Results 1 to 3 of 3

Thread: Simplifying Ugly Radical Expressions from Cardano's Method

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1

    Simplifying Ugly Radical Expressions from Cardano's Method

    I asked a similar question to this some time ago, but as we have gained several new specialists since then, I thought I'd try again.

    The problem comes from solving the equation
    $\displaystyle x^3 - 7x + 6 = 0$
    by Cardano's method.

    I get an expression for x of the form:
    $\displaystyle x = 2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right )$

    As it happens,
    $\displaystyle 2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right ) = 3$

    but how can I prove that? Is there any way to simplify this monster without knowing the answer ahead of time?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    I am not sure what you are asking.
    But sometimes I get expressions which look like,

    $\displaystyle \sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}$.

    And in general there is no way to simplify it, you just leave it like that. Part of the difficulty in the word "simplify" is that it is not well-defined, and hence we cannot come up with a theory which tells us when we can "simplify" and when we cannot. Perhaps you can create a definition.

    And, I also do not understand how you got trigonometric functions in your answer. I never remember learning it that.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    I am not sure what you are asking.
    But sometimes I get expressions which look like,

    $\displaystyle \sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}$.

    And in general there is no way to simplify it, you just leave it like that. Part of the difficulty in the word "simplify" is that it is not well-defined, and hence we cannot come up with a theory which tells us when we can "simplify" and when we cannot. Perhaps you can create a definition.

    And, I also do not understand how you got trigonometric functions in your answer. I never remember learning it that.
    The answer is that I would like to see if there is a way I can know that
    $\displaystyle 2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right ) = 3$
    without resorting to a decimal approximation.

    I did make a mistake in my derivation. (That's what I get for not writing it out in full.)

    As to how I got that form:
    $\displaystyle x^3 - 7x + 6 = 0$

    Let $\displaystyle x = p - q$ and $\displaystyle q = -\frac{7}{3p}$

    Then

    $\displaystyle p^3 + \frac{343}{27p^3} + 6 = 0$

    $\displaystyle 27p^6 + 162p^3 + 343 = 0$

    $\displaystyle p^3 = -3 + i \frac{10\sqrt{3}}{9}$ is one solution.

    Now put $\displaystyle p^3$ into complex polar form. (This will "simplify" the eventual answer for x.)

    $\displaystyle p^3 = \frac{7\sqrt{21}}{9} \cdot e^{i \cdot atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{ \pi}{2}}$

    So
    $\displaystyle p = \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}}$


    Thus
    $\displaystyle q = -\frac{7}{3p} = -\frac{7}{3} \sqrt[3]{\frac{9}{7\sqrt{21}}}
    e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}$

    $\displaystyle q = -\sqrt[3]{\frac{7\sqrt{21}}{9}} e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}$

    Thus
    $\displaystyle x = p - q = \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}} + \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}$

    $\displaystyle x = 2 \sqrt[3]{\frac{7\sqrt{21}}{9}} \left ( \frac{ e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}} +
    e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}}{2} \right )$<-- This is the reason for the complex polar form

    $\displaystyle x = 2 \sqrt[3]{\frac{7\sqrt{21}}{9}} cos \left (
    \frac{1}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + \frac{\pi}{6} \right )$

    $\displaystyle x = 2$

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Oct 15th 2011, 06:21 AM
  2. Simplifying radical expressions.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 19th 2009, 04:31 AM
  3. Simplifying Radical Expressions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jul 30th 2008, 10:31 AM
  4. Simplifying radical expressions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 21st 2008, 12:37 AM
  5. Simplifying radical expressions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 18th 2008, 05:26 PM

Search Tags


/mathhelpforum @mathhelpforum