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July 23

complex number

if 1,a and a².Find the values of 1+a² and 1+2a+2a² —Preceding unsigned comment added by 212.100.68.103 (talk) 11:27, 23 July 2008 (UTC)[reply]

if 1, a and a² what? If this is homework, please see the notice at the top of the page. AndrewWTaylor (talk) 11:32, 23 July 2008 (UTC)[reply]

If a is i then

1+a^2 is nothing

1+2a+2a^2 is one less than two imaginary things

But surely you know this and need no help from us. 122.107.182.1 (talk) 10:45, 24 July 2008 (UTC)[reply]

Expectation of heads and tails

N: number of tosses. $N_{H}$ : number of heads. < $N_{H}$ >: expected heads.

We also expect an actual $N_{H}$ to deviate from ${\frac {N}{2}}$ by about ${\frac {\sqrt {N}}{2}}$ or the fraction to deviate by:

${\frac {1}{N}}{\frac {\sqrt {N}}{2}}={\frac {1}{2{\sqrt {N}}}}$

The larger N is, the closer we expect the fraction ${\frac {N_{H}}{N}}$ to be to one-half.

And the question is: how do we arrive at these values of ${\frac {\sqrt {N}}{2}}$ and ${\frac {1}{N}}{\frac {\sqrt {N}}{2}}={\frac {1}{2{\sqrt {N}}}}$

83.40.249.61 (talk) 11:41, 23 July 2008 (UTC)[reply]

The results of a coin toss follow a Bernoulli distribution, with

p=q={\tfrac {1}{2}}

. The standard deviation of such a distribution is

\sigma ={\sqrt {pq}}={\sqrt {\tfrac {1}{4}}}={\tfrac {1}{2}}

, so the expected deviation in a set of

N

samples is

{\sqrt {N\sigma ^{2}}}={\sqrt {Npq}}={\sqrt {\tfrac {N}{4}}}={\tfrac {\sqrt {N}}{2}}

. The second part of your question is just simple multiplication and cancellation of terms; try rewriting the

{\tfrac {1}{N}}

part as

{\tfrac {1}{N}}={\tfrac {1}{{\sqrt {N}}{\sqrt {N}}}}

and you should be able to see it then. --_{tiny plastic} Grey Knight ⊖

The standard deviation of the head count is

{\begin{array}{ll}&{\sqrt {\sum _{k=0}^{N}(k-{\frac {N}{2}})^{2}\,p(k{\mbox{ heads}})}}\\=&{\sqrt {{\frac {1}{2^{N}}}\sum _{k=0}^{N}\left(k-{\frac {N}{2}}\right)^{2}{\binom {N}{k}}}}\\=&{\sqrt {\left({\frac {N}{2}}\right)^{2}-{\frac {1}{2^{N}}}\sum _{k=0}^{N}k(N-k){\binom {N}{k}}}}\\=&{\sqrt {\left({\frac {N}{2}}\right)^{2}-{\frac {N}{2^{N}}}\sum _{k=0}^{N-1}k{\binom {N-1}{k}}}}\\=&{\sqrt {\left({\frac {N}{2}}\right)^{2}-{\frac {N}{2}}({\mbox{expected heads in N-1 tosses}})}}\\=&{\sqrt {\frac {N}{4}}}.\end{array}}

-- BenRG (talk) 14:44, 23 July 2008 (UTC)[reply]

NB: in that,

p(k{\mbox{ heads}})

is "the probability of getting

k

heads", which is expressed in terms of the binomial coefficient as

{\tfrac {1}{2^{N}}}{\tbinom {N}{k}}

, and of course

({\mbox{expected heads in N-1 tosses}})

is just

\langle N-1\rangle

in your notation above.

\Sigma

is for summation if you didn't already know that. --_{tiny plastic} Grey Knight ⊖ 15:03, 23 July 2008 (UTC)[reply]

Summing sinusoids

A QPSK radio carrier can be represented as $a=sine(W+m*pi/2)$ where $m$ = 0, 1, 2 or 3. Clearly the peak amplitude of this carrier is | $a$ | = 1.

Suppose we have n such carriers with angular frequencies $W=1,2,....n.$ Sum them all together in a transmitter. I have two questions:

What peak amplitude must the transmitter handle?
How much can the peak amplitude be reduced by introducing any arbitrary constant phase offset(s) $P$ _n on particular carrier(s) so

$a$ _n $=sine(W$ _n $+m*pi/2+P$ _n $)$ ? Cuddlyable3 (talk) 12:34, 23 July 2008 (UTC)[reply]

This sounds horribly like homework. Anyway, if the peaks all line up then the amplitude will be maximum, as will happen if the valleys line up. If you can phase things so the peaks line up with the valleys then you get less amplitude. If you want to see the effects, calculate a couple of cycles in excel with the phases being variable. -- SGBailey (talk) 23:09, 23 July 2008 (UTC)[reply]

My question concerns the crest factor of OFDM and is a real-world engineering challenge. It is true as SGBailey observes that if peaks of all the n carriers should coincide then the peak amplitude is n (times that of one carrier). That requires a transmitter to handle n² power which is very expensive. I want ("the world needs") the nearest way to "phase things so the peaks line up with the valleys", hopefully given as a formula rather than an invitation to experiment in excel. To keep a lid on this problem note that a) the carrier frequencies must be 1, 2, ... n, and b) each carrier is independently modulated in 90 degree phase steps i.e. QPSK. These phase steps correspond to different time intervals on each carrier. (There are some ideas afoot of breaking rule a) and/or b) to suppress the peak but that may be the subject of patenting and therefore not for discussion here.)Cuddlyable3 (talk) 20:16, 27 July 2008 (UTC)[reply]

Can you clarify your question? You call W a frequency, but it looks more like a phase shift. Are you evaluating the

\sin

only at finitely many points, or are you interested in the maximal value of the sum over the entire period? A question that makes sense to me is to minimize

\operatorname {max} {\Bigl \{}\sum _{m=1}^{n}\sin(m\,x+p_{m}):x\in [0,2\,\pi ]{\Bigr \}}

over all choices of numbers

p_{1},p_{2},\dots ,p_{n}

. I can say a few things about that problem. Is that what you mean? Oded (talk) 22:11, 27 July 2008 (UTC)[reply]

Sorry I meant $W_{n}=2\pi f_{n}t$ . Hopefully clarifying in your better notation Oded my questions are:

Evaluate

$\operatorname {max} {\Bigl \{}\sum _{k=1,2,..}^{N}\sin(k\,x+.5m\pi ):x\in [0,2\,\pi ];m=[0,1,2,3]{\Bigr \}}$

Here the variable $m$ is information that can take 4 discrete values independently on each carrier $k$ and $x$ is continuous.

Minimise

$\operatorname {max} {\Bigl \{}\sum _{k=1,2,..}^{N}\sin(k\,x+.5m\pi +P_{k}):x\in [0,2\,\pi ];m=[0,1,2,3]{\Bigr \}}$ by well chosen constants $P_{k}$ .

For a satellite link that is strictly power limited, values of N = 2 to 16 are actual while some terrestrial broadcasting uses much higher values. An attraction of OFDM is that all the carriers can be simultaneously modulated or demodulated by a single FFT process. Cuddlyable3 (talk) 13:04, 30 July 2008 (UTC)[reply]

real analysis

what is the use of real analysis in real life ? —Preceding unsigned comment added by 202.140.53.156 (talk) 14:48, 23 July 2008 (UTC)[reply]

Real analysis is the basis of calculus, which has many applications. Algebraist 15:05, 23 July 2008 (UTC)[reply]

You can use it to look inside bodies without cutting them. This is tomographic reconstruction, which relies on the Radon transform. Similar ideas can be applied to select candidates for mines or oil wells. Michael Slone (talk) 03:13, 24 July 2008 (UTC)[reply]

Numerical analysis relies fundamentally on real analysis. If you are trying to solve a partial differential equation by (for instance) a finite element method, then real analysis is required in order to establish convergence of the method. Typically, it is essential to know convergence of the method as well as its limitations, at least in some crude sense, before it can properly be used for real-world problems. siℓℓy rabbit (talk) 03:42, 24 July 2008 (UTC)[reply]