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September 9[edit]

Atmospheric carbon scenario difference projections again[edit]

I'm moving this back from the archives in hopes that someone else can help balance it against File:Extreme-weather-cost.gif: What is the optimal amount of money to spend on climate change mitigation which will minimize total financial losses?

There is a calc problem that I don't understand at Wikipedia:Reference_desk/Archives/Science/2010 August 27#What is the wind-water-solar climate change mitigation scenario atmospheric carbon projection? [copied below] Why Other (talk) 22:22, 27 August 2010 (UTC)[reply]

From what I can tell, since there seems to be a lot of jargon there that would be better understood by a atmospheric scientist than a mathematician, the model being used is that the atmosphere is a giant tank of some fluid where some impurity is being added at a given rate while at the same time it's being removed at a rate proportional to its concentration. The tank is assumed to be well mixed, meaning you don't have to worry about the concentration not being the same in different parts of the tank. This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables. I hope that helps with the mathematical aspect of the model at least.--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)[reply]

Thanks! Why Other (talk) 03:24, 30 August 2010 (UTC)[reply]

In Jacobson, M.Z. (2009) "Review of solutions to global warming, air pollution, and energy security" Energy and Environmental Science 2:148-73 doi 10.1039/b809990c and Jacobson, M.Z. and Delucchi, M.A. (November 2009) "A Plan to Power 100 Percent of the Planet with Renewables" (originally published as "A Path to Sustainable Energy by 2030") Scientific American 301(5):58-65 what is the projected atmospheric carbon over time for their preferred wind-water-solar program?

What year do they start subtracting carbon and when do they reach 350 ppm?

I have asked also here, but I have been having better luck here at WP:RDS. Why Other (talk) 02:04, 27 August 2010 (UTC)[reply]

Per Dr. Jacobson, this is related to Eqn. 3 in http://www.stanford.edu/group/efmh/fossil/ClimRespUpdJGR%201.pdf

"[calculate] the time-dependent change in CO2 mixing ratio from a given anthropogenic emission rate, [and with that] the time-dependent difference in mixing ratio resulting from two different emission levels by subtracting results from the equation solved twice. Note that chi in the equation is the anthropogenic portion of the mixing ratio (this is explained in the text) and units of E need to be converted to mixing ratio. The conversion is given in the paper."

This almost might be ready for the math reference desk. Why Other (talk) 22:18, 27 August 2010 (UTC)[reply]

Maybe something like the scenario proposed by James Hansen's Alternative Scenario paper? ~A H 1^(TCU) 18:38, 28 August 2010 (UTC)[reply]

Interesting, but the assumptions in that paper don't involve adjusting the CO2 emissions rate, which is what I am trying to do.

This is from a response to a pointer to this question I put on the Math Reference Desk:

"... This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables....--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)"[reply]

I wish I had more experience with ODEs. Why Other (talk) 03:27, 30 August 2010 (UTC)[reply]

The separation of variables article describes it some, and a little more explanation is here. Basically the ODE is something like

{dy \over dx}=ay+b

Separating variables gives you

{dy \over ay+b}=dx

Integrating both sides,

{1 \over a}{\log(ay+b)}+C_{1}=x+C_{2}

Exponentiating both sides, rearranging terms, and renaming some constants,

y={ke^{ax}-b \over a}

where $k=e^{a(C_{2}-C_{1})}$ or something like that. It's possible that I messed up that calculation somewhere (it's late and I haven't done this in a while) but it's a basic ODE technique.

That said, that linear model sounds oversimplistic for something as complicated as the earth/atmosphere system. See for example clathrate gun hypothesis. 67.119.3.248 (talk) 08:50, 30 August 2010 (UTC)[reply]

Can anyone take it further than that? I know compared to File:Extreme-weather-cost.gif this is "merely" an optimization problem in probability distributions, but I'm sure it has few enough degrees of freedom to have a singular optimum solution.

If you want to consider a less tractable, but still important problem: how much is knowing the answer worth? Why Other (talk) 05:40, 8 September 2010 (UTC)[reply]

Fitting a parabola and Euler's theorem[edit]

I have two questions. I know that to write an equation for a parabola given three distinct points A, B, and C (which do not include any special points like the vertex) you substitute the x and y of A, B, and C for x and y in $y=ax^{2}+bx<c$ , then solve a system of equations. Is there a faster way to do this, using more sophisticated maths? I was think the mean value theorem and Rolle's Theorem could be used, if you find the slope of the lines AB, BC, and AC and set them all equal to $2xa+b$ , but I got hung up because the 'x' is not the same in each so you have x₁ etc. And #2 Euler's method goes like: $y_{n+1}=y_{n}+\Delta x_{step}\cdot f(x_{n},y_{n}),\$ where f is the derivative of some function y(x). Why the $f(x_{n},y_{n})$ ? What are (xn, yn) there for, since f is presumably a function only in x and is ther any difference to just f(xn)? Thanks a lot. 24.92.78.167 (talk) 00:21, 9 September 2010 (UTC)[reply]

Lagrange polynomial may be useful here. Euler's method is a method for solving differential equations and, unless I misunderstood the question, is not applicable.--RDBury (talk) 06:51, 9 September 2010 (UTC)[reply]

Euler's method is a separate question entirely; as I understand it, it's why might f depend on y. Usually it's for solving dy/dx = f(x,y) and f really could depend on y (eg dy/dx = y). Of course, if you can solve for y analytically then it only really depends on x (and the initial conditions), but if you can solve analytically you're not going to use Euler's method. 95.150.23.63 (talk) 15:28, 9 September 2010 (UTC)[reply]

Underline notation for n!/(n-k)![edit]

Binomial coefficient#Multiplicative formula states:

{\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}

Clearly $n^{\underline {k}}={\frac {n!}{(n-k)!}}$ , but I've never seen that notation before. Is it standard, and what is it called? -- 124.157.218.142 (talk) 02:42, 9 September 2010 (UTC)[reply]

Falling factorial.--203.97.79.114 (talk) 03:58, 9 September 2010 (UTC)[reply]

Thank you! And from Pochhammer symbol#Alternate notations, spoken "n to the k falling". -- 124.157.218.142 (talk) 04:07, 9 September 2010 (UTC)[reply]

SVD and Polysemy[edit]

I've read in a few papers on latent semantic indexing that singular value decomposition solves problems with polysemy in data. This claim is made, but not backed up. I understand clearly how synonymy and data sparsity are solved (the other two commonly noted data problems), but I do not clearly see how polysemy is solved. Any ideas? -- kainaw ™ 04:07, 9 September 2010 (UTC)[reply]

Probably because of the occurrence of words of related meaning tend to be correlated with each other. Say w1 and w2 have meanings m1 and m2 (that are related, say both are in medicine), and w1 and x2 have meanings n1 and n2 (i.e. w1 is polysemous) where n1 and n2 are related to finance. You'll see {w1,w2} in medical documents and {w1,x2} in financial documents. 75.57.241.73 (talk) 07:53, 9 September 2010 (UTC)[reply]

Radom number generation[edit]

Hi there. If I wanted to generate lognormal random numbers (from the distribution LN(μ,σ²)) can I just generate normal numbers (from the distribution N(μ,σ²)) and use the expenential of the second set? Thanks. --Mudupie (talk) 14:48, 9 September 2010 (UTC)[reply]

The second sentence of lognormal says yes! Note of course that the mean of the result won't be

\mu

or even

e^{\mu }

. --Tardis (talk) 16:24, 9 September 2010 (UTC)[reply]

Truncated lognormal distribution[edit]

Hello. Say I have a variable X that I know to be lognormally distributed. I would like to parametrise the distribution Y = X|X>k given some observations that I have. I only know the observations that are more than k but I know the number of observations that are smaller than k (although I don't know what those observations are). How do I parametrise Y? Thanks. --Mudupie (talk) 15:41, 9 September 2010 (UTC)[reply]

Consider log(X) which has a normal distribution. Find the number m such that the number of observations above m is equal to the known number of observations below k. Look up in a table of the normal distribution to find the number of standard deviations between log(m) and log(k), and you are done. Bo Jacoby (talk) 20:47, 9 September 2010 (UTC).[reply]

I corrected the above explanation several times, because it was unclear and incorrect, so let me try to be a little more precise. Let the total number of observations be N. Let the number of observations below k be K. The variable log(X) has normal distribution with mean value (log(m)+log(k))/2, and the K/N quantile is log(k). Bo Jacoby (talk) 21:03, 10 September 2010 (UTC).[reply]

Finding Remainder using Scientific Cal?[edit]

Is it possible for me to find out the remainder (long division?) using a scientific calculator?

My Scientific Calculator is Sharp EL-520W (EL-520WBBK).

The manual is here: http://ec1.images-amazon.com/media/i3d/01/A/man-migrate/MANUAL000031630.pdf

Say for example: 28,925.5 / 3,600 = 8.03486...

I have a fraction button which outputs either ${\tfrac {57,851}{7,200}}$ or $8{\tfrac {251}{7,200}}$ .

The remainder is supposed to be 125.5.

How can I find that using my calculator?

--33rogers (talk) 18:23, 9 September 2010 (UTC)[reply]

Either repeatedly subtract the divisor from the quotient until the result is less than the quotient, or once you have the $8{\tfrac {251}{7,200}}$ solution, take the integer part (8 in this case), multiply that by the divisor, and subtract that from the quotient.

28,925.5 - (8 x 3,600) = 125.5

--Rojomoke (talk) 19:46, 9 September 2010 (UTC)[reply]

The manual doesn't (appear to) list all the keys, so it's hard to be sure: what you're looking for is the modulo operation. Be warned that some systems that provide such an operation define it only for integers, some systems define it in an odd way for negative numbers, and that the operators modulo and remainder are closely related but not identical. It's confusing; if you'd like to know more, just ask. --Tardis (talk) 15:19, 10 September 2010 (UTC)[reply]

Look at the menu the MATH key brings up to see if you've got a floor function, probably called INT, or, failing that, a truncate function, called perhaps IPART or FIX. You can then compute the remainder of the division like this: 28925.5-3600*INT(28925.5/3600) where / means the division (÷ key). (This requires using both numbers twice, so you may want to store them in registers with the STO function instead.) You can alternately use the formula 3600*FPART(28925.5/3600) which is easier to type but may have more problems with precision.

To Tardis: if you want to see all the keys, either google for the name of the calculator, or look it up in V. T. Toth's calculator museum.

Hmm, after some more googling I suspect this calculator does not have a floor function, so this won't work. Sorry. – b_jonas 22:31, 12 September 2010 (UTC)[reply]

e-values and e-vectors of a transition matrix[edit]

I am trying to find all the eigenvalues and eigenvectors of this matrix: $\left({\begin{array}{cccc}0&1&0&0\\p&0&(1-p)&0\\0&p&0&(1-p)\\0&0&1&0\end{array}}\right)$

I know that one of the eigenvalues is 1 with an eigenvector of all 1's...

but when I try to expand out the polynomial taken from $|P-\lambda I|=0$ , it's not something I can work with. Are there actually four eigenvalues/eigenvectors, and is there a trick to getting the other three? —Preceding unsigned comment added by 130.102.158.15 (talk) 23:03, 9 September 2010 (UTC)[reply]

What is the polynomial you get? Rckrone (talk) 03:34, 10 September 2010 (UTC)[reply]

I got it to work out alright. Rckrone (talk) 03:50, 10 September 2010 (UTC)[reply]

my polynomial is

\lambda ^{4}+\lambda ^{2}(1-2p)+p(1-p)

and when I try and solve this I get a very ugly thing...

\pm {\sqrt {\frac {-(1-2p)\pm {\sqrt {(1-2p)^{2}-4p(1-p)}}}{2}}}

—Preceding unsigned comment added by 130.102.158.15 (talk) 07:09, 10 September 2010 (UTC)[reply]

That is not the correct characteristic polynomial. One way to see this is by checking if

\lambda =1

is a root. (And you know it's an eigenvalue, so it has to be a root of the characteristic polynomial.)

Once you have the correct polynomial, you can try to find the roots of that degree 4 polynomial. (We're actually lucky in this case, and can solve it using the quadratic equation-- often degree 4 is much harder.) But since you already know that 1 is an eigenvalue, you can factor out

\lambda -1

and get just a degree 3 polynomial. There is one other root you should be able to guess, since the characteristic polynomial is actually a function of

\lambda ^{2}

; and if you factor that out too, you have just a degree two polynomial to solve to find the final two eigenvalues.140.114.81.55 (talk) 09:27, 10 September 2010 (UTC)[reply]

It should be noted that finding the characteristic polynomial is easier if you already know some of the eigenvectors. In this case you have the all 1's vector and the alternating ±1 vector are eigenvectors. If you use these to create new columns in

|P-\lambda I|=0

you can factor out the

\lambda ^{2}-1

and reduce the problem to a 2×2 matrix. The other factor comes out as

\lambda ^{2}-p(1-p)

.--RDBury (talk) 20:25, 10 September 2010 (UTC)[reply]

got it, thanks —Preceding unsigned comment added by 130.102.158.15 (talk) 23:42, 11 September 2010 (UTC)[reply]