Wikipedia:Reference desk/Archives/Mathematics/2007 January 30

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January 30

Radicals

Just a quick question: is it correct to have a radical inside of another radical, or should I attempt to work the second out of the first radical? I guess "correct" isn't the best word. I guess I'm asking is it more formal format to have it out? 75.18.9.71 03:48, 30 January 2007 (UTC)

It depends on the context, but I would certainly say that there is nothing wrong with having square roots inside other square roots. Do you mean something like ${\sqrt {2+{\sqrt {2}}}}$? –King Bee ^{(T • C)} 04:05, 30 January 2007 (UTC)

Yes, that's exactly what I mean. Thanks :-) -75.18.9.71 04:11, 30 January 2007 (UTC)

That's perfectly ok. In fact, there are some numbers defined by an infinite amount of nested radicals. Check out that article. --Ķĩřβȳ♥ŤįɱéØ 07:13, 30 January 2007 (UTC)

If there is an unnested version, it will be considered simpler, so by the unwritten rule that (in the absence of other rules to the contrary) the simpler expression is preferred, it is the better form. For example, rather than

${\sqrt[{3}]{-2+{\sqrt {5}}}}\,,$

use

${\frac {-1+{\sqrt {5}}}{2}}\,.$

 --Lambiam^Talk 14:13, 30 January 2007 (UTC)

But from there I would prefer ${\sqrt {5}}/4-1/2$ -Arch dude 03:23, 31 January 2007 (UTC)

How do you go about to simplify such an expression? I tried the expression in Maxima (which I've only just begun to get acquainted with). It confirmed, of course, that both expressions evaluate to 0.6180339887499. But I was unable to make it simplify the expression (tried procedures ratsimp, radcan, trigsimp and trigreduce, i.e. all the buttons that had the words "simplify" or "reduce" on them). It was no more successful at simplifying

${\frac {{\sqrt {5}}-1}{2\,{\left({\sqrt {5}}-2\right)}^{\frac {1}{3}}}}$.

How does one approach the problem "by hand"? --NorwegianBlue ^talk 22:08, 30 January 2007 (UTC)

If you mean removing the bottom denominator - I'd suggest multiplying the bottom by (sqrt(5) -2 )^2/3 to give:

1/2 (sqrt(5)-1)(sqrt(5)-2)2/3 / (sqrt(5)-2)1 

then multiply top and bottom by (sqrt(5)+2) to give:

1/2 (sqrt(5)-1)(sqrt(5)-2)2/3(sqrt(5)+2) / (sqrt(5)-2)(sqrt(5)+2)

which evaluates to

1/2 (sqrt(5)-2+5)(sqrt(5)-2)2/3 / (sqrt(5)sqrt(5) - 4)
1/2 (sqrt(5)+3)(sqrt(5)-2)2/3 / (5 - 4)
(sqrt(5)+3)(sqrt(5)-2)2/3 / 2

Now (sqrt(5)-2)^2/3 = ((sqrt(5)-2)²)^1/3 = ( 9-4sqrt(5) )^1/3

(sqrt(5)+3)(9-4sqrt(5))1/3 / 2

Also (x³)^1/3 = x so using (sqrt(5)+3)³ = 72+32sqrt(5) gives

(32sqrt(5)+72)1/3 (9-4sqrt(5))1/3 / 2

And a^xb^x = (ab)^x so multiplying (32sqrt(5)+72) and (9-4sqrt(5)) gives 288sqrt(5) - 128x5 + 648 -288sqrt(5) = 648-640 = 8 so

81/3/2

Disclaimer - it's unlikely that I haven't made numerous obvious mistakes along the way - but the method is realistic - however you need to take into account the signs and different values that the roots can take - I've assumed sqrt(4) = 2 (not -2) so be careful, and get a second opinion.87.102.2.51 22:46, 30 January 2007 (UTC)

So the answer in the reals is 1!!. Was that right?87.102.2.51 22:50, 30 January 2007 (UTC)

One possible approach, which can have varying degrees of success, it to conjecture that ${\sqrt[{3}]{-2+{\sqrt {5}}}}$ will be of the form $a+b{\sqrt {5}}$ with a and b rational. This gives $(a+b{\sqrt {5}})^{3}=-2+{\sqrt {5}}$, or $a^{3}+3a^{2}b{\sqrt {5}}+3ab^{2}5+5b^{3}{\sqrt {5}}=-2+{\sqrt {5}}$ which gives $a^{3}+15ab^{2}=-2$ and $3a^{2}b+5b^{3}=1$. Solving this (the substitution $t={\tfrac {a}{b}}$ helps) should give the result. -- Meni Rosenfeld (talk) 22:26, 30 January 2007 (UTC)

Thanks for the quick reply! --NorwegianBlue ^talk 22:45, 30 January 2007 (UTC)

If you need to simplify radicals and you have a TI graphing calculator, I wrote a killer program that simplifies radical expressions automagically. ~ Flameviper ^{Who's a Peach?} 23:58, 30 January 2007 (UTC)

Simplifying nested radicals is generally tough. To get some idea of the issues and methods, try

• Susan Landau (1991). How to Tangle with a Nested Radical.
• Susan Landau (1990). Simplification of Nested Radicals.

Richard Fateman, who worked on MIT's Macsyma (the source for Maxima), made some early progress for computer algebra, but I'm not sure what's in Maxima today. --KSmrq^T 00:29, 31 January 2007 (UTC)

Thanks for the info and the links. I skimmed through the papers, but lack the knowledge of abstract algebra that's needed to understand them. Nevertheless, it's fascinating to see how complicated this seemingly simple problem is, and it's a nice illustration of a concrete and practical application of very abstract theory. --NorwegianBlue ^talk 19:09, 31 January 2007 (UTC)

Secondary school mathematic problem

i have a question that i can't answer. this is from the Indonesian secondary school math text book (equivalent to year 9)

If: 2^x = 3^y = 6^-z

Count : X^-1+Y^-1+Z^-1

Thank you for your detailed explanation and answer.

222.124.48.110 04:18, 30 January 2007 (UTC)SenZ

We aren't allowed to give answers to homework questions, but I'll help you by giving you some hints. First of all, the negative exponent in the 6^-z should be evaluated. Separate -z into -1 * z, and of course multiplying exponents is the same as raising a number to a power to a power:

$C^{mn}=(C^{m})^{n}\!$ So... $6^{-1*z}=(6^{-1})^{z}={({\frac {1}{6}})}^{z}\!$ And this simplifies your original given to: $2^{x}=3^{y}={({\frac {1}{6}})}^{z}\!$

That makes all three of your exponents nonnegative. Now you can substitute values which make all three equal. Find a common denominator among your bases.

I hope this helps. Also, please give more information. "equivalent to year 9" doesn't tell us what subject of math you're studying. Is this Calculus, Statistics, Algebra, Trigonometry, ? Thanks. --Ķĩřβȳ♥ŤįɱéØ 07:24, 30 January 2007 (UTC)

I find the question a bit tricky. If 2^x = 3^y = 6^−z, these three expressions are all equal to the same value, say v. So then 2^x = v, 3^y = v, and 6^−z = v. Solving these equations gives you expressions for x, y and z, which depend on (the unknown value of) variable v. Now you can also form the expression x⁻¹+y⁻¹+z⁻¹, and – lo and behold! miracle of miracles! – the dependence on v vanishes in a puff of smoke.  --Lambiam^Talk 13:57, 30 January 2007 (UTC)

Thx guys....anyway....it is not a homework, just an exercise that i want to do it alone...about the subject of math...it would be translated into : "Exponential number"....hmmmm....202.57.4.232 03:05, 31 January 2007 (UTC)SenZ

Have you figured out the trivial answer to the question? --Gerry Ashton 05:56, 3 February 2007 (UTC)

Counting problem

If x 'things' are distributed randomly over n 'holes', and I count every possible combination I would like to know the number of times a hole has 'z' things in it - (taking all the combinations or an average).

I can solve this problem when the 'holes' are distinguishable eg Hole 1 has 4, hole 2 has 0 (4,0) is different from Hole 1 has 0, hole 2 has 4 (0,4) and so each is counted once.

However I cannot as yet solve this problem when the 'holes' are indistinguishable.(eg Hole 1 has 4, hole 2 has 0 is the same as Hole 1 has 0, hole 2 has 4, and so the two are counted only as one combination. (4,0)=(0,4) )

I can't even find an expression for the total number of combinations..

Can anyone point me in the right direction (page) or explain how to begin.. I have no idea if this is even solvable (except by brute force). Thanks for any help you can give.87.102.13.207 16:05, 30 January 2007 (UTC)

My mind is blank on what Wiki article to look for, but I wanted to point out this question is equivalent to asking for the number of ways to sum n non-distinct integers greater than or equal to zero with total sum X. For example, if n=3 and X=5, there are five ways to add three non-negative integers to a total of five (5+0+0, 4+1+0, 3+2+0, 3+1+1, 2+2+1). Dugwiki 18:36, 30 January 2007 (UTC)

Yes - that's the same thing - at least what I'm asking is clear.87.102.13.207 18:50, 30 January 2007 (UTC)

It seems that you're describing what could be called a weak partition, where 0s are allowed and order doesn't matter, but the total number of addends is also fixed. I don't know a closed form for this, but I can write a reasonable recursion for you: define ${}_{x}D_{n}^{m}$ as the number of ways x objects can be distributed into n indistinguishable holes at at most m per hole. If there are 0 holes, then there is precisely 1 way if there are 0 objects and 0 ways otherwise: ${}_{x}D_{0}^{m}=\delta (x)$ (see Kronecker delta); if there are $x>nm$ objects, they cannot all fit and so ${}_{x}D_{n}^{m}=0$. Being allowed to place more objects than you have in one hole and being allowed to use more holes than you could fill do not help: ${}_{x}D_{n}^{m}={}_{x}D_{n}^{x}\;\forall \,m\geq x$ and ${}_{x}D_{n}^{m}={}_{x}D_{x}^{m}\;\forall \,n\geq x$; at the beginning there are no per-hole restrictions so ${}_{x}D_{n}:={}_{x}D_{n}^{n}$. Finally, we sort the holes by the number of objects placed in them (destroying their identity), and consider allotting i objects to the first hole (the one with the most objects); the important bit is that at most i objects can be placed into any hole thereafter. As such, ${}_{x}D_{n}^{m}=\sum _{i=\left\lceil x/n\right\rceil }^{\min(m,x)}{}_{x-i}D_{r-1}^{i}$. With some dynamic programming this can be evaluated quite efficiently. I don't know how to work out your $f(z)$ probability from this, except to say that whenever you use $i=z$ you've got a hit. (It also depends on whether you're counting the probability in terms of arrangements or in terms of holes.) However, if you just need numerical results, the case of a hole getting z objects should not be hard to notice during the algorithm doing the sums (just make sure that that data is memoized appropriately). I've calculated values for all $n\leq 100$ (${}_{100}D_{100}=190569292$, to give a sense of the growth rate which appears to be slightly subgeometric), and can provide those if it's helpful. --Tardis 20:59, 30 January 2007 (UTC)

Thanks for that - it's clearer at least now how to write an 'efficient' computer program to count the combinations - effectively I loop e₁ from x to (x/n) (rounding up) for the first hole - the residue is x-e₁ = e₂ and have another loop nested inside the first - and so on until I run out of 'things' or 'holes'.. Every time I get to the end of a list of loops I increment my count of holes with 'things'=f ( count_f ) ie count_e(n) is increase by one for each n, and obviously increase my count of the number of combinations. At least that's what I think you said! Cheers.

Not sure how to memoise this but I get the idea.87.102.2.51 23:18, 30 January 2007 (UTC)... Yes - I could store the results for when the residual 'things' is 'small' and pluck them out avoiding a few loopss - more complex to do - but would definately save time.87.102.2.51 23:51, 30 January 2007 (UTC)

87.102.2.51 23:14, 30 January 2007 (UTC)

[edit conflict] The memoization is that you cache the result for, say, ${}_{5}D_{3}^{3}$ because you'll need it when calculating ${}_{8}D_{4}^{7}\;(i=3)$ as well as ${}_{9}D_{4}^{6}\;(i=4)$, both of which you'll need when calculating, say, ${}_{15}D_{5}^{9}\;(i=6,7)$. So whenever you finish a recursive evaluation (your "nested loop", which should probably be a nested function call), you store into an array accessible from all the calls; if you then end up needing the same value again, you can just look it up in the array. It's quite important for an algorithm like this if you want to compute any non-trivial values in a reasonable amount of time. --Tardis 23:57, 30 January 2007 (UTC)

Yes - memoisation now understood.87.102.2.51 00:35, 31 January 2007 (UTC)

Isn't this just a Combinatorics#Combination with repetition? You are trying to choose the holes that have balls in them; and there can be more than one ball per hole (repetition)? --Spoon! 22:53, 30 January 2007 (UTC)

Er no. It's not the same because I'm ignoring combinations that look the same but have the same amount of things in different holes eg 5 things 2 holes - (5,0) and (0,5) count as just one, not two...at least I think it's not..87.102.2.51 23:14, 30 January 2007 (UTC)

For the record, I can't imagine what physical process would obey the mathematics you're setting up here; not to doubt unkindly, but perhaps you should describe your overall problem (if there is one) to have that checked too? --Tardis 23:57, 30 January 2007 (UTC)

Well yes - originally I was looking at counting x 'things' over n 'holes' which has meaning - it actually took me a long time to work out that it was indeed a combination with repetition problem - I now find the solution to this obvious. (It was a physics model to test - and turned out to have some 'interest' as a model of energy/particles distributed over very, very large volumes of space - amongst all the other things it can apply to.)

Having solved this (with an easy equation and various approximations for big n and x) I began to wonder if I could solve this similar problem... and if it did have any meaning.

I too can't think of anything as yet that matches this situation.. But if I could get an equation for it, I could at least compare the behaviour of the distribution vs n and x - maybe it would approximate something..

Looking at it I doubt more and more that a discrete solution will be findable (by me) - the round up(x_residue/n_remaining) part looks like it messes up simple counting techniques of the type I used to solve the previous problem eg figurate numbers.

As an aside I would be very interested if anyone could supply an example of a system that is relevant to this counting method - I probably should ask another question for that. So yes you're right when you question what could it could be possibly any use for - I too would like to know!87.102.2.51 00:35, 31 January 2007 (UTC)