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1 February 27

[edit] February 27

[edit] The washer method (disk integration)

I need to find the volume of a solid of revolution. The 2-dimensional area is bounded by y = √x, y = 0, and x = 4, and it is spun around the axis x = 6. I know that my outer radius is 6 - x, but since I'm integrating in terms of y (since the axis is parallel to the y-axis), I should write that as R(y) = 6 - y². But I can't figure out how to express the inner radius r(y) as a function, so that I can plug it into the formula V = π∫_a^b[R²(y) - r²(y)]dy! (I believe a and b = 0 and √6 respectively.) Can anyone offer any help? Thanks, anon. —Preceding unsigned comment added by 70.19.34.80 (talk) 01:10, 27 February 2008 (UTC)

Since the inner wall is a vertical line, the inner radius is constant with respect to y, and in this case is 2. Black Carrot (talk) 03:51, 27 February 2008 (UTC)

Also, if you're integrating with respect to y, it should run from 0 to 2. You should draw some diagrams if you haven't already. Black Carrot (talk) 03:54, 27 February 2008 (UTC)

[edit] Friend challenged me

IF A box of nutmeg ways 1 1/3 ounces and there is 20 scoops how much is each scoop? —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 02:51, 27 February 2008 (UTC)

Are you asking fot the ounces per one scoop where you are told that there are 1.333... ounces per 20 scoops? --hydnjo talk 03:05, 27 February 2008 (UTC)

It has to end in as a fraction like 1/20 —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 03:09, 27 February 2008 (UTC)

Well you know that 1 1/3 equals 4/3, right? And you know that dividing by 20 is the same as multiplying by 1/20, right? And how to reduce fractions, right? So, what more help do you need? --hydnjo talk 03:42, 27 February 2008 (UTC)

it is a pain in the ass... im not really understanding it it would be 0.065.. but thats not it! damn this is upseting —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 03:45, 27 February 2008 (UTC)

You're close, but not there yet. Come on, I would expect a 12 year old to know this, so why a 14 year old like you don't? --antilived^{T | C | G} 05:01, 27 February 2008 (UTC)

Shiβe i Fell stupid...i Got it. do simple things ever Kick your ass? —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 05:27, 27 February 2008 (UTC)

Nope. I've never made a mistkske in my life. Black Carrot (talk) 06:42, 27 February 2008 (UTC)

Eh? well im gonna have to raise the bullshit flag on that one. >.< —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 07:05, 27 February 2008 (UTC)

Look carefully at Black Carrot's spelling, and see Sarcasm. —Keenan Pepper 13:07, 27 February 2008 (UTC)

[edit] Chain Rule and Higher Derivative

Let

z = f (x, y)

x = g (s, t)

y = h (s, t)

Then

$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)z$		(1)

$\frac{\partial^2 z}{\partial t^2}=\frac{\partial}{\partial t}\frac{\partial z}{\partial t}=\frac{\partial}{\partial t}\left[\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)z\right]=\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)\frac{\partial z}{\partial t}$		(2)

Replace (1) into (2)

$\frac{\partial^2 z}{\partial t^2}=\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t}\right)z$		(3)

$\frac{\partial^2 z}{\partial t^2}=\left[\frac{\partial^2}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2}{\partial x\partial y}\frac{\partial x\partial y}{\partial t^2}+\frac{\partial^2}{\partial y\partial x}\frac{\partial y\partial x}{\partial t^2}+\frac{\partial^2}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2\right]z$		(4)

$\frac{\partial^2 z}{\partial t^2}=\frac{\partial^2 z}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2 z}{\partial x\partial y}\frac{\partial x\partial y}{\partial t^2}+\frac{\partial^2 z}{\partial y\partial x}\frac{\partial y\partial x}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2$		(5)

Is every step above correct? - Justin545 (talk) 07:07, 27 February 2008 (UTC)

No.

$\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}$

is not the same as

$\left(\frac{\partial}{\partial x}\frac{\partial x}{\partial t}\right)z.$

For example, if z = x = t, the former evaluates to 1·1 = 1, and the latter to 0·z = 0. Think of $\tfrac{\partial}{\partial x}$ as an operator, and abbreviate it as D. Also abbreviate $\tfrac{\partial x}{\partial t}$ as U. Then in your first line of equations you replaced Dz × U by DU × z. --Lambiam 09:46, 27 February 2008 (UTC)

Reply to Lambiam: You are right indeed. According to your answer, does that mean (5) is also an incorrect result? What is the correct answer if (5) is incorrect? (i.e. what is $\frac{\partial^2 z}{\partial t^2}$ equal to?) Thanks! - Justin545 (talk) 12:15, 28 February 2008 (UTC)

If you take f(x,y) = x and g(s,t) = t², then z = t², and the result should be 2. However, the right-hand side of (5) evaluates to 0. I think two terms have gone AWOL. Applying the product rule,

$\frac{\partial}{\partial t}\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}\right) = \frac{\partial}{\partial t}\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t} + \frac{\partial z}{\partial x}\cdot\frac{\partial}{\partial t}\frac{\partial x}{\partial t}.$

The second term seems to be missing, and likewise with x replaced by y. Taking them together, the following should be added to the right-hand side of (5):

$+ \frac{\partial z}{\partial x}\cdot\frac{\partial^2 x}{{\partial t}^2} + \frac{\partial z}{\partial y}\cdot\frac{\partial^2 y}{{\partial t}^2}.$

--Lambiam 21:59, 28 February 2008 (UTC)

According to your reply

$\frac{\partial}{\partial t}\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}\right) = \frac{\partial}{\partial t}\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t} + \frac{\partial z}{\partial x}\cdot\frac{\partial}{\partial t}\frac{\partial x}{\partial t}.$		(6)

Then

$\frac{\partial}{\partial t}\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}\right)=\frac{\partial}{\partial t}\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}+\frac{\partial z}{\partial x}\cdot\frac{\partial^2 x}{\partial t^2}$		(7)

But what does $\tfrac{\partial}{\partial t}\tfrac{\partial z}{\partial x}\cdot\tfrac{\partial x}{\partial t}$ evaluate to? Does it evaluate to 0 or does it evaluate to the rest of terms on the right-hand side of (5)? Thanks! - Justin545 (talk) 03:56, 29 February 2008 (UTC)

The latter (or, more precisely, half of these terms – there is also the same with x replaced by y). --Lambiam 00:50, 1 March 2008 (UTC)

According to our discussion above, we may conclude that

$\frac{\partial^2 z}{\partial t^2}=\frac{\partial^2 z}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2 z}{\partial x\partial y}\frac{\partial x\partial y}{\partial t^2}+\frac{\partial^2 z}{\partial y\partial x}\frac{\partial y\partial x}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2 +\frac{\partial z}{\partial x}\frac{\partial^2 x}{{\partial t}^2}+\frac{\partial z}{\partial y}\frac{\partial^2 y}{{\partial t}^2}$		(8)

where

$z = f (x, y)$		(9)

$x = g (s, t)$		(10)

$y = h (s, t)$		(11)

The conclusion seems to be correct since I can now verify our conclusion by a related problem in my text book of quantum mechanics:

Show that (12) is true

$\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}=\frac{\partial^2\psi}{\partial r^2}+\frac{1}{r}\frac{\partial\psi}{\partial r}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial \phi^2}$		(12)

where

$ψ = Ψ(x, y)$		(13)

$x = r cosφ$		(14)

$y = r sinφ$		(15)

Eq. (14) and (15) also implies

$x 2 + y 2 = r 2$		(16)

Fortunately and finally, I can solve the problem above with your great help! Although I don't understand how does $\tfrac{\partial}{\partial t}\tfrac{\partial z}{\partial x}\cdot\tfrac{\partial x}{\partial t}$ become half of the rest of terms on the right-hand side of (5) at the moment, I think I would figure it out later or open a new question here. Our discussion may end here. The following is just how I verify our conclusion with the problem. Just want to make the thread more complete. And you could simply skip the stuff below. Thanks for your help :-)

My proof of (12) is as below:

Replace

z

ψ

t

r

s

φ

f

Ψ

in (8), (9), (10) and (11). We have

$\frac{\partial^2\psi}{\partial r^2}=\frac{\partial^2\psi}{\partial x^2}\left(\frac{\partial x}{\partial r}\right)^2+\frac{\partial^2\psi}{\partial x\partial y}\frac{\partial x\partial y}{\partial r^2}+\frac{\partial^2\psi}{\partial y\partial x}\frac{\partial y\partial x}{\partial r^2}+\frac{\partial^2\psi}{\partial y^2}\left(\frac{\partial y}{\partial r}\right)^2 +\frac{\partial\psi}{\partial x}\frac{\partial^2 x}{{\partial r}^2}+\frac{\partial\psi}{\partial y}\frac{\partial^2 y}{{\partial r}^2}$		(17)

$ψ = Ψ(x, y)$		(18)

$x = g (φ, r)$		(19)

$y = h (φ, r)$		(20)

Evaluate derivatives

$\frac{\partial x}{\partial r}=\cos\phi$ , $\left(\frac{\partial x}{\partial r}\right)^2=\cos^2\phi$ , $\frac{\partial^2 x}{\partial r^2}=0$ ,

$\frac{\partial y}{\partial r}=\sin\phi$ , $\left(\frac{\partial y}{\partial r}\right)^2=\sin^2\phi$ , $\frac{\partial^2 y}{\partial r^2}=0$ ,

$\frac{\partial x\partial y}{\partial r^2}=\frac{\partial y\partial x}{\partial r^2}=\sin\phi\cos\phi$

(21)

Replace (21) into (17)

$\frac{\partial^2\psi}{\partial r^2}=\frac{\partial^2\psi}{\partial x^2}\cos^2\phi+\frac{\partial^2\psi}{\partial x\partial y}\sin\phi\cos\phi+\frac{\partial^2\psi}{\partial y\partial x}\sin\phi\cos\phi+\frac{\partial^2\psi}{\partial y^2}\sin^2\phi+\frac{\partial\psi}{\partial x}\cdot 0+\frac{\partial\psi}{\partial y}\cdot 0$		(22)

$\frac{\partial^2\psi}{\partial r^2}=\cos^2\phi\frac{\partial^2\psi}{\partial x^2}+2\sin\phi\cos\phi\frac{\partial^2\psi}{\partial x\partial y}+\sin^2\phi\frac{\partial^2\psi}{\partial y^2}$		(23)

By similar way from (17) to (23), we have

$\frac{\partial^2\psi}{\partial\phi^2}=r^2\sin^2\phi\frac{\partial^2\psi}{\partial x^2}-2r^2\sin\phi\cos\phi\frac{\partial^2\psi}{\partial x\partial y}+r^2\cos^2\phi\frac{\partial^2\psi}{\partial y^2}-r\cos\phi\frac{\partial\psi}{\partial x}-r\sin\phi\frac{\partial\psi}{\partial y}$		(24)

Multiply (23) by

r 2

$r^2\frac{\partial^2\psi}{\partial r^2}=r^2\cos^2\phi\frac{\partial^2\psi}{\partial x^2}+2r^2\sin\phi\cos\phi\frac{\partial^2\psi}{\partial x\partial y}+r^2\sin^2\phi\frac{\partial^2\psi}{\partial y^2}$		(25)

Add (24) and (25) together

$r^2\frac{\partial^2\psi}{\partial r^2}+\frac{\partial^2\psi}{\partial\phi^2}=r^2\left(\sin^2\phi+\cos^2\phi\right)\frac{\partial^2\psi}{\partial x^2}+r^2\left(\sin^2\phi+\cos^2\phi\right)\frac{\partial^2\psi}{\partial y^2}-r\cos\phi\frac{\partial\psi}{\partial x}-r\sin\phi\frac{\partial\psi}{\partial y}$		(26)

$r^2\frac{\partial^2\psi}{\partial r^2}+\frac{\partial^2\psi}{\partial\phi^2}=r^2\frac{\partial^2\psi}{\partial x^2}+r^2\frac{\partial^2\psi}{\partial y^2}-r\left(\cos\phi\frac{\partial\psi}{\partial x}+\sin\phi\frac{\partial\psi}{\partial y}\right)$		(27)

Divide (27) by

r 2

$\frac{\partial^2\psi}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\phi^2}=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}-\frac{1}{r}\left(\cos\phi\frac{\partial\psi}{\partial x}+\sin\phi\frac{\partial\psi}{\partial y}\right)$		(28)

By chain rule, we know

$\frac{\partial\psi}{\partial r}=\frac{\partial\psi}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial\psi}{\partial y}\frac{\partial y}{\partial r}=\cos\phi\frac{\partial\psi}{\partial x}+\sin\phi\frac{\partial\psi}{\partial y}$		(29)

Replace (29) into (28)

$\frac{\partial^2\psi}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\phi^2}=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}-\frac{1}{r}\frac{\partial\psi}{\partial r}$		(30)

$\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}=\frac{\partial^2\psi}{\partial r^2}+\frac{1}{r}\frac{\partial\psi}{\partial r}+\frac{1}{r^2}\frac{\partial^2\psi}{\partial \phi^2}$		(31)

Q.E.D. - Justin545 (talk) 07:19, 4 March 2008 (UTC)

The dy/dx notation is not exactly a fraction, although often you can treat it as a fraction (chain rule, integration by substitution) --wj32 ^t/c 05:47, 28 February 2008 (UTC)

Reply to wj32^t/c: That is ture. I think (well, I'm just a layman) Leibniz notation is somewhat confusing to me. I didn't see (yes, I am not experienced enough) any formal introduction which states when can I treat it as a fraction and when can't I after I learned the derivative for many years. It seems to be a mystery to me! :) - Justin545 (talk) 12:15, 28 February 2008 (UTC)

[edit] Recursively defined sets

I want to generate the first several iterations of this set, but I feel I'm not doing it correctly; my notes are sparse on certain details:

Defintion of $C \,$ .

$C \subseteq \Sigma$ where $\Sigma \,$ is the set of all strings over $\lbrace 1, 2, 3 \rbrace \,$

Basis step:

$\lbrace 1, 2 \rbrace \subseteq C$

Recursive step:

$w \in C$ $\rightarrow$ $3 \cdot w \in C$

$w_1, w_2 \in C$ $\rightarrow$ $w_1 : w_2 \in C$

$w \cdot 1 \in C$ $\rightarrow$ $2 \cdot w \in C$

The dot operator and the colon (:) operator represent character concatentation and string concatentation respectively.

It would follow, in my mind that the first few iterations of this set would go:

{1, 2}

{1, 2, 31, 11, 21}

{1, 2, 31, 11, 21, 32, 12, 22}

e.t.c

My main worry is this:

I assume w(1) and w(2) in the second part of the recursive step start at 1, 1 and go up 1, 2 then 2, 1 then 2, 2 and so on? Or do they have to be seperate values each time? Damien Karras (talk) 08:03, 27 February 2008 (UTC)

If I understand your last question correctly, the answer is that

w 1

and

w 2

don't have to be different. I'll guide you through the first iteration. You know that 1 ∈ C and 2 ∈ C. By rule 1, from 1 ∈ C it follows that 31 ∈ C, and from 2 ∈ C it follows that 32 ∈ C. We now apply rule 2: From 1 ∈ C and 1 ∈ C it follows that 11 ∈ C; from 1 ∈ C and 2 ∈ C it follows that 12 ∈ C; likewise 21 ∈ C and 22 ∈ C. Rule 3 doesn't give us anything new now. So the first iteration will be $\{1, 2, 31, 32, 11, 12, 21, 22\} \subseteq C$ . -- Meni Rosenfeld (talk) 09:51, 27 February 2008 (UTC)

(after edit conflict) I would interpret your rule #2 of the "recursive step" as being universally quantified over all w₁ and w₂ that are in C. So if 2 is a member of C_n in iteration n, then 22 is a member of the next generation C_n+1. Is that an answer to your question (which I do not fully understand)?

Because of rule #3, you can start the whole process by taking C₀ = {1}. Then the first few iterations give this:

C₀ = {1}

C₁ = {1, 31, 11, 2}

C₂ = {1, 31, 11, 2, 331, 311, 32, 131, 111, 12, 3131, 3111, 312, 1131, 1111, 112, 21, 231, 211, 22, 23}

--Lambiam 10:10, 27 February 2008 (UTC)

I think what I meant my overall question to be is this: For each iteration, do I pick the value of w, w1 and w2 just once? So the first iteration would be: w = 1, w1 = 1 and w2 = 1 and then I would move to the next iteration.

OR do I consider all possible values of w, w1, w2 for each iteration? In which case I would do, for instance, step one for w = 1 and w = 2, step two would be done for w1 = 1, w1 = 2, w2 = 1, w2 = 2 e.t.c

Furthermore, are the new elements of the set generated on each step? Because if so that would mean that w1 and w2 take on even more values before the end of the total recursive step (unless of course the former point is true). Damien Karras (talk) 10:41, 27 February 2008 (UTC)

I am still confused by your questions, but I have a better idea what it is you are asking. To find

C n + 1

, you apply every possible rule to the items in

C n

and only them. So if you start with

C 0 = {1,2}

, applying every possible rule to

C 0

gives you

C 1 = {1,2,31,32,11,12,21,22}

. You then apply every possible rule to

C 1

to get

C 2

, and so on. In each step, you only apply rules to elements you have had in the previous step, not to those you have generated recently. -- Meni Rosenfeld (talk) 11:44, 27 February 2008 (UTC)

Thanks both of you! That's exactly the answer I was looking for, sorry the question was phrased poorly. Just to check will I be right in saying that the start of $C_2 = \lbrace1, 2, 31, 32, 11, 12, 21, 22, 331, 332, 311, 312, 321, 322, 3131, 3132, 3111, 3112, 3121 ... \rbrace \,$ ? (I won't write the whole thing out because it appears to be massive!) Damien Karras (talk) 12:27, 27 February 2008 (UTC)

This looks correct. Don't forget that

C 2

also has elements such as 131, the concatenation of 1 and 31 which are both in

C 1

. -- Meni Rosenfeld (talk) 14:58, 27 February 2008 (UTC)

This is what I get starting from {1, 2}:

C₀ = {1, 2}

C₁ = {1, 2, 31, 32, 11, 12, 21, 22}

C₂ = {1, 2, 31, 32, 11, 12, 21, 22, 331, 332, 311, 312, 321, 322, 131, 132, 111, 112, 121, 122, 231, 232, 211, 212, 221, 222, 3131, 3132, 3111, 3112, 3121, 3122, 3231, 3232, 3211, 3212, 3221, 3222, 1131, 1132, 1111, 1112, 1121, 1122, 1231, 1232, 1211, 1212, 1221, 1222, 2131, 2132, 2111, 2112, 2121, 2122, 2231, 2232, 2211, 2212, 2221, 2222, 23}

--Lambiam 23:22, 27 February 2008 (UTC)

[edit] modular algorithm

how can i solve this problem?

[ x1= a (mod 100) , a= 20 (mod 37) ]

[ x2= b (mod 100) , b= 15 (mod 37) ]

[ x3= c (mod 100) , c= 18 (mod 37) ]

must be ; x2= a.k + y (mod100)

and

x3= b.k + y (mod100)

i need find b and c.. thank you best regards.. Altan B. —Preceding unsigned comment added by 85.98.230.220 (talk) 10:20, 27 February 2008 (UTC)

x1 does not play a role. Eliminating x2 by combining x2 ≡ b (mod 100) with x2 ≡ ak + y (mod 100) gives us b ≡ ak + y (mod 100). Likewise eliminating x3 results in c ≡ bk + y (mod 100). We can combine these to eliminate y and obtain b − ak ≡ c − bk (mod 100), or, equivalently,

b(k + 1) ≡ c + ak (mod 100).

Using the (mod 37) congruences given for a and b, and c, we rewrite these as a = 37α + 20, b = 37β + 15, and c = 37γ + 18. Substituting that in the equation brings us to:

(37β + 15)(k + 1) ≡ 37γ + 18 + (37α + 20)k (mod 100).

We now multiply both sides by the multiplicative inverse of 37 (mod 100), which is 73 (since 73·37 = 2701 ≡ 1 (mod 100). This results in

(β + 95)(k + 1) ≡ γ + 14 +(α + 60)k (mod 100).

Bringing γ to one side,

γ ≡ (β + 95)(k + 1) + (99α + 40)k + 86 (mod 100).

Now pick any values for α, β, k, and a new variable n, and compute

γ = (β + 95)(k + 1) + (99α + 40)k + 86 + 100n,

b = 37β + 15,

c = 37γ + 18.

For example, choosing α = β = 0, k = 12, n = −18 gives the solution b = 15, c = 55. --Lambiam 12:32, 27 February 2008 (UTC)

thank you for your cares, but this solution is not that i want.. 'cause i also know this solving system but, this system isnt exact solution. i will use this formula in the big machine ( this machine has 17 billions as modular ) as you know i cant expect(or according to your solution choosing) all value from my brain. thank you best regards Altan B. —Preceding unsigned comment added by 85.98.230.220 (talk) 14:51, 27 February 2008 (UTC)

Affedersiniz ama yukarıda yazdıklarınızı pek anlayamıyorum. --Lambiam 19:22, 27 February 2008 (UTC)

bu formul uzerine makina yapacagım. her seferinde nasıl tek tek su rakam olsun bu rakam olsun diye secme yaptıracagım makinaya? bana net kesin formul lazım , cunku makinanın ilk mod alması 17 milyar uzerinden olacak. yani 37 moda inince ikincisinde her rakam icin 435 milyon alternatif olacak. kesin net formul olması illa ki sart. ilginiz icin simdiden tesekkurler Altan B —Preceding unsigned comment added by 85.98.230.220 (talk) 09:01, 28 February 2008 (UTC)

Verdiğim cevap, "b = 15, c = 55" , yazdığınız probleme çözümdür, ve bu kesin bir formüldür. Acaba yazdığınız problem sadece daha büyük bir gruptan tek bir örnek mi? Bu durumda, genel problem nedir? --Lambiam 22:27, 28 February 2008 (UTC)

[edit] shapes and projections

I was wondering about something and I need abig help to correct my understanding to this issue.I am going to put my questions in two parts.part1:assume that we have ashape in two dimensions,like asphere,obviously we can determine the projections of that sphere on (x-y,y-z,z-x)planes, which they are going to be circles no matter how we rotate the sphere, the projections will be always circles.If we apply now the same thing on acone then we would have adifferent projections depend on how we rotate or fix the cone in two dimensions,we can get for example,acircle in(x-y)plane,atriangle in(y-z)plane and another triangle in(x-z)plane.we can determine the projections depend on the shape and its position or you can say the shape and the position vector between the origin and achosen point belong to that shape.Now my question is,can we determine the shape by looking at its projections?if yes,then what is the shape in two dimensions that has the projections,(circle,square and apoint)?p.s:you can improvise another projections.some times we can describe different shapes in 2 dimentions by looking at the projections,for example,assume we have 3identical circles as aprojections,one may say that the shape is atwo dimnensional sphere,or the shape could be 3 identical circles perpendicular to each other and sharing the same center. Part2:if we are able to say that the sphere in three dimensions has aprojection in two dimensions which is asphere too then for the cone in three dimensions what its projection in two dimensions going to be?IF THE ANSWER IS ACONE TOO NO MATTER HOW WE ROTATE OR CHANGE ITS POSITION IN 3 DIMENTIONS THEN IT SHOULD BE ASPHERE, because the sphere is the only shape that maintains its projections in all dimensions>2

Further discussion: Assume we have ashape in 3dimensions where its projections in 2dimensions as follows, (x,y,z)=s1,(x,y,w)=s2,(x,z,w)=s3&(y,z,w)=s4.The projections in one dimension would be,

[s1:(x,y)=a1,(x,z)=a2,(y,z)=a3],[s2:(x,y)=a1,(x,w)=a4,(y,w)=a5],[s3:(x,z)=a2,(x,w)=a4,(z,w)=a6],[s4:(y,z)=a3,(y,w)=a5,(z,w)=a6].We have 6 one dimensional projections .Now assume we have acone in 3 dimentions,we can finde some conditions to make that cone describable,for example ,if the con`s peak is located at the origin and the position vector between the peak and the center of the cone`s makes an angle=pi\4 with each of(x,y,z,w),then a1=a2=a3=a4=a5=a6 and each one of the six projections seems to be having ashape of sector of acircle.on the other hand if,a1=a2=a3=a4=a5=a6 ,shouldnot we get asphere in 3dimensions?my point here is,can we say that all different shapes in 3 dimensions with no preconditions have ashape of sphere???Husseinshimaljasimdini (talk) 12:05, 27 February 2008 (UTC)

Part 1 - you seem to have made a sort of mistake eg if I project a set of 3 circles orthogonal (at right angles) to each other onto the three planes I get a circle with a cross inside - not a filled circle as you might expect from the sphere.(Maybe this is worth thinking about?)

As for your question 'can we determine a shape from it's projections' - it might depend - the answer seems to be yes if we assume the shapes are all quadratics, ... but much more complicated shapes might project as for example 3 circles along one set of coordinate axis (suggesting a sphere) - but produce different shapes along a different set of coordinate axis.87.102.93.245 (talk) 12:39, 27 February 2008 (UTC)

Your question "(if yes),then what is the shape in two dimensions that has the projections,(circle,square and apoint)" - seems impossible - because if the shape projects onto a point it can only be a point or a line - therefor a circular or square projection is impossible...

(I removed a space from your question 2 since the text was going way off the screen..)87.102.93.245 (talk) 12:39, 27 February 2008 (UTC)

(edit conflict) I would guess in "3 identical circles perpendicular to each other and sharing the same center", he means *discs* not circles. Then the projections are the same as for a sphere. An obvious example of shapes that can't be distinguished by their projections are a sphere and a sphere containing another shape - you can't "see through" the sphere, so you'll never detect the contained shape. With suitable restrictions on the shapes (for example, 245's idea of restricting to quadratics) might work, but with general shapes there are all kinds of things that can go wrong. --Tango (talk) 13:21, 27 February 2008 (UTC)

In general, you cannot reconstruct the 3D-shape from three orthogonal 2D-projections. One shape that has three unit disks as projections is the union of these three disks:

{(x,y,z) | x = 0 ⋀ y²+z² ≤ 1 ⋁ y = 0 ⋀ x²+z² ≤ 1 ⋁ z = 0 ⋀ x²+y² ≤ 1}.

Among the convex shapes, one solution is the unit ball {(x,y,z) | x²+y²+z² ≤ 1}. The largest solid object giving the same three projections is given by the intersection of three cylinders:

{(x,y,z) | x²+y² ≤ 1 ⋀ x²+z² ≤ 1 ⋀ y²+z² ≤ 1}.

This is strictly larger than the unit ball: it contains the point x = y = z = ¹/₂√2, which is outside the unit ball. --Lambiam 13:18, 27 February 2008 (UTC)

[edit] Linear equations

Hello =], I'm just having some problems with a few linear equations and this text-book is terrible example wise. Ive had no trouble with the equations preceding it:
$5(y - 2) = 12$
$5 y - 10 = 12$ (Expand Brackets)
$5 y = 22$ (Add 10 to both sides)
$\frac{5y}{5}=\frac{22}{5}$ (Divide both sides by 5)
$y=\frac{22}{5}$

Can the same method be applied to the equation below? As far my answers have been incorrect:
$\frac{x}{2} + 3=5$
$\frac{x}{2}=2$ (Subtract 3 from both sides)
This looks finished but it is evidential that I have done something wrong, a point in the right direction will be appreciated.

Also something like this, is my answer correct?
$5x+ 4 \le 0$
$5x \le -4$ (subtract 4 from both sides)

Then this is where i get lost do i divide or perform some other operation? —Preceding unsigned comment added by 58.165.62.73 (talk) 13:06, 27 February 2008 (UTC)

You are right as far as you've gone in part 2, but you haven't finished. I assume you want to know what x on its own is, and you have x/2. Double each side. As for part 3, yes, you can just divide through, but be careful - if you divide or multiply through by a negative number, you have to invert the inequality. -mattbuck (Talk) 13:13, 27 February 2008 (UTC)

Mattbuck, note than he only subtracted 4 from each side of the inequality. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 14:50, 27 February 2008 (UTC)

Yes, but the end result presumably asks for $x \leq \frac{-4}{5}$ , which requires dividing through, and it's a note about multiplying by negative numbers generally. -mattbuck (Talk) 15:04, 27 February 2008 (UTC)

If you think about what these equations and expressions mean, you'll figure out how to solve these questions. Doing the same thing to both sides is just a simplification of what you're doing to two quantities--like other people have pointed out with multiplying/dividing by negative numbers, it's not enough. Imagine Reason (talk) 23:51, 27 February 2008 (UTC)

What you must remember about these types of problems is that 1.) your solving for a single variable (x in the above, and y above that) and 2.) That means isolating that variable so you get something like x=(whatever) or y=(whatever) etc. . To do that, you need to apply the inverse operations to those that act on the variable, take $\frac{x}{2}+3=5$ for example, for this problem, our unknown (variable) is x, and our equation says that, x is divided by 2 and then 3 is added to it to give us 5 as an "answer" our question however is what is x (our unknown) so we need to get the equation above to look like x=(whatever). First of what operation happens to x last? Remember the Order of operations? Try rereading the part in bold letters. It should be noted that when solving for a variable you apply the reverse of the order of operations to remove those terms from which ever side your variable is on to isolate the variable. A math-wiki (talk) 11:43, 1 March 2008 (UTC)

[edit] Sorting a list of points for "in order of closest points"

Hi, suppose I have a list of 3 dimensional points eg (x_n,y_n,z_n) in random order and I want to try to sort them so that closer points are nearer to each other in the list.

The only idea I had was to (after probably writing the numbers in binary form) would be to covert each set of numbers into a single 'string' in which the most significant digits are group together

eg in decimal (1234,6789,-5500) would convert to ( (1,6,-5) , (2,7,-5) , (3,8,0) , (4,9,0) ) and then sorting that in order from left to right... This would put points with 1000<x<=1999 close together in the list..

This works a bit, (but still gives more priority to x than y etc - working in binary reduces this effect a bit)

Can anyone suggest any improvements to this method, or a better method? Note that I only require the list be 'improved' so that nearer points be nearer in the list - it doesn't have to give a perfect sort in perfect order or nearness (that seems impossible anyway) (You can assume I don't need any help with the sort in order of linear magnitude algorhytm)

Any ideas welcome. Just the concept would do.. Thanks87.102.93.245 (talk) 16:58, 27 February 2008 (UTC)

Imagine that your points are uniformly distributed within a cube. How are you going to "walk" around the cube and visit each point? One way is to walk along y=0, z=0 from x=0 to x=N and then back along y=1 z=0 from x=N to x=0 etc. After a while you have visited all of the z=0 plane in a passably efficient manner, but the rest of z hasn't been started. so x=0, y=0, z=1 which is close to x=0, y=0, z=0 is N^2 points away from the that coord. In other words, unless the data has some known clustering that you can take advantage of, there isn't going to be a simple solution to your question. -- SGBailey (talk) 17:56, 27 February 2008 (UTC)

Is your goal to be able to quickly find the nearest points to a given point? If so, see kd-tree. --Trovatore (talk) 18:02, 27 February 2008 (UTC)

right thanks - I already had a sort of octree (if I used binary number) - this 'kd-tree' looks like a possible improvement (though a little more analysis of the data is require - it should give a better list)Thanks.87.102.93.245 (talk) 18:37, 27 February 2008 (UTC)

If you really need a single ordering rather than a data structure, you could try sorting them in their ordering along a Space-filling curve. For 3d variants of the Hilbert curve each comparison of the sorting algorithm can be done by finding the most significant bit at which any of the coordinates of the two compared points differ. —David Eppstein (talk) 18:24, 27 February 2008 (UTC)

Thanks, looks like a good possibilty. I've often wondered if I'd ever find a use for those 'peanut curve' things . Today was that day. I'm not convinced that the hilbert curve is better (than a bsp tree) in terms of points that are just on either side of 0.5 (ie halfway accross) - probably averages out to similarily as good (please correct me if you are more expert than me - which is likely..) Still very interesting though.87.102.93.245 (talk) 18:44, 27 February 2008 (UTC)

Assuming you use in-order depth-first traversal, a bsp tree does not tell you by itself in what order the children of a node should be put. If you partition by cutting up the cube at each level in eight equally sized cubes, it is possible to order the children at each node in such a way that the in-order traversal is the same as traversal along a 3D Hilbert curve, which is the best you can get with a uniform recursive method. But other orderings are not dramatically worse, at most a factor 2 or so in the path length, which is O(n^2/3). --Lambiam 20:25, 27 February 2008 (UTC)

This sounds very much like the Travelling Salesman Problem, which has been extensively studied. -- BenRG (talk) 16:04, 28 February 2008 (UTC)

It's only vaguely similar - the difference here amoungst others is that the distances are the square root of sums of squares rather than a linear sum.. given that I read that computer times for the solution are measured in years, I'll make do with an approximation!87.102.84.112 (talk) 17:44, 28 February 2008 (UTC)

If the distance metric used is the Euclidean metric, then finding the optimal solution is known as the Euclidean Travelling Salesman Problem, which is a special case of the general TSP, but lso hard to solve. (Christos H. Papadimitriou. "The Euclidean travelling salesman problem is NP-complete". TCS 4:237–244, 1977) The essential difference with TSP is that the questioner does not ask for the optimal solution, but only for improvements to a method. In terms of order of magnitude, the O(n^2/3) bound cannot be improved upon, as can be seen by considering N×N×N points placed in a completely regular cubic grid pattern. --Lambiam 22:53, 28 February 2008 (UTC)

Actually what I asked for was very vague, if I was being more specific to the problem I'd have to say that I wanted samples of points in order to be as close together as possible (L2 space), in fact what I really wanted was to group the set of points into n sets of m points so that the volume of the sets was minimised (I suppose some sort of 'sum of squares of volumes measure' would be applicable here..) further more I prefer that the volumes are close to cubic (perhaps some sum of distances^4? measure would work)..

The kdtree (or similar methods) look good enough, and I could work with the octree/peano/hilbert method too with a little extra algorhythm on the end.. I wouldn't try a exhaustive search. Thanks to everyone who replied. Feel free to give more ideas/information. Cheers.87.102.38.45 (talk) 12:02, 29 February 2008 (UTC)

Have a look at Cluster analysis. For a distance measure between two non-empty sets of points you may consider the Hausdorff metric, which is a proper metric, instead of the wishy-washy measures invented by social scientists. --Lambiam 00:58, 1 March 2008 (UTC)

[edit] Rounding conundrum?

I disagree with former teaching colleagues over the correct answer to a rounding problem I posed a few years ago to some 13-year-old pupils, and for my own personal satisfaction I would be interested in people's views to the following simple question:

What is 0.00999 rounded to 2 Significant Figures (2 S.F.)?

The two answers to consider, from myself as head of department and a few of my maths teachers, are 0.010 and 0.0100 (I'm not revealing who claimed which! Yet!). Answer PLUS REASON please if you think you know for sure which is the correct answer! Thanks! AirdishStraus (talk) 18:18, 27 February 2008 (UTC)

It's clearly 0.010, at least according to the convention in our article. The two significant figures are 1 and the subsequent 0. -- Meni Rosenfeld (talk) 18:38, 27 February 2008 (UTC)

As far as I remember from Physics and Chemistry, it should be 0.010. The first two zeros (the one before the decimal and the one right after the decimal) are just place holders. They appear in all the numbers being operated on and I was told that they are not considered a significant value in a problem like this. So we have the one and then the second zero to have two significant digits.A Real Kaiser (talk) 18:48, 27 February 2008 (UTC)

Mmm... obviously the number 0.010 has two significant figures and the number 0.0100 has three, but the issue, I believe, is in the actual process of taking 0.00999 and applying the 'rules'...

'Chop' the number off after the second significant digit and round if the third digit is 5 or greater; write down as your answer the digits you now have up the the 'chop off line'.

So... 0.00999 $\rightarrow$ 0.0099|9 $\rightarrow$ 0.0100| $\rightarrow$ 0.0100

Where, in any article does it say that the resultant answer is further truncated to 2 significant figures? This would mean that a small percentage of all decimals would have a different rule: anything from 0.00995 to 0.00999999... etc. for example for those numbers starting in the third places of decimals. AirdishStraus (talk) 19:47, 27 February 2008 (UTC)

Now I see the problem - which is that significant figures are stupid. For a number that starts with 1, a given number of significant figures represent less accuracy than for a number that starts with 9. For the number in your example, we are faced with the dilemma of treating it as starting with 9, which it does, or starting with 1, like its rounded value. I say the rounded value wins, since we measure the significant figures for it, not for the real value. The quirk is that 0.00999 rounded to 2 sf is 0.010, and rounded to 3 sf it's 0.00999 which is two orders of magnitude more accurate. Again, sfs are stupid. It makes much more sense to discuss relative error, and sfs were never meant to deal with such subtleties. -- Meni Rosenfeld (talk) 20:18, 27 February 2008 (UTC)

I'd think of it this way: the "numbers with two significant figures" are ..., 0.0097, 0.0098, 0.0099, 0.010, 0.011, 0.012, ... and rounding to two significant figures consists in choosing the one of those that (considered as an exact real number) is closest to your number, in this case 0.010. A rounding algorithm, like the one you gave, which produces two significant figures in most cases and three in a few boundary cases I would describe as buggy. Think of it also from the perspective of someone seeing the answer 0.0100; they'll most likely assume that it was selected from the list ..., 0.00998, 0.00999, 0.0100, 0.0101, ... and not from the list ..., 0.0098, 0.0099, 0.0100, 0.010, 0.011, ... (which doesn't even make mathematical sense, since two of the entries are the same). Not that there's anything particularly sensible about the standard convention, it's just standard. As Meni said, the whole concept is rather badly broken, and if you really care about accuracy you have to do a proper error analysis and give explicit error bars. -- BenRG (talk) 20:24, 27 February 2008 (UTC)

(outdent) Just a thought. While strictly speaking, rounding to 2 sig figs does give 0.010, a partial way to sidestep the noted discontinuity of accuracy is to look at the unrounded value and note at which place (tenths, hundredths, etc.) the number loses reliability, then round so as to remove all digits at and past that place. So if the third "9" is not reliable, quote the result as 0.0100, and so on.

The above posting has a good point which I reflected in my advice: rounding should not comply to rote symbolic manipulation rules of digit strings but should reflect the precision and reliability in obtaining the value to be rounded.

This discussion also relates to Benford's law. Baccyak4H (Yak!) 20:38, 27 February 2008 (UTC)

Guys... thanks for the contributions (others still welcome). I'm the one who believes 0.0100 is the correct answer as opposed to the rest of my (now ex-) department who favoured 0.010. The reasons are that it follows the given rule with no deviation (i.e. no 'special cases') and that it gives an indication of the accuracy of the original number. 0.0100 declared as having been rounded to 2sf means something that HAD TO BE originally between the bounds of 0.00995 and 0.009999999.... etc. whereas 0.010 could have come from between bounds of 0.00995 and 0.0105 Once again, many thanks for the various inputs. AirdishStraus (talk) 22:06, 27 February 2008 (UTC)

The correct answer is definitely 0.010. If you round to 2sf, you have to have 2sf at the end, otherwise it's nonsense. 0.0100 clearly has 3sf. Your reason for rounding to 0.0100 sounds to me like simply "3sf is more precise than 2sf", which is obviously true, but irrelevant, since you asked for 2sf. --Tango (talk) 22:30, 27 February 2008 (UTC)

You're probably right, rounding in such a way as to accurately reflect the precision of the measurement is best, but that only works if people know how to interpret the data. If I see "0.010" I know that means the real value is between 0.0095 and 0.0105. Conventions have to be followed whether they are good conventions or not, because otherwise people have no idea how to interpret what you say (you could devote a paragraph at the beginning of your paper to explaining the obscure rounding system you've used, but your readers would probably prefer you just to use the standard method). —Preceding unsigned comment added by Tango (talk • contribs) 22:35, 27 February 2008 (UTC)

HI! my rule here would be to convert to a mantissa/exponent form eg 9.99 x10^-3 then convert the mantissa.. giving ~~1.0 x10^-3~~ 1.0 x10^-2.. this method is always consistent... 87.102.84.112 (talk) 11:19, 28 February 2008 (UTC) see Significant_figures#Scientific_notation:

particular, the potential ambiguity about the significance of trailing zeros is eliminated

Also consider the reverse - how many sig figs in 0.0100 - the answer looks like 3 87.102.84.112 (talk) 11:21, 28 February 2008 (UTC)

I think you mean that 9.99 x10^-3 gives 10.0 x10^-3 and not 1.0 x10^-3 as you may have written in error. This, however, has the digits 1 0 0 supporting the 0.0100 version of the answer. I know that 0.0100 appears to be 3sf but the conundrum/ambiguity is in applying the commonly supported algorithm that I described above (don't forget that this rounding topic is aimed at 12- to 13-year-olds): 0.00999 $\rightarrow$ 0.0099|9 $\rightarrow$ 0.0100| $\rightarrow$ 0.0100

All text books in my possession (a considerable number!) avoid the issue completely and have no examples or questions of this nature whatsoever; nor does any Wiki article on rounding or sig figs. I'm still at home to any good-natured contributions or suggestions! AirdishStraus (talk) 11:38, 28 February 2008 (UTC)

NO! (but I made a typing error -see above) 2 sig figs means 2 digits in the mantissa - so 9.99 x10^-3 expressed to 2sig figs gives 1.0 x10^-2 to 2 sig fig..?87.102.84.112 (talk) 11:59, 28 February 2008 (UTC)

If you want an algorhythm that 'works' then how about this

0.0999 to 2sf

STEP 1

0.099|9 (ignore zeros on left, separate after 2 digits) ::gives 0.1 (this is rounding)

That is 0.1

STEP 2

Write the answer from STEP 1 to 2 significant figures

This is 0.10

This will give consistent and the answers will always be to the same number of significant figures as requested.

Whether or not a 12 year old can be expected to perform a 2 step process isn't something I can know. But I hope that the answer will always be yes. Clearly they must have a perfect understanding of rounding numbers before they can begin to do sig figs reliably.87.102.84.112 (talk) 12:19, 28 February 2008 (UTC)

I see what you're driving at... you are saying that after the 'chop off' to however many sf you choose, you should always ignore all trailing zeros and then write whatever decimal that gives you to your required degree of accuracy. So, as you imply... 0.00999 becomes 0.01 which then becomes 0.010 written to 2sf, because after the first step all trailing zeros are ignored. If only that rule existed in print somewhere!!! It still doesn't indicate the initial bounds for the original number in the same way though as my own 'logical' method does... but I guess you can't have everything! AirdishStraus (talk) 14:48, 28 February 2008 (UTC)

I agree that I too can't find examples for numbers that round up more than one digit.. there's also (if you're teaching this) the question of sematics in grammar - "round this number to 2 significant digits' can be different from 'write this number rounded so that your answer has 2 sig. digits' as you have shown. ie I could even write "round this number to 2 s.ds and write the answer obtained to 4 s.ds" (though that would be wrong for other reasons - ie inferred accuracy). If I was a teacher (which I'm not) I'd try to approach this topic having first covered scientific notation.. For-warned is for-armed.87.102.84.112 (talk) 16:16, 28 February 2008 (UTC)

It indicates bounds on the original number, just not as precisely as you get when using 3sf, which is obvious. Your method gives you 3sf, so of course it's more precise, it's just not what you're asking for. --Tango (talk) 14:59, 28 February 2008 (UTC)

I see the ambiguity in this case, it has to do really with where to draw the line as to when a number has moved it's leading significant digit to another place value. There is no clear 'right way' either, I much prefer notation like .00999±.00005 to sig. figs. A math-wiki (talk) 12:03, 1 March 2008 (UTC)