 05:50, 16 January 2006 (UTC)

Another very interesting article about large numbers and infinities can be found here. It's a long article, so be sure to check out all the pages. --DannyZ 04:39, 11 January 2006 (UTC)

January 9

Standard Deviation

If I recall correctly, in a normal distribution, each point of inflection is located one standard deviation away from the mean. Why isn't this in the normal distribution and standard deviation articles? --JianLi 01:30, 9 January 2006 (UTC)

It's mentioned here, at the bottom of the section: Normal distribution#Probability density function. —Keenan Pepper 02:34, 9 January 2006 (UTC)

I didn't see that, thanks :) I also added the info to Standard deviation.

BTW, something has to be done about this "standard deviation" illustration .

It doesn't seem correct. --JianLi 05:45, 9 January 2006 (UTC)

It looks fine to me at first glance. What specifically is wrong with it? —Keenan Pepper 17:22, 9 January 2006 (UTC)

There is a total that does not amount to 1OO% ? --Harvestman 19:17, 9 January 2006 (UTC)

no, the points of inflection seem to be off. though it is hard to judge, because it is pretty flat near those points

They add up to 100.4%, which isn't bad. Black Carrot 18:44, 11 January 2006 (UTC)

Are you sure? They seem to add up to 99.8, which is more accurate as - theoretically - we can never account for 100%. Typically, this is further illustrated by never completely closing the tails.

Yeah, I think it's 99.8%. But this is not because we can never account for 100%, as the "0.1%" takes into account everything past 3 st dev. Rather, the reason it is not perfectly 100% is just roundoff error -JianLi 23:48, 15 January 2006 (UTC)

pi

Can someone estimate how many years until those super-computers calculate the full extent of pi?

$\pi /0$ years. StuRat 04:07, 11 January 2006 (UTC)

Pi has infinite digits , they will never calculate it. helohe (talk) 19:55, 9 January 2006 (UTC)

What is the point, anyway? 100 digits are enough to circumscribe a circle around the visible universe accurate to the Planck length. —Keenan Pepper 22:29, 9 January 2006 (UTC)

The points of calculating many digits are as far as I can tell 1) prestige, 2) investigation of the digits' statistical properties, 3) testing/benchmarking computer hardware and software. Fredrik | tc 22:36, 9 January 2006 (UTC)

For a while in the 80s and early 90s it yeilded some interesting results in parallelization, but not as much as things like factoring or protein folding. —James S. 09:13, 11 January 2006 (UTC)

Contact by Carl Sagan told me that there are secret messages in pi. Proto t c 09:41, 12 January 2006 (UTC)

But since pi is conjectured to be a normal number, any message imaginable is contained on its digits. (hence this joke). ☢ Ҡieff⌇↯ 00:26, 14 January 2006 (UTC)

So you're saying there are secret messages, right? :) Superm401 | Talk 04:35, 14 January 2006 (UTC)

Pearson r

Can the Pearson r be used to analyze a univariate distribution? I say no, but my friend says yes. --Neutrality^talk 22:56, 9 January 2006 (UTC)

Pearson's ρ is a measure of correlation which is used to describe the linear dependence of two distributions. I suppose you could correlate a univariate distribution against an ideal uniform, normal, lognormal, or other distribution, if you really wanted a univariate measure, but I've never seen that done. Although the more I think about it, the more I believe that you might want to compare a distribution to the uniform distribution with Pearson's ρ. So, I say you are both right, but you are more right than your friend. —James S. 09:34, 11 January 2006 (UTC)

January 10

how much?

How much wood can a woodchuck chuck if a woodchuck could chuck wood?

A woodchuck would chuck as much wood as a woodchuck could chuck, if a woodchuck could chuck wood. (In other words, he would do his very best.) StuRat

ZOT! —Ilmari Karonen (talk) 17:49, 10 January 2006 (UTC)

---

As much wood that a woodchuck can chuck wood!

Mathematically, infinitely large. Through contradiction

Suppose a woodchuck can chuck, say x, amount of wood. then the woodchuck can after chucking that chuck a small amount, say p, of wood. so that he chunks x+p amount of wood. this is a contradiction to the satement that he can chuck only x amount. Thus Proved ;-) —Rohit_math 1:45 11 January 2006 (IST)

Not really. Suppose, as a counterexample, that the woodchuck drops dead of a heart attack after chucking x amount of wood. He can't very well chuck any amount of wood, however small, after that, now can he?

Besides, even if your proof was correct, you would only have proven the nonexistence of a maximum amount of wood. A supremum, however, could still exist. —Ilmari Karonen (talk) 20:42, 10 January 2006 (UTC)

Two cords. Night Gyr 20:40, 10 January 2006 (UTC)

Definitely less than two chords 'cause a woodchuck aint that good at playin' music. And besides which, this mere nursery rhyme would seem to agree 'bout that ('cause they lack opposin' thumbs) which adds to their being musically challenged. So, whereas a 'chuck might get away with an Em^(open) there's no way he could do a Cm⁷. hydnjo talk 01:22, 11 January 2006 (UTC)

Aach... I underestimated our pet 'chuck who at this very moment is riffin' the opening chords of Stairway To Heaven (geesh, who knew?) ;-) I guess a woodchuck can... hydnjo talk 03:01, 11 January 2006 (UTC)

If he had a saw in his little paw, a ton of wood he could! Black Carrot 18:47, 11 January 2006 (UTC)

And what if another woodchuck came chucking wood ?

Two woodchucks would chuck twice as much wood as a woodchuck could chuck, if woodchucks could chuck wood. --Harvestman 21:57, 11 January 2006 (UTC)

How much wood could two woodchucks chuck if two woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could three woodchucks chuck if three woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could four woodchucks chuck if four woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could five woodchucks chuck if five woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could six woodchucks chuck if six woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could seven woodchucks chuck if seven woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could eight woodchucks chuck if eight woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could nine woodchucks chuck if nine woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could ten woodchucks chuck if ten woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could eleven woodchucks chuck if eleven woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could twelve woodchucks chuck if twelve woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could thirteen woodchucks chuck if thirteen woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How much wood could fourteen woodchucks chuck if fourteen woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...

How many roads must a man walk down? – b_jonas 18:18, 12 January 2006 (UTC)

What is six times nine? Night Gyr 09:56, 14 January 2006 (UTC)

Alligation

I've placed requests for an article on Alligation (alligation medial and alligation alternate) [1]. It is all explained in that source, but with the archaic language, I'm likely to mess it up. Someone with more experience in mathematics or the history of mathematics would be better for the job. If you use that source in creating the article, please add {{1728}} to the page. Thanks! — ₀₉₁₈^BRIAN • 2006-01-10 21:23

I'm working on an article. I never knew what this was called. It's hard thinking of examples... I keep thinking of ones that involve drugs for some reason. =P —Keenan Pepper 03:41, 11 January 2006 (UTC)

Well, thank you KP for making that red link blue. We never heard of it either, that's one of the wonders of the WP:RD. hydnjo talk 04:10, 11 January 2006 (UTC)

Thanks for all your help! — ₀₉₁₈^BRIAN • 2006-01-11 04:23

What's fun is that with that latin origin word alligation one could generate an english word like adlinking, which is of widespread use over the net. --Harvestman 21:54, 11 January 2006 (UTC)

Huh? hydnjo talk 07:30, 12 January 2006 (UTC)

January 11

derivation-circumcenter

can anyone give me the derivation for finding the coordinates of the circumcenter of a triangle?

(Please sign your questions with ~~~~.) Did you read the explanation at circumcircles of triangles? If there is something additional you need, please let us know. --KSmrq^T 15:56, 11 January 2006 (UTC)

Let's see. Knowing about barycentric coordinates makes this proof really easy. If you have a triangle $ABC$ and a point $P$ anywhere on the plane, and the area of the triangles $PBC,PCA,PAB$ are $\lambda _{a},\lambda _{b},\lambda _{c}$ respectively than you can get the vector $\mathbf {p}$ pointing to $P$ can be calculated from the vertices by the formula $\mathbf {p} =(\lambda _{a}\mathbf {a} +\lambda _{b}\mathbf {b} +\lambda _{c}\mathbf {c} )/(\lambda _{a}+\lambda _{b}+\lambda _{c})$. Here, $\lambda _{a}$ is considered to be negative if the line $BC$ separates the points $A$ and $P$, and the sign of $\lambda _{b}$ and $\lambda _{c}$ is assigned similarly.

Knowing this, it's easy to get the coordinates of the circumcenter. Let $P$ be the circumcenter, the radius of the outscribed circle be $r$, and the angles of the triangle $\alpha ,\beta ,\gamma$. Now the angle $BPC$ is $2\alpha$, so its area is $\lambda _{a}=r\cos \alpha$. Similarly, $\lambda _{b}=r\cos \beta$ and $\lambda _{c}=r\cos \gamma$. Thus, $\mathbf {p} =(\cos \alpha \mathbf {a} +\cos \beta \mathbf {b} +\cos \gamma \mathbf {c} )/(\cos \alpha +\cos \beta +\cos \gamma )$. – b_jonas 18:02, 12 January 2006 (UTC)

what is the formula for the diameter of a circle f

What information is given? —Keenan Pepper 15:34, 11 January 2006 (UTC)

Is the radius is given? The area? The circumference? Three points on the circle? Each produces a different formula. --KSmrq^T 15:51, 11 January 2006 (UTC)

In any case, see Circle. —Ilmari Karonen (talk) 22:46, 11 January 2006 (UTC)

Let diameter be d, radius r, circumfrence c, and area a.

d=2r

d=c/pi

d=2*sqrt(a/pi)

--Superm401 | Talk 04:39, 14 January 2006 (UTC)

17/23 Correlation

Has anyone ever heard of the above, supposedly related to the Illuminati? What is it? --Black Carrot 23:23, 11 January 2006 (UTC)

Hmm.. well apparently, 23 is a significant number among crackpots numerologists, I mean.. and that page says 17 is popular too, so it's not surprizing if some conspiracy nut Illuminati researcher, thinks there's some connection. --BluePlatypus 07:50, 12 January 2006 (UTC)

Anybody know something more specific than that? Black Carrot 01:06, 17 January 2006 (UTC)

January 12

lines that converge

—The preceding unsigned header was added by 68.82.121.86 (talk • contribs) .

If I recall correctly, Myrna Minkoff was opposed. --George 01:57, 13 January 2006 (UTC)

• WTF? --- Charles Stewart 02:09, 13 January 2006 (UTC)

What do you call lines that are not parallel?

—The preceding unsigned header was added by 68.82.121.86 (talk • contribs) .

I'm confused. Did this person answer their question before they asked it? Black Carrot 00:49, 12 January 2006 (UTC)

(Hmm, I see what you mean, Nah, just jumpin' in before anybody had a chance (I think)) hydnjo talk 04:39, 12 January 2006 (UTC)

In any event, lines that are not parallel can either intersect (share a point in common) or be skew lines. —Keenan Pepper 01:12, 12 January 2006 (UTC)

To be more specific, if the lines are in the same plane, then they must either intersect or be parallel. If the lines are not coplanar, then they must be skew. StuRat 01:55, 12 January 2006 (UTC)

Parabolas

• How can you determine the equation of a parabola when given only a point on the parabola and the x-intercepts?
• When given a point on the parabola and the vertex?

Thanks, anon.

In general, it is not possible to determine a parabola given only three points, or given a single point and the vertex. You probably want to know how to find the parabola whose axis is perpendicular to the x axis.

• If the two x intercepts are x₁ and x₂, the equation of the parabola must have the form y = k(x-x₁)(x-x₂) for some k. Find k using the third point on the parabola.
• If the vertex is at (x₁,y₁), the equation must have the form y = k(x-x₁)² + y₁. Again, find k using the other point. —Keenan Pepper 01:22, 12 January 2006 (UTC)

I could be mistaken, but your preamble assertion is not correct. The Lagrange interpolating polynomial always gives you a parabola when given three points (not all collinear). If you know the vertex of the parabola, the other point gives you some idea of its concavity. By an argument appealing to the parabola's symmetry, you have three points and thus you can determine the parabola exactly. Dysprosia 09:50, 12 January 2006 (UTC)

... but by using Lagrange polynomials you are implicitly assuming that the axis of the parabola is perpendicular to the x axis, so its equation takes the form y = <second degree polynomial in x>. This is probably what the original question meant, but it was not explicitly stated. Without this assumption, three points or one point + vertex is not sufficient to define the parabola. For example, the parabolas y = x² and x = y² both pass through (1,1) and have their vertex at (0,0). Two general parabolas can intersect in up to 4 different points (see Bézout's theorem) so you need 5 points to define a unique parabola. Assuming a given direction for the axis of the parabola implicitly provides two of these points, because the direction of the axis determines where the parabola is tangent to the line at infinity. Gandalf61 15:35, 12 January 2006 (UTC)

If one knows the orientation of the parabola in advance, one could simply change variables when necessary to obtain a description in Lagrange polynomials. However, you're right, you need to know this orientation ahead of time. Dysprosia 22:52, 12 January 2006 (UTC)

You might also want to see Wikipedia:Reference_desk_archive/Mathematics/December_2005#Two_parabola_questions which is a generalization of this question. – b_jonas 16:26, 14 January 2006 (UTC)

January 13

decade

how long is a decade

I know that one. A decade is ten years long. Dmharvey 02:33, 13 January 2006 (UTC)

(Or a factor of ten, as in "100 is two decades away from 1," but probably not in the context above. Sdedeo (tips) 05:04, 13 January 2006 (UTC))

Some decades seem to last longer than others. The 1970's, for example, seemed to refuse to end. :-) StuRat 11:13, 13 January 2006 (UTC)

That's a good point. Technically a decade is ten years, but when referring to generational/cultural decades, they can be anywhere from, say, 6 to 15 years. Black Carrot 07:33, 16 January 2006 (UTC)

number theoretic transform

I have a prime p, say around 10⁷ or so. I have a vector of values $x_{k}\in \mathbb {Z} /p\mathbb {Z}$ for $k=0,\ldots ,p-2$. (i.e. the vector has length $p-1$). I would like to efficiently compute the number theoretic transform,

$y_{j}=\sum _{k=0}^{p-2}\omega ^{jk}x_{k},$

where $\omega$ is some primitive $(p-1)$th root of unity in $(\mathbb {Z} /p\mathbb {Z} )^{\times }$. I've been reading about the fast fourier transform, number theoretic transform etc on and off wikipedia but not having much luck finding a clean, fast way to do it. It would be really nice to be able to do this without leaving the world of arithmetic mod p. Any suggestions? Dmharvey 12:27, 13 January 2006 (UTC)

If you can get on IRC, I would recommend asking that one on efnet's #math channel. You could send them the URL to this section and see who answers. --James S. 21:49, 13 January 2006 (UTC)

January 14

characterictic function

I have such a problem. Decide whether the following teorem is true and explain why. Let |φ(t)| ≤ φ(0) = 1 and |Ψ(t)| ≤ Ψ(0) =1 for each t, t is an element of R. If (φ+Ψ)/2 is a characteristic function of a random variable, then φ and Ψ are also characteristic functions of some random variables. Please help me. Thank you in advance Julia

Our article on Characteristic function is fairly incomplete, and sadly I forgot the details of probability theory as well. I guess it exists by the inversion theorem, which is mentioned in the article, but in general case it's somewhat hairier. The right-side integral is additive in regards to φ, so I guess you'll just get F_Z=(F_X+F_Y)/2, which should be a cumulative function of some random variable Z.  Grue  11:26, 16 January 2006 (UTC)

convergence in distribution

Help me please to solve my homework please. Let X1,X2,...Xn be independent random variables such that P(Xk = k) = P(Xk = -k) = 1/(2k^2) and P(Xk = 1) = P(Xk = -1) = 1/2(1-1/k^2). Let Sn=∑Xk (k goes from 1 to n). Prove, that (1/√n)Sn converges in distribution to N(0,1) and var((1/√n)Sn) converges to 2. (n goes to infinity) Thanks;o)

You'll notice at the top of this page, it says "please do not post entire homework questions and expect us to give you the answers". If you want to ask homework questions, you have to be more devious. But you might want to look at convergence in distribution. Not sure exactly what "convergation" is, but it probably has something to do with "convergence". Dmharvey 23:34, 14 January 2006 (UTC)

Most-used math functions/symbols?

Will anyone please go to the list of math codes and give me a list of any they would like to see added to the Special characters box on edit pages? I'm looking for the top 25-50 characters/functions/formula/etc. Thanks! — ₀₉₁₈^BRIAN • 2006-01-14 13:47

• ≠ ≤ ≥ < > ≡ ≈ ≅ ∝   − × ÷ ± ⊥ ⊕ ⊗ ∗   …   ¼ ½ ¾ ¹ ² ³ °   ∂ ∫ ∑ ∞ ∏ √ ∇   ← → ↔ ⇐ ⇒ ⇔   ⌈ ⌉ ⌊ ⌋ 〈 〉   ¬ ∧ ∨ ∃ ∀   ∈ ∉ ∋ ∅ ⊆ ⊇ ⊃ ⊂ ⊄ ∪ ∩ ℵ (mostly copied from Wikipedia:Mathematical symbols) -- Jitse Niesen (talk) 16:04, 14 January 2006 (UTC)
  • I'm specifically looking for TeX/LaTeX expressions, but this is a start. — ₀₉₁₈^BRIAN • 2006-01-14 16:11
    • I see, I should have answered the question you actually asked instead of the question I thought you asked. Well, let's see. \nabla \partial \forall \exists \emptyset \in \not \subset \cap \cup \setminus \sqrt \sim \approx \cong \equiv \le \ge \ll \gg \pm \to \mapsto \iff \ldots \times \cdot \circ \sum \prod \int \frac \choose \begin{matrix}\end{matrix} \begin{bmatrix}\end{bmatrix} \begin{cases}\end{cases} \mbox \pi \mathbb \mathbf \left \right \langle \rangle \| \quad \qquad \infty \hbar. But, at least for me, I'd prefer to have the first list, as I know the TeX/LaTeX commands rather well. -- Jitse Niesen (talk) 17:51, 14 January 2006 (UTC)
      • Alright, if you try editing a page, and go to "Math/TeX" under "Special characters", you should see what I've put in so far. You might have to press CTRL+F5 on the edit page to see the latest version. — ₀₉₁₈^BRIAN • 2006-01-14 18:25

I have a large table of Unicode mathematics characters at User:KSmrq/Chars. I'm fond of infty ("∞"), mapsto ("↦"), frasl ("⁄", as in ⁵⁄₇), lang ("⟨"), and rang ("⟩"), among others. --KSmrq^T 22:02, 14 January 2006 (UTC)

Special characters is a bit of a travesty. In non-Unicode compatible clients, it mucks the edit process up rather seriously. Dysprosia 11:07, 16 January 2006 (UTC)

January 15

(no questions today)

January 16

Partial harmonic sum

Greetings:

Does anyone know how to compute the following expression exactly without using a computer? :

1 + 1/2 + 1/3 + 1/4 + ... + 1/2004 + 1/2005

note that there are only 2005 terms in the above question. It's a question taken out of a certain 2005 math competition.

Regards,

129.97.252.63 04:31, 16 January 2006 (UTC)

Without using a computer? Wikipedia is not paper

http://www.macalester.edu/aratra/chapt2/chapt2_4_1.html

--James S. 06:14, 16 January 2006 (UTC)

Okay, let's see...each rational in the sequence is ${\frac {1}{x}}$, where $1\leq x(\in \mathbb {N} )\leq 2005$. Now, we must get a common denominator—we'll use $2005!$ for simplicity—and each fraction become ${\frac {\frac {2005!}{x}}{2005!}}$. This thus makes the equation $\sum _{x=1}^{2005}{\frac {\frac {2005!}{x}}{2005!}}$...which in reality is not much improvement from where we started.

In reality I don't think there is a non-computer way other than by brute force, which will take a while. However, the computer (yes, I cheated) brought it to 8.18086436047126000000 (to 20dp). --JB Adder | Talk 06:33, 16 January 2006 (UTC)

But if it's a maths competition, then presumably you can do it by hand. enochlau (talk) 08:31, 16 January 2006 (UTC)

Well, it is not much of a simplification, but

1 + 1/2 + 1/3 + 1/4 + ... + 1/2004 + 1/2005 = H(2005) = ψ(2006) + γ

Was the task really to compute the expression exactly? Do you have the exact formulation of the question?

Rasmus (talk) 10:55, 16 January 2006 (UTC)

I strongly suspect the solution here is not by a general method, but rather some clever tricks of algebraic manipulation. --137.205.236.45 11:32, 16 January 2006 (UTC)

I plugged it into Mathematica and it was not a pretty result, so I don't think there's any trick that'll help us. Perhaps they wanted an approximate answer only - in that case, upper and lower Riemann sums would do the trick? enochlau (talk) 11:40, 16 January 2006 (UTC)

Thanks, I also suspected that this question might not be attainable by traditional means after checking with Maple. 129.97.252.63 13:56, 16 January 2006 (UTC)

Last night's Maple (c) results for the above problem:

825419062412938618065172093089947889397149613536805433233293531037345374983607845
607168571303751022983376044774946356625609868037459266524315881348303251988897297
264702267356771192306405854938327434861393197790709922461870578674170520718259858
709208867298356740872930086641532477167774390192178199597628909379208113940191580
795063697454416576692622505998280631510167317360912683033791803968444942384758990
858234347547510716553741330866608178304475580453652608327809900363947607113897164
452225510715716886988749112270863058179358139930700335029925139304914516700451064
337264806479832238026346350115626220638452291756962627332973452014220212662317566
133773747505598340212718763148981187341324675165293976681254362615682179088701625
329089664066770381767365995112667658279228474516706783621733174277607689316898498
294088334946622546427438165210457801275834481235079191000093
/
100896314379840321186998408533386120937997942211239805942055164219806651357887170
338069896071979300241604627960528226608198826833104802676509391191233622864697953
746131105044797407517764321203308711546496890490543206638668744705597392728340013
118518193575666739999190087664589823668755903919618121564163432457507043319012050
368704876735996000856162502069906637807469380429233657356927717928175747560331309
853759376077421934183746322086962398625735054481422211469583634354868579997531326
324730964497861549892549888791063696999263322606486002338059407329131457462819758
458471125333723038281994099496140309993163410926769370241232610047663184932185829
917494388279088327072506676178218231921632353872338260712580851072644683225000157
132703283102083111875042687135934903664056637859033901806260787677436158789532946
406020520294300450258958671116562100185384730286114583680000

now, if that was the final fraction that could not be reduced further, then it could rightly be argued that no clever algebraic trick exists for miraculously solving the above problem. However, if it could be shown that the above fraction does reduce to a far simpler fraction then the disturbing prospect remains that a clever algebraic method exists that would circuitously produce the same result.

So I guess the grand question is this:

Does Maple automatically reduce all of its fractional outputs? Does it even have the capability of reducing humongous fractions like this??????

Thanks.

129.97.252.63 17:50, 16 January 2006 (UTC)

Yes, the above fraction is reduced. Reducing a fraction simply amounts to cancelling the numerator and denominator by their greatest common divisor, and the GCD is inexpensive to calculate even for humongous numbers. Fredrik Johansson - talk - contribs 17:59, 16 January 2006 (UTC)

AHA! Via the Euclidean Algorithm!!! In fact I hit upon that insight while going to lunch. So the fraction Maple displays MUST have been reduced to the simplest form. Thank you so much Fredrik Johansson! and everyone else for your responses. 129.97.252.63 18:37, 16 January 2006 (UTC)

I think that the original person must bave misread the question, it actually asks for one to show that the sum is greater than a certain number. But inexperienced ppl might think that in order to prove the sum to be greater than some number the sum itself must be sought of in the first place, which is not necessarily the case. 129.97.252.63 18:37, 16 January 2006 (UTC)

Hello, and thank you for lending your time to help improve Wikipedia! If you are interested in continuing to edit, I suggest you make an account to gain a bunch of privileges. Happy editing! 129.97.252.63 21:38, 16 January 2006 (UTC)

Is it just me?

Or are these questions getting much harder compared to last week? I guess school is back in session. --James S. 08:38, 16 January 2006 (UTC)

mathematics

what is the value of root4 is it only +2 or +2 and -2

Both +2 and -2 are square roots of 4. However, in certain circumstances, only the positive root may apply, say if the height of an object launched vertically from the ground varies with time acording to H = T^2. Here the time could only be positive. The positive root is sometimes called the principal root. StuRat 15:17, 16 January 2006 (UTC)

If you want to define a square root function, it can only have one value for each argument (that's the definition of a function). The convention is to define it to be nonnegative. —Keenan Pepper 17:50, 16 January 2006 (UTC)

January 17

Popes

What is the chance of four popes having died on the same day ?

There has been 265 popes. Each having died a different year.

Since a new pope is elected only after the current one dies, I don't see how that's even theoretically possible. (Therefore the probability would be exactly zero.) —Keenan Pepper 01:17, 17 January 2006 (UTC)

My guess is, the questioner wants to know the probability that four have died on the same day of the year, a routine but tedious calculation I don't personally feel like making at the moment. (Of course it's "routine" only under the assumption that all dates except February 29 have the same probability and that the probabilities are independent; I suspect neither assumption is exactly right but that they're close enough for the purposes in question.) --Trovatore 01:36, 17 January 2006 (UTC)

Actually, the "routine" answer completely ignores February 29th, assuming all years have 365.0 days. Bah! On that basis alone, I refuse to offer anything more than the link given. --James S. 01:45, 17 January 2006 (UTC)

Questions at this desk enjoy an extremely w-i-d-e latitude but a bit of context would help me to feel a bit less like I was being quized (or trolled). hydnjo talk 02:40, 17 January 2006 (UTC)

Sorry, it's very similar to the birthday problem to which I linked, but as a matter of policy I won't do anything more than link to the page because whenever I've tried to get people to consider February 29th in the past, they tell me the best answer is just to ignore it. Maybe that's true. If someone else wants to do the whole problem for the questioner, fine, and I don't care whether they accommodate Feb 29 or not. --James S. 02:56, 17 January 2006 (UTC)

Sorry James S., no offense intended. With or without the inclusion of February 29, I thought that the question lacked context. I realize that the numerical odds would be affected by the inclusion (or exclusion) of that particular date but not so much as to alter my comment (a generalization) as to the nature of the question. hydnjo talk 03:20, 17 January 2006 (UTC)

Oh, you were asking for the questioner's context instead of my comment's context. I'm guessing it's either a homework problem or someone trying to decide whether a coinsidece is uncanny. Indentation modified. --James S. 03:35, 17 January 2006 (UTC)

Exactly. :-) hydnjo talk 03:45, 17 January 2006 (UTC)

Coterminal angles

What are coterminal angles? Maybe someone should make a page on this?--Urthogie 08:51, 17 January 2006 (UTC)

Coterminal angles are angles that coincide (when placed in standard position). For example 20°, 380° and -340° are coterminal angles. I am not sure we could make much more than a dictionary definition out of this. Rasmus (talk) 09:14, 17 January 2006 (UTC)

OK thanks that answers my question...whats a reference triangle, though?--Urthogie 09:16, 17 January 2006 (UTC)

Euler characteristic

Wikilutations!
I wanted to know how one could find out the Euler characteristic of a given surface. Say atleast for simple surfaces like Torus, Disc, etc.
Rohit_math 18:24, 17 January 2006 (UTC)

Divide the whole surface to triangles, and use F − E + V formula to calculate the characteristic. Also, if the surface in question is orientable and has g holes, Eurer's characteristic is 2-2g.  Grue  18:43, 17 January 2006 (UTC)

Smarties lids

File:Smarties (Candy).jpg

Smarties. Good to eat, but how many tubes before I get the entire alphabet?

Before the new, horrendous "hexatube" packaging, tubes of Nestlé Smarties used to contain a letter of the alphabet under the cap. Assuming the cap letters are chosen uniformly at random, how many Smarties tubes would I expect to have to buy before I had collected the entire alphabet? — Matt Crypto 18:33, 17 January 2006 (UTC)

I believe we had an equivalent question some time ago. Ah, here it is. The probability of finding all the n = 26 letters of the alphabet in m tubes is

$P(m,n)=\sum _{k=0}^{n}(-1)^{k}{n \choose k}\left({\frac {n-k}{n}}\right)^{m}$.

This is the number of surjections from M = {1 … m} to N = {1 … n} divided by the total number of functions from M to N (= n^m). Using the Perl code on the page I linked to above, I find that 94 tubes is enough to give you just barely more than even odds of getting all the letters. I even made a graph of the probabilities for 0 ≤ m ≤ 200. —Ilmari Karonen (talk) 19:33, 17 January 2006 (UTC)

I am not sure about the wording of the problem. I think that the question is not how many tubes would guarantee (with some probability) that I have all 26 letters in them. You start buying the tubes, and in X-th tube you get the last needed letter (you have 25 letters in X-1 tubes and 26 in X tubes). What is expected value of such X?

Is this the right question?(Igny 22:11, 17 January 2006 (UTC))

Smarties? Isn't that just Communist for M&Ms?'--George 19:41, 17 January 2006 (UTC)