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ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}\]I'm trying to do this problem.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3one way is to write it as:\[x=\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+...}}}}}\]and then square both sides to get:\[x^2=2+\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+...}}}}}=2+x\]then just solve the quadratic

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3unless I have mis understood your question?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Actually, I'd just want to know how to denest \(\sqrt{2 + \sqrt2}\)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3I guess you could use the same technique, i.e:\[x=\sqrt{2+\sqrt{2}}\]\[x^2=2+\sqrt{2}\]\[x^22=\sqrt{2}\]then square both sides and solve the resulting quartic equation

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3which should end up being just a quadratic in \(x^2\)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Ah,\[(x^2  2)^2 = 2\]\[x^4 4x^2 +4 = 2 \]\[x^4  4x^2  2 = 0\]Assume \(y = x^2\).\[y^2  4y  2 = 0\]\[y = 2\pm \sqrt{6}\]\[x = \sqrt{2 \pm \sqrt6}\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0@asnaseer But that again brings up a nested radical.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3I'm not sure what exactly you mean by "denest" here?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Well, a radical expression containing another radical expression.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3btw: you have a small sign error in your calculation above

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Oh, that was supposed to be a +... not .

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3are you trying to simply \(\sqrt{2+\sqrt{2}}\) further?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, I want to remove the inner radical.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0http://math.stackexchange.com/questions/196155/strategiestodenestnestedradicals I can't get Bill Dubuque's technique to work here. The motivation behind this question is... I want to learn something new. :)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3so you are trying to express this as:\[\sqrt{2+\sqrt{2}}=a\sqrt{b}\]and what to find a and b?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, I want this in the form \(a + b\sqrt n \), not \(\sqrt{a + b\sqrt n }\)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3ok, I see now  let me think over this. I am at work now so will probably get back to this after work  looks like an interesting problem. :)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you!\[2 + \sqrt 2\]Here a = 2, b = 1, n = 2. So,\[\sqrt{\text{parth}} = \sqrt{a^2  nb^2 } = \sqrt{4  2} = \sqrt{2}\]Subtracting the above,\[2 + \sqrt{2}  \sqrt{2} = 2\]Now, find \(\sqrt{\text{kohli} } = \sqrt{2a} = \sqrt{4} = 2\). But that yields \(1\).

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3ok, after some research, I found out that this is actually quite a difficult problem. however, there are some special cases that can be solved for quite easily. see this wiki article for details: http://en.wikipedia.org/wiki/Nested_radical the article mentions: "In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested. Earlier algorithms worked in some cases but not others."

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0WELL:\[\sqrt{2+\sqrt{2}} \] cannot be expressed in the forms dw:1348223973261:dw

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Thank God! I was so confused.

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0But I think it can be expressed in some form likedw:1348224135337:dw
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