imagionary numbers and proving

**mahadeva** · January 25th 2006, 01:23 PM

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.

**ThePerfectHacker** · January 25th 2006, 02:31 PM

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.

**mahadeva** · January 26th 2006, 03:25 AM

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

**ThePerfectHacker** · January 26th 2006, 06:26 PM

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.

**CaptainBlack** · January 26th 2006, 11:47 PM

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

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