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May 1[edit]

Gene DNA GC content against whole genome DNA GC content.[edit]

Hi guys, I have a simple question concerning GC (Guanine-Cytosine) content in DNA sequences. Suppose I have a bacterial genome, with a total GC content of 60%. And one of the genes in the genome is 1000 base pairs long, and has a GC content of 70%. I have been trying to do a bootstrap analysis to see if the 70% gene GC content value is significant, but I'm not sure about my steps:

- I create 100 random sequences, each 1000bp long, with each position in every sequence having a 60% chance to be G or C, and 40% to be A or T.

- For every sequence I calculate the resulting GC content, and add all values in an array.

- I calculate the Mean and StDev in the array.

- If the GC content of my gene is higher than Mean + StDev*4.47 or lower than Mean - StDev*4.47 i have a 95% probability chance of significance.

- If the GC content of my gene is higher than Mean + StDev*10 or lower than Mean - StDev*10 i have a 99% probability chance of significance.

I feel I'm wrong somewhere, and this produces the same results as if I had simply used the genome GC content for Mean and 1 for StDev. Any help would be much appreciated. PervyPirate (talk) 10:54, 1 May 2010 (UTC)[reply]

Are the base pairs of a genetic sequence independent of one another ? If not, it's a bit like saying "the average number of Ford logos per average car on the road is 0.3, yet my car has 4, is that significant ?". In the case of DNA, while they may initially be random, the selection process may tend to favor certain (non-random) combos. StuRat (talk) 11:23, 1 May 2010 (UTC)[reply]

I didn't want to complicate things, but yes, let's assume that we are talking about synonymous positions (the ones that have no effect on the encoded amino acid, and therefore they are independent from one another as well as there is no selective pressure on them). PervyPirate (talk) 11:35, 1 May 2010 (UTC)[reply]

Help in Solving Tensor Equation[edit]

Lets say we have a tensor equation $2A_{\alpha \beta \gamma }={B^{\lambda }}_{\gamma }C_{\alpha \beta \lambda }+{B_{\beta }}^{\mu }C_{\alpha \gamma \mu }$ . How do we solve it? Telling me how to find the tensor ${D^{\alpha }}_{\beta }$ or ${D_{\beta }}^{\alpha }$ given $E_{\beta \gamma }={D^{\alpha }}_{\beta }F_{\alpha \gamma }$ or $E_{\beta \gamma }={D_{\beta }}^{\alpha }F_{\alpha \gamma }$ respectively would be great help too. P.S. This is not a homework problem. Thanks in advance.~~The Successor of Physics~~ 13:35, 1 May 2010 (UTC)[reply]

Give all the possible values for the free indices and you get a (linear) system of equations for the components of B, C or D (whichever is unkown). In the last two equations it may be possible (always, if F is symmetric) to first diagonalize F with a coordinate transformation, then the equation system becomes trivial to solve; afterwards you just transform the obtained D back. — Pt _(T) 18:37, 2 May 2010 (UTC)[reply]

Thanks.~~The Successor of Physics~~ 14:07, 4 May 2010 (UTC)[reply]

(I've taken the liberty to fix the staggering of your indices.) You could also look at the last two as matrix equations, if you arrange the components into matrices appropriately and carefully. In two dimensions, the second equation would then read

E=DF

, where

E={\begin{pmatrix}E_{11}&E_{12}\\E_{21}&E_{22}\end{pmatrix}}{\text{,}}\quad D={\begin{pmatrix}{D^{1}}_{1}&{D^{2}}_{1}\\{D^{1}}_{2}&{D^{2}}_{2}\end{pmatrix}}{\text{,}}\quad F={\begin{pmatrix}F_{11}&F_{12}\\F_{21}&F_{22}\end{pmatrix}}{\text{.}}

Thus

D=EF^{-1}

and the solution exists if and only if F is invertible, that is, the determinant of F is non-zero. — Pt _(T) 12:35, 5 May 2010 (UTC)[reply]

Operator norm and isomorphisms[edit]

Let A, B be Banach spaces and L(A,B) be the vector space of the linear operators between A and B.

The natural norm for L(A,B) is the operator norm:

|T|_{L(A,B)}:=\sup _{|x|_{A}=1}|Tx|_{B}

now my problem is the following: there can be isomorphisms between these operator spaces, for example

L(A,L(B,C))\cong L(B,L(A,C))

and if we think the space in the first way rather than the second isomorphic way the definition of the operator norm changes.

My question is: is the operator norm independent on these isomorphism or do we get different operator norm if we look the space in a different way?

In other word is it true that

|T|_{L(A,L(B,C))}:=\sup _{|x|_{A}=1}|Tx|_{L(B,C)}=\sup _{|y|_{B}=1}|Ty|_{L(A,C)}=:|T|_{L(B,L(A,C))}

?

Thank you for your help.--Pokipsy76 (talk) 14:21, 1 May 2010 (UTC)[reply]

I think I've the answer. The equality holds just because if we explicit the meanings of the terms we have

|T|_{L(A,L(B,C))}:=\sup _{|x|_{A}=1}|Tx|_{L(B,C)}=\sup _{|x|_{A}=1}\sup _{|y|_{B}=1}|Txy|_{C}=\sup _{|y|_{B}=1,|x|_{A}=1}|Txy|_{C}

and the same happens for

|T|_{L(B,L(A,C))}

.--Pokipsy76 (talk) 16:14, 1 May 2010 (UTC)[reply]

Correct, and note that both spaces are also isometrically and linearly isomorphic with the space of bounded bilinear maps, L²(A×B, C). The norm of a bilinear map u:A×B→C is by definition

\|u\|_{L^{2}(A\times B,C)}:=\sup _{|x|_{A}\leq 1,\,|y|_{B}\leq 1}\|u(x,y)\|_{C}

,

and of course does depend on the choice of the norms on the spaces A B C. But the other equivalent norms on A, B resp. C, would produce an equivalent norm on L²(A×B, C).--pm a 17:45, 1 May 2010 (UTC)[reply]

games that do not end[edit]

What games do not have moves that progress through time, i.e., what games are not temporal? 71.100.1.71 (talk) 16:44, 1 May 2010 (UTC)[reply]

I don't know if this is what you mean, but some games consist of the players each making a single move all simultaneously. Those games don't progress through time in any meaningful sense since they happen at a single moment. Prisoner's dilemma is one example of such a game. Rckrone (talk) 16:55, 1 May 2010 (UTC)[reply]

These are known as Normal-form games. -- Meni Rosenfeld (talk) 20:01, 1 May 2010 (UTC)[reply]

There are games that can get into draw states, where no player can win. Chess is one. Conway's Game of Life can also get into repeating loops. StuRat (talk) 17:07, 1 May 2010 (UTC)[reply]

Many social games do not progress in any meaningful way. They are just there to maintain society. Dmcq (talk) 19:15, 1 May 2010 (UTC)[reply]

Do you mean games like rock, paper, scissors? 71.100.1.71 (talk) 09:11, 2 May 2010 (UTC)[reply]

No I mean like the banter between James Bond and Miss Moneypenny. Though I guess playing a game of rock scissors paper every day with someone would be roughly equivalent. Dmcq (talk) 15:49, 2 May 2010 (UTC)[reply]

Asians, Indians, and Eastern Europeans[edit]

Why are Asians, Indians, and Eastern Europeans so much better at math (in modern times, at least)? --75.33.219.230 (talk) 17:35, 1 May 2010 (UTC)[reply]

See Race and Intelligence. By the way Eastern Europe isn't part of Asia. Dmcq (talk) 19:08, 1 May 2010 (UTC)[reply]

That's why I listed Asians and Eastern Europeans separately. --75.33.219.230 (talk) 19:36, 1 May 2010 (UTC)[reply]

Assuming that the assertion in your question is indeed true, I think the reason is mostly cultural—how math is taught and how much emphasis is placed on math education, whether math is commonly perceived as a hard subject, whether great mathematicians and their achievements are seen as a source of national pride, existence of lucrative career possibilities not requiring much math, ... --98.114.146.58 (talk) 08:22, 2 May 2010 (UTC)[reply]

If you're talking about research in mathematics, France is probably the top country, in terms of Fields medallists. If not in absolute terms, per capita certainly. The US and Russia are also high.

On the other hand, if you're talking about the ability of schoolchildren, Eastern Europe no longer ranks highly. Since the fall of the Soviet Union, their education systems have gone down the tubes. France isn't great either. Have a look at the PISA results (for 15-year olds), ranking countries in the OECD as well as some others: [1]. Japan is tops in math, South Korea second, but then come New Zealand, Finland, Australia, Canada, Switzerland, the UK... Two Canadian provinces, Quebec and Alberta, would rank above South Korea if they were countries. The Czechs, Russians, Hungarians and Americans are all below average, and Latvia near the bottom. France is slightly above average. I would suspect that India wouldn't have ranked high if they had participated. 82.124.97.111 (talk) 16:16, 2 May 2010 (UTC)[reply]

Sorry, the latest results from 2006 (in which more countries participated) have the following top ten in math: Taiwan, Finland, Hong Kong, South Korea, Netherlands, Switzerland, Canada, Macao, Liechtenstein, Japan. Estonia 14th, Czech Republic 16th, Slovenia 20th, France 23rd, Poland 24th, the UK 25th, Slovakia 26th, Hungary 27th, Latvia 30th, Lithuania 31st, Russia 34th, the US 35th. Many Eastern European countries are ranked lower as well. Clearly, it's possible for Western countries to match Asia, since five of the top ten are Western. We don't need to try and match Japan. Let's copy whatever Finland is doing instead. —Preceding unsigned comment added by 82.124.97.111 (talk) 16:45, 2 May 2010 (UTC)[reply]

A better way to measure how different countries compare would be to let university students do the problems assigned to students from other universities in other countries. E.g. consider these problems assigned to first year students at Novosibirsk State University . These are way too hard for most first year students at Western universities. Quite a few of these problems would be challenging even for most fourth year students. Count Iblis (talk) 14:36, 4 May 2010 (UTC)[reply]

Is the OP (Mr. 75.33) an American? If so, you must realize that the Asians, Indians, and Eastern Europeans that you are exposed to are not representative of all Indians, Asians, and Eastern Europeans. It will be precisely those who have a good mind for numbers who will be more likely to immigrate here for a technical job, or similar. Buddy432 (talk) 15:29, 4 May 2010 (UTC)[reply]

Derivatives[edit]

Hey guys, I'm having trouble with some derivatives; maybe someone could help me out? I don't get how ${\frac {d}{dx}}a^{x}=\ln(a)a^{x}$ or how ${\frac {d}{dx}}\ln(x)={\frac {1}{x}}$ . Could someone show me a proof? What I'm not getting is where the natural logarithm is coming from. Thanks! PS: In the former, why is it that ${\frac {d}{dx}}a^{x}=\ln(a)a^{x}$ , and not ${\frac {d}{dx}}a^{x}=(x)a^{x-1}$ ? 76.229.195.134 (talk) 19:49, 1 May 2010 (UTC)[reply]

{\frac {d}{dx}}a^{x}=\ln(a)a^{x}

follows from

a^{x}=e^{xlna}

and the chain rule.

{\frac {d}{dx}}\ln(x)={\frac {1}{x}}

follows from the derivative of

e^{x}

and the inverse function theorem. The former is not

xa^{x-1}

because that rule is for when x is the base, not the exponent. -- Meni Rosenfeld (talk) 20:09, 1 May 2010 (UTC)[reply]

The erroneous proposed derivative xa^x−1 would imply that the slope is negative when x is negative, and obviously that's not what you see when you look at the graph. If you go back to the definition of the derivative, you've got

{\begin{aligned}{d \over dx}a^{x}&=\lim _{h\to 0}{a^{x+h}-a^{x} \over h}\\[10pt]&=\lim _{h\to 0}a^{x}{a^{h}-1 \over h}\\[10pt]&=a^{x}\lim _{h\to 0}{a^{h}-1 \over h}{\text{ (since }}a^{x}{\text{ does not depend on }}h)\\[10pt]&=\left(a^{x}\cdot {\text{constant}}\right){\text{ (since the limit does not depend on }}x).\end{aligned}}

so next you need to know what the "constant" is. To be continued...... Michael Hardy (talk) 20:48, 1 May 2010 (UTC)[reply]

Note that the derivative of a function with respect to x looks at how the function changes when we change x. In that regard a^x behaves very differently from xⁿ because x is the variable we care about. It is true that ${\frac {d}{da}}a^{x}=xa^{x-1}$ , but that's something different. Rckrone (talk) 05:59, 2 May 2010 (UTC)[reply]

Here is an explanation. Michael Hardy (talk) 19:14, 2 May 2010 (UTC)[reply]

long differential equations problem[edit]

This was a problem in my homework a couple of days ago.

"Use the method of variation of parameters to determine the general solution of the given differential equation: $y^{iv}+2y''+y=\sin(t)$ "

The characteristic equation is $k^{4}+2k^{2}+1=0$ . That has roots of {i, i, -i, -i} so the general solution of the corresponding homogeneous equation is $y=c_{1}\cos(t)+c_{2}\sin(t)+c_{3}t\cos(t)+c_{4}t\sin(t)$ .

To find the particular solution, I have to evaluate five Wronskians:

$\left|{\begin{array}{cccc}\cos(t)&\sin(t)&t\cos(t)&t\sin(t)\\-\sin(t)&\cos(t)&\cos(t)-t\sin(t)&\sin(t)+t\cos(t)\\-\cos(t)&-\sin(t)&-2\sin(t)-t\cos(t)&2\cos(t)-t\sin(t)\\\sin(t)&-\cos(t)&t\sin(t)-3\cos(t)&-3\sin(t)-t\cos(t)\end{array}}\right|$

$\left|{\begin{array}{cccc}0&\sin(t)&t\cos(t)&t\sin(t)\\0&\cos(t)&\cos(t)-t\sin(t)&\sin(t)+t\cos(t)\\0&-\sin(t)&-2\sin(t)-t\cos(t)&2\cos(t)-t\sin(t)\\1&-\cos(t)&t\sin(t)-3\cos(t)&-3\sin(t)-t\cos(t)\end{array}}\right|$

$\left|{\begin{array}{cccc}\cos(t)&0&t\cos(t)&t\sin(t)\\-\sin(t)&0&\cos(t)-t\sin(t)&\sin(t)+t\cos(t)\\-\cos(t)&0&-2\sin(t)-t\cos(t)&2\cos(t)-t\sin(t)\\\sin(t)&1&t\sin(t)-3\cos(t)&-3\sin(t)-t\cos(t)\end{array}}\right|$

and so forth. How can I quickly calculate these by hand? I also need some help evaluating the resulting integrals. I know the problem can be done pretty easily using the method of undetermined coefficients but the problem says to use the method of variation of parameters. —Preceding unsigned comment added by Metroman (talk • contribs) 21:37, 1 May 2010 (UTC)[reply]

This assignment looks like torture to me. I would suggest simplifying things as follows. You can write the diff. equation as:

(D^2 + 1)^2 y = sin(t)

This means that if you first solve the equation:

(D^2 + 1) y_1 = sin(t)

then the solution of

(D^2 + 1)y = y_1(t)

Will also be a solution of the original diff. equation. You can use variation of constants here too, so this should be allowed. You can simplify things further by replacing sin(t) by exp(it) and taking the imaginary part of the solution. Also, you can factor D^2 + 1 = (D+i)(D-i) and solve each of the two equations in two steps. Count Iblis (talk) 22:43, 1 May 2010 (UTC)[reply]

As to the issue of computing by hand the determinants: note that, in the first matrix, if you sum the first row into the third, and the second row into the fourth, you get a matrix with a null 2x2 block, so its determinant is easily computed as the product of the determinants of the two principal 2x2 blocks (at a first glance: isn't -2?). Analogous simplifications work with the other matrices. --pm a 07:07, 2 May 2010 (UTC)[reply]

Romanian army / vehicle volume[edit]

In What Am I Doing in New Jersey?, comedian George Carlin mentioned, "People who don't know how wide their cars are...'Well, I don't know if I can fit in there!' - You can get the f*****g Romanian Army in there!"

I have been unable to find data on the approximate volume of the largest commercially-available cars, nor the average body capacity of Romanian military personnel. I wonder if anyone could help prove or disprove his hypothesis? Chzz ► 23:53, 1 May 2010 (UTC)[reply]

The hypothesis is that the Romanian army will fit in there? I think it would depend on where. Rckrone (talk) 05:51, 2 May 2010 (UTC)[reply]

As far as I can ascertain, it is the space in which one could comfortably fit a car. Chzz ► 06:05, 2 May 2010 (UTC)[reply]

check Hyperbole. --pm a 08:09, 2 May 2010 (UTC)[reply]

As an ex-resident of the Manchester (UK) area, the corresponding expression I'm familiar with is "You could get a Ferranti transformer through there". As it says in the article: "High voltage power transformers became an important product for Ferranti; some of the largest types weighed over a hundred tons". Moving the largest by road required a low-loader with motive power fore and aft, and skilled driving to pass through built-up areas.→81.131.162.203 (talk) 13:44, 2 May 2010 (UTC)[reply]

More common versions of the expression include "you could get a jumbo jet/supertanker through that gap". I would imagine that the same goes for any substantial object. 194.205.143.136 (talk) 05:54, 3 May 2010 (UTC)[reply]