Wikipedia:Reference desk/Archives/Science/2016 September 11

= September 11 =

Rotation and poles of the Earth, Solar System, and galaxy
Can someone verify that I'm thinking of this correctly?

Using the right hand rule, the Earth spins counter clockwise with the north pole on top. The Solar System also spins in this same direction and thus the north pole of the Solar System is on the same side of the ecliptic as the north pole of the Earth. The galaxy on the other hand (hehe) spins in the opposite direction, thus the galaxy's south pole is on the same general side of the Solar System's ecliptic as the Solar System and Earth's north pole.

Right? † Dismas †|(talk) 03:46, 11 September 2016 (UTC)


 * The Sun is currently moving towards Hercules therefore the direction of rotation is towards the nearer 90° galactic longitude 0° lat, not the 0°, 270°. That means you're right about it being backwards (the north galactic pole (NGP) is the one in the northern hemisphere) That redirect might satisfy some curiosity on the chirality of the galactic coordinate system. Sagittarian Milky Way (talk) 04:32, 11 September 2016 (UTC)


 * Note btw that the galactic pole and the ecliptic pole are 60&deg;20&prime; apart; that is, the Solar System's plane is tilted that much from the Galactic plane. —Tamfang (talk) 04:52, 11 September 2016 (UTC)


 * Yes it's very random. The north celestial pole and north galactic pole are also a bit over 60° apart. Sagittarian Milky Way (talk) 05:00, 11 September 2016 (UTC)
 * A small correction - the Earth, of course, spins counterclockwise looking from the North pole i.e. from West to East. Ruslik_ Zero 19:19, 11 September 2016 (UTC)
 * What is this a correction to? Sagittarian Milky Way (talk) 19:38, 11 September 2016 (UTC)

These answers make things murkier for me. What if I phrased it this way? Imagine I shoot a gun straight up from the north pole of the Earth. Will that bullet travel into the southern side of the galactic ecliptic? † Dismas †|(talk) 04:16, 12 September 2016 (UTC)


 * Galactic ecliptic is incorrect terminology. You mean galactic equator. It will go to northern galactic latitudes. I think if you're a bit but not super lucky it'll orbit for an eight of a billion years before entering the south of the galaxy (because all two body orbits that aren't perfectly equatorial have a descending node). Also the bullet will halt a mile or so in the air before orbiting our galaxy and you have like 2 seconds to run. If you can aim perfect enough to make it drop like a stone back first instead of an arc it could only hit you at like 65 mph instead of way faster. But that's not what you meant. Sagittarian Milky Way (talk) 05:35, 12 September 2016 (UTC)

Follow up questions: other solar systems
What about other solar systems that we know about? If we assume that the planets that we know of, in these systems, rotate along their systems' ecliptic, can we create statistics about how many systems have their plane as the same plane as the Galaxy, and how many are roughly rotating in the same direction as the galaxy? Or do we already know there is no correlation for either?--Lgriot (talk) 13:00, 13 September 2016 (UTC)
 * I've been told, in reply to a similar question, that there is no correlation between galactic axis of rotation and the axes of individual stars and solar systems. In other words, as far as we know, the axis and direction of rotation of a random star system is random. --Stephan Schulz (talk) 13:59, 13 September 2016 (UTC)
 * The galaxies' orientation is not random though. They prefer to align their planes with the surface of the nearest bubble of cosmic foam. i.e. The nearest non-dwarf galaxy to us looks close to edge-on and is only 21.5729606 degrees from Earth's galactic equator. There's even a supergalactic coordinate system and supergalactic plane that thousands of galaxies correlate to. Sagittarian Milky Way (talk) 23:24, 14 September 2016 (UTC)
 * Thank you, good people!--Lgriot (talk) 11:23, 15 September 2016 (UTC)

Spiraling of the Moon
Does the Moon spiral out of its orbit? Can it leave the Solar System? What is the quantitative description of this phenomenon?--82.137.11.181 (talk) 07:04, 11 September 2016 (UTC)
 * See Orbit of the Moon, Tidal acceleration and Tidal locking. Angular momentum is transferred from the Moon to the Earth through the action of tides, causing the radius of the Moon's orbit to increase and the rotation of the Earth to slow down.  The radius of the orbit is currently increasing at about 40 mm per year.  Eventually (after approximately 50 billion years), the length of the day on the Earth would increase to be the same as the period of the Moon's orbit - the bodies would then be tidally locked, and the system would be stable.  However, the Sun will have become a red giant before then (approximately 5 billion years), expanding so as to destroy both the Earth and the Moon.  Tevildo (talk) 08:12, 11 September 2016 (UTC)
 * Quibble: actually the angular momentum is transferred from Earth to Moon. Wnt (talk) 12:33, 14 September 2016 (UTC)

Ceres (dwarf planet)
What would be the effect of Ceres clearing the rest of the asteroid belt and gaining all that mass? Would it have disastrous effects on the rest of the planets (or at least Mars or Jupiter) due to all that mass now being concentrated in one place instead of spread throughout the belt? † Dismas †|(talk) 13:05, 11 September 2016 (UTC)


 * 1. Lighter than the Moon. 2 No. Mars gets closer to Earth than to Ceres. Sagittarian Milky Way (talk) 13:43, 11 September 2016 (UTC)


 * According to asteroid belt "gravitational perturbations from Jupiter imbued the protoplanets with too much orbital energy for them to accrete into a planet. Collisions became too violent, and instead of fusing together, the planetesimals and most of the protoplanets shattered. As a result, 99.9% of the asteroid belt's original mass was lost in the first 100 million years of the Solar System's history." So, if it could somehow accrete into a planet now, which would seem to require the removal of Jupiter, it would be too small to have much effect.


 * Also according to our article: "The total mass of the asteroid belt is approximately 4% that of the Moon, or 22% that of Pluto, and roughly twice that of Pluto's moon Charon (whose diameter is 1200 km)". If we multiply by 1000 as if Jupiter wasn't there and 99.9% of the mass wasn't scattered away, then we get a mass of 220 times Pluto, or about 2 0.5 Earths.  So, we could have had a significant planet there.   I doubt if that would be enough to have much effect on Jupiter (if we didn't need to remove it to make this planet in the first place), but it might have had a noticeable effect on Mars, depending on the exact orbit, orbital resonance, etc. StuRat (talk) 13:49, 11 September 2016 (UTC)


 * The asteroid belt may have had a significant effect on Jupiter in it's current form. According to, Jupiter has moved towards the sun by about 18 million miles, as a result of ejecting asteroid belt objects into the outer solar system.   StuRat (talk) 13:49, 11 September 2016 (UTC)


 * 220 Pluto mass is more like 0.5 Earth mass. Sagittarian Milky Way (talk) 14:03, 11 September 2016 (UTC)


 * Thanks, math error there (I had the inverse). StuRat (talk) 14:06, 11 September 2016 (UTC)


 * I think the question was what would happen if all current asteroid belt mass is concentrated in Ceres. The answer is nothing worthy of attention. The mass of Ceres is about 1/3 of the total asteroid belt mass and if the mass of Ceres increases by three times it will be still a very small mass. Ruslik_ Zero 19:08, 11 September 2016 (UTC)


 * I agree, but the only way the asteroid belt could accrete into a planet was if there was no Jupiter there, and that would have led to the accretion of the original, much larger mass of the asteroid belt. StuRat (talk) 22:47, 11 September 2016 (UTC)

Thanks! † Dismas †|(talk) 04:17, 12 September 2016 (UTC)

cooking times
I bought a package of frozen kale. The sole listed ingredient is kale. These are the cooking instructions: 1.) Pour frozen kale in saucepan and cover with water. Boil for 3 minutes. 2.) Reduce heat, cover and simmer 20 minutes. My question concerns division of cooking time between step one and step two. Why not just cook it at a certain heat level for a certain period of time, for instance why not boil it for 10 minutes or simmer it for 25 minutes? Bus stop (talk) 16:40, 11 September 2016 (UTC)


 * The rate of cooking is determined almost entirely by the temperature (roughly doubling in rate for every 10 °C increase), so once the food is heated to boiling temperature, simmering keeps it there and any extra heat applied increases the bubbling (stream generation) and wastes energy without any increase in speed of cooking. So your "boil for 10 minutes" would simply leave it undercooked and with much of the water boiled off.  Cooking it more slowly would work efficiently, but the time taken to heat the food through initially might be quite long, and indeed variable according to exactly how well heat can flow around, and just how high the heat is set.  Most of the cooking happens only near boiling, so this could lead to very variable results.  Boiling initially ensures that the food is heated through (via agitation of the water through boiling, and possibly through steam condensation) in a short time, reaching a temperature throughout for cooking quickly and at a fairly well-determined time.  —Quondum 17:32, 11 September 2016 (UTC)


 * Note that properly, Simmering should be below boiling temperature (not just a slow boil), although precisely how much depends on what you're doing (and since most people aren't using thermometers or other measurements and also it isn't uncommon to have elements with limited digital thermostat controls, somewhat random). The lower temperature is significant enough to affect how stuff particularly meat (edit:) or other proteins cook. See e.g.   Nil Einne (talk) 17:53, 11 September 2016 (UTC)


 * All true. Boiling water is essentially a crude way to take a temperature, and the "simmer" bit is just a downward tweak of the reading.  Preferably you'd simply type in 85 C for 23 minutes and get the temperature you want at the pot.  Ideally, you'd type in the internal temperature you want... one technology I miss in the kitchen is the terahertz scanner for measuring internal meat temperature during cooking, but we can't have everything. Wnt (talk) 12:40, 14 September 2016 (UTC)


 * I cook kale for about 30 seconds. If the recipe calls for 20 minutes, was this a recipe book that came with your 1950s time machine?  For a dish named "kale slurry" or somesuch?
 * Steamed dark green leaf vegetable a la Madhur Jaffrey
 * Place generous oil in a lidded heavy cast iron pan and add a teaspoon or more of whole spices - black onion seed is a good one.
 * Heat the pan.
 * Shred the leaves finely, or roughly tear them. Remove any thick stalks.
 * Mix up a tablespoon each of water, light rice vinegar (acidic but not strong tasting) and a teaspoon of honey.
 * Heat until the spices start to pop.
 * Turn the heat off. Throw in the leaves. Throw in the cold liquid. Put the pan lid on and shake enthusiastically.
 * 30 seconds to a minute later (this is why it needs a heavy cast-iron pan), it's ready to eat.
 * Andy Dingley (talk) 18:29, 11 September 2016 (UTC)


 * They started from frozen, while your recipe seems to start from room temperature. Also note that, like spinach, kale can be eaten raw, or cooked to the consistency of slime, so highly variable cooking times are to be expected.  (I never liked spinach until I started eating it raw.) StuRat (talk) 22:45, 11 September 2016 (UTC)


 * Myself, I would adapt the recipe to cook the kale in the microwave, since they always have a timer, unlike most stoves. However, microwaves heat rather unevenly, especially those which lack a rotating plate, and this would necessitate heating at low power for a long time, so the heat would have time to transfer evenly.  But the chances of burning the kale are drastically reduced, say if distracted by a phone call during the cooking process.  Thawing the frozen kale first would speed up the cooking process, whether on the stove or in the microwave. StuRat (talk) 23:14, 11 September 2016 (UTC)


 * Thank you to everybody. All helpful responses. Bus stop (talk) 01:14, 12 September 2016 (UTC)


 * One of my favourite kitchen gadgets (UK) is an electric steamer. It's like having an extra ring or two on the stove. Like the microwave it also has a timer. I boil root vegetables, but anything above ground gets steamed by some means. Andy Dingley (talk) 10:39, 12 September 2016 (UTC)


 * I assume you boil root vegetables (carrots, potatoes, etc) because they require more cooking, and also boiling eliminates the need to clean them. However, some of the nutrients leach into the water.  The fix for this is to drink the water, that is, make soup out of it.  Of course, if you haven't washed the root vegetables, you may have some sand in your soup, so you may want to a least rinse them first.  As for where to boil them, I'd go with the microwave, because of the timer, so less chance of burning it.  Also, using a stove, especially a gas stove with an uncovered pot, can result on burnt food at the waterline. StuRat (talk) 14:59, 13 September 2016 (UTC)


 * "I assume you boil root vegetables (carrots, potatoes, etc) because they require more cooking"
 * Then you would be assuming wrongly. Andy Dingley (talk) 15:07, 13 September 2016 (UTC)


 * LOL, then may I ask why you don't like to steam them ? StuRat (talk) 16:02, 13 September 2016 (UTC)


 * Can we agree on this. The OP bought a package of frozen kale. The purveyors wants to keep cooking instructions simple. As Andy Dingley suggested, it does seem out of the 1950's in an age when wives stayed at home and got the meal ready for table by simply stewing all veg for 20 minutes - regardless of its final palatability. My COI on this, is I am a supertaster, not by choice but I was born with a now recognized generic trait. No member of the Brassica family ought to be simmered for 20 minutes. Beet root, etc, OK but not Brassica. After all, The OP may have friends round for diner that are too polite to mention that he  over-cooks and they may not accept a second invitation -even if he  follows exactly what it says on the packaged cooking destructions. For some tough old greens -which now are unavailable in the mall but you can grow  at home. Simmer to bring out the bitter oils, then change the water for the final simmer. As the say -Grandma always knows best. Think the problem may well have occurred during the second World War when wives went to work in the factories and  came home to simmer everything for 20 minutes, because that is all they had time for.--Aspro (talk) 16:26, 14 September 2016 (UTC)


 * Generic trait ? Are those less expensive than name-brand traits ? :-) StuRat (talk) 16:52, 14 September 2016 (UTC)

Why is 5 petals most common?
At least where I'm at. But I think 5 is most common for a lot of the Earth. 6 seems to be next commonest of the low numbers (i.e. lily). 3 isn't too common I don't think (maybe in the tropics) I think I 've seen 4 petal flowers only once. Is there a fitness advantage to 5 petals vs 4 or is it just the difficulty of evolving such different anatomies? (similar to how no one ever invents a 6-legged mammal) Then there's "many petals", i.e. rose, sunflower, many others but that mutation seems easy for some reason. Sagittarian Milky Way (talk) 17:49, 11 September 2016 (UTC)
 * Dogwood has four petals. ←Baseball Bugs What's up, Doc? carrots→ 20:53, 11 September 2016 (UTC)
 * That might be it. It's kind of a Southern tree isn't it? Sagittarian Milky Way (talk) 21:05, 11 September 2016 (UTC)
 * Working on the assumption that you're from the Americas, Pacific Northwest Wildflowers - Browse Plants by Type: 4 petals runs to six pages. You may be right about cultivated flowers, where the more petals the better. Alansplodge (talk) 21:12, 11 September 2016 (UTC)
 * Ah, I was forgetting those (usually small?) wildflowers where 4's not unusual. I don't see nature much. Sagittarian Milky Way (talk) 21:20, 11 September 2016 (UTC)
 * This page lists the families of plants (growing wild in the USA) with four petals, which include the Onagraceae (evening primroses), Gentianaceae (gentians), Hydrangeaceae (hydrangeas) and probably the best known, Papaveraceae (poppies). But you're right, poppies aside, they're mostly rather small. Perhaps bigger flowers benefit from more petals? Alansplodge (talk) 21:28, 11 September 2016 (UTC)


 * There are also flowers with a lot more petals, like chrysanthemums or sunflowers (or, more broadly, the asteraceae). I could see how a small number of large petals could be problematic, catching too much wind and damaging the flower instead of passing between the petals.  For comparison, there are some trees with very large leaves, but I believe most of those are down in the jungle, where strong winds are rare. StuRat (talk) 22:52, 11 September 2016 (UTC)


 * The most primitive flowering plants have a highly variable number of petals. Having evolved from plants like the conifers, and Welwitschia it is not odd that the most primitive known, true flowering plant, Amborella vary in form from having bunches and whirls, to lacking petals entirely.  (I am simplifying a bit for those who may complain about my terminology, but the essence is correct.)


 * The movement of direction in plant evolution seems to have been towards having a flat target with a bull's eye in the middle for the flying pollinator. The magnoliids accomplished this early on, and they have petals that come in multiples of three.  Then, ignoring a few small outliers, the great bulk of all the remaining plants fell into two separate branches, the monocots, which have petals in multiples of three, a unique leaf blade with parallel veins, a single pore in their pollen, and a single seedling leaf (think onions, lilies, tulips, grass and coconuts); and the dicots, which have petals normally in mulitiples of 5 on a single flower, three pores in their pollen and seedlings (think soy) with two leaves when they sprout.


 * Of course everyone had thought the magnolias and water lillies were dicots, and that the dicot-monocot split was primary. Only with the development of genetic analysis and the recognition of Amborella's significance has the plant family tree taken shape.


 * Now, as to the 4/5 eudicot vs 3/6 monocot difference, and the suite of other difference, it might simply be the chance of a founder effect, or there might be a single pleiotropic gene, given there are other effects, such as with stem development.


 * Finally, flowers like mums and sunflowers are actually comprised of scores of separate little flowers, each with only one petal In the sunflower, for example, if a flower develops in the middle of the head, it produces only sex organs, but no petals.  The opposite applies to those that find they have no neighbor at the edge.


 * μηδείς (talk) 03:05, 12 September 2016 (UTC)




 * Good call on bringing up Amborella. Arabidopsis also used to be thought more basal than it is currently understood to be. The compound flower of the Sunflowers and many other Asteraceae we see around are inflorescences, and they do indeed have highly variable petal numbers. Magnolids evolved before pollination was widely performed by Hymenoptera, so that's why they don't necessarily look much like other more recent flowers. SemanticMantis (talk) 15:29, 12 September 2016 (UTC)


 * I came across, however the journal International Journal of Pure and Applied Mathematics is published by Academic Publications, Ltd. who appear in this list . Nil Einne (talk) 05:55, 12 September 2016 (UTC)
 * BTW I had a slightly more careful skim through the paper and I'd say 2 things. One even assuming the statistics are accurate, it sounds like it's only looking in one area, in the addition there's the risk as mentioned above of that's we're biased in favour of flowers that are more important to use for whatever reason. And even if the statistics hold true worldwide and aren't simply biased by flowers we've looked at, it could simply be random for various reasons as also mentioned above. And I have to admit, I'm sceptical how well someone in the Department of Business Information actually understands the biology even assuming the modelling is good. (And getting back to the earlier point, it's often easier to come up with many reasons why something may be so, but it's generally difficult to know which ones may actually be the reason or even if it's just random.) On the other hand, if we treat this as a more popsci explaination rather than quality academic research, perhaps it's still interesting. Nil Einne (talk) 11:12, 12 September 2016 (UTC)
 * I wonder how no real journal had copied the International Union of Pure and Applied Chemistry and International Union of Pure and Applied Physics with the International Journal of Pure and Applied Mathematics before. That name sounds awesome.  Is there really no applied mathematics?   Sagittarian Milky Way (talk) 14:04, 12 September 2016 (UTC)
 * The first two are unions, not journals. There is of course plenty of applied mathematics. For whatever reason, the main two professional associations in the USA are the AMA and SIAM, which cleave along pure/applied lines. The international organization more analogous to IUPAC and IPAP is the International Mathematics Union, which includes math broadly, including pure, applied, and educational. SemanticMantis (talk) 15:34, 12 September 2016 (UTC)


 * If you want a mechanistic look you might like stuff like this . I think the premise has been sufficiently challenged, but I'll note that even within species, petal number can be highly variable . SemanticMantis (talk) 15:37, 12 September 2016 (UTC)

Nobody seems to have questioned the question. Are 5 petals the most common? If not, we perhaps need to edit the thread title to avoid misleading readers. DrChrissy (talk) 16:31, 12 September 2016 (UTC)


 * It's not clear what the title even means. It's certainly true that five-petaled flowers are most common in one part of my yard. I have no reason to believe that five-petaled flower species have a plurality among flowering species worldwide. I also have no reason to believe that flowers with five petals make up a plurality of all flowers found worldwide.
 * Note the paper linked by Nil Einne has all kinds of bontanical mis-statements, e.g. "Flowers that belong to the same family have the same number of petals" - that is not even true within the same species, as illustrated by third link above. Now, if we ignore that, and trust that the author properly tabulated Makino(1989), and the families have not been significantly rearranged since then, then we can see from Table 1 that "five-petaled families" have a plurality among the 159 plant families occurring in Japan that are considered. Honestly, that paper is maybe OK for a student project, but I'd reject it from any journal I review for.
 * Anyway, If we ignore within-family variation, and note the plurality of five-petaled families in Japn, I still don't think the question can really be answered, not with extant references. You can get fascinating discussion and guesses and intuition, and invoke some general principles like pollinator services, resource competition, conserved traits, resource allocation, r/K selection, mutualism, coevolution, etc. Floral display is a good keyword to find relevant literature petal number/color/size, flower number/size, etc. E.g. here is a paper discussion how petal number and many other factors interact with pollinator behavior. Here  is another nice paper about evolution and petal number, this one restricted to the Asteridae.  SemanticMantis (talk) 19:49, 12 September 2016 (UTC)
 * Thanks for this. So do we need to do something about the title of this thread?  This is a reference list and, in the future, people scanning archives might be mislead.  We could ask the OP to clarify what they meant and form it as an open question. DrChrissy (talk) 19:56, 12 September 2016 (UTC)
 * I don't know- I see your point, but changing titles can also be problematic. I've started a discussion on the talk page about how to deal with this type of issue because misinformation in the the title does come up occasionally. As a stopgap, I've given your followup a header, and that will also show up in archives.  SemanticMantis (talk) 20:20, 12 September 2016 (UTC)
 * Good solution. Thank you. DrChrissy (talk) 20:37, 12 September 2016 (UTC)
 * But why start with the word "Question"? Isn't it apparent that the rest of it is a question? Normally I change headers called "Question" into something more revealing, but this already has the required detail, so again, why the word "Question"?  --   Jack of Oz   [pleasantries]  21:38, 12 September 2016 (UTC)


 * Changing titles is not a problem if you use the "anchor" template as I have now done here. ←Baseball Bugs What's up, Doc? carrots→ 22:56, 12 September 2016 (UTC)


 * I just spent an hour writing a long response my browser erased, so, short version:


 * The IJPAM article is rationalistic, even quasi-mystic, with its reference at the end to Daphne from Greek Mythology and the five-fingered hand. (The first tetrapods has 6 or 7 fingers.)


 * Most importantly, the article fails to deal with the many bundles of features that distinguish 3/6 monocots (which lack wood/secondary growth, largely have leaves with grass-blade like shapes and parallel veins, and different developmental morphologies) while 5/4 eudicots have pollen with three pores, two seedling leaves (cotyledons) and largely bear leaves with a net of veins in a palmate structure. Nothing about the number five explains these suites of structures.


 * The author did, howver, mention something interesting, the parallel in the development of pentagonal symmetry in early echinoderms Echinoderm ancestors were bisymmetrical, their ancestors experimented with spiral-shaped and  trimerous forms (Helicoplacoids).  See this diagram of early echinoderms and relatives from this article on their evolution


 * Once fully modern echinoderms developed with fivefold-symmetry it was the best way to tile what in essence is a spherical plane. Try to tile a soccer ball with only hexagons.  It can't be done.  (I am not sure if it was Gould, Penrose, or someone else who pointed this out in my reading, so no source.)  But this article does describe the development of the pentaradial form from the triradial one in the fossil records.  Perhaps Pentaradial symmetry is simply a more stable variant.


 * Oh, and BTW, it turns out mum flowers are indeed originally pentaradial, but one petal grows to form the spike and the others fail to develop.  In any case, each of the spikes is a digenerate pentaradial flower. μηδείς (talk) 01:15, 13 September 2016 (UTC)


 * Pentagonal symmetry is the basis for the Dodecahedron, one of the platonic solids. -- Jayron 32 01:35, 13 September 2016 (UTC)

Feynman Lectures. Lecture 22
Link: http://www.feynmanlectures.caltech.edu/I_22.html Can someone explain me please last phrases, when Feynman tries to find $$\lg(i)$$.

First, why did Feynman decide that by changing sigh of $$i$$ he can replace division by multiplication? And what's the matter with $$x^2+y^2=1$$? Username160611000000 (talk) 19:04, 11 September 2016 (UTC)

Why should we divide by $$0.99996+0.00900i$$ but not any other row of table? And why divide but not multiply? — Preceding unsigned comment added by Username160611000000 (talk • contribs) 19:14, 11 September 2016 (UTC)


 * A complex number multiplied by its complex conjugate is equal to the square of the norm of the original number. That is, $$\begin{align}  z\overline{z} &= {\left| z \right|}^2\end{align}$$.

If x^2 + y^2 = 1 as Feynman says, then the norm is 1, so $$\begin{align} z\overline{z} &= 1\end{align}$$. Multiply this equation by any number a, and divide by $$\overline{z}$$, and you get $$\begin{align} a{z} &= a / \overline{z}\end{align}$$. So multiplying any number by a complex number whose norm is 1 is the same as dividing by its conjugate. CodeTalker (talk) 20:48, 11 September 2016 (UTC)
 * (edit conflict) :The answer to the first question is that $$1/(x+iy) = (x-iy)/(x-iy) * 1/(x+iy) = (x-iy)/(x^2+y^2) = (x-iy)/1$$--Wikimedes (talk) 20:55, 11 September 2016 (UTC)


 * But $$0.99996+0.00900i \neq x+iy$$. And $$(0.99996+0.00900i)(0.99996-0.00900i)=0.99996^2+0.00900^2 \neq 1$$.
 * I don't understand anything.
 * Feynman tries to find $$\lg(i)$$. So he thinks he can write $$10^{is}=i$$ and try to expand $$is$$ into summands: $$10^{is}=10^{is_1+is_2+is_3+...}=10^{is_1}\cdot 10^{is_2}\cdot 10^{is_3}\cdot ...$$.
 * He finds:
 * $$x+iy=10^{is}=(?)\cdot(0.40679+0.91365i)\cdot(0.95885+0.28402i)\cdot(0.98967+0.14349i)=(?)\cdot(-0.01308+1.00008i)$$
 * Can you explain every step from this moment?
 * This lecture is working through a constructive proof of the properties of the complex number system. This is actually a quite advanced mathematical technique.  Although this methodology is very satisfying to people who have formal mathematical training, it is absolutely not the easiest way to learn complex numbers for most people!
 * What has happened in the steps leading up to the point where you are confused?
 * Well, the author has created a numerical method (and glossed over where it came from). He uses mathematical implication to say that if you accept his numerical method, then you must also accept his consequent steps.  He never formally proves his premise (although, as it turns out, his numerical method is valid and correct - it is a truncated Taylor series - he does not ever demonstrate this.  You, the reader, must simply recognize this by inspection because you are already intimately familiar with that method and have used it a hundred zillion times for the last five years ... right?  Are the undergraduates still following the material?)
 * If you can't follow the Feynman lectures, you aren't alone. In fact, Richard Feynman wrote about this topic in his memoirs: he wrote those lectures to teach undergraduate material in a way that advanced research physicists loved - but undergraduate students got lost in the details!
 * If you haven't already learned to think like a mathematician, a good book is A Transition to Advanced Mathematics.
 * I do not think we can explain "every step" any better than the lecture you have linked: as I read it, every single step is outlined by the author. But every single step is an advanced mathematical step.
 * If you are not already intimately familiar with the rules for algebra as they pertain to exponentiation, you need to go back to those fundamental mathematical definitions. Feynman is taking the definition of exponentiation, and applying it to an unknown entity (the complex number), and using it to demonstrate numerous logical consequences.  From these consequences, he expresses the elementary properties of the complex number, by construction.  Constructive proof is one of the more difficult techniques for new students; it's less intuitive, because it can be hard to see where you are going with the proof until you get there.
 * In this case, where you are going is a formal, provable set of properties about the complex number system. Those properties are necessary consequences of a simple definition of an unknown quantity, the imaginary unit.
 * As it turns out, these properties are exactly the same properties that most easier textbooks just tell you. Those books sometimes simply define these properties as part of the complex number system.  Fundamentally, it's not any better or worse to use mathematical definition in that way; but as you get deeper into your study of higher mathematics, you will find that purists prefer to keep the total number of definitions as small as possible, and to derive properties from that tiny set of definitions.  This is what mathematicians mean when they use the adjective "elegant."  In fairness to some people who do not see how this extra difficulty is "elegant," ... for those readers, just accept that "elegant" must be interpreted as a technical term of mathematical jargon, and not interpreted according to its normal meaning as an English-language adjective.  Feynman's method is elegant because it reduces the number of requisite premises to one - not because it's easy!
 * Engineers usually don't care: they are just as happy to memorize thirty complicated definitions of various properties and rules; they have little patience (or necessity!) for learning a difficult method that reduces all thirty complicated rules into a common thread. That's one reason why Feynman's lectures aren't great introductions of physics for engineering students.
 * Nimur (talk) 05:14, 12 September 2016 (UTC)
 * OP is correct that $$0.99996^2+0.00900^2 \neq 1$$ (though it is close). Feynman asks how one divides by a complex number.  The answer is that one can't directly, so one changes it to a different form so that one can multiply by a complex number instead.  Instead of dividing by $$a + bi$$, one multiplies by $$a/(a^2+b^2)-bi/(a^2+b^2)$$ which equals $$1/(a+bi)$$.  I'm not sure why Feynman thinks $$a^2+b^2$$ is close enough to 1 to ignore it.  For an introduction to complex arithmetic, the first chapter of "Complex Variables and Applications" by Churchill and Brown is a pretty good source.--Wikimedes (talk) 14:57, 12 September 2016 (UTC)


 * Thanks. I think I get it. The whole 3rd column of Table 22–3 consist of $$10^{is_{_{1,2,3...}}}$$ so every that complex number satisfies the condition $${x_{1,2,3...}}^2+{y_{1,2,3...}}^2=1$$.
 * $$x+iy=10^{is}=(?)\cdot(0.40679+0.91365i)\cdot(0.95885+0.28402i)\cdot(0.98967+0.14349i)=(?)\cdot(-0.01308+1.00008i)$$.
 * Next multiplier must have imaginary part close to real part of previous result (-0.01308). He should simply say "we must multiply by 0.99996-0.00900i" . And if $$0.99996+0.00900i = 10^{is_{_3}}$$ then $$0.99996-0.00900i$$ is simply equal $$10^{-is_{_3}}$$. In addition $$(-0.01308+1.00008i)\cdot(0.99996+0.00900i)=-0.02208+0.99992i$$ and $$(-0.01308+1.00008i)\cdot(0.99996-0.00900i)=-0.00408+1.00016i$$ . So second is much better.
 * $$(?)\cdot(-0.01308+1.00008i)=(?_{II})\cdot(-0.01308+1.00008i)\cdot(0.99996-0.00900i)=(?_{II})\cdot(-0.00408+1.00016i)$$.
 * $$(?_{II})\cdot(-0.00408+1.00016i)=(?_{III})\cdot(-0.00408+1.00016i)\cdot(1 -0.0045i)=(?_{III})\cdot(0.00042+1.00018i)$$
 * Next multiplier or $$(?_{III})$$ must have imaginary part close to 0.00042. We can suppose imaginary part is equal to 0.00042. Using $$10^{im/1024}=1+0.0022486im$$: $$0.0022486im=0.00042i \implies m=0.19$$.
 * $$(?_{III})=10^{0.19i/1024}$$.
 * Is it correct?
 * Feynman got $$10^{0.20i/1024}$$ instead of $$10^{0.19i/1024}$$. Is it just error by rounding? Username160611000000 (talk) 18:22, 12 September 2016 (UTC)