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November 24

[cis (x)]^n = cis (nx)?

I've recently proved that [cis (x)]^n = cis (nx) (where cis(x) = cos(x)+isin(x)) in class. Does the equation [f(x)]^n = f(nx) hold for any other trigonometric function? What is the point of doing the proof anyway? How is this helpful? 24.92.78.167 (talk) 03:56, 24 November 2010 (UTC)[reply]

It is helpful in that exponential identities are algebraically much simpler than trigonometric identities. You'll see this if you study Fourier series in any depth. The equation holds for (for example) f(x) = cis(6x), etc. But whether there are any others besides these trivial modifications is a subtler question. I think one can show that there are no others that are continuous, and the ones that are not continuous are freaks of no particular interest. Michael Hardy (talk) 04:12, 24 November 2010 (UTC)[reply]

You need the axiom of choice to exhibit a noncontinuous linear function. Robinh (talk) 08:24, 24 November 2010 (UTC)[reply]

The fact that cis(x)^n = cis(nx) is de Moivre's theorem, and is of use when considering powers and roots of complex numbers. For example, computing (1 + i√3)²¹ is simple with de Movire's theorem, but a nightmare with the binomial theorem. (In answer to your question "How is this helpful?") —Anonymous Dissident^Talk 08:44, 24 November 2010 (UTC)[reply]

A generalization also allows you to interpret complex number multiplication geometrically. Take two complex numbers in the complex plane. Multiplying them is the same as adding their angles from the x-axis, assuming their lengths are both 1. In the special case that the complex numbers in question are equal, induction gives your theorem immediately from this interpretation. 67.158.43.41 (talk) 20:06, 24 November 2010 (UTC)[reply]

algebra

if a^2+2b=7, b^2+4c=-7, & c^2+6a=-14 then value of (a^2+b^2+c^2)is ?????????// —Preceding unsigned comment added by Shubham81095 (talk • contribs) 05:23, 24 November 2010 (UTC)[reply]

Can you tell us what you've tried so far, or where you're stuck? Also, I'm curious what the source of the problem is, I've seen it online but with no attribution. Eric. 82.139.80.252 (talk) 08:22, 24 November 2010 (UTC)[reply]

I get a trial and error solution of a=-3, b=-1, c=-2 giving a²+b²+c²=14. I don't know if that's the only solution though.--RDBury (talk) 09:43, 24 November 2010 (UTC)[reply]

Wolfram Alpha thinks it is (assuming we're staying within the real numbers). —Anonymous Dissident^Talk 09:54, 24 November 2010 (UTC)[reply]

The above is the only solution in real numbers, there should be up to 6 complex solutions though.--RDBury (talk) 09:58, 24 November 2010 (UTC)[reply]

Never mind, I see Alpha gives the complex roots as well. Looks like a²+b²+c² is different in each case, three pairs of conjugate values. So the answer appears to be 14 but only if you assume real numbers.--RDBury (talk) 10:18, 24 November 2010 (UTC)[reply]

I found the answer immediately by googling "a^2+2b=7". As our OP seems unlikely to return, I'll just say that if one adds the equations and completes the square, one gets

(a+3)^{2}+(b+1)^{2}+(c+2)^{2}=0

. So that is the only real solution. Eric. 82.139.80.252 (talk) 15:49, 24 November 2010 (UTC)[reply]

Eliminating b and c from a^2+2b−7=b^2+4c+7=c^2+6a+14=0 gives the following equation of degree 8.

a^8−28a^6+350a^4−2156a^2+1536a+9513 = 0

Alpha factors this to be

(a+3)^2 ((a-2)^2+3) (a^4−2a^3−16a^2+10a+151) = 0

The real solution a=−3 is a double root. The second factor has a pair of complex conjugate roots. The third factor has two pairs of complex conjugate roots. Bo Jacoby (talk) 22:55, 24 November 2010 (UTC).[reply]

Probability distribution for gestational length in human pregnancy

My wife is due to give birth today...

What is the probability distribution for gestational length? They give you an expected date of delivery and they tell you that "normal" births are regarded as being between 37 and 42 weeks (with births normally being induced at 42 weeks so that will affect the distribution). But how does that relate to the statistics? Is the due date actually the mean?

And then, just to make things more complicated, how are things affected by previous births? Our first child was a week late. Does that change the probabilities for my wife?

I promise not to talk to my wife about this once she's in labour.

Darkhorse06 (talk) 20:05, 24 November 2010 (UTC)[reply]

I googled "gestational duration distribution", and the fifth hit was a pdf "A re-look at the duration of human pregnancy" which states that the mean is 272.1 days and the standard deviation is 9 days. Yes, this will be affected by previous births, but probably not by much. If you want more information you should try a biologist at the science desk.... Eric. 82.139.80.252 (talk) 21:58, 24 November 2010 (UTC)[reply]