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Nested radicals

  1. Jan 22, 2010 #1
    I've come across this before, but for the life of me can't seem to get anywhere with it. I can plug this into technology and get an answer (and then move on...), but I want to know how to do it without.

    An example problem:

    [tex]4\sqrt{\sin(\pi/4)}[/tex]

    I get as far as

    [tex]2\sqrt{2\sqrt{2}} = 2(2(2)^\frac{1}{2})^\frac{1}{2}[/tex]

    but then I run out of steam and don't see the next step. Surely it is something simple that I am missing, but I just don't see it.

    Thanks for any tips. Cheers!
     
  2. jcsd
  3. Jan 22, 2010 #2

    Char. Limit

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    Gold Member

    I don't know what you're asking, but since [tex]2\sqrt{2}=\sqrt{8}[/tex], it equals two times the quartic root of 8...
     
  4. Jan 22, 2010 #3
    I have [tex]\sqrt{8}[/tex] as an intermediate step in my work.

    [tex]

    4\sqrt{\sin(\pi/4)} = \sqrt{\frac{16\sqrt{2}}{2}} = \sqrt{8\sqrt{2}} = \sqrt{4*2\sqrt{2}} = 2\sqrt{2\sqrt{2}}

    [/tex]

    When I plug into technology, I get the answer:

    [tex]

    2*2^\frac{3}{4}

    [/tex]

    but I don't know how to get there. That's what I'm asking; sorry for being unclear.
     
  5. Jan 22, 2010 #4

    Char. Limit

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    Gold Member

    Well, yeah.

    Going with my two times quartic root of 8 solution, since 8 is 2 cubed, the quartic root of 8 is 2 to the three-fourths power, and the number simplifies to your number.
     
  6. Jan 22, 2010 #5
    Of course. Thank you very much! I definitely should've seen that! :)

    Cheers.
     
  7. Jan 22, 2010 #6
    When you multiply a two numbers with the same base, the exponents add. So,

    [tex] 2(2^{1}*2^{\frac{1}{2}})^{\frac{1}{2}} = 2(2^{1+\frac{1}{2}})^{\frac{1}{2}} = 2(2^{\frac{3}{2}*\frac{1}{2}}) = 2*2^{\frac{3}{4}} [/tex]
     
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