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ThudnBlunder
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Nested Radicals
« on: May 27th, 2003, 11:35am » |
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Evaluate:
1] sqrt(1 + sqrt(1 + sqrt(1 +...
2] sqrt(4 + sqrt(16 + sqrt (64 + ...
3] sqrt(2 + sqrt(8 + sqrt(32 + ...
4] sqrt(x2 + sqrt(4x2 + sqrt(16x2 + sqrt(64x2 + ... (where x > 0)
5] sqrt(1 + 2sqrt(1 + 3sqrt(1 + 4sqrt(1 + ...
6] cbrt(1 + cbrt(1 + cbrt(1 + ... (cbrt = cube root)
7] sqrt(2^(21) + sqrt(2^(22) + sqrt(2^(23) + ...
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towr
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Re: Nested Radicals
« Reply #1 on: May 27th, 2003, 12:32pm » |
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...
1] f = sqrt(1+f) => f = (sqrt(5)+1)/2
2] f = sqrt(4 + 2f) => f^2 = 4 + 2f => f= 1+sqrt(5)
3] f = sqrt(2 + 2f) => f = sqrt(3)+1
4] f = sqrt(x^2 + 2f) => f = sqrt(x^2 + 1) + 1
...
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ThudnBlunder
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Re: Nested Radicals
« Reply #2 on: May 27th, 2003, 2:09pm » |
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Quote:| 2] f = sqrt(4 + 2f) => f^2 = 4 + 2f |
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While I agree with your answer to 1], I don't think that multiplying f by 2 changes the exponents as required.
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Icarus
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Re: Nested Radicals
« Reply #3 on: May 27th, 2003, 5:16pm » |
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Here are the results I found. [2],[3],[5], and [7] were obtained by spread sheet. [4] is just the generalization of [2] and [3]. I have not deduced them mathematically yet.
[1] as towr has already said, if f = sqrt(1+...), f=sqrt(1+f) so f2 - f - 1 = 0.
The positive root is f = (1+sqrt(5))/2 - the golden ratio.
[2] f = 3
[3] f = 1 + sqrt(2)
[4] f = 1 + x
[5] f = 2
[6] f ~ 1.324718. Using the same trick as with [1], f3 - f - 1 = 0.
The "exact value" is: f = cbrt(1/2 + sqrt(23/108 )) + cbrt(1/2 - sqrt(23/108 )).
[7] f ~ 3.236.
I particularly found the behavior of [2] and [3] interesting. They converge surprisingly fast.
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towr
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Re: Nested Radicals
« Reply #4 on: May 27th, 2003, 11:31pm » |
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hmm.. you're right..
now if it was 256 instead of 64..
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wowbagger
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Re: Nested Radicals
« Reply #6 on: May 30th, 2003, 7:57am » |
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Icarus,
I agree with your numerics except for number five. I come up with f = 3 .
Consider
f > sqrt( 1 + 2*sqrt( 1 + 3*sqrt( 1 ) ) ) = sqrt (5) > 2
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Icarus
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Re: Nested Radicals
« Reply #7 on: May 31st, 2003, 9:00am » |
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Sorry - I took the sequence one too many steps back. The result I found was actually:
2 = sqrt(1+1*sqrt(1+2*sqrt(1+3*sqrt(...
More generally, n+1 = sqrt(1+n*(sqrt(1+(n+1)*sqrt(...
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ThudnBlunder
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Re: Nested Radicals
« Reply #8 on: May 31st, 2003, 11:39am » |
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8] sqrt(1 + sqrt(2 + sqrt(3 + sqrt(4 + .......
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Icarus
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Re: Nested Radicals
« Reply #9 on: Jun 3rd, 2003, 7:34pm » |
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I've got a spread-sheet number for (8) as well, but I don't see much point in posting it, since I haven't found a derivation.
Instead I want to see if I can't make progress on proving (4) is x+1:
let fn(y) = sqrt(4nx2 + y) and gn(y) = y2 - 4nx2 is the inverse function. Restrict all variables to >= 0, so both functions are 1-1 and increasing.
Fix R>0
Let Ak = f0of1of2of3o...fk (R).
The claim is that limk Ak = x+1, regardless of R.
If n >= log2 (R/x), then fn(R) > R, and since each of the f functions is increasing, An > An-1. Thus {A} is eventually increasing.
fk(2k+1 + 1) = sqrt( 22kx2 + 2(2kx) + 1 ) = 2kx + 1
If k >= log2((R-1)/2x), then R < 2k+1x + 1, and
Ak = f0o...fk(R) < f0o...fk(2k+1x + 1) = x + 1
So the sequence {A} is eventually increasing and bounded above by x + 1. Therefore limk Ak exists and is <= x+1, for any positive R. (Negative R works as well, except that Ak is not well-defined until k is large enough. But this does not affect the limit.)
Now I need to show that limk Ak >= x+1 as well. But I am out of time, so maybe tomorrow.
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SWF
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Re: Nested Radicals
« Reply #10 on: Jun 3rd, 2003, 8:54pm » |
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For the ones that have not been done mathematically here are some forms that help in finding mathematical solutions.
4] f(x)2=x2+f(2x)
This leads to f(x)=1+x (which is not immediately obvious, because f(x)=1-x also fits the above form).
5] Let f(x) be the form given by Icarus: f(x) = sqrt(1+x*(sqrt(1+(x+1)*sqrt(1+...
f(x)2=1+x*f(x+1)
And f(x)=1+x is a solution. For case 5], x=2 and f(2)=3.
8] Let f(x)=sqrt(x+sqrt(x+1+sqrt(x+2+sqrt(x+3+...
f(x)2=x+f(x+1)
I don't see a simple solution to this, and the numerical result for x=1 looks irrational.
Here is another one that is a little different because there are a finite number of sqrt()'s:
9] sqrt(2+sqrt(2+sqrt(2+sqrt(2...+sqrt(3)))...))
There is an exact solution in terms of N (the number of sqrt's) and yes the final sqrt contains a 3 instead of a 2.
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Icarus
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Re: Nested Radicals
« Reply #11 on: Jun 4th, 2003, 6:27pm » |
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I considered the functional equation approach, but the problem you hint at for [4] is broader still (besides which, it assumes a priori that the limit exists, which is anathema to a mathematician like me  ). Are you sure that f(x) = x+1 and f(x) = 1-x are the only solutions to f(x)2 = x2 + f(2x) ?
Before you can be sure that you have found the correct limit, you need to eliminate all other possible solutions, and functional equations may have infinitely many (though this can be cut down considerably if you can prove continuity). So I passed them up to try a more analytic attack.
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| « Last Edit: Jun 4th, 2003, 6:30pm by Icarus » |
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Re: Nested Radicals
« Reply #12 on: Jun 4th, 2003, 6:49pm » |
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[9] The 3 is a giveaway: 2cos (PI/3*2N)
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Re: Nested Radicals
« Reply #13 on: Jun 5th, 2003, 10:36pm » |
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Very good, Icaraus. The argument of cos() gets small rapidly with increases in N, which means cos(x) is approximatly equal to 1-x2/2. Solving for x and rearranging gives a good approximation to pi that gets even better as N increases.
Quote:| Are you sure that f(x) = x+1 and f(x) = 1-x are the only solutions to f(x)2 = x2 + f(2x) ? |
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That is why I said, the answer is not immediately obvious. And maybe that functional equation will lead to a dead end in trying to prove the nested radicals equal 1+x, but it sure helps finding a candidate to try and prove. It looks like there are an infinite number of solutions to that functional equation, for example:
f(x) = 1 - x2/2 + x4/56 - x6/3472 + ...
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SWF
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Re: Nested Radicals
« Reply #14 on: Oct 26th, 2003, 1:56pm » |
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Now that Icarus has convinced my of the importance of being more complete, I will try give a better reason that that the answer to [4] is 1+x. But I sure regret ever starting to type all these equations.
Let Ak = [surd](x2 + [surd](4x2+ ...+ [surd](22kx2)...)) and will try to show that Ak approaches 1+x as k grows larger.
From the expression for Ak keep squaring and rearranging to get:
(...((Ak2 - x2)2 -4x2)2...)2 - 22(k-1)x2 = [surd](22kx2)=2kx
The steps involved above give a sequence of functions I'll call h():
h(y,0)= y
h(y,1)= h(y,0)2-1*x2=y2-x2
h(y,2)= h(y,1)2- 4*x2=(y2-x2)2- 16*x2
h(y,k)= h(y,k-1)2- 22(k-1)*x2
Therefore h(An,n)=2kx
To get bounds, start with some value of n large enough relative to x such that 4/(2nx)<1. For this value of n,
h(An,n)=2nx
Adding 22(n-1)x2 to both sides, and using relation between h(An,n) and h(An,n-1)2:
h(An,n-1)2 = 22(n-1)x2+2*2n-1x
Adding 1 to the right hand side makes it a perfect square and an inequality. I will simultaneously get a lower bound, and the first expression for h2 below is factored as shown to aid in getting a bound:
22(n-1)x2(1+4/(2nx)) = h(An,n-1)2 < (2n-1x+1)2
Defining u=4/(2nx) (and by the choice of n, u<1), take square root and use the fact that [surd](1+u)>1+u/2-u2/8 to get the lower bound:
2n-1x+1-1/(2nx) < h(An,n-1) < 2n-1x+1
Repeat to get bound for h(A,n-2), except on lower bound use [surd](1-w)>1-w (for 0<w<1):
2n-2x+1-1/(2nx(2n-2x+1)2) < h(An,n-2) < 2n-2x+1
The same thing for h(A,n-3) gives:
2n - 3x + 1 - 1/(2nx(2n-2x+1)2(2n-3x+1)2) < h(An,n-3) < 2n-3x+1
Keep repeating until the bound is on of h(An,0), at which point it equals An and gives:
1+x-e < An < 1+x
where the expression for e clearly goes to zero as n grows.
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ThudnBlunder
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Re: Nested Radicals
« Reply #15 on: Dec 30th, 2003, 1:20pm » |
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10] sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + ......
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Re: Nested Radicals
« Reply #16 on: Jan 21st, 2004, 2:30pm » |
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on Dec 30th, 2003, 1:20pm, THUDandBLUNDER wrote:10] sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + ......
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10] Nobody's had a go at this yet, so...
Let
x = sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + ......
y = sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 +......
z = sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - ......
w = sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 -......
Then
x2 + y = 2......(1)
y2 - z = 2......(2)
z2 - w = 2.....(3)
w2 + x = 2.....(4)
Eliminating for x we get,
([{(2 - x2)}2 - 2]2 - 2)2 = 2 - x.........(5)
FWIW, Mathematica expands this as
x16 - 16x14 + 104x12 - 352x10 + 660x8 - 672x6 + 336x4 - 64x2 + x + 2 = 0
According to Mathworld the answer is 2sin{[pi]/18} = 0.3472963554... and this value satisfies (5).
But Mathematica's Solve function gives only 6 (of 16) roots involving radicals, none of which equal 0.3472963554... Now it's a fact that sin{[pi]/9} and cos{[pi]/9} cannot be expressed in terms of radicals of real numbers. So I guess this means sin{[pi]/18} can't either, except with infinitely nested radicals. But I would expect to see some solutions involving i. Perhaps I'm misreading the output. I will check it again.
Unfortunately, I can't give the Mathworld link (search for 'Trigonometry Values [pi]/18') because I - and this whole subnet - have been blocked for more than two weeks. Reason: I downloaded too many Mathematica notebooks in a short space of time and their bots thought I was another bot! I had to laugh at the time but now it's getting ridiculous. Not to worry, too bad for those breadheads at CRC Press (remember how they treated Weisstein a few years back?) that I already have a bootleg version of their encyclopaedia. However, my colleagues are also effected. http://mathworld.wolfram.com/docs/faq.html
By the way, there is a new section at http://functions.wolfram.com/.
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| « Last Edit: Jan 22nd, 2004, 6:39pm by ThudnBlunder » |
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Barukh
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Re: Nested Radicals
« Reply #17 on: Jan 22nd, 2004, 7:12am » |
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Reverse engineering: using the identity (it follows from de Moivre’s theorem)
sin(3[theta]) = -4sin3 [theta] + 3 sin[theta]with [theta] = [pi]/18, we get that sin( [pi]/18 ) is a root of the cubic 8x3 – 6x + 1 = 0. Therefore, 2 sin( [pi]/18 ) is a root of the cubic x3 – 3x + 1, which is much simpler than Mathematica’s expansion.
I wonder if it’s possible to derive by manipulating radicals in 10]. My intuition tells it is…
on Jan 21st, 2004, 2:30pm, THUDandBLUNDER wrote:| Now it's a fact that sin9[pi]/90 and cos9[pi]/90 cannot be expressed in terms of radicals of real numbers. So I guess this means sin9[pi]/180 can't eithe, except with infinitely nested radicals. But I would expect to see some solutions involving i. Perhaps I'm misreading the output. I will check it again. |
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I really don’t understand what are the angles your consider. 2 sin( [pi]/18 ) cannot be expressed in finite “real” radicals, since the discriminant of its cubic polynomial (1/4 – 33/27) is negative.
Quote:| Reason: I downloaded too many Mathematica notebooks in a short space of time and their bots thought I was another bot! I had to laugh at the time but now it's getting ridiculous. Not to worry, too bad for those breadheads at CRC Press (remember how they treated Weisstein a few years back?) that I already have a bootleg version of their encyclopaedia. |
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Wow! Have you downloaded all the 11,599 entries, 109,439 cross-references, 5,804 figures, 276 animated graphics, 1,028 live Java applets? 
Bottom line: awesome radicals, THUD&BLUNDER!
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ThudnBlunder
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Re: Nested Radicals
« Reply #18 on: Jan 22nd, 2004, 7:36am » |
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Quote:| I really don’t understand what are the angles your consider. |
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Those angles should have been sin{[pi]/9} and cos{[pi]/9}.
Hmm...I didn't think of using that multiple-angle formula.
In fact, by inspection 2sin{[pi]/18} satisfies
[(2 - x2)2 - 2]2 - 2 = x.......(6)
This factorises to
(x + 1)(x - 2)(x3 - 3x + 1)(x3 + x2 - 2x - 1) = 0
(6) implies that z2 = x + 2 and that x = w
But I still can't see why (5) doesn't give the x3 - 3x + 1 factor.
Quote:| Wow! Have you downloaded all the 11,599 entries, 109,439 cross-references, 5,804 figures, 276 animated graphics, 1,028 live Java applets? |
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No, I downloaded a 100Mb zipped file from a Chinese student's website.
There are 62,205 files (mainly small formula gifs) and they take about an hour to burn onto a CD!
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| « Last Edit: Jan 22nd, 2004, 6:36pm by ThudnBlunder » |
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Barukh
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After a careful inspection, I think the problem is here:
on Dec 30th, 2003, 1:20pm, THUDandBLUNDER wrote:| 10] sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 - sqrt(2 - sqrt(2 + ...... |
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That’s not the infinite sequence for 2sin( [pi]/18 )! Look at the sequence of signs in the nested radicals: +-++--++--… - it has a period of 4. The correct one has a period of 3 – here’s the Mathworld’s page for those who can see it and the attached snapshot for those who cannot  . That leads to your formula (6) and the factorization:
on Jan 22nd, 2004, 7:36am, THUDandBLUNDER wrote:| (x + 1)(x - 2)(x3 - 3x + 1)(x3 + x2 - 2x - 1) = 0 |
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The positive roots of this are: 2; 0.347… and 1.246… (from the first cubic); 1.532… (from the second cubic). Simple considerations show that the sought result must be less than 1, leaving the only possible solution.
Want to guess who is "responsible" for this problem in the first place? 
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| « Last Edit: Jan 25th, 2004, 10:49am by Barukh » |
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ThudnBlunder
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Re: Nested Radicals
« Reply #20 on: Jan 25th, 2004, 7:10pm » |
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Yes, very careless of me.  At least my name means something round here.
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Barukh
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Re: Nested Radicals
« Reply #21 on: Jan 31st, 2004, 10:39am » |
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on Jan 25th, 2004, 10:47am, Barukh wrote:| Want to guess who is "responsible" for this problem in the first place? |
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As no guesses arrived: in 1914, Ramanujan posed this problem (in a more general form) to the Journal of the Indian Math Society.
(TenaliRaman, this time it's not Euler, but you should be proud!  )
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Sameer
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Re: Nested Radicals
« Reply #22 on: Feb 12th, 2004, 10:45am » |
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on Jan 31st, 2004, 10:39am, Barukh wrote:
(TenaliRaman, this time it's not Euler, but you should be proud! )
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So should be I and I am...
Btw isn't the answer to first problem a golden ratio? hmm...
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TenaliRaman
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Re: Nested Radicals
« Reply #23 on: Feb 18th, 2004, 9:29pm » |
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on Jan 31st, 2004, 10:39am, Barukh wrote:
(TenaliRaman, this time it's not Euler, but you should be proud! )
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You bet i am!! Ramanujan is my childhood hero! I could not read up on any of his works unfortunately. The first time i came to know him was through a small article on my maths textbook way back in 4th std. His life history sort of inspired me and maths being my fav subject , i was determined to make it my very own. Alas, i was destined to do something else but that has not stopped him from inspiring me.
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