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March 24[edit]

Large Powers[edit]

I am trying to get some sense of perspective as to what 10 to the power of 2860 means in real terms, but alas this computation seems to be beyond the power of my humble calculator. Can anyone care to enlighten me as to what this number might be in real terms?--Flaming Ferrari (talk) 00:00, 24 March 2011 (UTC)[reply]

Forgive me if this is too elementary an answer, but 10 to the 2860 would be written as a 1 followed by 2860 zeros. This looks like:

10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

That's pretty big, and much larger than any number that you are likely to encounter in your everyday life. It's much bigger than the number of particles in the observable universe, which is about 10⁸⁰. See Large numbers for some other comparisons. Staecker (talk) 01:13, 24 March 2011 (UTC)[reply]

Thanks, I didn't realise it was just a case of 1 followed by 2860 zeros, but there we go, I have learnt something today. Anyhow in case you're wondering where I got this number from, according to evolutionary theorists it is 10²⁸⁶⁰ times more likely that all life on earth is decended from a common ancestor, than multiple ancestors. See Last universal ancestor. --Flaming Ferrari (talk) 02:08, 24 March 2011 (UTC)[reply]

That must be for every species having developed independently. Having two source species on Earth shouldn't be nearly as unlikely. StuRat (talk) 04:33, 24 March 2011 (UTC)[reply]

The source is here, in case you're curious:

Saey, Tina (5 June 2010). "Life has common ancestral source". Science News. 177 (12): 12. doi:10.1038/465168a. PMID 20463725.

--Sławomir Biały (talk) 13:07, 24 March 2011 (UTC)[reply]

10 to the power of 2860 is approximately the number of different ways that 1097 objects can be arranged (the number of ways to shuffle a deck of 1097 cards, for example), which may make it seem more accessible. 86.181.169.87 (talk) 04:28, 24 March 2011 (UTC)[reply]

It's still much smaller than the hypothetical number of monkeys you need to get the text of Hamlet. – b_jonas 13:12, 24 March 2011 (UTC)[reply]

Dang! How many more monkeys will I need? Sławomir Biały (talk) 13:27, 24 March 2011 (UTC)[reply]

One, but his name has to be Francis Bacon.

Infinite_monkey_theorem gives a ballpark figure of about 10¹⁸³⁹⁴⁶. The more precise figure will depend on how precise a reproduction you want, and how suitable the typewriters are for the monkeys (eg. don't use one of these modern mobile phones with such a small keyboard that a monkey can't press only one button at a time, use a real keyboard). – b_jonas 20:34, 24 March 2011 (UTC)[reply]

My wife isn't going to like this. She's already barely coping with our

10^{2860}

simian guests. Where will they sleep? Sławomir Biały (talk) 20:39, 24 March 2011 (UTC)[reply]

They don't have to sleep at all. They are required to type the manuscript in a single day, and you let them away after that. The requirement for speed only increases the number of monkeys by a few orders of magnitude. – b_jonas 08:01, 25 March 2011 (UTC)[reply]

Oh, so they have to be

10^{183946}

motivated monkeys. Unfortunately, our own experience more closely resembles the unpleasant scenario described here. Sławomir Biały (talk) 11:27, 25 March 2011 (UTC)[reply]

Let the goat in, too. (Why do all online versions of this story involve buying a goat? I believe in the original version they already had a goat in the yard.) -- Meni Rosenfeld (talk) 08:37, 25 March 2011 (UTC)[reply]

What's the best way to prevent a long text such as in Staecker's reply from creating a horizontal scroll bar? I thought about using Template:Collapse, but it places the hide button on the right which is very inconvenient. -- Meni Rosenfeld (talk) 15:04, 24 March 2011 (UTC)[reply]

I suppose converting it into a not-so-long text, e.g. by putting a space every ten or twenty decimal positions, will do... --CiaPan (talk) 17:21, 24 March 2011 (UTC)[reply]

I wrote a script to insert zero-width spaces after every character so that it wraps properly. « Aaron Rotenberg « Talk « 19:15, 24 March 2011 (UTC)[reply]

Nice touch- I should've done that in my original script. But actually I thought that the scrolling right might give a better impression about just how big the number is. Looks pretty big either way, I guess. Staecker (talk) 00:07, 25 March 2011 (UTC)[reply]

I agree. But the scrolling should be collapsible rather than bothering everyone on the page. If we could put a scrollbar embedded in a text box it would be optimal. Splitting to multiple lines is the next best thing. -- Meni Rosenfeld (talk) 08:31, 25 March 2011 (UTC)[reply]

Latex-amsproc-paperwidth[edit]

I used amsproc documentclass to prepare a manuscript in Latex. Now the publishers have given me a specific page size ( height and width) which does not match with the standards letterpaper, a4, etc. I inserted \paperwidth = *.**in and \paperheight = *.**in but it seems they are ignore while compiling ( I don't see any change in the total numbers of pages). Any suggestions? 14.139.128.14 (talk) 03:21, 24 March 2011 (UTC)[reply]

Unless specified, I don't think \textheight and \textwidth will update based upon \paperwidth and \paperheight. So you'll have to fiddle with the former parameters too. Perhaps \textwidth=0.9*\paperwidth might help. SemanticMantis (talk) 14:59, 24 March 2011 (UTC)[reply]

Differential Equation[edit]

Given differential equation

${\frac {d^{2}a}{dt^{2}}}+a(t)k(t)=0$

And conditions

k(t)>0∀t,

a(0)=1

a'(0)=a₀>0

What is the simplest way to prove that there is at least one root of a(t)=0 for t<0? —Preceding unsigned comment added by 92.21.95.193 (talk) 13:59, 24 March 2011 (UTC)[reply]

In the ta-plane, the graph of a(t) is concave down wherever a>0. Sketch a graph. A rigorous argument follows easily. Sławomir Biały (talk) 14:41, 24 March 2011 (UTC)[reply]

Mathematical morphology notation in research paper[edit]

The following question is with reference to the following research paper which uses some mathematical morphology.

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.84.4751&rep=rep1&type=pdf

Equation (6) uses the following notation:

f1 = γ Φ

What does this notation (no operator between the two inputs) mean? Do I apply one after the other? Multiply them? Ordinarily I'd assume element-wise multiplication but in this case I can't see the motivation for that, and the authors go on to show what I assume is element-wise multiplication later in the paragraph using the * symbol (f * fm).

If anyone thinks they can help but can't access the source PDF let me know and I'll type the equations out. Cheers! --94.0.185.139 (talk) 17:28, 24 March 2011 (UTC)[reply]

It tells you what γ_l and φ_l are in equations (8) and (9). They're minima and maxima of certain sequences of values. So, I would say, that ƒ₁ is indeed a product. — Fly by Night (talk) 18:34, 24 March 2011 (UTC)[reply]

(ec) It says right after equation (1) that * means convolution, so (6) could still be pointwise multiplication, though as you say that appears to make little sense when nothing else seems to depend on the choice of zero point for sample values. Have you sought out the reference [12] that the authors say this comes from? –Henning Makholm (talk) 18:43, 24 March 2011 (UTC)[reply]

Sadly I couldn't find reference 12 anywhere. All other references to ASF in literature appear to be quite different to the one in this paper, and none use pointwise multiplication. I'll probably just have to implement it then play around with the various possibilities until I get something that looks like the figure. Thanks --94.0.185.139 (talk) 18:58, 24 March 2011 (UTC)[reply]

I think it is simple composition. I.e apply the Φ operator to the image and then apply the γ operator to the resulting image. You can't really consider this as pointwise as Opening, Closing (morphology) operate on a small regions around each point.--Salix (talk): 00:19, 25 March 2011 (UTC)[reply]

But γ and φ here are not operators; they are entire images. And f₁ itself is also used as if it were an image, rather than an operator -- for example f₁ is use as the left operand to a dilation in equation (7). –Henning Makholm (talk) 13:21, 25 March 2011 (UTC)[reply]

You can regard γ₁ and φ₁ as operators which act on images and return images. If you define

\gamma _{1}(f)=\max(f\circ b_{0,l};f\circ b_{45,l};f\circ b_{90,l};f\circ b_{135,l})

and similarly for φ₁. Our expression is then

f_{1}(G)=\gamma _{1}(\phi _{i}(G))

where G is the image being acted on. Eq 7 is actually defining a filter/operator rather than an actual image.--Salix (talk): 15:01, 25 March 2011 (UTC)[reply]

Sure, but what the source's equation (8) actually says does not have

\gamma _{1}(f)

on the LHS, only

\gamma _{l}

. And then afterwards you need to take the convolution of an image with an operator in the paragraph following (9), the meaning of which is not clear either. –Henning Makholm (talk) 15:23, 25 March 2011 (UTC)[reply]

After some playing (6) almost certainly is composition. I was confused by a) the ambiguous notation and b) the fact (8) and (9) are defined only in terms of the original image function f, rather than with a parameter. Like Henning I have no idea what the authors mean by the "applying the morphological function" part at the end, but the star operator certainly can't mean convolution in that context. f_m itself, on visual inspection, appears to contain the desired edges, so i'm unsure what's intended in applying this in some way to the image f. Maybe this is just how filters are conventionally defined, I'm kind of unfamiliar with the field, or maybe the paper is just a bit sloppy. Thanks both for your help anyway --94.0.185.139 (talk) 19:06, 25 March 2011 (UTC)[reply]

Simultaneous equation[edit]

7p-12d=15

-5p+d=10

Izayki (talk) 20:51, 24 March 2011 (UTC)[reply]

Can you show us what you've tried, rather than just have us do your homework for you ? Do they want you to do it by substitution or some other way ? StuRat (talk) 21:04, 24 March 2011 (UTC)[reply]

Standard approach is to call the unknown quantities x and y rather than p and d. It is also common to have zero on the right hand side of the equality sign. So your problem is 7x−12y−15 = −5x+y−10 = 0. If 12 times the second expression is added to the first expression you get −53x−135 = −5x+y−10 = 0. If then 5 times the first expression is subtracted from 53 times the second expression you get −53x−135 = −53y−145 = 0. This process of elimination has transformed the two equations in two unknowns into two equations, each having a single unknown. Such equations are solved numerically: x=-2.55 and y=-2.74. Bo Jacoby (talk) 11:20, 25 March 2011 (UTC).[reply]

Bo, these solutions are incorrect. For example: [7×(−2.55)] − [12×(−2.74)] = 15.03 ≠ 15. — Fly by Night (talk) 13:01, 25 March 2011 (UTC)[reply]

Round-off errors cannot be avoided. The solutions are correct to two decimal places. Bo Jacoby (talk) 13:22, 25 March 2011 (UTC).[reply]

... i.e. incorrect :-) — Fly by Night (talk) 17:24, 25 March 2011 (UTC)[reply]

Rewrite in terms of matrices. So, we have 7p − 12d = 15 and −5p + d = 10. We can express that as

\left[{\begin{array}{cc}7&-12\\-5&1\end{array}}\right]\left[{\begin{array}{c}p\\d\end{array}}\right]=\left[{\begin{array}{c}15\\10\end{array}}\right].

The two-by-two matrix on the left has a non-zero determinant, and so has an inverse matrix. Calculate the inverse matrix, multiply both sides by the inverse matrix and that will give

\left[{\begin{array}{c}p\\d\end{array}}\right]=\left[{\begin{array}{cc}?&?\\?&?\end{array}}\right]\left[{\begin{array}{c}15\\10\end{array}}\right].

The two-by-two matrix on the right with a question mark in each place is the inverse matrix which you will need to calculate. Multiply out the right hand side to give you the values for p and d. When you do this, you will see that p = −135/53 and d = −145/53. You will need to calculate the inverse matrix if you're to get any marks; don't just write the answer! You could do it by direct calculation, e.g. if −5p + d = 10 then d = 10 + 5p. Substitute that into the first equation and solve for p to give p = −135/53. Put that value for p back into the second equation and solve for d to give d = −145/53. I don't use that method because it's easy to make a mistake and it gets very messy; expecially if you have lots of equations in many variables. — Fly by Night (talk) 12:54, 25 March 2011 (UTC)[reply]

Note that the matrix inversion method only solves linear equations, and that the subtitution method only works where you can isolate a variable algebraically, while the elimination method also works for equations of high order. Bo Jacoby (talk) 13:32, 25 March 2011 (UTC).[reply]

Note that the OP only presented linear equation! Also, note that the method above can be used for higher order equations. — Fly by Night (talk) 17:22, 25 March 2011 (UTC)[reply]

Matrix inversion seems way over the top for the very simple pair of equations given by the OP. If he (?) can't solve those by simple algebraic manipulation then inversion is likely to be beyond his capabilities. AndrewWTaylor (talk) 17:25, 25 March 2011 (UTC)[reply]

That's why I supplied the simple solve and substitute solution in small text as an after thought. — Fly by Night (talk) 17:27, 25 March 2011 (UTC)[reply]

P.S. We learned about matrices and simultaneous equations round about the same time in school. It is possible that he knows how to invert a two-by-two matrix (after all, you only need to take negatives, multiply and swap some numbers) but hadn't realised the connection to simultaneous equations. Even if he doesn't, I supplied link to the relevant articles and, hopefully, he'll follow those links and learn a bit more mathematics. — Fly by Night (talk) 18:15, 25 March 2011 (UTC)[reply]

Solving an algebraic equation of degree >4 cannot be done algebraically, and solving an algebraic equation of degree 1, 2, 3 or 4 algebraically involves fractions and roots, and the numerical representation of a solution as a decimal fraction involves an unavaidable round-off error. If we cannot accept this round-off error we better leave the equation unsolved. The 'solution' x=-135/53 says nothing more than the equation 53x+135 = 0, and an approximate solution like x≃-2.55 is the useful answer. I don't see that "the method above can be used for higher order equations". Matrix inversion does not do the trick, and algebraic substitution is not practical for equations of degree 3 or 4, and not generally possible for equations of degree >4. Bo Jacoby (talk) 01:41, 26 March 2011 (UTC).[reply]

First of all, are you saying that we can't solve x⁶ − 1 = 0 algebraically? It has degree six, I know I can solve it. Second of all, how is an approximate, i.e. incorrect, solution of more use than an exact rational solution? With the rational solutions, anyone that can do long division by hand can calculate p and d to arbitrary degrees of accuracy. In fact, using the fact that the decimal expansion starts to repeat when the remainders in the division repeat, we can show that p = −135/53 = −2.5471698113207 and d = −145/53 = −2.7358490566037.

If you went back to high school and wrote x = −2.55 and y = −2.74 as your solution to the OP's problem then you'd be going home with no marks, because they're wrong! Finally, you can use the matrix method to solve systems of simultaneous polynomial equations. No-one said you have to solve for linear variables. Take, for example, x⁵ + ax³ + b = x⁵ + cx³ + d = 0, you can write this as

\left[{\begin{array}{cc}1&a\\1&c\end{array}}\right]\left[{\begin{array}{c}x^{5}\\x^{3}\end{array}}\right]=\left[{\begin{array}{c}-b\\-d\end{array}}\right].

Provided a ≠ c we can invert the matrix on the left to give

{\frac {1}{c-a}}\left[{\begin{array}{cc}-c&a\\1&-1\end{array}}\right]\left[{\begin{array}{c}b\\d\end{array}}\right]=\left[{\begin{array}{c}x^{5}\\x^{3}\end{array}}\right].

Expanding give the compatibility condition that (b − d)⁵ = (ad − bc)³(c − a)². If a = c then we don't even need the matrix method; we just subtract the equations. This method has given us the conditions for x⁵ + ax³ + b = x⁵ + cx³ + d = 0 to have a solution and, provided it has a solution, it has given us the solution. The moral of the story here is that we should always use our intelligence and try not to blindly apply mathematical machinery; using the right method for the right problem. That's why mathematics is part art and part science. — Fly by Night (talk) 18:30, 26 March 2011 (UTC)[reply]

A generally applicable method should be preferred to methods that only applies in special cases. The general equation of degree >4 cannot be solved algebraically even if special cases such as x⁶ − 1 = 0 can.
"how is an approximate, i.e. incorrect, solution of more use than an exact rational solution?". Because decimal numbers are easier to compare to one another than fractions. It is easier to compare −2.55 to −2.58 than to compare $-{\frac {135}{53}}$ to $-{\frac {67}{26}}$
The problem 7x−12y−15 = −5x+y−10 = 0 was two equations with two unknowns. The elimination method allows the unknowns to be separated: 53x+135 = 53y+145 = 0. Your example x⁵ + ax³ + b = x⁵ + cx³ + d = 0 is two equations with one unknown. Bo Jacoby (talk) 04:30, 27 March 2011 (UTC).[reply]

I agree. You should be more careful with your statements. You said that "Solving an algebraic equation of degree >4 cannot be done algebraically. That implies that no equations of degree greater than four are solvable by algebraic methods; which is clearly false. I provided a counter example.
I have already shown you that −135/53 = −2.5471698113207 and −145/53 = −2.7358490566037. The rational solutions are better than you solutions in two ways: firstly they are correct, while yours are not. Secondly they lead to more accurate decimal expansions than yours.
My example was an example to show you that the matrix method can solve non-linear equations. After all, you said you don't see how, so I provided an example to demonstrate.

If you still have any questions or queries then please drop me a line on my talk page. This thread is becoming farcical, and we should not waste any more space on the reference desk discussing what is and what isn't a correct solution to a high school problem. — Fly by Night (talk) 06:19, 27 March 2011 (UTC)[reply]

Farcical indeed. The original post wasn't a question, just a pair of linear equations which it could be presumed were to be solved simultaneously. The second post was a sensible refusal to solve the problem without something more from the OP. Then things escalated into way over the top point-scoring from two new posters who ignored the embargo on doing someone's presumed homework. There was a pointless complaint about using unknowns other than x and y and having non-zero RHSs, the ludicrous appearance of matrices, a debate about non-exact answers and various completely different kinds of equations - just what do you two expect the OP to make of all this? Is it likely that he/she will be back?←86.132.167.61 (talk) 19:15, 27 March 2011 (UTC)[reply]

Fly by Night and especially Bo Jacoby are far from being new posters (having posted in WP:RD/math since early 2010 and mid 2007, respectively). You can criticize their posts, but there's no need to make stuff up. (Or did I misunderstand what you meant by "new posters"?) -- Meni Rosenfeld (talk) 20:22, 27 March 2011 (UTC)[reply]

Matrices are standard practise in solving linear equations. My first ever class at university was about matrices and expressing systems of linear equations in terms of matrices equations, reducing to row echelon form and then solving. There was no braking of any "embargo": I did tell the OP to calculate the inverse matrix themselves. Maybe you would also like to mention the equally ludicrous appearance of matrices on our System of linear equations article's talk page? Meni: that feedback tool is awesome. I haven't seen anything like that before. That's for adding the link. — Fly by Night (talk) 01:24, 28 March 2011 (UTC)[reply]

Yes Meni, you misunderstood - I meant new posters to the thread, after the OP and StuRat, who made the standard "no homework" response. Bo Jacoby then ignored this and gave a solution by elimination. I think that what followed was a pointless escalation of things, given the apparent level of the OP's question, and their total absence from the developing debate. The later question on percentage shows a much more sympathetic approach.←86.132.167.61 (talk) 16:20, 28 March 2011 (UTC)[reply]

See the article Simultaneous equations that describes several methods you can use to find the variables p and d. The two methods shown below differ in the intention of their third equations (3).
Substitution method
Equation (3) is to isolate one variable in terms of the other so it can be put back into the given equations.

Given
            7p - 12d = 15    (1)
           -5p +   d = 10    (2)
From (2)
            d = 10 + 5p      (3)
(3) in (1)
            7p - 12 ( 10 + 5p ) =15
          -53p = 15 + 120
         
            p = -135/53
              = -2.54717 to 6 sig. fig.
                --------
From (3)
            d = 10 + 5p = 10 - 12.7358
              = -2.73585 to 6 sig. fig.
                --------

Elimination method
Equation (3) is made by scaling one of the given equations to have the same number of one of the variables so that that variable can be eliminated by combining equations.

Given
            7p - 12d = 15    (1)
           -5p +   d = 10    (2)
(2) times 12
          -60p + 12d = 120   (3)
(1) + (3)
          -53p = 15 + 120 = 135
            p = -135/53
              = -2.54717 to 6 sig. fig.
                --------
From (3)
            d = 10 + 5p = 10 - 12.7358
              = -2.73585 to 6 sig. fig.
                --------

A System of linear equations can also be solved in matrix form which is useful when there are a large number of variables and equations, see the linked article. Cuddlyable3 (talk) 10:49, 29 March 2011 (UTC)[reply]