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October 6

I understand Complex Numbers

Complex number used to be really hard. But I understand it now. It's is just ScaleRot. Scaling and Rotation.
4+3I is just PolarCoor( 5 , 36.86 deg ) or ScaleRot( 5 , 36.86 deg )
So the square of (4+3i) is just scale the unit number (1) by 5 twice and rotate by 36.86 degrees twice.
1*5*5 = 25
0 degree + 36.86 degree + 36.86 degree = 73.72 degree

So the answer is just ScaleRot( 25 , 73.72 degree ) or PolarCoor( 25 , 73.72 degree ) or 7.008 + 23.997i

Why does the maths teacher at school makes it seems so difficult??? 220.239.51.150 (talk) 02:51, 6 October 2013 (UTC)[reply]

We don't know how your teacher taught it, or why this way seemed difficult to you.

But while it's true that multiplying by a complex number scales and rotates, I wouldn't say that complex numbers are just that, it's deeper than that. Perhaps the teacher was trying to present a more complete view of complex numbers. -- Meni Rosenfeld (talk) 06:26, 6 October 2013 (UTC)[reply]

Is it deeper than that? Complex numbers is just a convenient way of expressing the 2 by 2 matrix group whose Lie algebra is given by the generators

I_{2}

(1 in complex numbers) and

\sigma _{z}

(i in complex numbers) which, as pointed out above, is exactly the same group of continuous scaling and rotation operations in 2D. They are one and the same thing. — Preceding unsigned comment added by 109.144.255.136 (talk) 14:40, 6 October 2013 (UTC)[reply]

That captures only the multiplicative structure. They also have an additive structure and various sorts of topological/differential structure, and these interact in deep ways. --Trovatore (talk) 22:49, 6 October 2013 (UTC)[reply]

Should we say that the square root of 4+3I is just ScaleRot(sqrt(5), 18.43 degrees) ? Pldx1 (talk) 17:02, 10 October 2013 (UTC)[reply]

4 + 3 I = ScaleRot( 5 , 36.87 deg)

squareroot of (4 + 3 I) is ScaleRot( A , B deg )

Therefore

1 * A * A = 5 so A=sqrt(5) as A>= 0 (A cannot be negative)

For B , we have to consider all possibilities

0 + B + B = 36.87 deg so B = 18.44 deg

0 + B + B = 360 deg + 36.87 deg = 396.87 deg so B = 198.44 deg

0 + B + B = 360 deg + 360 deg + 36.87 deg = 756.87 deg so B = 378.44 deg = 360 deg + 18.44 deg = 18.44 deg

0 + B + B = 360 deg + 360 deg + 360 deg + 36.87 deg = 1116.87 deg so B = 558.44 deg = 360 deg + 198.44 deg = 198.44 deg

and so on and so forth.

You will notice that B can only have two possible values 18.44 deg or 198.44 deg

So the squareroot of (4 + 3 I) is ScaleRot( sqrt(5) , 18.44 deg ) or ScaleRot( sqrt(5) , 198.44 deg )

202.177.218.59 (talk) 04:36, 11 October 2013 (UTC)[reply]

Symbol in the center of a commutative diagram?

In the article Banach bundle, all the commutative diagrams have what look to be small copyright symbols in the centers of them. Is this meant to convey that the diagrams commute? If so, is this notation standard? It is new to me. --Mark viking (talk) 03:46, 6 October 2013 (UTC)[reply]

While not a copyright sign, this [1] mentions putting a symbol in the center to indicate the diagram commutes. However, I didn't find this notation anywhere else- and I don't think it's that good of an idea- so it may be unrelated, but it's the only like thing I could find. A lot of the things I work on, and read, involve heavy use of category theory, and I don't think I've ever seen diagrams marked like that. At the very least, if it means something, it should be explained in the article, maybe we can track down who put the diagrams in?Phoenixia1177 (talk) 05:58, 6 October 2013 (UTC)[reply]

It was User:Sullivan.t.j, and I've left a note on his/her talk page. Rojomoke (talk) 10:55, 6 October 2013 (UTC)[reply]

Yes, the c-in-a-circle denotes commutativity, not copyright. I think that there mathematically needs to be some indication that a diagram commutes, whether that's done (a) globally, by saying "all diagrams on this page / in this article commute", (b) locally in the text, by saying "the following diagram commutes", or (c) locally in the diagram itself, as done by the "copyright sign". I'm not particularly wedded to any one way of achieving that aim, and I've seen all three used in practice. Sullivan.t.j (talk) 16:16, 6 October 2013 (UTC)[reply]

I'm used to a little circle, not a copyright sign. That's what Serge Lang uses, if I recall correctly. I think this is less confusing, because the copyright sign might be interpreted as an assertion of copyright. (But only a little less confusing, as it still has to be explained, and anyway in practice one never draws a non-commutative diagram.) --Trovatore (talk) 21:55, 6 October 2013 (UTC)[reply]

I would like to thank you all for explaining the notation and giving insight about its context. The discussion is ongoing, but my particular questions are well resolved. --Mark viking (talk) 22:44, 6 October 2013 (UTC)[reply]

Hopefully nothing went amiss on this wind energy assignment...

Without access to a tutor or the instructor this late at night, I figured that the only way to have the assignment checked was by placing it here. The rules of the RD asks to do our own homework. I already did. I'd just now like for someone to check whether they're all correct, or if I forgot to plug something into a formula here and there, or whether I used the right formula. I do feel iffy on the airplane problem (Problem #3.)

Problems 1-4; P. 45

1. The blade of a wind turbine experiences a force of 100 lbs. when the wind speed is 10 ft/sec. To what value does the force change if under the same conditions the wind speed increases to 15 ft/sec?

a. First force = 100 lbs., First wind speed = 10 ft / sec, Second wind speed = 15 ft / sec
b. 15 ft/sec = Force of 10 ft/sec * (15/10)^2
c. 15 ft/sec = (100)(15/10)^2 = (100)(2.25)
d. = 225 lbs of force at the wind speed of 15 ft/sec

2. If the lift and drag coefficients in an airfoil are 0.998 and 0.05, respectively, determine by how many times the lift force is larger than the drag force in a blade with that airfoil profile.

a. Lift coefficient = .998, Drag coefficient = .05
b. .998 / .05
c. 19.96
d. The lift force is 19.96x larger than the drag force in a blade with this airfoil profile.

3. If a small 3200-lb airplane has two wings with the airfoil profile in problem 2, and the drag force on each wing at a given wind speed is 100 lbs., what is the net lift force on the plane?

a. Drag coefficient = .05, Lift coefficient = .998
b. Lift coefficient is 19.96x larger than the Drag coefficient
c. 100 lbs * 19.96 = 1996 lbs lift force
d. 1996 lbs lift force – 3200 lbs plane weight = 1204 lbs shortfall (Net lift force = -1204 lbs)

4. If a turbine has three blades and the center of pressure (where the lift and drag forces apply) of each blade is 10 ft from the axis of the turbine and the lift force on each blade is 100 lbs., how much is the torque that rotates the turbine?

a. Lift force = 100 lbs; distance from center of gravity on each blade = 10 feet. (Blade lengths = ~20 feet each)
b. 100 lbs * 10 ft = 1000 lb-ft of torque for each blade.
c. 1000 lb-ft of torque per blade * 3 blades = 3000
d. 3000 lb-ft of total torque rotates the turbine

But hopefully everything checks out. --2602:30A:2EE6:8600:9C3C:99E3:B76F:BDBA (talk) 05:30, 6 October 2013 (UTC)[reply]

In 3, I think you have done the lift force calculation for one wing - but we are told the plane has two wings ... Gandalf61 (talk) 07:31, 7 October 2013 (UTC)[reply]

Data distribution

Assuming that data is normally distributed, what percentage of that data would be expected to lie within three standard deviations of the mean (i.e., the mean +/– three standard deviations)? Thanks. Joseph A. Spadaro (talk) 17:04, 6 October 2013 (UTC)[reply]

See Normal distribution#Standard deviation and tolerance intervals. Duoduoduo (talk) 17:59, 6 October 2013 (UTC)[reply]

Perfect. Thanks! Joseph A. Spadaro (talk) 21:05, 6 October 2013 (UTC)[reply]

Diffusion Model

Hello. I have a diffusion equation modelled by the Gaussian function. $f(x)={\frac {Ae^{-{\frac {x^{2}}{4Dt}}}}{8({\sqrt {\pi Dt}})^{3}}}$ A and D are constants. I want to find the distance from the source where the density of particles is the greatest. Given the time when the density is greatest, why do I take the partial derivative with respect to t instead of x? Thanks in advance. — Preceding unsigned comment added by Mayfare (talk • contribs) 22:45, 6 October 2013 (UTC)[reply]

The gaussian function is a solution to a diffusion equation without source, so your question is confusing. Bo Jacoby (talk) 18:30, 7 October 2013 (UTC).[reply]