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        riddles >> hard >> Denesting Nested Radicals
        
(Message started by: Barukh on Dec 29^th, 2003, 5:08am)

Title: Denesting Nested Radicals
Post by Barukh on Dec 29^th, 2003, 5:08am

Simplify the following doubly nested radicals to a combinations of singly nested radicals (and thanks towr for his amazing formula generator! :D)

P.S. The usual question: Does it belong to the hard section? Not all the expressions are hard, but I believe some of them are.

Title: Re: Denesting Nested Radicals
Post by towr on Dec 29^th, 2003, 7:28am

::[hide]
1) sqrt(3)+sqrt(2)
6) 5^(3/4)/2 + 5^(1/4)/2 + sqrt(5)/2 + 3/2
[/hide]::

Title: Re: Denesting Nested Radicals
Post by towr on Dec 29^th, 2003, 8:30am

::[hide]
4) 98^(1/3)/3 - 28^(1/3)/3 - 1/3
[/hide]::

Title: Re: Denesting Nested Radicals
Post by Barukh on Dec 29^th, 2003, 8:40am

towr, so far so good :D! Although 6) also has a nicer looking expression.

I am curious: how do you get the answers?

Title: Re: Denesting Nested Radicals
Post by towr on Dec 29^th, 2003, 9:26am

I'm mostly just trying things out in my math program, the first one rolled out immediately, the other two took a bit more work.
In 6 I just tried to work out the denominator, by multiplying numerator and denominator with first (3+2(5)^(1/4)) and then (9 + 4 sqrt(5)). After that my mathprogram simplified it further..
#4 took most work from my part..
sqrt(28^(1/3) - 3) = x
(28^(1/3) - 3) = x^2
x^6 + 9·x^4 + 27·x^2 + 27 = 28
(x^3 + x^2 + 5·x - 1)·(x^3 - x^2 + 5·x + 1) = 0
solving the first factor then gives the answer (though I wouldn't know how to do that without my trusty math program either..)

http://mathforum.org/library/drmath/view/52657.html
http://mathforum.org/library/drmath/view/52660.html
helped a bit as well..
I wish I could find some better tools to solve these kind of problems though..

Title: Re: Denesting Nested Radicals
Post by THUDandBLUNDER on Dec 30^th, 2003, 12:36pm

Quote:

Although 6) also has a nicer looking expression.

6) [hide](5^1/4 + 1)/(5^1/4 - 1)[/hide]

Surprisingly, 5) has a simpler answer than 2).

Title: Re: Denesting Nested Radicals
Post by TenaliRaman on Jan 13^th, 2004, 4:43am

Is it possible to give any hints or clues for the above problems (except maybe the first one!!) :)

Title: Re: Denesting Nested Radicals
Post by Barukh on Jan 14^th, 2004, 1:28am

on 01/13/04 at 04:43:20, TenaliRaman wrote:

Is it possible to give any hints or clues for the above problems (except maybe the first one!!) :)

Here are some hints (I try not to give in too much – please ask for more if you need):
[sqrt]-[hide] Start with 6). The answer will have the same form: it’s a fraction ;), and the numerator is a^1/4 + b^1/4.
Then try 5). As THUD&BLUNDER said, it has a simple form – namely, it’s a linear combination of two terms.
All the other cases may be represented as a linear combination of 3 terms[/hide] -[sqrt].

Unfortunately, an algorithm for denesting radicals in the general case is still unknown (or is it fortunately?), so don’t get disappointed. The radicals presented here (except the first one) are due Ramanujan who gave just the answers, not the explanations of the solutions.

Title: Re: Denesting Nested Radicals
Post by Barukh on Jan 14^th, 2004, 1:32am

on 12/30/03 at 12:36:35, THUDandBLUNDER wrote:

Surprisingly, 5) has a simpler answer than 2).

Correct. Do you know the answers to all the cases?

Title: Re: Denesting Nested Radicals
Post by John_Gaughan on Jan 14^th, 2004, 6:59am

Powers and radicals like these fascinate me, but I am unlearned in this corner of math. Could someone please explain how they get their answers? I made it as far as calculus and discrete math in college, so please don't confuse me too much :-)

Title: Re: Denesting Nested Radicals
Post by TenaliRaman on Jan 14^th, 2004, 10:43am

Yeeehaw!!!!!!!!!!!!!!!
ok i haven't solved 6 yet!
But i have solved 3 and if my hunch is right i have a method for 4 as well.

I will hide the answer for 3 but i am not hiding the method i used to solve it.

Ok,
first of all lets look at a general expression,
sqrt(1-cubrt(b/a)) with a>b
let me write b/a as w so w<1 and our general expression is,
sqrt(1-cubrt(w))

expand this binomially and soon it will be noticed that we have a period 3 cycle which implies that, i can write
sqrt(1-cubrt(w)) = x + y*cubrt(w) + z*cubrt(w²) .... (*)
(provided x,y and z are finite converging limits .. i am assuming it for now)
I can see 2 methods for getting x,y and z,
method 1>
Enumerate the series which equals to x , series which equals to y and series which equals to z. If it is possible to find the value of the series then we would have found x,y and z.

method 2>
square both sides of (*) and equate the coefficients of 1,w and w² and solve the simultaneous equation. (This gets pretty tedious and i used this to do third question and it took me pretty long time.)

The above method can be used to solve the 4th question as well.

Answer to 3rd :
::[hide](1/3)*(cubrt(2)+cubrt(20)-cubrt(25))
Note: If you see 25 = 5^2, 20 =5*4
25 = 25/16 * 16 = (5/4)^2 * 16
20 = (5/4) * 16
2 = (2/16) * 16

It seems there is a pattern running here which means a generalised solution for sqrt(1-cubrt(w)) is possible given some convergence conditions.[/hide]::

[edit]
Giving it some more thought,
this idea will work for any nthroot(1-cubrt(w)), since each one of this will be a period 3 cycle. It is just that evaluating x,y and z will become hectic. That's not very encouraging to me :D (however this implies we can solve 2 by this method)

Also, nthroot(1-mthroot(w)) will be a period m cycle , so we will have m variables. This implies this could be used to solve 7th as well ,only problem is that we will have 5 variables to contemplate with!! phew!!
[/edit]

All that is left now is 6 and 5.

Title: Re: Denesting Nested Radicals
Post by THUDandBLUNDER on Jan 15^th, 2004, 1:11pm

Quote:

Giving it some more thought,
this idea will work for any nthroot(1-cubrt(w)),

TenaliRaman, for 3) there is a solution to [smiley=surd.gif](³[smiley=surd.gif]a + ³[smiley=surd.gif]b) iff there exist integers p,q such that

a = (4p + q)q³
b = 4(p - 2q)p³

In which case,
[smiley=surd.gif](³[smiley=surd.gif]a + ³[smiley=surd.gif]b) = [smiley=pm.gif][{³[smiley=surd.gif]2(p - 2q)²} - {³[smiley=surd.gif](4p + q)²} - {³[smiley=surd.gif]4(p - 2q)(4p + q)}]/3

For [smiley=surd.gif](³[smiley=surd.gif]5 - ³[smiley=surd.gif]4), p = q = 1 and the result easily follows.
Finding p,q in general, given they exist, is not easy and requires some familiarity with Galois field theory. :(

But the above assumes that neither a nor b are zero and that a/b is not a cube in [smiley=bbq.gif].
That is, a/b does not equal c³ for any c in [smiley=bfcq.gif].
If c [smiley=in.gif] [smiley=bbq.gif] and a/b = c³ then [smiley=surd.gif](³[smiley=surd.gif]a + ³[smiley=surd.gif]b) = ⁶[smiley=surd.gif]b[smiley=surd.gif](c + 1)
Unfortunately, we cannot use this form to solve 5) as 20 is not a perfect cube.

For some reason (to do with the expressions for a and b not being relatively prime for certain p,q),
we sometimes need to swap them around and have

a = 4(p - 2q)p³
b = (4p + q)q³

Applying this form to 4), where a = 28 and b = -27, we get p = -1 and q = 3.

This gives [smiley=surd.gif](³[smiley=surd.gif]28 - ³[smiley=surd.gif]27) = [smiley=pm.gif](³[smiley=surd.gif]98 - ³[smiley=surd.gif]28 - 1)/3

(The above formulae are due to Ramanujan.)
============================================

A simple method for 1)

5 + 2[smiley=surd.gif]6 = (10 + 4[smiley=surd.gif]6)/2 = (2 + [smiley=surd.gif]6)²/2

So [smiley=surd.gif](5 + 2[smiley=surd.gif]6) = [smiley=pm.gif](2 + [smiley=surd.gif]6)/[smiley=surd.gif]2 = [smiley=pm.gif]([smiley=surd.gif]2 + [smiley=surd.gif]3)

Title: Re: Denesting Nested Radicals
Post by SWF on Jan 15^th, 2004, 5:34pm

For 2) cbrt( cbrt(2)-1) = [hide] ( cbrt(3)-cbrt(6)+cbrt(12) )/3[/hide]

Solution method: [hide]Tried to write cbrt(2)-1 as the cube of something. After failing to find a two term expression that worked, tried a three term expression, which I assumed would have the form
A + B*2^(1/3) + C*2^(2/3)
under the assumption that constants A, B, C could be chosen to zero out any 2^(2/3) term. That left equations for A, B, and C, which weren't difficult to solve since it turned out that A=C=-B=9^(-1/3). Then rearranged to get the 9^(1/3) out of the denominator[/hide].

Title: Re: Denesting Nested Radicals
Post by THUDandBLUNDER on Mar 21^st, 2004, 1:29am

Solve [smiley=surd.gif][x - 2[smiley=surd.gif](x-1)] + [smiley=surd.gif][x + 2[smiley=surd.gif](x-1)] = 1/(x-1)

Title: Re: Denesting Nested Radicals
Post by Barukh on Mar 21^st, 2004, 6:35am

on 03/21/04 at 01:29:51, THUDandBLUNDER wrote:

Solve [smiley=surd.gif][x - 2[smiley=surd.gif](x-1)] + [smiley=surd.gif][x + 2[smiley=surd.gif](x-1)] = 1/(x-1)

Are we looking at the real solutions?

[smiley=blacksquare.gif][hide]1.5[/hide][smiley=blacksquare.gif]