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March 5

What's the symbol that means "repeat this function"? And what's the process called?

Resolved

– ± Lenoxus (" *** ") 20:23, 6 March 2011 (UTC)[reply]

It's kind of a jagged E. I think of it as "Greek-looking" but I don't think it's part of the Greek alphabet. Underneath this symbol, you put something like "1 > n > 50", or if you only want to evaluate the function for a single term, say 20, you write "n = 20". To the right of this symbol you put the function. If the part under the function is the infinity symbol, then you are to find the function's limit. Correct any of that which I got wrong.

I want to call this process "cocentating", but I don't think I'm spelling (or pronouncing) that correctly. And I feel that I'm getting that word from some other process that involves adding all the numbers from 1 to x, but that's related to what I'm looking for…

See, I'd like to know what it's called when you have a function, say, f(n)=7n, and you add together the outputs of this function for every integer from A to B. (If A is 2 and B is 5, then the final result in this very simple example would be 98).

Ultimately, I'm seeking an online calculator that can do that, with a rather more complex function: namely, ((-1)^(n+1)) * (16 choose n) * ((106-n) choose (10-n)).

To clarify, even though I just linked to Binomial coefficient, I'm really only familiar with the choose operation and what it does, and only from extracurricular study, so don't assume that I've taken a math course that would normally come "before" what I do know. (As you can tell, I'm also rusty on the stuff I did learn, anyway.) ± Lenoxus (" *** ") 01:00, 5 March 2011 (UTC)[reply]

I think you may be refering to capital Sigma Σ which is often used to represent summation so

\sum _{n=1}^{10}n

is the sum of numbers from 1 to 10.--Salix (talk): 01:24, 5 March 2011 (UTC)[reply]

Looks like it! Thanks so much. (Apparently I didn't look hard enough at that list of Greek letters.) Now all I'm still wondering is what my brain meant by "cocentation"… I swear the word is very close to an actual mathemathical term… ± Lenoxus (" *** ") 04:02, 5 March 2011 (UTC)[reply]

Concatenation juxtaposes two strings against each other. --COVIZAPIBETEFOKY (talk) 04:16, 5 March 2011 (UTC)[reply]

For examples of number concatenation you may have been thinking of, see oeis:A007908 and Champernowne constant. PrimeHunter (talk) 15:38, 5 March 2011 (UTC)[reply]

That's right. Thank you both! 20:23, 6 March 2011 (UTC)

Maths with Intensive and extensive properties

When trying to construct a valid mathematical formula in physics using intensive and/or extensive properties, would some mathematical operations with them be invalid? For example would it be valid to add an intensive property to an extensive property, or divide an intensive property by an extensive property? Which operations would be valid, which invalid?

Similarly with the Level of measurement - see http://people.math.sfu.ca/~cschwarz/Stat-301/Handouts/node5.html - would some operations (such as add, divide, multiply etc) be invalid, others valid, when done with the different scales?

I'm wondering if the two things above could give clues to the correct form of a formula in a similar way to how Dimensional analysis does? Thanks 92.15.6.227 (talk) 19:54, 5 March 2011 (UTC)[reply]

I don't think you can get much out of the "intensive/extensive" idea here that dimensional analysis won't already give you. If you can find an intensive and an extensive quantity of the same dimension, one could probably concoct a situation where it would make sense to add them. Of course their sum would itself be neither intensive or extensive -- it would say something about this particular system, and we could not derive from it the corresponding number for one of the halves if we cut the system in twain and considered each separately.

The level of measurement article does try to explain what you can do with various kinds of measurements, through links to the underlying mathematical structures. –Henning Makholm (talk) 00:34, 6 March 2011 (UTC)[reply]

I am tempted to say: No you cannot add an intensive property to an extensive property, but there's no reason not to multiply or divide them. Michael Hardy (talk) 01:36, 9 March 2011 (UTC)[reply]

Numerical Integration problem

I have two points in 2D with position vectors ${\vec {x}}=(x_{1},x_{2})$ and ${\vec {y}}={(y_{1},y_{2})}$ . We define ${\vec {r}}={\vec {x}}-{\vec {y}}$ and $r=|{\vec {r}}|$ . Another function $G$ is defined as $G_{ij}({\vec {x}},{\vec {y}})=-\mathrm {ln} (r)+{\frac {(x_{i}-y_{i})(x_{j}-y_{j})}{r^{4}}}$ , where $i=1,2$ and $j=1,2$ . How would I find $\int _{{\vec {x}}_{1}}^{{\vec {x}}_{2}}G_{ij}({\vec {x}},{\vec {y}})\mathrm {d} {\vec {x}}$ over the straight line joining ${\vec {x}}_{1}$ and ${\vec {x}}_{2}$ , if the interval $({\vec {x}}_{1},{\vec {x}}_{2})$ contains ${\vec {y}}$ . As we notice, we have singularity and normal numerical methods cannot be applied. Any help would be appreciated. (For curious readers, I am trying to solve Stokes problem using Boundary Elements and $G$ is the Green's function for velocity.) Thanks. - DSachan (talk) 23:13, 5 March 2011 (UTC)[reply]

Assuming I understand the notation and question, it looks as though the integral (of, say,

G_{ii}

) diverges to infinity, since the integrand is

O(1/r^{2})

. Could it be that you have the wrong integrand? The logarithmic singularity should be easier to cope with. Sławomir Biały (talk) 23:52, 5 March 2011 (UTC)[reply]

The logarithmic term does look suspicious for another reason, namely that it does not seem to depend on i and j -- this means that the entire construction is not symmetric under rotations of the coordinate system (or at least

G_{ij}

and its integral do not transform as tensors). Can that really be intended? I would expect something like a factor of

\delta _{ij}

there. –Henning Makholm (talk) 00:10, 6 March 2011 (UTC)[reply]

Also, what is the notation

\textstyle \int G_{ij}\,d{\vec {x}}

supposed to mean here? I'm guessing it's a line integral

\textstyle \int G_{ij}\,|d{\vec {x}}|

, but a case could also be made for

\textstyle \int (\sum _{j}G_{ij}dx_{j})

-- or even that the integral evaluates to a vector for each of the four i,j-combinations. –Henning Makholm (talk) 00:49, 6 March 2011 (UTC)[reply]

The first thing I'd do is to translate it into a one-dimensional definite integral, to avoid muddying the matter with irrelevant multidimensional details.

Now, suppose you want to evaluate

\int _{a}^{b}f(x)\ dx

, and f has a simple pole at c. You consider the integral over a pole to cancel out. Then what you can do is:

Find the residue $R=\lim _{x\to c}f(x)(x-c)$ .
Numerically evaluate $\int _{a}^{b}\left(f(x)-{\frac {R}{x-c}}\right)\ dx$ .

-- Meni Rosenfeld (talk) 10:55, 6 March 2011 (UTC)[reply]

Thanks Sławomir, Henning and Meni. Yes, I goofed up a bit. There is

\delta _{ij}

next to

-\mathrm {ln} (r)

and it is a line integral

\textstyle \int G_{ij}\,|d{\vec {x}}|

. Regarding divergence of the second term in the integral, I took one particular term out considering it constant over that element, but taking that term variable probably I can make a coupling with this term to give a more meaningful converging term. I am trying to manipulate it. If I have doubts, I will ask again. Thanks. - DSachan (talk) 08:04, 7 March 2011 (UTC)[reply]

Ok, I think, finally, I got the full expression making a correction in the second term, which now seems to converge. I have to numerically compute the line integral of

G_{ij}({\vec {x}},{\vec {y}})=-\mathrm {ln} (r)\delta _{ij}+{\frac {(x_{i}-y_{i})(x_{j}-y_{j})}{r^{2}}}

(notice now the

r^{2}

term in the denominator). Another expression to be computed the line integral of is

T_{ijk}({\vec {x}},{\vec {y}})={\frac {(x_{i}-y_{i})(x_{j}-y_{j})(x_{k}-y_{k})}{r^{4}}}

, which is the Green's function for stress tensor. I am pretty sure now that these expressions are correct but not sure how to incorporate and implement Meni's ideas in my problem. Any other help would be highly appreciated. Thanks very much - DSachan (talk) 16:57, 7 March 2011 (UTC)[reply]