Wikipedia:Reference desk/Archives/Mathematics/2007 November 13

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November 13

fridaythethirteenthnumbersequencethingy

Hi. I've discovered a possible repeating pattern of numbers, which depicts the number of Friday the thirteenths in a given year. I start the sequence when there is four of two of Friday the Thirteenths in the years. Here's the sequence: 2222112213112212122131132131. This comes from listing the number of Friday the Thirteenths from 1920 to 2019. The only year that breaks the pattern in these years is 1926, which has the wrong number of F13s from the number that it should have. Do anomalies such as these increase as you move farther forwards or backwards in time from this set of years? Can this be added to the article to provide the user with a good idea for how to calculate the number of F13s? What are some other years that have this anomaly? Or, could either I or the printer of the book have made a mistake? Oh, and please don't use this as your password or hash chain. Do leap years, especially centuarinary leap years, influence this number? Thanks. ~AH1^(TCU) 01:11, 13 November 2007 (UTC)

Well the number of F13 depends on whether January 1st is Monday and whether this year is a leap year. If we didn't have centuarinary leap years, I would expect the period of this sequence to be 7*4=28. With centuarinary leap years, I would expect the period to be 7*400. (Igny 03:19, 13 November 2007 (UTC))

The Gregorian calendar repeats every 400 years. So the period of the sequence must be a factor of 400. --Spoon! 09:10, 13 November 2007 (UTC)

I meant to say that there are a whole number of weeks in the Gregorian cycle of 400 years. --Spoon! 09:42, 13 November 2007 (UTC)

Are you sure that 1926 was an exception ? Since 2000 was a leap year, there is a leap year every four years between 1901 to 2099, and you would expect the cycle of Friday 13th counts to repeat every 28 years between these dates. Our Friday the 13th article gives a 28 year cycle from 2001 to 2028 that is essentially the same as your cycle, with an offset - your cycle starts in 2017 - and it says that this cycle repeats every 28 years between 1901 and 2099. I wouldn't expect the 28 year cycle to fail until 2100. Gandalf61 13:02, 13 November 2007 (UTC)

The sequence starting 2222 and ending 2131 applies to the years from 1933 to 1960 in your interval from 1920 onwards. Counting back, the year 1926 should have just one Friday the 13th, which it did, in August. So, no anomaly.→81.154.106.140 11:55, 14 November 2007 (UTC)

I suspect you've coined a new word "centuarinary" there, Igny. Congratulations. -- JackofOz 23:44, 14 November 2007 (UTC)

This sequence from OEIS [1] gives the pattern for the 400-year cycle, where the values represent the 14 different annual calendars (7 for non-leap years, 7 for leap years). Each year number corresponds consistently to a particular number (1, 2 or 3) of Friday 13ths.81.153.220.176 22:30, 15 November 2007 (UTC)

Functional Analysis

Okay, I was reading this grad level functional analysis book and came across this statement.

"For a sequence $(x_{n})$ in an inner product space (real or complex), the conditions $\|x_{n}\|\rightarrow \|x\|$ and $\langle x_{n},x\rangle \rightarrow \langle x,x\rangle$ imply convergence $x_{n}\rightarrow x$. This statement has a special case in the complex plane."

My questions is how do we prove the first part and then what is this special case in the complex plane?

Another statement was that,

Let $T:X\rightarrow X$ be a bounded linear operator on a complex inner product space $X$. If $\langle Tx,x\rangle =0$ for all $x\in X$, then $T=0$.

Is this true (a proof perhaps)? The reason I ask is because there is no proof in the book and I know that this is false in the case of a real inner product space with rotation. Rotation is a linear (non-zero) operator in $\mathbb {R} ^{2}$. So if I take two orthogonal vectors and rotate them, they are still orthogonal.
A Real Kaiser 04:12, 13 November 2007 (UTC)

For the first question: remember that $x_{n}\rightarrow x$ means that $\|x-x_{n}\|\rightarrow 0$. Using the two givens, this can be shown by expanding out $\|x-x_{n}\|^{2}=\langle x-x_{n},x-x_{n}\rangle$

For the second, note that if $\langle Tx,y\rangle =0$ for all x,y, then T = 0. Over the complex numbers we can pull a trick to show that this is implied by $\langle Tx,x\rangle =0$ for all x. To do this, consider that $0=\langle T(x+\lambda y),x+\lambda y\rangle =\langle Tx+\lambda Ty,x+\lambda y\rangle =\ldots$ (expand this out). If you do this for $\lambda =1$ and for $\lambda =i$, you'll see that $\langle Tx,y\rangle =0$ (you'll also show $\langle Ty,x\rangle =0$ at the same time). kfgauss 09:27, 13 November 2007 (UTC)

Partial deriviative equation

I have an equation which is function of two variables c and phi. I need to find a c that will make both equations equal to 0 at a certain phi value.

Or in math terms:

Fk(phi,c) = ... dFk(phi,c) = ... // dFk = diff(Fk,phi)

Fk(phi,c) = dFk(phi,c) = 0 => c = ?

The actual equation is this one: Fk1 = c1*hmax*(1 - (phi-pi)/beta1 + delta/hmax) - (hmax/2*pi) * ((((2*pi)^2*m1*omega^2)/beta1^2) - c1) * sin(2*pi*(phi-pi)/beta1);

Note that c1 (unknown), hmax, beta1, delta, m1 and omega are known constants. phi >= pi && <= 73deg. c1 is around 470E3.

Although, I can find a suitable value of c numerically, I'd like to try an analytical solution. Any insight? Thanks! --83.233.147.22 15:24, 13 November 2007 (UTC)

Let's get that in LaTex: $F_{k1}(c_{1},\phi )=c_{1}h_{max}\left(1-{\frac {\phi -\pi }{\beta _{1}}}+{\frac {\delta }{h_{max}}}\right)-{\frac {h_{max}}{2\pi }}\left({\frac {(2\pi )^{2}m_{1}\omega ^{2}}{\beta _{1}^{2}}}-c1\right)sin\left({\frac {2\pi (\phi -\pi )}{\beta _{1}}}\right)$. (Feel free to edit if I've made a mistake). Looks like a lot of work to me. Have you tried a system like Mathematica, to do it automatically? risk 16:06, 13 November 2007 (UTC)

If I understand correctly you are looking for a $c_{1}$ that makes $F_{k_{1}}$ and ${\frac {\partial {}F_{k_{1}}}{\partial \phi }}$ vanish. Start by calculating the derivative. You must then solve for $c_{1}$ a system of two linear equations. From there you obtain a solution for $c_{1}$ as well as a condition on the constants which guarantees the existence of a solution.

I obtain

$c_{1}={\frac {4\pi ^{2}m_{1}\omega ^{2}}{{\beta _{1}}^{2}\left(1-\sec \left({\frac {2\pi (\phi -\pi )}{\beta _{1}}}\right)\right)}}$

I haven't calculated the existence condition. Morana 18:39, 13 November 2007 (UTC)

Thanks for your help, although your final result does not seem to work once you put in the constants. I've tried to solve it using Maple but it gives me junk result as well.

restart;

Fk1 := c*hmax*(1-(phi-Pi)/beta+delta/hmax)-(1/2)*hmax*Pi*((2*Pi)^2*m*omega^2/beta^2-c)*sin(2*Pi*(phi-Pi)/beta);

dFk1 := diff(Fk1, phi)

a := dsolve({Fk1, dFk1}, [c, phi])

Strangely, if I solve using [phi, c] instead, it tells me that c = c... or if I use solve instead of dsolve. If I try to substitute the constants for their real values, Maple gives me "Solutions may have been lost a:=0".

$c=4\,{\pi }^{4}\cos \left(2\,{\frac {\pi \,\left(\varphi -\pi \right)}{\beta }}\right)m{\omega }^{2}{\beta }^{-2}\left(-1+{\pi }^{2}\cos \left(2\,{\frac {\pi \,\left(\varphi -\pi \right)}{\beta }}\right)\right)^{-1}$ (sorry, Maple is not exporting right)

I'm curious as to why we have different results, how did you find c?

Here's an image of the function: [[2]] Note that phi has been converted to degrees for display purposes and c is at 470E3. There is also an absolute function to make the graph symmetric. Just assume that we want to find a c value that will make the minimum on the right coincide with the x axis.

This should be simple...Argh! --83.233.147.22 20:17, 13 November 2007 (UTC)

I have obtained this solution by solving ${\frac {\partial {}F_{k_{1}}}{\partial \phi }}(c_{1},\phi )=0$ for $c_{1}$. If you then substitute this solution into $F_{k_{1}}(c_{1},\phi )=0$ you obtain a condition for the existence of a solution. As I said, I haven't calculated this condition but I have the impression that it is non-trivial, and that it probably doesn't hold for many choices of constants.

It would help if you gave the value of all the constants, and an example of a $(\phi ,c_{1})$ pair, where $c_{1}$ is a solution. Morana 23:44, 13 November 2007 (UTC)

$m:=0.22;hmax:=0.003;beta:=37*Pi/90;omega:=120*Pi$.

At $\phi =0.3615$ and $c=464000$, $F_{k_{1}}\approx 0$ and ${\frac {\partial {}F_{k_{1}}}{\partial \phi }}\approx 0$.

Thanks for your help. --83.233.147.22 08:22, 14 November 2007 (UTC)

Unknown f(x)

What sort of function could this be? it appears to be exponential on both sides, but I don't see how to join them both in a single function of x. Takk Rós 21:05, 13 November 2007 (UTC)

Finding a function that fits these data is not the problem. for example, if all else fails, you can use Lagrange interpolation, or guess that it is a polynomial of some degree and find the coefficients that minimize the error. Or, if you have other reasons to believe it should be exponential, you can try to fit a function of the form ${\frac {e^{\alpha x}}{\gamma e^{\beta x}+\delta }}$ or something similar. The problem is that there are too many possibilities, and there doesn't seem to be enough data to be sure which one is better.

If you can provide a numerical representation of the data (as will as any additional information or predictions you might have), I can try to find a good fit. -- Meni Rosenfeld (talk) 21:42, 13 November 2007 (UTC)

I see two possiblities that might be reasonably correct. One it is a skewed version of a Normal distribution or two it is an algebraic function with a horzontal asymptote as $x\to +\infty$ and a slant asymptote as $x\to -\infty$ A math-wiki 23:57, 13 November 2007 (UTC)

Do you have the data available in numeric form?  --Lambiam 01:20, 14 November 2007 (UTC)

Some context would also help tremendously. Where does this data come from? Why do you want to fit a curve to it? risk 02:13, 14 November 2007 (UTC)

Poisson distribution or maybe Rayleigh distribution can have shapes similar to that. - Rainwarrior 07:32, 14 November 2007 (UTC)

Or maybe a black body spectrum - see Planck's law. Gandalf61 10:18, 14 November 2007 (UTC)

The data don't seem to be discrete, so the Poisson distribution is probably inappropriate. If we do assume the function starts at (0,0) and only defined on $[0,\infty ]$, an obvious simple choice is $\alpha x^{\beta }e^{-\gamma x}\;\!$. -- Meni Rosenfeld (talk) 13:53, 14 November 2007 (UTC)

Math help

What is(-x^2)(y)(2xy)(-4y) in simplest form? Sseballos 21:47, 13 November 2007 (UTC)

What have you tried thus far to solve it?—Cronholm¹⁴⁴ 21:55, 13 November 2007 (UTC)

Is it (8)(x^3)(y^3), (-8)(-x^3)(y^3) or (8)(x^3)(-y^3)? Sseballos 21:58, 13 November 2007 (UTC)

Your original expression has two negative values, and the product of two negatives is always positive. --LarryMac | Talk 22:05, 13 November 2007 (UTC)

So is it my first answer? Sseballos 22:06, 13 November 2007 (UTC)

Yes. Jix2 22:11, 13 November 2007 (UTC)

I'm curious, is (-x^2)

(a) -(x^2)

or

(b) (-x)^2

202.168.50.40 23:27, 13 November 2007 (UTC)

It would be A since the exponent has priority over the minus. See Order of operations. --83.233.147.22 23:35, 13 November 2007 (UTC)

It can be ambiguous. binary and unary minus have different priorities and unary minus has priority over multiplication, so 1+2*-3 is 1+(2*(-3)). Quite how the priority of unary minus and power is not particularly well defined. The normal mathematical interpretation is (a), different programming languages have different standard, notably Excel has -x^2 = (-x)^2, matlab has -x^2 = -(x^2). There is a certain logic to excels choice as then all unary operations have the highest priority. --Salix alba (talk) 19:36, 14 November 2007 (UTC)

I think it would be best to review the idea of simplest form first and then avoid skipping steps if your unsure of how it calculates out. First off there is always many (in fact an infinite number) of ways to write a given expression (this can be proved by showing that $1={\frac {x-1}{x-1}}={\frac {x-2}{x-2}}={\frac {x-3}{x-3}}...$ etc. etc. if the denominators are not zero). So what is simpler, well in laymans terms were trying to write it with the minimum number of mathematical symbols (includes numbers). Technically there are some specific rules such as (in elementry (high school) algebra) you must remove any (square) roots from the bottom of any fractions. Now for our specific case, since - in a product is shorthand for the factor -1. Pull out the two -1's, $-1\cdot -1=1$ so with the negatives out of the way it's a simple algebraic product. we can drop the parenthesis (and in doing so multiply the 2 and (now positive) 4 together and then add exponents for x and y. (Do note that $xy^{3}$ is $x\cdot y^{3}$ not $x^{3}y^{3}$) A math-wiki 00:09, 14 November 2007 (UTC)

Actually to be pedantic, it is not the case that $1={\frac {x-1}{x-1}}={\frac {x-2}{x-2}}={\frac {x-3}{x-3}}\ldots$ Since each expression is equal to 1 for all x except for a single point, which varies with each expression. Donald Hosek 00:14, 14 November 2007 (UTC)

And a much simpler way to get an infinitude of expressions of equal value is 0 = 0 + 0 = 0 + 0 + 0 = ... .  --Lambiam 01:11, 14 November 2007 (UTC)

Indeed those are better examples, but mine is true for all x not an integer. I wanted to show how algebraic expressions can change through many different forms without changing the value of the expression for a given x value. (non-integer x's in this case) A math-wiki 00:03, 15 November 2007 (UTC)