January 29

So I have five different elements (A,B,C,D,E) and 2 spots they can fill. The question is how many unique pairs can I make, where the elements in the spots can be the same (non-unique). For this simple case the possible combos are (AA)(AB)(AC)(AD)(AE)(BB)(BC)(BD)(BE)(CC)(CD)(CE)(DD)(DE)(EE) or 5+4+3+2+1 = 15 different unique combos allowing repeated elements. My question is how can you find the answer for n elements and s spots? I'm sure there is a name for this kind of combination, but I don't know what it is. Am I missing a clever trick or a way to manipulate permutations and combinations which don't allow repeating elements? 24.106.205.226 02:43, 29 January 2007 (UTC)[reply]

You want the formula for combinations with repetition, where n is the number of elements and r is the number of spots. As an alternative, you could use the approach you did to work out that it was 5+4+3+2+1, and then generalise with this summation identity: ${\displaystyle {\sum _{i=1}^{n}i}=((n(n+1)} \over {2))}$. Maelin (Talk | Contribs) 03:23, 29 January 2007 (UTC)[reply]

Thanks for the help Maelin. I solved it formulized it, and then promptly found it on the combinatorics page. It's

${\displaystyle (((n+s-1)!} \over {s!(n-1)!))=((n+s-1} \choose {s))=((n+s-1} \choose {n-1))}$

where n is the number of objects from which you can choose and s is the number to be chosen.

Even though it took me a while, I was right in my hunch that it did involve a combinatoric, and it was a well wasted hour or two in terms of learning! 24.106.205.226 04:00, 29 January 2007 (UTC)[reply]

Can

${\displaystyle f(x,y)={\frac {\sum _{i=x+1}^{max(x,y)+19}[i-20(1-max(min(\left({\frac {x-i+21}{20))-1\right),1),0))(1-max(min(\left({\frac {y-i+21}{20))-1\right),1),0))-x]}{max(x,y)+19-x))}$

be rewritten without the summation into something that can be evaluated in a single spreadsheet cell? NeonMerlin 05:37, 29 January 2007 (UTC)[reply]

I have serious doubts whether you're going to find a closed form expression for that sum because it involves choosing maxes and mins.

Try

${\displaystyle f(x,y)=\mathrm {If} \left(y-x<0,-10,\mathrm {If} \left(y-x<20,{\frac {(y-x)^{2}-190}{y-x+19)),{\frac {(y-x)^{2}+39(y-x)-760}{2(y-x+19)))\right)\right)}$

Also note that this only depends on ${\displaystyle t=y-x}$, which can be used to simplify the writing. -- Meni Rosenfeld (talk) 20:25, 29 January 2007 (UTC)[reply]

Is it true that the expected values of a sorted sample, size N, are the 1st, 3rd, 5th, ... and (2N-1)th 2N-quantiles? NeonMerlin 07:35, 29 January 2007 (UTC)[reply]

I believe that would qualify as a systematic sample, although I'm not sure if you'd want them to be sorted for that. I'm not an expert on statistics anyway, so would someone else like to clarify? Curtmack of the Asylum 18:37, 29 January 2007 (UTC)[reply]

good aftenoon.... may i please know about the different mathematical formulas in solving problems which involved more than 7 digits? like for example finding the square root of a 7 digit figure in just a couple of second by just using the speed of my mental computation? thank you very much...your kind answer to this question would be a priceless help for my profession.... —The preceding unsigned comment was added by Never17fade (talk • contribs).

While there are definitely algorithms for calculating things like square roots, and there are undoubtedly tricks which can make the calculation easier, I sincerely doubt that there is any way for a human without an extraordinary gift to calculate the square root of a 7-digit number mentally, let alone in a few seconds. The obvious question I should ask is: What is wrong with using a calculator for whatever purpose you need this calculation? -- Meni Rosenfeld (talk) 18:45, 29 January 2007 (UTC)[reply]

Perhaps both Never17fade and Meni Rosenfeld would benefit from reading the article on mental calculation. --KSmrq^T 19:33, 29 January 2007 (UTC)[reply]

Having read the relevant sections, I now know that ${\displaystyle 15-15.21\approx -0.02}$ and that 3.8975 is closer to 3.8729833... than 3.875 is (I have taken the liberty of sofixiting it), but alas, I still cannot calculate square roots of 7-digit numbers (under the assumption that 4 significant digits are required) mentally in a few seconds. I agree, though, that I have overestimated the difficulty of this problem - but not if a greater precision is required. -- Meni Rosenfeld (talk) 20:48, 29 January 2007 (UTC)[reply]

This is not a direct answer to the question, but perhaps the following will impress your audience. It is a way to calculate the 17th root of a 32-digit number. The large number must be a natural number raised to the 17th power, otherwise the following does not work. Let F be the first digit of the 32-digit number, and let L be its last digit. Then if F = 1, the 17th root of the number equals 60+L. Otherwise, if F > 1, the 17th root is equal to 70+L. Example 1: find the 17th root of the number 11 04769 42366 68359 04801 61345 93027. F = 1 and L = 7, so the 17th root is 60+7 = 67. Example 2: find the 17th root of the number 59 83278 73799 50079 88913 23447 05024. F = 5 and L = 4, so the 17th root is 70+4 = 74. For maximum effect, do not give the answer immediately, but stand there mumbling silently for some time, while occasionally glancing at a corner of the ceiling.  --Lambiam^Talk 14:25, 30 January 2007 (UTC)[reply]

To elaborate on a question asked a few days ago, I have a formula to determine the length of the angle bisectors, from vertex to the opposite side, in terms of the sides of the triangle. The length of the bisector from vertex A is 2*b*c*cos(A/2)/(b+c), the others having an obvious cyclic symmetry. cos(A/2) can be found in terms of a, b and c, of course.

But can it be done the other way round - is there an expression to get each side of the triangle from given values of the angle bisectors?…86.132.237.140 21:07, 29 January 2007 (UTC)[reply]

The simple answer is of course - yes - you have three equations (as supplied above) and three unknowns (a,b,c). You can of course get angles A,B,C from a,b, and c. So a solution can be found. Whether there will be multiple solutions is another question. (It looks a bit tricky and time consuming though Good Luck)87.102.13.207 15:36, 30 January 2007 (UTC)[reply]

(ok I thought I had an easy method, it's easy for side bisectors..)87.102.13.207 15:51, 30 January 2007 (UTC)[reply]

My question, "can it be done ... " was shorthand for "please show me, as it looks rather non-invertible to me and I'd appreciate having the answer". I wasn't the asker of the numerical question a few days ago, by the way. If by side bisectors you mean medians, yes the sets of sides and medians are easy to get from each other.…81.153.220.25 16:33, 30 January 2007 (UTC)[reply]

Well I get cos (A/2)= sqrt( (b+c)²-a² / 4bc )

and so A (length) = bc/(b+c) x sqrt( (b+c)²-a² / bc )

giving a² = (b+c)² (1- a²/bc)

Is that right so far? (If so I could substitute 'a' into the related expressions for B and C (length) - giving after rearrangement and squaring two hexic equations in b and c... I give up sorry. It's unsolveable by me.87.102.13.207 17:01, 30 January 2007 (UTC)[reply]

OK try again - I've derived (C²-a²)b = (C²-b²)a where C is the angle bisector length of the line that crosses triangle side c of length c.. The rest can be got by "obvious cyclic symmetry" - this looks simpler. For now I leave the rest to you.87.102.13.207 17:39, 30 January 2007 (UTC)Apologies for not having been of any use.87.102.13.207 18:19, 30 January 2007 (UTC)[reply]

I'd love to know how you got that, because I didn't see any useful way forward myself. However, isn't what you have equivalent to ${\displaystyle C^{2}(b-a)=a^{2}b-b^{2}a}$ and so ${\displaystyle C^{2}=-ab<0}$? Perhaps there's merely a typo; please elucidate. --Tardis 18:46, 30 January 2007 (UTC)[reply]

Mmm, grr. ok I divided the triangle length c into c_A and c_B such that c_A + c_B = c

Also c_A / c_B = b/a (is this wrong perhaps??)

So then using c_A = C² + b² -2Cbcos (angle C) and c_B = C² + a² -2Cacos (angle C) eliminating "angle C" - I got the relationship above - something may have gone wrong here - though for the moment my mistake elludes me

Found it (missing square) (IGNORE the following)

cA = C2 + b2 -2Cbcos (angle C) and cB = C2 + a2 -2Cacos (angle C)
therefore 
(cA - C2 - b2)/ -2Cb =  cos (angleC) = ( cB - C2 - a2)/ -2Ca
(cA - C2 - b2)/b = ( cB - C2 - a2)/a
cAa - C2a - b2a =  cBb - C2b - a2b   (equation 1)
but cB / cA = a/b see Angle bisector theorem
so cB = cAa/b
and cB + cA = c 
so  cA +  cAa/b =c 
cA (1 + a/b) =c
cA ( (b+a)/b ) =c 
so cA  =cb/(b+a) and cB  =ca/(b+a) (equations 2)
Combining equation1 and two equations 2 gives 
cba/(b+a) - C2a - b2a =  cab/(b+a) - C2b - a2b 
C2a + b2a = C2b + a2b 

I can't for the life of me see were I've gone wrong but I am getting tired so will come back tommorow.87.102.13.207 19:41, 30 January 2007 (UTC)[reply]

(A,B,C are the bisector lengths, a,b,c are the lengths of the sides, and c_A is the length of triangle side c from a corner opposite b to the bisection)??87.102.13.207 19:11, 30 January 2007 (UTC)[reply]

Ignore below - made mistake..

C2 = ab
So if I've not made a mistake (which is often the case) I get
C2=ab  B2=ac  A2=bc87.102.13.207 19:13, 30 January 2007 (UTC) 
So that gives the ratio of side lengths - another bit of info would probably finish off this problem - anyone?87.102.13.207 19:22, 30 January 2007 (UTC)[reply]

Sorry, but this just isn't working; your assumptions (probably about proportionality of "corresponding" parts) are wrong somehow. Consider an equilateral triangle, where ${\displaystyle a=b=c}$ and ${\displaystyle \{A,B,C\}<a}$ (the altitudes are less than the side lengths): your formulas give ${\displaystyle \{A,B,C\}=a}$. --Tardis 19:30, 30 January 2007 (UTC)[reply]

I tried to show how I got there above - I used the angle bisector theorum - I accept that something has gone wrong here - but at present can't spot my mistake - If you could check the above and point it out I'd appreciate it..87.102.13.207 19:47, 30 January 2007 (UTC)[reply]

OK I found it - a missing square - the result is...

c_A² = C² + b² -2Cbcos (angle C) and c_B ² = C² + a² -2Cacos (angle C)

therefore

(c_A² - C² - b²)/ -2Cb = cos (angleC) = ( c_B² - C² - a²)/ -2Ca

(c_A² - C² - b²)/b = ( c_B² - C² - a²)/a

c_A²a - C²a - b²a = c_B²b - C²b - a²b (equation 1)

but c_B / c_A = a/b see Angle bisector theorem

so c_B = c_Aa/b

and c_B + c_A = c

so c_A + c_Aa/b =c

c_A (1 + a/b) =c

c_A ( (b+a)/b ) =c

so c_A =cb/(b+a) and c_B =ca/(b+a) (equations 2)

Combining equation1 and two equations 2 gives

(cb)²a/(b+a)² - C²a - b²a = (ca)²b/(b+a)² - C²b - a²b

Rearranging gives

abc = (C²-ab)(a+b)²/c and two other equations for A and B.. Not as simple.. SORRY.87.102.13.207 20:08, 30 January 2007 (UTC)[reply]

This http://www.cut-the-knot.org/triangle/TriangleFromBisectors.shtml

is a reference to an article claiming to prove that, in general, a triangle cannot be constructed from given values of its angle bisectors. I presume this is in the classical sense of constructability by ruler and compasses, but does this mean also that the sides cannot in general be calculated?.....81.154.108.56 01:21, 2 February 2007 (UTC)[reply]

Yes - not by ruler and compasses - but as a calculation - I think that would all depend on your ability to solve complex algebraic equations - for me the answer is no..87.102.4.6 10:43, 2 February 2007 (UTC)[reply]

By the way I got

(C2-ab)(a+b)2/c = (B2-ac)(a+c)2/b = (A2-bc)(c+b)2/a

Where capitals are bisector lengths, lower case are side lengths - solvable yes, easy no, multiple solutions don't know)87.102.4.6 10:47, 2 February 2007 (UTC)[reply]

See Wikipedia:Reference_desk/Mathematics#triangle angle bisectors287.102.4.6 11:24, 2 February 2007 (UTC)[reply]