Wikipedia:Reference desk/Archives/Mathematics/2007 May 22

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May 22

Can this be done without using a computer

A company opens a bank account with a fixed interest of 10%. At the start of each 
year the company deposits X dollars, X[0] = $10 . The bank gives a simple annual 
interest of 10% . Also the amount the company deposits at the start of each year 
grows by 10% as well.

Year   X      1stJan    31stDec
 0    $10      $10       $11
 1    $11      $22       $24.20
 2    $12.10   $36.30    $39.93
 
 How much money is in the bank account after 5000 years (aka 31stDec year 4999)?

This question drove me nuts because I not allow to use a computer program to generate the answer (considered as cheating). Is there a way to get the answer without using an electronic computer or programable calculator?.

202.168.50.40 00:45, 22 May 2007 (UTC)

This problem is indeed solvable without a computer. Let T(n) be the total amount the company has in its account at the end of year n. Then $T(n)=1.1(T(n-1)+X(n))$. You should be able to figure out what X(n) equals. Once you plug that in, you will need to find a closed-form expression for this recurrence relation. The geometric series article should be useful for that. nadav (talk) 01:56, 22 May 2007 (UTC)

A good way to find such a closed-form expression involves 4 steps:

1. Calculate T(n) for n = 1, 2, 3, ... (how many values should you calculate? enough for you to do the next step).
2. Try and find a pattern in the T(n)s - try rearranging them so they look similar.
3. Guess what the formula might be for a general n.
4. Prove your guess via mathematical induction (depending on circumstances, you may not have to do the full, formal proof, just enough to justify to yourself that it's correct).

Having done that, you will have your expression for T(n) and can just pop in the appropriate value of n. Confusing Manifestation 02:55, 22 May 2007 (UTC)

$T(n)=1.1(T(n-1)+X(n))$. You should be able to figure out what X(n) equals.

$T(n)=1.1(T(n-1)+10*1.1^{n})$ where T(-1)=0

T(0) = 1.1 (T(-1) + 10 * 1.1 ^ 0) = 1.1 (0 + 10) = 11

T(1) = 1.1 (T(0) + 10 * 1.1 ^ 1) = 1.1 (11 + 11) = 24.20

This is not helping me because I don't want to do this 5000 times. 202.168.50.40 02:58, 22 May 2007 (UTC)

Damn it, someone else in class claims to have the answer already. He says

f(n) = 11 * n * 1.1 ^ (n - 1)

f(1) = 11.00

f(2) = 24.20

f(3) = 39.93

f(5000) = 11 * 5000 * 1.1 ^ 4999

How the hell did he get the answer so quick?202.168.50.40 03:02, 22 May 2007 (UTC)

Damn it! That smartarse is going to the front of the class to explain how he got the answer. What is he saying? He use the dollar, euro and pound method.

At the end of 1st year, the amount is 11 dollars.

He converted it to euro using a fake conversion rate of 1 euro = 1.1 dollar. So 11 dollars = 10 euro

At the second year, starts with 10 euros and deposit 10 euros for the end of year amount of 1.1 * (10 euro + 10 euro) = 22 euros (or 24.20 dollars)

He converted it to british pound using a fake conversion rate of 1 pound = 1.1 euro. so 22 euros = 20 pounds.

At the third year, starts with 20 pounds and deposit 10 pounds for the end of year amount of 1.1 * (20 pounds + 10 pounds) = 33 pounds (or 36.30 euros or 39.93 dollars)

In sumary he says

1st year 11 dollars

2nd year 22 euros

3rd year 33 pounds

so f(n) = 11 * n * 1.1 ^ (n - 1)

202.168.50.40 03:12, 22 May 2007 (UTC)

Lol, I really hope he is joking. Further suggestion: When you follow Confusing's method above, try to use variables to replace some of the numbers. For example, replace 1.1 with r, and 10 with X(0), say. This should help elucidate the pattern. My guess is that you will be able to find the pattern without going beyond T(4). nadav (talk) 03:52, 22 May 2007 (UTC)

Yeah, I would point out that you don't want to put it into a number, because then you almost certainly won't see the pattern. Like Nadav suggests, replace a few of the numbers with variables, or at least write it out in as much detail as possible - I'll give you the first couple of steps to start you off, just this time :)

T(0) = 11

T(1) = 11 * 1.1 + 10 * 1.1 = (11 + 10) * 1.1

T(2) = ((11 + 10) * 1.1) * 1.1 + (10 * 1.1 * 1.1) = (11 + 10 + 10) * 1.1 * 1.1

T(3) = ?

Does that give you an idea of what the pattern is? Confusing Manifestation 04:02, 22 May 2007 (UTC)

Why does he have to be joking? let's say unit(n) = 1.1^n dollars, so unit(0) = dollar, unit(1) = euro, unit(2) = pound etc. This way, if you have how many of unit(x) in year(x), you can calculate dollars with the inverse, which is dollars=unit(n)*(1.1^n). Get it? By the way, it would be esier to use 1st year = 10 euros, 2nd year = 20 pounds ect.— Daniel 19:17, 22 May 2007 (UTC)

So you are saying that

f(n) = 10 * n * 1.1 ^ n

f(1) = 11.00

f(2) = 24.20

f(3) = 39.93

f(5000) = 10 * 5000 * 1.1 ^ 5000

202.168.50.40 21:53, 22 May 2007 (UTC)

That's about right, and since it fits your first calculations, it's quite definitely right (for December 31st). — Daniel 23:44, 22 May 2007 (UTC)

Rotating a sine/cosine graph

Hey guys, how would I rotate a sine or cosine graph by, say, 45 degrees? I'm trying to make it tangent to y=x+1 and y=x-1. I don't want to use rotation matrices. Thanks, Fbv65edel / ☑t / ☛c || 03:11, 22 May 2007 (UTC)

$f(x)=\sin {x}+x$

Does that work? --0rrAvenger 03:33, 22 May 2007 (UTC)

That is a shear rather than a rotation, but it may be suitable for the desired tangents. - Rainwarrior 03:41, 22 May 2007 (UTC)

Whether or not you want to use rotation matrices, you will still have to do all of the math involved in the rotation (perhaps you should look at the Rotation (mathematics) article, which might be easier to understand?). If you are trying to rotate: $y=\sin(x)$, the usual way to rotate (x,y) looks like:

$x'=x\cos \theta -y\sin \theta \,$

$y'=x\sin \theta +y\cos \theta \,$

So from here, substitute these into your equation:

$x\sin \theta +y\cos \theta \ =\sin(x\cos \theta -y\sin \theta \ )$

Now the trick is to pull y back out and put it on the left side by itself. I'll leave that for you. (Careful! Depending on the angle there may be multiple values for y.) - Rainwarrior 03:38, 22 May 2007 (UTC)

Well, it turns out a shear what I was looking for, but the rotation using sin and cos of theta is something for me to explore as well. Thanks a lot! --Fbv65edel / ☑t / ☛c || 03:47, 22 May 2007 (UTC)

Actually, a much easier suggestion than trying to find y in terms of x is to define two orthogonal vectors, such as (1,1) and (1,-1), and then map the original function's x and y onto these, so: y = sin(x) gives you x and y, and then to get your coordinate, take: y*(1,1) + x*(1,-1). (Which is: y' = y-x, and x' = y+x.) - Rainwarrior 03:57, 22 May 2007 (UTC)

And if you use generic orthonormal vectors (cos θ, −sin θ) for x and (sin θ, cos θ) for y, you actually get rotation over an arbitrary angle – which you'd need if you want the rotated curve to be tangent to a given line.  --Lambiam^Talk 08:44, 22 May 2007 (UTC)

xkcd

This is based on [1]. (Hover your mouse above the comic for a moment to see it.) Is there such a thing as a "tightly closed nonorientable space," and if so, what would it be like? Would "nonorientable" in such a context include something like the Klein bottle? Black Carrot 06:52, 22 May 2007 (UTC)

See Projective space.  --Lambiam^Talk 08:38, 22 May 2007 (UTC)

(Note visibility of the popup comment is browser-dependent.) Since "tightly closed" suggests a metric, perhaps it is meant as a colloquial substitute for "compact". A Klein bottle is a compact nonorientable surface, but his suggestion would seem to require a three-dimensional space. Euclidean 3-space is not compact, and a 3-ball is orientable, so Lambiam has suggested the most obvious simple example. One way to model the projective space RP³ is to begin with a 4-sphere, such as {(w,x,y,z):w²+x²+y²+z²=R}, for some fixed radius R. (Make it a "tight" sphere, if desired.) Now identify (w,x,y,z) with (−w,−x,−y,−z). An equivalent, slightly more complicated approach, begins with a 3-ball. As for the rest of his description, I think it refers to a homomorphic operator on curvilinear forms, but this goes beyond my expertise. --KSmrq^T 09:54, 22 May 2007 (UTC)

You could be right, he might mean compactness. He's usually pretty careful about that kind of thing, though. What would tight closure do to it? If I'm understanding the beginning of the tight closure article correctly, the space would have to be a specialized type of ring. Is it possible to define addition and multiplication such that a real projective plane or 3-space is tightly closed? Since it seems like that would effect measure and movement, would that change what it was like to live inside it? Black Carrot 00:00, 23 May 2007 (UTC)

Problem on Pearson Correlation

I encountered a problem on correlation that is rather strange to me. I have got two sets of time-series data (N = 115) which have correlation coefficient of -0.0636 (Significance 0.01 - two tailed). However if I divide the data into two halves (N1 = 58 and N2 = 57) the correlation coefficient of first half becomes +0.6134 (Significance 0.01 - two tailed)and the second half +0.2317 (Significance 0.08 - two tailed). Previous studies say that the correlation coefficient between the data should be positive and significant. I am not a very statistics person. Can anyone help me what might be the reason for these results, in non-technical terms? How to interpret such results and relate it to the previous studies?

The data sets are stock market trading volume and volatility, if it helps you to answer more effectively.

Thanks in advance.--202.52.234.141 08:45, 22 May 2007 (UTC)

Plotting the data with one set on the x-axis and the other on the y-axis is always a good first step. You may find that you have data which looks like

                  x
                 x                     x
                x                     x
               x                     x
                                    x

Taking as a whole you could fit a slightly downward pointing line. But each group individually would fit an upwards pointing lines. Is there a reason for the two halves, say if the data comes from two seperate days there might be an explination. --Salix alba (talk) 09:05, 22 May 2007 (UTC)

Square root of pi

What is the square root of pi? —The preceding unsigned comment was added by Amleth (talk • contribs).

Approximately 1.77245385? Something tells me you want something a little bit more than just a number. x42bn6 Talk Mess 20:18, 22 May 2007 (UTC)

(after edit conflict) It is the positive number whose square equals the number π. Just like π, it is a transcendental number, so any numeric values we give are only approximations. Then

${\begin{array}{rcl}\pi &=&3.14159265358979323846264338327950288419716939937510...\\{\sqrt {\pi }}&=&1.77245385090551602729816748334114518279754945612238...\end{array}}$

Does that answer your question?  --Lambiam^Talk 20:28, 22 May 2007 (UTC)

positive *real* number ^^ —The preceding unsigned comment was added by 84.187.55.83 (talk • contribs) 20:41, May 22, 2007 (UTC) - Please sign your posts!

? Is there a positive *not-real* number whose square equals π? If not, what is the point of this addition?  --Lambiam^Talk 21:54, 22 May 2007 (UTC)

If I'm remembering my complex algebra correctly, there are only two numbers that, when squared, equal π, and neither one has an imaginary component. Thus, the qualifier "real" is unneccessary. --Carnildo 21:58, 22 May 2007 (UTC)

Yes, it was a fun addition, but you should at least think a moment about the possibility of additional complex roots. You can prove that there is no additional root by considering the polar form. Sqaring is doubling the angle and it becomes obvious, that there can be no additional square root. There are however additional fourth roots.

I just want to point out that there are two real numbers whose square is pi. I'm quite positive about one of them but I am negative about the other one. 202.168.50.40 22:02, 22 May 2007 (UTC)

Yes, if you are talking about solutions to the equation x² = π. However, that was not what was asked, and it is convention to consider sqrt, ie., the square root, as positive and single-valued.

Can you clarify this for me? My memory (such as it is) tells me that the words "the square root of <whatever>" means both positive and negative roots, and in an exam you'd be wrong to provide only the positive one. But if the question is "What is ${\begin{array}{rcl}{\sqrt {\pi }}\end{array}}$", then the use of the symbol means that all you're being asked about is the positive root, and in an exam you'd be wrong to provide the negative one as well. I know this is not an exam, but isn't this the technical difference between the words "square root" and the square root symbol? JackofOz 03:14, 23 May 2007 (UTC)

I use "a square root" if I mean any number whose square is equal to the argument, and "the square root" for the principal square root. I don't recollect ever seeing or hearing a mathematician use the locution "the square root of" in another meaning.  --Lambiam^Talk 10:13, 23 May 2007 (UTC)

If that's true, what use is letting $x={\sqrt {2}}$ when, using the definition you gave, leads to ambiguity? This notation almost always implies the positive root for people who are familiar with the notation, and reads as "let x equal the square root of 2". There is a reason we have "$\pm$". However, $x=\pm {\sqrt {2}}$ can be rather confusing since x actually stands for two values. I suspect this is where most peoples confusion lies. However, I cannot say that in some context (possibly where people are ignorant of convention or are abusing notation) that ${\sqrt {x}}$ may mean something else other than the principle square root of x. Root⁴(one) 03:46, 23 May 2007 (UTC)

(edit conflict) Nod, the fourth root of one (yes, I'm spamming myself!) is generally only considered to be one, although there are 4 roots to $x^{4}=1$. To generalize the convention to the complex plane, if a number x be expressed as $(r,\theta )$, you calculate the nth root of x to be $({\sqrt[{n}]{r}},\theta /n)$. I believe this concept is called the "principle root". One reason for this name may be that this is the only root from which you can calculate the rest of the n-1 roots by incrementally adding $2\pi /n$ to the angle and not have to calculate the angle additions modulo $2\pi$. That is, you don't have to subtract $2\pi$ from any resulting angle so that $0\leq \theta <2\pi$. Root⁴(one) 03:30, 23 May 2007 (UTC)

To JackofOz: I think using what's likely to be marked correct on an exam is a silly metric to use for almost anything. On an exam, you should put down whatever the convention established in your class by your professor, whether right or wrong -- that doesn't make it mathematically correct. To Root4: I don't know any context in which the fourth root of 1 would be only considered to be 1 (of course there is no argument that 1 is a fourth root of 1). In fact, the only time I can think of for preferring one nth root of unity over another is when you want to focus on primitive roots, and of course 1 is never an primitive nth root of unity. Tesseran 04:55, 23 May 2007 (UTC)

I realise that words and symbols are mathematically "correct" only to the extent that there is agreement as to what they mean in mathematical contexts. I was taught the above difference by a number of teachers, both in school and university, as an established convention that I understood was universally agreed on. But maybe this isn't the case. I don't understand what you mean by "a silly metric". Surely professsors don't just make up their own meanings for mathematical symbols and expect their students to comply. Do they? JackofOz 05:10, 23 May 2007 (UTC)

This is not a very deep topic. In formulas, there is never any confusion since the symbol always denotes the positive root as you said. When words are used, then it depends on the context. nadav (talk) 05:32, 23 May 2007 (UTC)

Thank you, that's all I wanted to know. (Sorry for dragging you into shallow waters, but they tend to be warmer and permit more illumination). JackofOz 05:39, 23 May 2007 (UTC)

It should be emphasized that the word "the" in mathematics can only be used when the object described is unique. For example, the expression "the real number whose square is 4" is meaningless, as there are two such numbers. Similarly, using "the square root of 4" to describe either the positive or the negative root is incorrect. -2 is a square root of 4, not the square root of 4. By convention, the phrase "the square root of x" is shorthand for "the principal square root of x", which is the nonnegative one, and denoted ${\sqrt {x}}$. I can understand high school teachers being confused about this point, but not university teachers, so you may want to discuss the matter with them (making sure you understood their convention correctly). -- Meni Rosenfeld (talk) 11:53, 23 May 2007 (UTC)

Excellent explanation, Meni. I parted company with my university mathematics teachers over 25 years ago, so I'm not inclined to hunt them down at this remove. Obviously I had the wrong impression. (Quite strange - this is only the second time I've ever been wrong. The first time was many years ago, when I thought I'd given a wrong answer to a question. As it turned out, I was right and my examiner was wrong; but I was wrong to have thought I was wrong.)  :) JackofOz 00:17, 24 May 2007 (UTC)

Can anyone think of any interesting properties of the square root of π? All I can think of is that it's the area under the Gaussian function. nadav (talk) 02:50, 23 May 2007 (UTC)

Silly me, I forgot about the gamma function. nadav (talk) 05:24, 23 May 2007 (UTC)

This has already been asked. — Daniel 18:43, 23 May 2007 (UTC)

Thanks. You have a good memory. But I'm somewhat disappointed because I expected a lot of interesting-looking Ramanujan type identities besides the two properties above and Stirling's approximation, nadav (talk) 19:13, 23 May 2007 (UTC)

${\sqrt {\pi }}=\pm {\sqrt {\pi }}$

but...

$\pi ^{\frac {1}{2}}=+{\sqrt {\pi }}$MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 22:25, 23 May 2007 (UTC)

Um, no, because if the square root sign implied both roots, then we wouldn't have to write the plus-or-minus sign. ${\sqrt {}}$, as a function over the positive reals, is defined as the positive square root. Even over the complex numbers, it's generally taken to be the principal root. Confusing Manifestation 22:50, 23 May 2007 (UTC)

OK, folks, drop the stick and slowly back away from the horse carcass.  --Lambiam^Talk 23:05, 23 May 2007 (UTC)

LOL! :-D --KSmrq^T 15:23, 24 May 2007 (UTC)

Me too. —Bromskloss 13:54, 25 May 2007 (UTC)