Results 1 to 5 of 5

Math Help - imagionary numbers and proving

  1. #1
    Newbie
    Joined
    Jan 2006
    Posts
    4

    imagionary numbers and proving

    This is the problem:

     (1+((1+i)/2)) (1+((1+i)/2)^2) (1+{((1+i)/2)^2}^2)... (1+{((1+i)/2)^2}^n) =  (1+i)(1-1/{2^2}^n)

    it is given that n must be >=2


    I tried to solve things by ofcourse using  i^2=\sqrt{-1} and that sorting and calculating to see what will I get but I just cant get anything usseful,so if u have some ideas please reply.
    Attached Thumbnails Attached Thumbnails imagionary numbers and proving-zad1.gif  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Where do you come up with such problems? I love them, you already posted a similar one in the trigonometry section.
    I did not prove it yet, but again I recommend mathematical induction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2006
    Posts
    4
    Thank you for trying to help.Your answer on my previous question was great.And yes,mathematical inuduction could probably help here,too.I've tried to solve it with different aproach,by trying to make some combinations of geometrical rows,but that didnt really help. I'l try again so if I come up with something i will post.

    You want to know where I find these problems?Ask my teacher
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let me just say something about your first post.
    The relationship i=\sqrt{-1} is not entirely correct. Let me explain, we define \sqrt{n}=x where x\geq 0 for n\geq 0 as a real number such as x^2=n. Now we can prove that such a real number always exists and is unique.

    Now the problem with defining \sqrt{-1} is that the complex-numbers have no sense of positive or negative. Mathematically we say that they lack ordering thus we cannot define what it means a positive complex or negative complex number as in the first paragraph.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by mahadeva
    This is the problem:

     (1+((1+i)/2)) (1+((1+i)/2)^2) (1+{((1+i)/2)^2}^2)... (1+{((1+i)/2)^2}^n) =  (1+i)(1-1/{2^2}^n)

    it is given that n must be >=2


    I tried to solve things by ofcourse using  i^2=\sqrt{-1} and that sorting and calculating to see what will I get but I just cant get anything usseful,so if u have some ideas please reply.
    The first thing you have to do is show that:

    \left(\frac{1+i}{2} \right)^{2^n}=\frac{1}{2^{2^{n-1}}},

    for n>2. We can do this by induction, starting from

    \left(\frac{1+i}{2} \right)^{2^3}=\left(\frac{1+i}{2} \right)^{8}=\frac{1}{16}}=\frac{1}{2^{2^{3-1}}},

    I will leave the rest of the induction for this result to the reader.

    Now the main problem:

     (1+((1+i)/2)) (1+((1+i)/2)^2) (1+{((1+i)/2)^2}^2)... (1+{((1+i)/2)^2}^n) =  (1+i)(1-1/{2^2}^n)

    is true for n=2. Now if it is true for some m \geq 2:

     (1+((1+i)/2)) (1+((1+i)/2)^2) (1+{((1+i)/2)^2}^2)... (1+{((1+i)/2)^2}^{m+1}) =  (1+i)(1-1/{2^2}^m) (1+{((1+i)/2)^2}^{m+1}).

    The RHS of this may be rewritten using our earlier result:

    (1+i)\left(1-\frac{1}{2^{2^m}}\right) \left(1+\frac{1}{2^{2^m}}\right)= (1+i)\left(1-\frac{1}{2^{2^{m+1}}}\right) ,

    Hence the identity to be proven is true for m+1, which is sufficient to complete the induction.

    RonL
    Last edited by CaptainBlack; January 27th 2006 at 05:40 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving complex numbers
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 9th 2010, 10:55 AM
  2. Proving numbers in sequences and summation
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 16th 2009, 03:13 PM
  3. proving for real positive numbers
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: July 23rd 2009, 12:34 AM
  4. Proving Lucas Numbers and Fibonacci Numbers
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 18th 2008, 11:33 PM
  5. Proving Bernoulli numbers eq'n please help
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: May 22nd 2007, 08:00 PM

Search Tags


/mathhelpforum @mathhelpforum