Results 1 to 4 of 4

Math Help - Little Question

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    36

    Little Question

    if x+\frac{1}{x}=-1.
    What's the value of x^{2007}+\frac{1}{x^{2007}}?

    I try using this:

    x^{2007}+\frac{1}{x^{2007}}=\left(x+\frac{1}{x}\ri  ght)^{2007}-\sum_{k=1}^{2006}\binom{2007}{k}x^{2k-2007}

    but i don't find any results.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    On the basis of the identity...

     x + \frac{1}{x} = e^{\ln x} + e^{-\ln x}= 2\cdot \cosh(\ln x) = 2\cdot \cos (i\cdot \ln x) (1)

    ... we can write...

    x + \frac{1}{x} = -1 \rightarrow \ln x = i\cdot \frac{4}{3}\cdot \pi (2)

    Since 2007 devides 3 is \ln x^{2007} = 2007\cdot \ln x = 4\cdot i\cdot k\cdot \pi so that is...

     x^{2007} + x^{-2007} = 2\cdot \cos (4\cdot k \cdot \pi)= 2 (3)

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by streethot View Post
    if x+\frac{1}{x}=-1.
    What's the value of x^{2007}+\frac{1}{x^{2007}}?

    I try using this:

    x^{2007}+\frac{1}{x^{2007}}=\left(x+\frac{1}{x}\ri  ght)^{2007}-\sum_{k=1}^{2006}\binom{2007}{k}x^{2k-2007}

    but i don't find any results.
    If:

    x+1/x=-1

    then x=-1/2\pm \sqrt{3}/2=e^{2\pi i/3}\text{ or }e^{4\pi i/3}

    Then as 2007=3\times 669 we have x^{2007}=e^{1338 \pi i} \text{ or }e^{2676 \pi i} =1 and so:

     <br />
x^{2007}+x^{-2007}=1+1=2<br />

    cb
    Last edited by CaptainBlack; August 4th 2009 at 02:13 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2009
    Posts
    36
    Then as we have and so:
    (e^{\frac{2i\pi}{3}})^{2007}=e^{i1338\pi}

    and

    (e^{\frac{4i\pi}{3}})^{2007}=e^{i2676\pi}
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum