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July 10

Is there a name for this pattern?

In high school I proved this by simple induction:

1+2=3, 4+5+6=7+8, 9+10+11+12=13+14+15, ...

Is it just a fun pattern? Thanks in advance. Imagine Reason (talk) 00:13, 10 July 2008 (UTC)[reply]

Well, each of these has the form

k^{2}+\sum _{i=1}^{k}(k^{2}+i)=\sum _{i=1}^{k}(k^{2}+k+i)

for some natural number

k

. This is equivalent to

k^{2}=\sum _{i=1}^{k}(k^{2}+k+i)-(k^{2}+i)=\sum _{i=1}^{k}k

which is obviously true. In addition you are using the fact that

(k+1)^{2}=k^{2}+2k+1

(e.g. 4 = 1+3 and 9 = 4+5). —Preceding unsigned comment added by Aenar (talk • contribs) 07:43, 10 July 2008 (UTC)[reply]

Cool. Given that S_n = 1+2+3+...+n = 1/2n(n+1) you probably could come up with a non inductive proof. eg Show S_n+a-S_n-1=S_n+2a-S_n+a (or very similar)

As for a name - I don't think so, some very ancient mathematician might be credited with being the first to spot it, but I doubt it's got a name.87.102.86.73 (talk) 10:46, 10 July 2008 (UTC)[reply]

There's a geometry way to look at it. Think of the two sums as stacks of x unit blocks. Note that the first term on the LHS of each equation is a perfect square, call it

n^{2}

. We can rearrange it as a square, which fits neatly under the remaining LHS terms and we have the RHS stacks. This is obvious algebraically as well. And you can see that you're going to end your sums getting to the next perfect square by noting that

(n+1)^{2}=n^{2}+2n+1

and

2n+1

is precisely the number of terms in the combined LHS and RHS of each equation. Donald Hosek (talk) 16:20, 10 July 2008 (UTC)[reply]

Absurdity

Absurdity, a terminology of discrete mathemetics. What is this? —Preceding unsigned comment added by 121.54.67.106 (talk) 03:12, 10 July 2008 (UTC)[reply]

Well, in logic, which is sometimes considered a mathematical discipline, there is reductio ad absurdum, or "reduction to the absurd". StuRat (talk) 04:53, 10 July 2008 (UTC)[reply]

5 digit number

Hi! I have just completed an exam question but have no way of checking that my answer if correct, which would be nice. I'd be grateful if someone could check it for me. It's not homework and I'll give you my answer so everything should be fine.

12345 is a five digit number whose digits sum to 15, i.e 1+2+3+4+5=15.

How many five digit numbers are there whose digits sum to 39?

I get an answer of 210; I'd write out my method but it'd take alot of effort and, if I'm right, would be unnecessary. Thanks in advance for doing this for me! 92.2.122.213 (talk) 14:04, 10 July 2008 (UTC)[reply]

Yes, that's correct. Algebraist 14:10, 10 July 2008 (UTC)[reply]

Cheers, really appreciate it. Can I ask, how did you do it? My method was noting that 39 differs from 45 by 6 and then using numbers that will create this difference, and using combinations to work out how many numbers there are using the same digits (poorly explained I know), but you did it much faster than I did and must have used a more efficient method. 92.2.122.213 (talk) 14:12, 10 July 2008 (UTC)[reply]

My method was essentially the same as yours. I listed all possible sets of digits summing to 39, and added up the number of orderings of each. Algebraist 14:29, 10 July 2008 (UTC)[reply]

Wow - that took you less than 6 minutes?! I obviously made a bit of a meal of it then. Thanks again. 92.2.122.213 (talk) 14:32, 10 July 2008 (UTC)[reply]

You didn't use a calculator, did you? That tends to slow down this sort of thing. Algebraist 14:40, 10 July 2008 (UTC)[reply]

Lol - no I didn't. This is from a STEP mathematics paper and you're not allowed to use calculators in them; often that seems like a bit of a pain but here I agree that it would delay everything. —Preceding unsigned comment added by 92.2.122.213 (talk) 15:14, 10 July 2008 (UTC)[reply]

I think there's a way to solve this kind of problem with a generating function, if you're interested in a more general case. I'll post if if I get it. --_{tiny plastic} Grey Knight ⊖ 16:32, 10 July 2008 (UTC)[reply]

inequalities for one-dimensional Schroedinger equation

Greetings, your question has been moved to the Science Reference Desk where hopefully you'll receive a better response. In the future, please try to post questions in the reference desk most relevant to the subject area so that we may better help you. EagleFalconn (talk) 15:58, 10 July 2008 (UTC)[reply]

See Wikipedia:Reference_desk/Science#inequalities_for_one-dimensional_Schroedinger_equation_(moved_from_Math_desk) (somtimes we even link to it!!)87.102.86.73 (talk) 17:39, 10 July 2008 (UTC)[reply]

about elementary operations in arithmatic

When we multiply a multiple digit number by another number we start multiply from the right side to the left side till we reach the last digit of the former number. On the other hand when we divide a multiple digit number from another number we start dividing from the left side till we reach the last digit of the former number in the right side. WHY? —Preceding unsigned comment added by 59.94.113.148 (talk) 16:26, 10 July 2008 (UTC)[reply]

Because that's the way the dependencies work. When multiplying, the tens column depends on the units column, because you can carry over, when dividing it works the other way round because you have remainders being carried over in the opposite direction. Since multiplication and division are inverses of each other, it makes sense that you perform them in opposite directions. Division is just multiplication backwards, really. --Tango (talk) 17:06, 10 July 2008 (UTC)[reply]

By "divide a multiple digit number from another number," do you mean something like 125/5 or 5/125? Concerning multiplication, you could start from the left and go to the right since addition is ~~associative~~ commutative (in the last step of a multi-digit multiplication calculation you add up the rows of numbers you got from digit by digit multiplication, if you know what I'm referring to). The only difference is that instead of getting a stack of numbers to add up that looks like a triangle like this /|, you get a triangle like this \|. The different "directions" for multiplication and division calculations are more results of convention than the two operations being inverses of each. 99.145.236.183 (talk) 17:18, 10 July 2008 (UTC)[reply]

I think you mean commutative (you can change the order), not associative (you can move brackets around). You have to go from right to left within each row, though - if I do 15x12, I can start with either the 1 or the 2 of the 12, but with each one, I have to start with the 5 of the 15, otherwise I'll have to go back on myself when I realise I need a carry a 1. --Tango (talk) 17:22, 10 July 2008 (UTC)[reply]

My bad -- as Tango pointed out, I meant to say commutative. 75.4.134.134 (talk) 06:37, 11 July 2008 (UTC)[reply]

Modelling Wikipedia's growth with exponential calculations - what am I doing wrong?

To model Wikipedia's growth, I made a fairly simple exponential formula: $ae^{2x_{n}}+be^{x_{n}}+c=y_{n}$ , where $x_{n}$ is the time spent in years, and $y_{n}$ is the number of articles (since a given epoch). By making three such equations with n varying from 1 to 3, I was able to come up with:

$a={\frac {e^{2x_{3}+x_{2}}-e^{2x_{3}+x_{1}}+e^{2x_{2}+x_{1}}-e^{2x_{2}+x_{3}}+e^{2x_{1}+x_{3}}-e^{2x_{1}+x_{2}}}{y_{3}e^{x_{2}}-y_{3}e^{x_{1}}+y_{2}e^{x_{3}}+y_{1}e^{x_{3}}-y_{1}e^{x_{2}}}}$
$b={\frac {y_{2}-y_{1}-a(e^{2x_{2}}-e^{2x_{1}})}{e^{x_{2}}-e^{x_{1}}}}$
$c=y_{1}-ae^{2x_{1}}-be^{x-1}$

I then set the x and y values as follows:

$x_{1}=0.9564198,y_{1}=500000$
$x_{2}=1.4808718,y_{2}=1500000$
$x_{3}=2.3036097,y_{3}=1941760$

Thus I got these values:

$a=-0.0000951$
$b=557284.84$
$c=-950257.29$

To verify the answers, I placed all numerical values calculated above back to the original equations. The first two of them were calculated correctly, but the third resulted in $y_{3}=4628303.7$ . Why did this happen? I must have calculated something wrong at some point, but where? JIP | Talk 17:30, 10 July 2008 (UTC)[reply]

Did you use a machine to do it? You could try replacing x1 with x2 y1 with y2 etc eg permuting the values and seeing if you get the same error..87.102.86.73 (talk) 19:34, 10 July 2008 (UTC)[reply]

I haven't check your equations so I'll assume they're right - you've got the solutions to a set of three linear equations with three unknowns right.? Maybe you should look at System_of_linear_equations#Consistency or somewhere else on that page.87.102.86.73 (talk) 19:37, 10 July 2008 (UTC)[reply]

I did all the symbolical calculations by hand, but used SciLab to do the numeric work. What puzzles me is that I got

y_{1}

and

y_{2}

right but

y_{3}

wrong. Shouldn't I either have got them all right, or all wrong? If necessary, I can post all the steps I did in the symbolic calculations. JIP | Talk 19:45, 10 July 2008 (UTC)[reply]

what I'm suggesting is that you try:

x_1=2.3036097, y_1=1941760
x_2=1.4808718, y_2=1500000
x_3=0.9564198, y_3=500000

...in your equation (it looks right but I haven't checked it for 'human error'). If you get different values of a,b,c by permuting the variables then this means that you have a model that just doesn't work (sorry) (seeSystem_of_linear_equations#Consistency ). If the values of a,b,c are similar then you can put that down to noise. (An alternative way would be to take the three equations and replace all x1 y1 with x3 y3 and vice versa..) you can also do this with x2 y2 and x1 y1 or x3 y3 (there are 6 combinations in total..) Did that make sense?87.102.86.73 (talk) 20:41, 10 July 2008 (UTC)[reply]

Quote:What puzzles me is that I got $y_{1}$ and $y_{2}$ right but $y_{3}$ wrong. if you've done this by hand you'll be aware that there is more than one way to eliminate the variables c,b to get an equation for a. eg I guess you had 3 equations, and picked two to eliminate c, did that again with another combination to elimate c agian.. and the combined these two equations in b to eliminate b and get a..? So in fact there are 3 different equations for 'a' which should (if the data fits) give the same 'a', I think you might find that these three equations (you can easily generate them by swapping variables as described above) will give different 'a's for the data you have.. This would mean that the data doesn't fit the model you've used..87.102.86.73 (talk) 20:48, 10 July 2008 (UTC)[reply]

I swapped

x_{1}

and

x_{3}

, and

y_{1}

and

y_{3}

around as suggested. This gave me different values of a, b, and c. Again, placing them back to the original formula, the first two (

y_{1}

and

y_{2}

) were correct but the third one (

y_{3}

) was wrong. So it looks like I have an inconsistent system. JIP | Talk 04:24, 11 July 2008 (UTC)[reply]

Or not - see post below - I think you've made a mistake in the formula for a. Sorry I didn't spot the formula's error earlier.87.102.86.73 (talk) 13:35, 11 July 2008 (UTC)[reply]

For your x's and y's, the correct result is a = -64605.3, b = 1.00947×10⁶, c = -1.68947×10⁶. I simply solved numerically, so I don't claim to know where your mistake is. Dragons flight (talk) 22:18, 10 July 2008 (UTC)[reply]

Do those values give the correct y for each x?87.102.86.73 (talk) 22:54, 10 July 2008 (UTC)[reply]

Yes, they apparently do (with a small rounding error). How exactly did you solve the equations? JIP | Talk 04:28, 11 July 2008 (UTC)[reply]

The on-line linear solver linked to in the article System of linear equations gives the above solution once I replace all $e^{2x_{n}}$ , $e^{x_{n}}$ and $y_{n}$ with their numerical values (which are known, it is a, b and c that are unknown). I tried solving the system by hand by first expressing a in terms of b, c, $x_{n}$ and $y_{n}$ , then b in terms of c, $x_{n}$ and $y_{n}$ , and finally c in terms of $x_{n}$ and $y_{n}$ . The first two steps gave simple enough formulas, but the formula for c started getting so big I would have had a headache trying to solve c. JIP | Talk 05:46, 11 July 2008 (UTC)[reply]

So you got an answer then - your equations looked ok - maybe there was just a simple error somewhere in them.87.102.86.73 (talk) 11:21, 11 July 2008 (UTC)[reply]

Actually I think you need to check the equation for a - having 5 terms in the denominator seems wrong.87.102.86.73 (talk) 12:33, 11 July 2008 (UTC)[reply]

HERE! : —Preceding unsigned comment added by 87.102.86.73 (talk) 13:36, 11 July 2008 (UTC)[reply]

a=

(y3-y1)(e^x₂-e^x₁)-(y2-y1)(e^x₃-e^x₁) 
----------------------------------------------------- 
(e^2x₁-e^2x₂)(e^x₃-e^x₁) - (e^2x₁-e^2x₃)(e^x₂-e^x₁)

(corrected once) does this work?

which should be the same as this?: (again note you can permute eg swap x1y1 for x3y3 etc):

y3e^x₂-y3e^x₁-y1e^x₂-y2e^x₃+y2e^x₁+y1e^x₃
-------------------------------------------------------------------------
e^x₃+2x₁-e^x₃+2x₂+e^x₁+2x₂-e^x₂+2x₁+e^x₂+2x₃-e^x₁+2x₃

It looks like you' inverted the formula for a? 87.102.86.73 (talk) 12:56, 11 July 2008 (UTC)[reply]

equations for b and c seem right.87.102.86.73 (talk) 12:59, 11 July 2008 (UTC) possibly b should be "-b" ie change sign.87.102.86.73 (talk) 13:01, 11 July 2008 (UTC)[reply]

Hi. Please note, however, that Wikipedia's growth is not always exponential. Sometimes it is linear, and other times just unpredictable. Also, see Modelling Wikipedia's growth. Thanks. ~A H 1^(TCU) 14:42, 11 July 2008 (UTC)[reply]

With my most recent sample data, the discriminant of the resulting quadratic equation ends up negative with $y_{n}$ equalling five million articles or more. This means that according to the model using a quadratic function of the exponential function, Wikipedia will never reach so many articles, but will instead start to decrease after a certain "ceiling" size. I am certain this is false, so I have to find a better calculation model. JIP | Talk 09:11, 12 July 2008 (UTC)[reply]

How to find out what the recurring part of the decimal expansion of a/b is

As we all know, 1/7 has a decimal expansion of 0.142857142857...., where 142857 recurs an infinite number of times. Is there any method to find out what that is given a fraction a/b? Like, without looking at the expansion and making an educated guess? --Oskar 21:24, 10 July 2008 (UTC)[reply]

Use long division until you get a remainder you've had already (at this point the whole process starts to repeat). Algebraist 21:26, 10 July 2008 (UTC)[reply]

We even have an article: repeating decimal#How a repeating or terminating decimal expansion is found. Algebraist 21:34, 10 July 2008 (UTC)[reply]

Imagines your number,(a simpler one) .181818..., is n. Maultiply it by 100 because you have 2 deciaml places that are repeationg. So 100n=18.1818... . Then subtract n, 99n=18.18...-.18...=18. divide. 18/99=n. 2/11=n. using the equivalent property or something, .181818...=n=2/11--Xtothe3rd (talk) 21:46, 10 July 2008 (UTC)[reply]

Conway's The Book of Numbers has an interesting section on this. He studied the cycles in the decimal expansion of a/n for a given n. For 13 there are three different cycles. The length of the repeating digits is the smallest number l such that

10^{l}\equiv 1\mod p

. --Salix alba (talk) 00:05, 11 July 2008 (UTC)[reply]

What's p? --Tango (talk) 01:37, 11 July 2008 (UTC)[reply]

p=n

, the denominator; the numerator can only reduce the cycle length, I believe. I suppose also that Salix alba meant the smallest positive integer l, but I wouldn't sink to that kind of nit-picking! --Tardis (talk) 16:12, 14 July 2008 (UTC)[reply]

Greek Letters

What do all these Greek letters and other symbols mean in algebra? I mean I know $\pi$ means pi which is ${\frac {C}{d}}\approx 3.14\approx {\frac {22}{7}}$ . But I still don't get what signs like $\Delta$ , $\varpi$ , $\Omega$ , or even $\vartheta$ mean. Can someone please help me w/ this, it would help me understand a lot of things on wikipedia. I would appreciate it and thank you in advance. Earthan Philosopher (talk) 21:52, 10 July 2008 (UTC)[reply]

Depends on context. Sometimes they are names for specific things, sometimes they're just names for variables, like x and y are used. Some of the standard usages are listed at Greek letters used in mathematics, science, and engineering. What articles have you been looking at? They ought to make it clear what's referred to. Algebraist 21:59, 10 July 2008 (UTC)[reply]

Though you can't always rely on good practice being followed, it is good practice to always mention clearly and in non-mathematical language exactly what each symbol means when it is first introduced. I'd venture that

\Delta

usually represents a change in something, and an expression like

\Delta x

represents a change in

x

, rather than "Delta times x". This is more the connotation of

\Delta

; in practice, it could denote many things. As for the other symbols, they could mean anything at any time — I'm unaware of any particular connotation associated with them.

One thing though. Symbols like

\varpi

,

\varphi

, and

\vartheta

are often used, confusingly, when

\pi

,

\phi

, and

\theta

are already being used for similar quantities, so pay close attention when reading mathematical formulae. —Preceding unsigned comment added by Gogobera (talk • contribs) 03:21, 11 July 2008 (UTC)[reply]