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# Invertibility of compact operator on infinite dimensional Hilbert space

I am trying to solve an exercise from the book Analysis for Applied Mathematics by Ward Cheney. Exercise 2.3.25 states: "Prove that if $$X$$ is an infinite-dimensional Hilbert space, then a compact operator on $$X$$ cannot be invertible."

I interpret the question to mean that the operator maps $$X$$ to a general normed linear space $$Y$$. I have been able to show that if the inverse exists, then it cannot be bounded/continuous. This is also covered in several questions on this site. This also takes care of the case where the operator maps to a Banach space because the interior/open mapping theorem then implies that if the inverse exists then it must be bounded/continuous.

Is the assertion in the exercise true if the operator maps $$X$$ to $$Y$$, where $$Y$$ is a normed linear space? After failing to prove this myself I have tried searching both this site, as well as google, but still have not found a proof. As I am learning functional analysis for the first time, I realize I might have overlooked something I did not quite understand. I have seen some mentions of spectral theory when searching, but this is not covered before the next chapter in the book.

Thanks in advance to anyone that took the time to read this post.

• You shall modify slightly the title of your question as it is not exactly what you ask at the end of your question... – Nicolas Jun 13 at 12:46

Let $$(x_n)$$ be a sequence in the closed unit ball of $$X$$. Then $$T(x_n)$$ has a convergent subsequence in $$Y$$, say $$T(x_{n_k})$$ by compactness of $$T$$. If $$T$$ has continuous inverse then $$(x_{n_k})=(T^{-1} (Tx_{n_k}))$$ would also be convergent. But this shows that close unit ball of $$X$$ is compact and this implies that $$X$$ is finite dimensional.

EDIT: It is customary in FA to say that a bounded operator is invertible if it has a bounded inverse. If you just want $$T$$ to be a bijection onto some normed linear space there are many examples. Let $$T:\ell^{2} \to \ell^{2}$$ be defined ny $$T(x_n)=(\frac 1 n x_n)$$. Then $$T$$ is a compact one-to-one operator . If $$Y$$ is the range of $$T$$ then $$T$$ is a bijection from $$X$$ onto $$Y$$.

• Thank you for your reply! If I understand correctly, this only shows that $T$ cannot have a continuous inverse, which as I wrote above I have been able to show myself in a similar fashion. This in turn implies that $T$ cannot have an inverse when$Y$ is a Banach space. What about the case where $T^{-1}$ is not required to be continuous? – tfjaervik Jun 13 at 12:30
• @tfjaervik I have edited my answer. – Kavi Rama Murthy Jun 13 at 12:45
• Thanks a lot! I was just trying to check whether the previous example you gave satisfied the conditions in the question (I am a little slow being new to functional analysis). Really appreciate it! – tfjaervik Jun 13 at 12:51
• My first example was defined in a Banach space which was not a HIlbert space. So I changed the example. @tfjaervik – Kavi Rama Murthy Jun 13 at 12:52

Alternatively, recall the following two facts:

(1) The identity operator $$\operatorname{id}_X: X \to X$$ is compact if and only if $$X$$ has finite dimension.

(2) If $$S$$ is compact and $$T$$ is any bounded operator, then $$ST$$ is compact.

Using this, we can easily prove your claim:

Assume that $$S$$ is compact and invertible. Then $$\operatorname{id}= SS^{-1}$$ is compact by $$(2)$$, and by $$(1)$$ it follows that $$X$$ has finite dimension.

Note: This does not rely on any properties of Hilbert spaces, i.e. the claim is true in a Banach space.

• Thanks! I had been able to show it in a similar fashion, but I was not aware that inverses of bounded operators usually means bounded inverses. Hence, I had been struggling to prove something that was not true, as the example in the accepted answer shows. – tfjaervik Jun 13 at 12:58
• @tfjaervik Usually, invertibility is meant to happen in some ring/algebra, here the algebra of bounded operators. But, in fact it doesn't matter in this context, because by the bounded inverse theorem, the set-theoretic inverse of a continuous linear operator between Banach spaces is automatic continuous again. See en.wikipedia.org/wiki/…. – QuantumSpace Jun 13 at 13:00
• Thanks for the input! Indeed I mentioned the last part of your comment in my original post. I was struggling to prove the assertion for general normed linear spaces $Y$ ($T: X\rightarrow Y$) – tfjaervik Jun 13 at 13:03
• @tfjaervik In that case, you are right. You need to require that $X \to Y$ has continuous inverse. – QuantumSpace Jun 13 at 13:05

I add an alternative to the (excellent) answer of Kavi Rama Murthy. This concerns infinite dimensional Hilbert spaces and assume as well that invertibility implies continuity of the inverse as it is generally required.

Let $$K$$ be a compact operator and let $$(K_n)_{n\geq0}$$ be a sequence of finite rank operators converging in norm topology to $$K$$. If $$K$$ is invertible, then $$K_nK^{-1}=(K_n-K+K)K^{-1}=(K_n-K)K^{-1}+\mathrm{Id}$$ is invertible for $$n$$ large enough by a Neumann series argument, and thus the finite rank operator $$K_n$$ is invertible, which is absurd.

• @KaviRamaMurthy The OP's background is that of infinite dimensional Hilbert spaces. – Nicolas Jun 13 at 12:42
• @KaviRamaMurthy Actually the end of the OP asks for something else indeed, so I will clarify what I am doing in my answer. – Nicolas Jun 13 at 12:47


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# Gauge (bore diameter)

Gauge

From left to right; a .45 ACP, a .410 bore shotshell, a 20 gauge shotshell, and a 12 gauge shotshell
Type Shotgun
Place of origin Various

The gauge of a firearm is a unit of measurement used to express the diameter of the barrel. Gauge is determined from the weight of a solid sphere of lead that will fit the bore of the firearm, and is expressed as the multiplicative inverse of the sphere's weight as a fraction of a pound (e.g., a 112th pound ball fits a 12-gauge bore). Thus there are twelve 12-gauge balls per pound (etc.). The term is related to the measurement of cannon, which were also measured by the weight of their iron round shot; an 8 pounder would fire an 8 lb (3.6 kg) spherical cast iron ball and had a bore diameter of about 91 mm (3.6 in).

Gauge is commonly used today in reference to shotguns, though historically it was also used in large double rifles, which were made in sizes up to 2 bore during their heyday in the 1880s, being originally loaded with black powder cartridges. These very large rifles, sometimes called elephant guns, were intended for use in India and Africa for hunting dangerous game.

Gauge is abbreviated "ga.", "ga", or "G". The space between the number and the abbreviation is often left out, as in "12ga".

## Calculating gauge

An n-gauge diameter means that a ball of lead (density 11.352 g/cm3 or 6.562 oz/in3) with that diameter has a mass equal to 1/n part of the mass of the international avoirdupois pound (453.59237 grams). Therefore an n-gauge shotgun or n-bore rifle has a bore diameter (in centimeters) of approximately

$d_n = \left(\frac{6 \times 453.59237~\mathrm{g}}{n \times \pi \times 11.352~\mathrm{g/cm}^3}\right)^{\frac{1}{3}} = 4.2416~\mathrm{cm} \times \frac{1}{\sqrt[3]{n}}$

Another source for a gauge size formula can be found in Shotgun shell.

It should be noted that the results of the calculations given above need not be carried out to further than one decimal place, as shotgun and rifle gauges are stated in integers. Furthermore, the density of the lead that was once used as the standard varied, since bullets and slugs are not made of chemically pure lead, but are instead made of lead alloyed with a variety of materials.

## Gauges in use

From left to right: 9×19mm Parabellum, .40 S&W, .45 ACP, 5.7×28mm, 5.56×45mm NATO, .300 Winchester Magnum, and a 2.75-inch and 3-inch 12 gauge

Since shotguns were not originally intended to fire solid projectiles, but rather a compressible mass of shot, the actual diameter of the bore can vary. The fact that most shotgun bores are not cylindrical also causes deviations from the ideal bore diameter.

The chamber of the gun is larger, to accommodate the thickness of the shotshell walls, and a "forcing cone" in front of the chamber reduces the diameter down to the bore diameter. The forcing cone can be as short as a fraction of an inch, or as long as four inches on some guns. At the muzzle end of the barrel, the choke can constrict the bore even further, so measuring the bore diameter of a shotgun is not a simple process, as it must be done away from either end.

Shotgun bores are commonly "overbored" or "backbored", meaning that most of the bore (from the forcing cone to the choke) is slightly larger than the value given by the formula. This is claimed to reduce felt recoil and improve patterning. The recoil reduction is due to the larger bore producing a slower acceleration of the shot, and the patterning improvements are due to the larger muzzle diameter for the same choke constriction, which results in less shot deformation. A 12-gauge shotgun, nominally 18.5 mm (0.73 in), can range from a tight 18.3 mm (0.72 in) to an extreme overbore of 20.3 mm (0.80 in). Some also claim an increased velocity with the overbored barrels, up to 15 m/s (49 ft/s), which is due to the larger swept volume of the overbored barrel. Once only found in expensive custom shotguns, overbored barrels are now becoming common in mass marketed guns. Aftermarket backboring is also commonly done to reduce the weight of the barrel, and move the center of mass backwards for a better balance. Factory overbored barrels generally are made with a larger outside diameter, and will not have this reduction in weight—though the factory barrels will be tougher, since they have a normal barrel wall thickness.

Firing slugs from overbored barrels can result in very inconsistent accuracy, as the slug may be incapable of obturating to fill the oversized bore.

## Sizes in use

left-to-right: .410, 28ga, 20ga, 12ga

Certain sizes are more common than others; 12 gauge is the most common size,[1] with up to 50% of the overall shotgun market in the United States. The 20-gauge shotgun is popular with shooters who are uncomfortable with the weight and recoil of a 12 gauge gun, and is popular for upland game hunting. The next most popular size is a tie between the 28 gauge and .410 bore shotgun, which is not actually a gauge but rather a caliber. 10, and 16 gauges, while less common, are still readily available.

Shotguns larger than the 10 gauge are rarely manufactured nowadays. Eight gauge is rare in the United States due to its prohibition in duck hunting. However, it is still used in many parts of the world (notably: Britain) for bird hunting. Its shells are usually black powder paper cartridges as opposed to the plastic/wax cartridge and smokeless powder of today.

The 11, 15, 18, 2, and 3 gauge shells are the rarest of them all;[2] people who own these types of rare shotguns will usually have their ammunition custom reloaded by a company that specializes in rare and custom bores for a high price. The 14 gauge has not been loaded in the United States since the early 1900s, although the 2-9/16 inch hull is made in France.[3] The very small 24 and 32 gauges are still produced and used in some European countries and Brazil. Punt guns and special purpose guns, such as the Russian 23 mm KS-23 (approximately 6 gauge), do exist, but are rarely encountered.

Also seen in limited numbers are smoothbore firearms in calibers smaller than .410, such as .22 Long Rifle, .22 WMR, and 9 mm rimfire, designed for short range pest control.[4][5]

To further complicate matters, special shot cartridges are available for typical handgun chamberings such as 9 mm Parabellum, .45 ACP, .38 Special/.357 Magnum, .44 Special/.44 Magnum, and .45 Colt. These rounds are either crimped in or in a plastic casing replacing the bullet. These are not generally considered "shot shells" by shotgun users, and the patterning performance is questionable since they are fired through rifled barrels. Thompson/Center makes special pistol barrels in .38/.357, .44 and .45 Colt that have "straight rifled" chokes in them to reduce the spin of the shot column and produce better patterns, but they are still suitable only for pest control at very short ranges.

### Gauge and shot type

The 10 gauge was headed into obsolescence until steel and other nontoxic shot became required for waterfowl hunting as the larger shell could hold the much larger sizes of shot needed to get the low-density steel shot to the ranges needed for waterfowl hunting. The move to steel shot reduced the use of 16 and 20 gauges for waterfowl hunting, and smaller 2.75" (70mm) 12 gauge shells as well. However, the advent of the 3.5 in (89 mm) 12 gauge shell, with its higher SAAMI pressure rating compared to standard 12 gauge guns, begin to approach the performance of the 10 gauge loads. Newer nontoxic shots, such as bismuth and tungsten-nickel-iron alloys, and even tungsten-polymer blends, regain much or all of the performance loss, but at much higher cost than steel or lead shot.[6]However, scientific research indicates that tungsten alloys can actually be quite toxic internally.[7]

### Sizes found in the UK

4 bore
8 bore 3¼” (83mm)
10 bore 2⅝”, 10 bore 2⅞”, 10 bore 3½” (89mm)
12 bore 2” (51mm), 12 bore 2½” (63mm), 12 bore 2¾” (70mm), 12 bore 3” (76mm), 12 bore 3½” (89mm)
14 bore 2½” (63mm)
16 bore 2½” (63mm), 16 bore 2¾” (70mm)
20 bore 2½” (63mm), 20 bore 2¾” (70mm), 20 bore 3” (76mm)
24 bore 2½” (63mm)
28 bore 2½” (63mm), 28 bore 2¾” (70mm)
32 bore 2½” (63mm)
.410 bore 2”, .410 bore 2½” (63mm), .410 bore 3” (76mm)
.360 bore 1¾” (45mm)
9mm (no.3 bore) rimfire [1.439" (9 x 36.56mm)], 9mm (no.3 bore) long rimfire
7mm (no.2 bore) rimfire
6mm (no.1 bore) short rimfire, 6mm (no.1 bore) rimfire

## Conversion guide

Portrait of Frederick Courteney Selous with his 4 bore single-shot Boer rifle and African hunting regalia, 1876.

A table showing the various gauge sizes with weights. The bores marked * are found in punt guns and rare weapons only. The .410 bore and 23mm caliber are exceptions; they are actual bore sizes, not gauges. If the .410 and 23mm were measured traditionally, they would be 67.62 gauge and 6.278 gauge, respectively.

Gauge
(Bore)
Diameter Weight of unalloyed (pure) lead ball
(mm) (in) grams ounces grains
AA* 101.60 4.000 6225.52 219.6 96,080
* 76.20 3.000 2626.39 92.64 40,530
0.25* 67.34 2.651 1814.36 64.000 28,000
0.5* 53.45 2.103 907.18 32.000 14,000
A* 50.80 2.000 778.19 27.45 12,010
0.75* 46.70 1.838 604.80 21.336 9328
1* 42.42 1.669 453.59 16.000 7000
* 38.10 1.500 328.3 11.58 5066
1.5* 37.05 1.459 302.39 10.667 4667
2* 33.67 1.326 226.80 8.000 3500
3* 29.41 1.158 151.20 5.333 2333
4* 26.72 1.052 113.40 4.000 1750
B* 25.40 1.000 97.27 3.43 1501
5* 24.80 .976 90.72 3.200 1400
6* 23.35 .919 75.60 2.667 1166
6.278 23.00 .906 72.26 2.549 1114
7* 22.18 .873 64.80 2.286 1000
8 21.21 .835 56.70 2.000 875
9* 20.39 .803 50.40 1.778 778
10 19.69 .775 45.36 1.600 700
11/C½* 19.05 .750 41.24 1.454 636
12 18.53 .729 37.80 1.333 583
13* 18.04 .710 34.89 1.231 538
14* 17.60 .693 32.40 1.143 500
15* 17.21 .677 30.24 1.067 467
16 16.83 .663 28.35 1.000 438
17* 16.50 .650 26.68 0.941 412
18* 16.19 .637 25.20 0.889 389
20 15.63 .615 22.68 0.800 350
22* 15.13 .596 20.62 0.728 319
24 14.70 .579 18.90 0.667 292
26* 14.31 .564 17.44 0.615 269
28 13.97 .550 16.20 0.571 250
32 13.36 .526 14.17 0.500 219
36 12.85 .506 12.59 0.444 194
C* 12.70 .500 12.16 0.429 188
40 12.40 .488 11.34 0.400 175
67.62 10.41 .410 6.71 0.237 104

Note: Use of this table for estimating bullet masses for historical large-bore rifles is limited, as this table assumes the use of round ball, rather than conical bullets; for example, a typical 4 bore rifle from circa 1880 used a 2,000-grain (4.57 oz) bullet, or sometimes slightly heavier, rather than using a 4 ounce round lead ball. (Round balls give progressively much worse external ballistic performance than conical bullets at ranges greater than about 75 yards.) In contrast, a 4-bore express rifle often used a 1,500-grain (3.43 oz) bullet wrapped in paper to keep lead buildup to a minimum in the barrel. In either case, assuming a 4-ounce mass for a 4-bore rifle bullet from this table would be inaccurate, although indicative.

## References

1. ^ Carter, Greg Lee (2002). Guns in American Society: An Encyclopedia. Santa Barbara, Calif.; Oxford: ABC-CLIO. p. 361. ISBN 978-1-57607-268-4.
2. ^ Frank C. Barnes (2009), Layne Simpson editor, Cartridges of the World, 12th Ed.
3. ^ Frank C. Barnes (2009), Layne Simpson editor, Cartridges of the World, 12th Ed.
4. ^ Clair Rees (March 2000). "Marlin's 'Garden Gun'—Model 25MG". Guns Magazine.
5. ^ Frank C. Barnes (2003). Stan Skinner, ed. Cartridges of the World (10th ed.). Krause Publications. ISBN 0-87349-605-1.
6. ^ Randy Wakeman (2007). "Why The 10 Gauge Shotgun is Obsolete". chuckhawks.com.
7. ^ John Kalinich et al. (2005). "Embedded weapons-grade tungsten alloy shrapnel rapidly induces metastatic high-grade rhabdomyosarcomas in F344 rats.". Environmental Health Perspectives.

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# The final parsec “problem”

Many and perhaps all galaxies seem to contain supermassive black holes of about $10^7 M_\odot$ at their centres. Determining their origins is of great astrophysical interest.

In what I understand to be the currently-favoured theory of "hierarchical merger", the supermassive black holes get larger over time via repeated mergers between smaller holes, during for example galactic mergers. This requires a natural mechanism dissipating orbital energy and angular momentum between the holes such that they will generically merge in much less than a Hubble time.

When the holes are very far apart (much further than a parsec) such a mechanism exists in the form of gas drag (EDIT: the actual mechanism here is dynamical friction; thanks Chris White for pointing this out). And when they are very close the gravitational radiation reaction becomes important. But at distances of about a parsec the holes don't seem to be driven together, and according to current theory should never merge. Thus the "final parsec problem": how do we get the holes close enough together for a merger to happen?

However as far as I'm aware there isn't actually any evidence that the black holes ever do merge. A large swarm of small black holes about a parsec apart would be equally compatible with observation. Or else some process other than hierarchical merger could account for the holes. So it seems this "problem" has more to do with the careers of gravitational wave scientists than with any actual difficulty in astrophysics. Or am I being unfair?

• Judging by the end of your question, you seem to be ranting against the name of the idea in question. I'm not sure whether or not you think that the problem is a problem or isn't a problem. It's unclear what you're trying to say. – HDE 226868 Nov 28 '15 at 20:09
• At very large distances it's dynamical friction, not gas drag, that brings black holes together. Also, a large swarm of black holes would be incredibly dynamically unstable -- very rapidly, some would be ejected while the rest merged. – user10851 Nov 28 '15 at 20:48
• Update also @JohnDuffield: There is now solid evidence from a direct measurement of black holes merging so the issue has been settled (experimentally). – Danu Feb 16 '16 at 16:18
• @Danu : note the "the" in my comment, and that this question concerns supermassive black holes. We have good evidence of decaying pulsar binaries, I don't think there's much of an issue re stellar black holes merging. – John Duffield Feb 16 '16 at 20:26
• @Danu From an astrophysical perspective, whether supermassive black holes merge is a qualitatively different question from whether stellar mass ones do: supermassive black hole binaries are (presumably) created during galactic mergers, while stellar mass ones (presumably) formed as binaries in the first place. This question concerns supermassive binaries; LIGO saw a stellar mass merger. – AGML Feb 20 '16 at 20:25

Great question.

1) There is indirect (and circumstantial) evidence that they do merge. While there are some famous examples of apparently 'binary' (or more accurately 'dual') AGN (e.g. Komossa+2003, or Rodriguez+2006) there seems to be a very conspicuous dearth of such systems --- suggesting that they don't spend very long at observable separations. Note that "observable" doesn't necessarily mean (optically) resolvable as VLBI has also found very few binaries, and there are also spectroscopic methods which should be more sensitive the closer the systems are. Additionally, people have suggested that the tightness of the M-sigma relation suggests that BH binaries merge effectively. Finally, all of the well constrained observations of BHs (e.g. through stellar velocity profiles) can fairly reliably rule out the presence of massive companions.

2) There are growing theoretical reasons to believe the 'final-parsec problem' isn't as bad as previously thought. One of the best reviews of the subject is given by Merritt and Milosavljević 2009.

On the scales of parsecs, the dominant 'hardening' mechanism (bringing the binary closer together) is through scattering of single-stars in the so-called 'loss-cone'. If you take the simplest analytic model for the loss-cone, of a spherically symmetric, isotropic distribution of stars, then the loss-cone is rapidly 'depleted' --- there is only a small number of stars which are able to scatter with the binary, and once they do the loss-cone is 'empty' and very slowly refilled. More realistic models, which take into account perturbations from anisotropic and nonspherical systems suggest that the loss-cone can be refilled much more rapidly and efficiently.

Additionally, the importance of local gas-drag on the binary on parsec and sub-parsec scales seems to be much more significant than originally thought. This is for two reasons: i) large amounts of additional, dense gas seem to get funneled towards the galactic center during mergers. ii) the formation of a circumbinary gaseous disk can be much more effective at extracting orbital energy / angular-momentum than was initially thought (e.g. Cuadra+2009, MacFadyen and Milosavljević 2008).

Right now, all of the theoretical models suggest that a very large fraction of MBH binaries should merge very effectively, i.e. on timescales less than about $10^8$ years. Even pretty substantial modifications to the knobs of these models, don't seem to change the basic conclusion that we are on the verge of detecting MBH Binary systems with Pulsar Timing Arrays (PTA). The only big caveat to this, in my opinion, is that the frequency range which PTA are most sensitive to - could actually be at the frequencies where MBH Binaries spend the least time. None of the models I've seen, however, suggest we should be too worried about it.

All of this is an extremely active area of research, and both the observations and the theoretical models are very difficult to perform --- but hopefully we'll be seeing some exciting progress on both fronts very soon.


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# Options and Volatility Smile

An equity option represents the right to buy (“call” option) or sell (“put” option) a unit of underlying stock at a pre-specified price (strike) at a predetermined maturity date (European option) or at any time up to the predetermined date (American option).

Option writer sells an option and option holder buys an option.

For a European call option on an index with strike 8,000 and index level of 8200 at maturity, the option holder receives the difference of $200 from option writer. This is called the instrinsic value or payoff of the option from the holder’s point of view. The payoff function for a call option is $$h_{T}(S,K) = max[S_{T}-K, 0] \tag{Eq. 1}$$ where T = maturity date,$\ S_T $is the index level at maturity and K is the strike price. In-the-money: a call (put) is in-the-money when S > K (S < K) At-the-money: call or put is at-the-money when$\ S \approx K $Out-of-the-money: a call is out-of-the-money when S < K (S > K) A fair present value (is different than payoff) of a European call option is given by Black-Scholes formula:$\ C_{0}^{*} = C^{BSM}(S_{0},K,T,r,\sigma) \tag{Eq. 2}\S_{0} $current index level (spot) K strike price of the option T time-to-maturity of the option r risk-free short rate$\ \sigma $volatility or the std dev of the index returns$\ C^{BSM} = S_{t} . N(d_{1}) – e^{r(T-t)} . K. N(d_{2})\tag{Eq. 3} $where$\displaystyle N(d) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{d} e^{-\frac{1}{2}x^{2}} dx\displaystyle d1 = \frac{\log\frac{S_{t}}{K} + (r + \frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}\displaystyle d2 = \frac{\log\frac{S_{t}}{K} + (r – \frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}$The present value of the option is always higher than the undiscounted payoff, the difference being the time value. In other words, the option’s present value is composed of payoff plus the time value. Time value indicates that there is always a chance of option going in-the-money or more in-the-money during that time. ### Simulating Returns The Geometric Brownian motion model of the BS equation is given by $$\displaystyle dS_{t} = rS_{t}dt + \sigma S_{t} dt dZ_{t}\tag{Eq.4}$$ The discretized version is $$\displaystyle S_{t} = S_{t – \Delta t} e^{(r – \frac{1}{2}\sigma^2) \Delta t + \sigma \sqrt{\Delta t} z_{t}}\tag{Eq.5}$$ where t$\in {(\Delta t, 2\Delta t,…..,T)}$Using the above discretized version, we will simulate the spot prices with$S_{0}$=100, T=10, r = 0.05 and$\sigma$=0.2 Implied Volatility is the value of$\sigma$that solves Eq. 2 given the option market quote$C_{0}^{*}\$

Volatility surface is the plot of the implied volatilities for different option strikes and different option maturities on the same underlying (an option chain).

Vol Surfaces exhibit :
Smiles: option implied volatilities exhibit a smile form, i.e. for calls the OTM implied volatilities are higher than the ATM ones; sometimes they rise again for ITM options
term structure:smiles are more pronounced for short-term options than for longer-term options; a phenomenon sometimes called volatility term structure

To demonstrate Vol Surface, I will use an option chain on AAPL stock as of 5/11/2017. I have downloaded this data from a reputable vendor, you can find this file here AAPL_BBG_vols

Once we have the implied volatilities, we will generate a grid of strikes and maturities and use Cubic interpolation to derive the missing implied volatilities needed for a smooth surface.


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# Find a set $A$ such that $A \notin \mathrm{Rng}{f}$

Let $f:\mathbb N\to \mathbb P(\mathbb N)$ be given by the following expression: $$f(n)=\{m\in\mathbb N\mid 3m-10>n\}$$ Find a set $A$ such that $A \notin \mathrm{Rng}{f}$.

I came up with $A=\{n\in\mathbb N\mid n\notin f(n)\}$, but I don't believe that it works because $6\in f(6)$.

• Where's the contradiction? Does that mean $A=f(6)$? No. It means that $6\notin A$ and therefore that $A\neq f(6)$. – Thomas Andrews Apr 19 '15 at 22:50
• Of course, you could just take $A=\{1\}$. Since $1\notin f(n)$ for any $n$, so this would mean that $A\cap f(n)=\emptyset$ for all $A$. Or you could take $A$ any finite set, since $f(n)$ is always infinite. Or you could take $A$ to be the set of all primes, since if $m\in f(n)$ then $m+1\in f(n)$, so $f(n)$ contains all numbers larger than its least element. – Thomas Andrews Apr 19 '15 at 22:54

The set $A=\{n\in\mathbb N\mid n\notin f(n)\}$ should work for every application $f$, for instance in your application it's clear that $n\in f(n)$ if and only if $3n-10> n$ which means that: $$A=\{0,1,2,3,4,5\} \tag 1$$ now assume that there exists some $x$ such that $f(x)=A$ then $A$ is infinte because always $f(x)$ is infinite for any $x$, this contradicts $(1)$.
Note that the set $A$ always works without any restriction on $f$, and it's a part of Cantor's argument that $\mathbb N$ and $\mathbb P(\mathbb N)$ don't have the same cardinal.


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# Decidability of rank-k polymorphism vs. System F

There's a paper by Kfoury from 1992, "Type Reconstruction in Finite Rank Fragments of the Second-Order $$\lambda$$-Calculus", that proves that type inference for Curry-style rank-$$k$$ polymorphic lambda calculus is undecidable if $$k \ge 3$$. My thought was that since there must be terms that can be typed in rank-3 but not rank-2, and since System F is rank-$$n$$ (i.e. the union of all finite rank-$$k$$), surely this implies that Curry-style System F type inference is undecidable as well. However, its undecidability isn't proven until 1996 by Wells in "Typability and Type Checking in the Second-Order $$\lambda$$-Calculus Are Equivalent and Undecidable", which explicitly mentions Kfoury's paper (along with others) and states that decidability of both type inference and type checking have not yet been proven until then.

So given that rank-3 polymorphic lambda calculus is a fragment of System F, why doesn't the former's undecidability imply the undecidability of the latter?

We prove that [...] for every $$k\ge 3$$ there is a typing of constants that assigns types in $$S(1)$$ such that the type reconstruction problem for $$\Lambda_k$$ extended by this typing is undecidable. Unfortunately, this result does not give any clue as to whether type reconstruction for $$F_2$$, with or without constants, is decidable. Our decidability/undecidability proofs use in an essential way the apriori information about the rank of types.


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5

# F' =fl V V + VsSolve for V,Determine X and y2x-123x+72y...

## Question

###### F' =fl V V + VsSolve for V,Determine X and y2x-123x+72y

f' =fl V V + Vs Solve for V, Determine X and y 2x-12 3x+7 2y

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# Better still, a Tramway to Space

1. Dec 28, 2004

### maglifter

Forget space elevators and pipelines to space, the tramway is far more realistic.

A big 1500 mile tube suspended with superconducting cables, floating in the air, reaching 70000ft, containing a maglev track.

http://www.startram.com/main.html [Broken]

"StarTram - Ultra Low-Cost To Orbit

The current cost of launching payloads to LEO orbit is very high, on the order of $10,000 per kilogram. The launch costs to GTO and GEO are considerably greater. Major research efforts are underway on advanced reusable launch vehicles that could potentially reduce launch costs to LEO to ~$1000 per kilogram. However, the technology for these reusable vehicles is very demanding and it is not clear whether this goal can be achieved.

The cost of the energy required to accelerate a kilogram of payload to orbital speeds, i.e., 8 km/sec., is actually very small, on the order of 50 cents. If there were no atmosphere on Earth, payloads could be accelerated to orbital speed on Earth’s surface, using a superconducting Maglev suspension to levitate and propel the payload. The resultant total cost (energy plus facilities) to launch payloads would be extremely small, on the order of a few dollars per kilogram.

The heating and air drag forces that would occur in the presence of Earth’s atmosphere prevent such a launch system from being practical. However, if the spacecraft were accelerated in a low pressure tunnel and then launched into Earth’s atmosphere at high altitude where the atmospheric density is very low, then it would be possible to launch payloads at a very low cost.

Plus Ultra has investigated a radically new concept for magnetically levitating and accelerating spacecraft to orbital speeds and launching them at high altitudes. The Plus Ultra concept, termed StarTram, accelerates the spacecraft to 8 kilometers per second in a long (~1000 km) low pressure tunnel on the surface. After the levitated spacecraft has reached orbital speed, it transitions into an upwardly curved low pressure launch tube, along which it travels to an altitude of ~70,000 feet. The cylindrical launch tube is magnetically suspended by the magnetic repulsion forces between a set of superconducting cables attached to it, and a second set of superconducting cables, located on the surface beneath the launch tube. The two sets of superconducting cables carry oppositely directed, zero loss supercurrents.

The launch tube is stabilized vertically and laterally by a set of high tensile strength, lightweight tethers that anchor the tube to the ground, and prevent sway or movement from wind forces (Sidebar 1). The spacecraft exits through a fast opening shutter at the upper end of the launch tube into the low-density atmosphere at 70,000 feet and rapidly ascends to orbit. At the launch altitude, the atmospheric heating rate and drag forces are low enough (Sidebar 2) that the spacecraft can safely ascend into Earth orbit.

StarTram has the potential to increase the present launch volume of a few hundred tons per year to orbit to many thousands of tons per year, at a unit cost that is well below $100 per kilogram." Last edited by a moderator: May 1, 2017 2. Dec 28, 2004 ### errorist Why not build it up the side a very tall mountain? It should get you out of the very heavy air near sea level? 3. Dec 28, 2004 ### hitssquad Mount Everest is only 29,035 ft. You might start there, but you would still have another 40,000 feet to go. 4. Dec 28, 2004 ### maglifter I've always wondered why it wouldn't be possible to use a hypersonic scramjet with maglev launch assist. You build a long maglev tube, make a vacuum of it, and then launch the scramjet. At the end of the tunnel, when the jet has reached mach 5 or 6, you open a diaphragma, and POP, the scramject sucks in air and propels itself to mach 10. 5. Dec 28, 2004 ### Astronuc ### Staff: Mentor A maglaunch tube inside a mountain still has the column of air above the launch point. Mount Everest is a bit impractical - one would has to bore into the mountain - and that requires getting heavy equipment (e.g. tunnel boring machine) to the site. Mountain climbers to Everest have to hike up glaciers to get to the highest base camps which are still quite aways from the summit. Itinerary - http://www.alpineascents.com/everest-dtd.asp Climb overview - http://www.alpineascents.com/everest.asp Everest Maps - http://www.alpineascents.com/everest-map.asp Besides 70000ft is no where close to space. On October 4, 2004, SpaceShipOne rocketed to an altitude in excess of 328,000 ft. Maybe the OP meant 700,000 ft. But then one might as well go with a space elevator. 6. Dec 28, 2004 ### cronxeh or.. or.. why not scotch tape a scramjet to a hot air balloon and then release it from 65000 feet? 7. Dec 28, 2004 ### NateTG I think that was the canuckistani X-prize project approach. Of course you'd have to get a working scramjet first. Regarding reusable vehicles: Everything that gets launched into space that has to come down again is really a wasted effort, doubly so for things like the space shuttle that go into orbit because they also need fuel to slow themselves out of orbit to come back down. 8. Dec 28, 2004 ### maglifter I didn't know Israel was called Canuckistan. But anyway, the maglauncher is not as crazy as you think, since it will be a practical means to get rid of the first stage of ordinary rockets. ESA and NASA are working on it. Their maglauncher will be 2.4 miles long, just to get rid of the first stage. The Startram, on the contrary, is pure fiction. And Cronxeh, you were kindly asked to stick to making your LPG powered airplane, remember? Last edited: Dec 28, 2004 9. Dec 28, 2004 ### Astronuc ### Staff: Mentor I haven't dismissed the mag-launch concept - but putting in Everest would be very difficult, if not impossible. Ideally - put it in a mountain as close to the equator - Andes perhaps - as possible. In the US, the mountains are pretty much limited to 14,000 range - from sea-level. But still there is the air column in the launch tube. The question is - how much acceleration in 14,000 ft (or there abouts). On the other hand, non-living cargo (supplies) can be shot at > 10 g. People cargo need lower g's, so may be an improved shuttle design is useful. 10. Dec 28, 2004 ### NateTG Any surface-launched object will initially be in an orbit that collides with the earth unless it is traveling at escape velocity. That means that any gun-launched satelite will have to have additional significant rocket boosters. Regarding G-forces and materials: In the 70's Gerrard Bull was using artillery to shoot weather probes to high altitude, and did some research into gun-to-space type technology, and apparently had little trouble hardening all of his equipment to handle 40G by encasing delicate components in resin. The pie-in-the-sky gun launch scenario would involve a very large light gas gun and 4000G's. At those forces, special electronics would be necessary. It's pretty easy to caculate the approximate acceleration that you'd need to get for a particular distance. If the length of the barrel is $d$ and the muzzle velocity is $v$ then the acceleration is going to be: $$a=\frac{v^2}{2d}$$ So, let's say you want to have a maximum acceleration of 10G's ($\approx 100 \frac{m}{s^2}$) and a muzzle velocity of 5000 m/s. Then you're looking at a length of about 120,000 meters. If you can handle 1000G's (which is reasonable for raw materials) you can get the same muzzle velocity in 1,200 meters. I remember going through and calculating things like power consumption for a device like this, and deciding that it would probably be worth building it near a good power source rather than trying to find a high altitude location since cable losses are significant. 11. Dec 28, 2004 ### hitssquad Chemical spin-traps to protect astronauts from high-g effects Not necessarily. If the problem is nerve death, spin-traps (such as the chemicals PBN and N-t-butylhydroxylamine that life-extensionists use today to preserve the youthfulness of their brains and other organs) and other mitochondrial protectors might allow rocket-passeners to survive unharmed without blood in their brains for a few minutes. Also, astronaut brains might be supercooled - again, to protect against high-g-induced brain death - on the launchpad with cold breathable-liquids such as perflourocarbons. 12. Dec 28, 2004 ### maglifter Lots of confusing messages I think some concepts are being confused here: 1. a maglauncher is a simple maglev track, a few miles long (2 - 3 miles) from which a spaceplane is launched, horizontally. We're talking 1 g here. Nothing big. Nasa is building it. (see this interesting presentation: http://science.ksc.nasa.gov/shuttle/nexgen/Nexgen_Downloads/Spaceport_Visioning_Final_Report.pdf ) Such a maglauncher is merely used to give the spaceplane a headstart, in order to reduce the hassle of the first stage of an ordinary rocket launch. 2. a stratram is a very long (near) vacuum tube, with a maglev track, on which the carrier gradually moves to extremely high speeds, before it pierces a diaphragm and enters the atmosphere at high altitude. 3. Bull's mega guns are: nonsense. Ok. -A maglauncher is feasible, cheap, and it's being built. -Startram is fiction, fun and will never be built, just like the space elevator. -Mega guns are not fun, not useful and have been built by hundreds of amateurs. One of which was assassinated in Iraq. 13. Dec 28, 2004 ### russ_watters ### Staff: Mentor Aerodynamic resistance is a surprisingly small piece of the energy required to get into orbit - and at 70,000 feet, you're above >95% of the atmosphere. Still, getting a ramp up to 70,000ft requires construction materials unlike anything even unlikely to be available in the forseeable future. A launch ramp up the rockies, however, puts you above half the atmosphere. Last edited: Dec 28, 2004 14. Dec 29, 2004 ### maglifter Russ, you seem to know quite a bit about this. I don't. Could you help me with this? Suppose you want to launch 3 people from a maglev track up to 100km altitude (merely a parabola with a top of 100km) (the X-prize criteria). -how long should such a track be? -would it make any considerable difference if you were to build it on a mountain slope with the track end at 5km altitude? -and what would be the advantage of building it on a mountain right on the equator ? (I've never quite understood why equatorial launches are best). Thx! 15. Dec 29, 2004 ### Astronuc ### Staff: Mentor Equatorial lauches take maximum advantage of the rotational speed of the earth's surface. And one would want to lauch eastward, as the shuttle and other manned craft do. One can do a simple trajectory calculation and figure with or without air resistance, what the initial velocity would need to be from some height H (~ 2-3 km) to 100,000 km, and then what the acceleration over H would have to be to get that velocity. Or suppose, one starts off on a level track, the acclerates through a curve to the final 2-3 km lauch tube. Then one has to calculate that tolerable accelerations in the three sections. Re: No thanks. 16. Dec 29, 2004 ### cronxeh its doable. not maglev.$4-6 million. got cash?

17. Dec 29, 2004

### Kenneth Mann

Difficult yes, impossible no. Actually, the whole idea of space access is difficult, and we're just looking for the best trade-offs. Mountain launch seems quite promising in this light. We just need to study it more carefully to determine if it is a winner. Also, why does it have to be Everest? There are many mountains over 25,000 feet high, (over a hundred, if my recollection is correct) most of them in the general area of the Himalayas, the Hindu Kush, etc.

Actually, the Equator is ideal only for Equatorial applications, such as Earth Synchronous insertions. Everest and those mountains around it are at about 28 degrees North, the same as Canaveral, and wouldn't you know, the inclination to the Moon is 28 degrees. (Ideally accessible from either North or South). To the Ecliptic, it's about 23 degrees. I'd be careful selecting Andean mountains. Some of them are in the mid-twenties range, but if I recall correctly some may be volcanic.

Nothing's perfect, but then do we really need a purely ballistic insertion? We really want to find an approach that is more economical over the very long term. For this, we need simply to appreciably increase our effective specific impulse, and if we cheat a bit by off-loading part of the job from Today's purely rocket-based approach, then we are ahead in the game. The drawback is, this will only work if we commit to a large-scale move into the space environment, thus amortizing the very large up-front costs of a mountain-based launcher. What might be investigated is, using something like a Scramjet, multiple-use booster with small rockets for the very upper ranges.

18. Dec 30, 2004

### Kenneth Mann

I agree with this. Why even consider 70,000 feet. All we really need is enough boost to get us to a point and velocity at which we can light off something like a Scramjet, and which will put the vehicle into a mode from which it can safely glide down in case of some sort of systems failure at launch. Mach 1, 2 or the like may be quite enough.

KM

19. Dec 30, 2004

### Kenneth Mann

PostScript

If we are really interested in using the Maglev launcher concept where it is of best advantage, we could consider setting up one on the Moon (after we have bases there). There we can simply run the track along the surface (covered or underground, if we don't want to continually have to clean off dust encrustation) wherever we can find a long enough stretch of relatively flat surface. (Actually, we'd probably have to do a lot of landscaping just to get a flat enough stretch of surface.)

Then we could use it to sling out "stuff" to the planets, etc. (We'd have to wait for a monthly window for each launch track. The Moon is only five degrees off the Ecliptic, so we might be able to fudge this a bit.)

KM


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# Thread: prove sup and inf

1. ## prove sup and inf

The question is:

assume A is a proper subset of B and both belongs to R^n dimention

show that this implies sup(A)< and= sup(B).

It looks quite obvious, but I don't know how to construct the proof formally.

Thanks for help.

2. Originally Posted by delhaize
The question is:

assume A is a proper subset of B and both belongs to R^n dimention

show that this implies sup(A)< and= sup(B).

It looks quite obvious, but I don't know how to construct the proof formally.

Thanks for help.
Actually, you have a serious problem right from the start. While there is a standard "order" on R, there is no standard order on R^n for n> 1. And without a specific order there is no "lim" or "sup".

The statement you give is easy to prove in R but makes no sense in R^n for n> 1.


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# "weak" or "holographic" vielbein framings?

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Here I am going to ask if there is any work on field theories with "weak" metric structure exhibited not by vielbein fields on the tangent bundle, but by the analog kind of "viel+1-bein fields" on the direct sum of the tangent bundle with the trivial line bundle.

A vielbein field configuration on a manifold $X$ is locally a choice of trivialization of the tangent bundle, and if it is also globally so then it is a choice of global parlallelism, a "framing". Traditionally in physics this receives attention in the corner of "teleparallel gravity", but this is not what I am after here.

In mathematics, notably in generalized cohomology,  a major role is played by the weaker concept of "stable framings" which are framings not necessarily of the tangent bundle $TX$ itself, but of the sum of that with a trivial vector bundle of any rank. In between that very weak concept and the ordinary strong concept of framing (global parallelism) is that of "$(d+1)$-framing", a choice of trivialization of $TX \oplus \underline{\mathbb{R}}$ on a $d$-dimensional manifold $X$.

There is some sense in which a $(d+1)$-framing is analogous to a conformal structure. A conformal structure, too, may be thought of as given by a local vielbein field but relaxing its conditions a bit -- by allowing a common rescaling of all the beins in the vielbein. Here for a $(d+1)$-framing we don't have rescalings, but we have the extra freedom of moving around one extra "bein" and of rotating all the "beins" in one more dimension than the manifold has.

A more precise way to say how $(d+1)$-framings play a role analogous to that of conformal structures in conformal field theory is the cobordism hypothesis-theorem, which says among many other things that if we produce a Chern-Simons-like topological field theory of dimension $(d+1)$,then its Wess-Zumino-Witten-like holographic dual theory in dimension $d$ has a flat bundle of "conformal blocks" on the moduli space of the "$(d+1)$-structures" on its spacetime $X$. If the dual CS theory is fully anomaly-free in that it is framed, then this means that the $d$-dimensional boundary theory has a flat bundle of "conformal blocks" on the moduli space of its $(d+1)$-framings. (More in this MO question which, it turns out, has a positive answer.)

Has any field theory of such kind ever surfaced anywhere in the physics literature? Maybe in discussion of generalized WZW-like theories or in discussion of variants of AdS/CFT. Or elsewhere?

Should this be tagged reference-request'' or the analogue of the same here? I find the editing process sort of confusing, probably because I did not sleep well last night.
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# Math Help - Frenet-Serret Formulas

1. ## Frenet-Serret Formulas

Let $\kappa=\frac{\delta \overline T}{\delta s}\cdot \overline N$ and $\tau=\frac{\delta \overline N}{\delta s}\cdot \overline B$. Assume in turn that each of the intrinsic derivatives of $\overline T, \overline N, \overline B$ are some linear combination of the unit vectors $\overline T, \overline N, \overline B$ and hence derive the Frenet-Serret formulas of differential geometry.

I am sure thus must be easy but I cannot see it!

If $\frac{\delta \overline T}{\delta s}=a \overline T + b \overline N + c \overline B$ then taking the dot product of both sides with N and using the first equation given produces $\frac{\delta \overline T}{\delta s}=a \overline T + \kappa \overline N + c \overline B$ If I claim to know that N is orthogonal to T then I can get the first equation, that is $\frac{\delta \overline T}{\delta s}=\kappa \overline N$

But when I move onto the next equation $\frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + c_2 \overline B$ then taking the dot product of both sides with B and using the second equation given produces $\frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + \tau \overline B$.

I cannot see how to get from here to the required equation:
$\frac{\delta \overline N}{\delta s}=-\kappa \overline T + \tau \overline B$

Or the third equation:
$\frac{\delta \overline B}{\delta s}=-\tau \overline N$

2. I am sure thus must be easy but I cannot see it!

What happens when you differentiate a unit vector?

3. Differentiating a unit vector we get:

$\frac{\delta \overline T}{\delta s}\cdot \overline T=\frac{\delta \overline N}{\delta s}\cdot \overline N=\frac{\delta \overline B}{\delta s}\cdot \overline B=0$

So this helps me to reduce my three equations to:

$\frac{\delta \overline T}{\delta s}=\kappa \overline N + c_1 \overline B$

$\frac{\delta \overline N}{\delta s}=a_2 \overline T + \tau \overline B$.

$\frac{\delta \overline B}{\delta s}=a_3 \overline T + b_3 \overline N$

But how do I show that $c_1=a_3=0$ and $a_2=-\kappa, b_3=-\tau$?

4. Originally Posted by Kiwi_Dave
Let $\kappa=\frac{\delta \overline T}{\delta s}\cdot \overline N$ and $\tau=\frac{\delta \overline N}{\delta s}\cdot \overline B$. Assume in turn that each of the intrinsic derivatives of $\overline T, \overline N, \overline B$ are some linear combination of the unit vectors $\overline T, \overline N, \overline B$ and hence derive the Frenet-Serret formulas of differential geometry.

I am sure thus must be easy but I cannot see it!

If $\frac{\delta \overline T}{\delta s}=a \overline T + b \overline N + c \overline B$ then taking the dot product of both sides with N and using the first equation given produces $\frac{\delta \overline T}{\delta s}=a \overline T + \kappa \overline N + c \overline B$ If I claim to know that N is orthogonal to T then I can get the first equation, that is $\frac{\delta \overline T}{\delta s}=\kappa \overline N$

But when I move onto the next equation $\frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + c_2 \overline B$ then taking the dot product of both sides with B and using the second equation given produces $\frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + \tau \overline B$.

I cannot see how to get from here to the required equation:
$\frac{\delta \overline N}{\delta s}=-\kappa \overline T + \tau \overline B$

Or the third equation:
$\frac{\delta \overline B}{\delta s}=-\tau \overline N$
Given $\frac{d\vec{T}}{ds} = \kappa \vec{N}$ and $\frac{d \vec{B}}{ds} = - \tau \vec{N}$,

if we consider

$\vec{N} = \vec{B} \times \vec{T}$ then

$\frac{d \vec{N}}{ds} = \frac{d \vec{B}}{ds} \times \vec{T} + \vec{B} \times \frac{d \vec{T}}{ds} = - \tau \vec{N} \times \vec{T} + \vec{B} \times (\kappa \vec{N})
= \tau \vec{B} - \kappa \vec{T}
$

as required.


text:
Fibration of the phase space for the Korteweg-de Vries equation.(English)Zbl 0731.58033

We prove that the fibration of $$L^ 2(S^ 1)$$ by potentials which are isospectral for the 1-dimensional periodic Schrödinger equation, is trivial. This result can be applied, in particular, to N-gap solutions of the Korteweg-de Vries equation (KdV) on the circle: one shows that KdV, a completely integrable Hamiltonian system, has global action-angle variables.

MSC:

 37J35 Completely integrable finite-dimensional Hamiltonian systems, integration methods, integrability tests 37K10 Completely integrable infinite-dimensional Hamiltonian and Lagrangian systems, integration methods, integrability tests, integrable hierarchies (KdV, KP, Toda, etc.) 37C85 Dynamics induced by group actions other than $$\mathbb{Z}$$ and $$\mathbb{R}$$, and $$\mathbb{C}$$ 37A30 Ergodic theorems, spectral theory, Markov operators
Full Text:

References:

 [1] [CL] , , Theory of Ordinary Differential Equations, McGraw-Hill, New York, 1955. [2] [DU] , On global action-angle coordinates, C.P.A.M., 33 (1980), 687-706. · Zbl 0439.58014 [3] [FIT] , , , An explicit solution of the inverse problem for Hill’s equation, SIAM J. Math. Anal., 18 (1987), 46-53. · Zbl 0622.34021 [4] [ChF] , , A remark on a paper by C. Fefferman, Proc. A.M.S., (Feb. 1990), 407-409. [5] [GT2] , , Gaps and bands of one dimensional periodic Schrödinger operators II, Comm. Math. Helv., 62 (1987), 18-37. · Zbl 0649.34034 [6] [GK] , , Introduction to the Theory of Linear Non Selfadjoint Operators, Transl. of Math Monogr., vol. 18, AMS, Providence, 1969. [7] The module $${\cal D}f^s$$ [8] [Kp] , On the periodic spectrum of the 1-dimensional Schrödinger operator, Comm. Math. Helv., 65 (1990), 1-3. · Zbl 0703.34085 [9] [Ma] , Sturm Liouville Operators and Applications, Birkäuser, Basel, 1986. [10] [MM] , , The spectrum of Hill’s equation, Inv. Math., 30 (1975), 217-274. · Zbl 0319.34024 [11] [MW] , , Hill’s Equation, Wiley-Interscience, New York, 1986. · Zbl 0594.34030 [12] [MT] , , Hill’s operator and hyperelliptic function theory in the presence of infinitely many branch points, CPAM, 24 (1976), 143-226. · Zbl 0339.34024 [13] [PS] , , Aufgaben und Lehrsätze aus der Analysis, vol. 2, 3rd ed., Grundlehren, Bd 20, Springer-Verlag, New York, 1964. [14] [PT] , , Inverse Spectral Theory, Academic Press, 1987.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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### Home > INT2 > Chapter 12 > Lesson 12.2.4 > Problem12-131

12-131.

Twelve students will be chosen at random from the $900$ students at Rolling Meadows High School to serve as the Judicial Board for minor student infractions.

1. How many different Judicial Boards are possible?

$_{900}C_{12}\approx 5.48 \cdot 10^{26}$

2. Mariko hopes to be on the board. How many possible boards include her?

$_{899}C_{11}\approx 7.30 \cdot 10^{24}$

3. What is the probability (in percent) that Mariko is chosen for the board?

Divide the number of boards containing Mariko by the number of possible boards.


text:
# Masters Thesis

## Researching resistive sensors

From 2010-2012, as graduate student in mechanical engineering at the University of Massachusetts Lowell, I researched resistive sensors. These Graphene sensors were being tested for thier ability to detect chemical traces of explosives. I fabricated and tested sensors coated with Graphene using photo-lithography and groove-and-fill techniques. Additionally, a battery powered micro-controller based device was created, employing a four point probe, to demonstrate the system's portability.

I wrote and defended my thesis obtaining my master's degree in 2012. My thesis title and abstract follows.


text:
# 5.E: Pedigrees and Populations (Exercises)

These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.

## Study Questions

1. What are some of the modes of inheritance that are consistent with this pedigree?

2. In this pedigree in question 1, the mode of inheritance cannot be determined unamibguously.What are some examples of data (e.g. from other generations) that, if added to the pedigree would help determine the mode of inheritance?

3. For each of the following pedigrees, name the most likely mode of inheritance (AR=autosomal recessive, AD=autosomal dominant, XR=X-linked recessive, XD=X-linked dominant).(These pedigrees were obtained from various external sources).

a)

b)

c)

d)

4. The following pedigree represents a rare, autosomal recessive disease. What are the genotypes of the individuals who are indicated by letters?

5. If individual #1 in the following pedigree is a heterozygote for a rare, AR disease, what is the probability that individual #7 will be affected by the disease? Assume that #2 and the spouses of #3 and #4 are not carriers.

6. You are studying a population in which the frequency of individuals with a recessive homozygous genotype is 1%. Assuming the population is in Hardy-Weinberg equilibrium, calculate:

a) The frequency of the recessive allele.

b) The frequency of dominant allele.

c) The frequency of the heterozygous phenotype.

d) The frequency of the homozygous dominant phenotype.

7. Determine whether the following population is in Hardy-Weinberg equilibrium.

 genotype number of individuals AA Aa aa 432 676 92

8. Out of 1200 individuals examined, 432 are homozygous dominant (AA)for a particular gene. What numbers of individuals of the other two genotypic classes (Aa, aa) would be expected if the population is in Hardy-Weinberg equilibrium?

9. Propose an explanation for the deviation between the genotypic frequencies calculated in question 8 and those observed in the table in question 7.

5.1 The pedigree could show an AD, AR or XR mode of inheritance. It is most likely AD. It could be AR if the mother was a carrier, and the father was a homozygote. It could be XR if the mother was a carrier, and the father was a hemizygote. It cannot be XD, since the daughter (#2) would have necessarily inherited the disease allele on the X chromosome she received from her father.

5.2 There are many possible answers. Here are some possibilities: if neither of the parents of the father were affected (i.e. the paternal grandparents of children 1, 2, 3), then the disease could not be dominant. If only the paternal grandfather was affected, then the disease could only be X-linked recessive if the paternal grandmother was a heterozygote (which would be unlikely given that this is a rare disease allele).

5.3 a) The mode of inheritance is most likely AD, since every affected individual has an affected parent, and the disease is inherited even in four different matings to unrelated, unaffected individuals. It is very unlikely that it is XD or XR, in part because affected father had an affected son.

b) The mode of inheritance cannot be AD or XD, because affected individuals must have an affected parent when a disease allele is dominant. Neither can it be XR, because there is an affected daughter of a normal father. Therefore, it must be AR, and this is consistent with the pedigree.

c) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It is not XR, because there are unaffected sons of an affected mother. It is therefore likely AR, but note that the recessive alleles for this condition appear to be relatively common in the population (note that two of the marriages were to unrelated, affected individuals).

d) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It could be either XR or AR, but because all of the affected individuals are male, and no affected males pass the disease to their sons, it is likely XR.

5.4 If a represents the disease allele, individuals a, d, f (who all married into this unusual family) are AA, while b, c, e, g, h, i, j are all Aa, and k is aa.

5.5 There is a ½ chance that an offspring of any mating Aa x AA will be a carrier (Aa). So, there is a ½ chance that #3 will be Aa, and likewise for #4. If #3 is a carrier, there is again a ½ chance that #5 will be a carrier, and likewise for #6. If #5 and #6 are both Aa, then there is a ¼ chance that this monohybrid cross will result in #7 having the genotype aa, and therefore being affected by the disease. Thus, the joint probability is 1/2 x 1/2 x 1/2 x 1/2 x 1/4 =1/64.

5.6 a) q = -0.01. = 0.1

b) 1-q = p; 1-0.1 = 0.9

c) 2pq = 2(0.1)(0.9) = 0.18

d) p2 = 0.81

5.7 First, calculate allele frequencies:

$\mathrm{p = \dfrac{2(AA) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(432) + 676}{2(432+676+92)} = 0.6417}$

$\mathrm{q = \dfrac{2(aa) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(92) + 676}{2(432+676+92)} = 0.3583}$

Next, given these observed allele frequencies, calculate the genotypic frequencies that would be expected if the population was in Hardy-Weinberg equilibrium.

$\mathrm{p^2 = 0.6417^2 = 0.4118}$

$\mathrm{2pq = 2(0.6417)(0.3583) = 0.4598}$

$\mathrm{q^2 = 0.3583^2 = 0.1284}$

Finally, given these expected frequencies of each class, calculate the expected numbers of each in your sample of 1200 individuals, and compare these to your actual observations.

 expected observed (reported in the original question) AA 0.4118 × 1200 = 494 Aa 0.4598 × 1200 = 552 aa 0.1284 × 1200 = 154 432 676 92

The population does not appear to be at Hardy-Weinberg equilibrium, since the observed genotypic frequencies do not match the expectations. Of course, you could do a chi-square test to determine how significant the discrepancy is between observed and expected.

5.8 If in this theoretical question, the frequency of genotype of AA is set at 432/1200 and we are asked what frequencies of the other classes would fit a Hardy-Weinberg equilibrium. So, given that p2 = 432/1200, then p=0.6, and q=0.4. Given these allele frequencies and a sample size of 1200 individuals, then there should be 576 Aa individuals (2pq × 1200 = 2(0.6)(0.4) × 1200=576) and 192 aa individuals (q2 × 1200 = 0.42 × 1200 = 192), if the population was at Hardy-Weinberg equilibrium with 432 AA individuals.

5.9 The actual population appear to have more heterozygotes and few recessive homozygotes than would be expected for Hardy-Weinberg equilibrium. There are many possible reasons that a population may not be in equilibrium (see Table 5.1). In this case, there is possibly some selection against homozygous recessive genotypes, in favour of heterozygotes in particular. Perhaps the heterozygotes have some selective advantage that increases their fitness.

It is also worth noting the discrepancies between the allele frequencies calculated in 5.8 and 5.7. In question 5.7, we calculated the frequencies directly from the genotypes (this is the most accurate method, and does not require the population to be in equilibrium). In 5.8, we essentially estimated the frequency base on one of the phenotypic classes. The discrepancy between these calculations shows the limitations of using phenotypes to estimate allele frequencies, when a population is not in equilibrium.

This page titled 5.E: Pedigrees and Populations (Exercises) is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Todd Nickle and Isabelle Barrette-Ng.


text: But they are not a linear pair because they are not connected, and they do not share a common side. The line through points A, B and C is a straight line. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. Linear Pair: Definition, Theorem & Example ... we can say that angle NSI and angle ISD also form a pair of adjacent angles. In the adjoining figure, ∠AOC and ∠BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line. True, if they are adjacent and share a vertex and one side. Definition: Two angles that add up to 180°, Adjacent Angle – Definition, Examples & More. The vertex of an angle is the endpoint of the rays that form the sides of the angle. Example: Find the pairs of adjacent angles in the following figure. In Other Words, It Is Between A Right Angle And A Straight Angle. Start studying M54.2c. E. Both angles … In this article, we are going to discuss the definition of adjacent angles and vertical angles in detail. (ii) Angles in a linear pair which are (iii) Complementary angles that do not form a not supplementary. When two lines intersect, the angles in a linear pair have sides that are opposite rays; but the vertical angles formed by the two intersecting lines also have sides that are opposite rays. Solution: ∠ 1 and ∠ 2 are adjacent to each other∠ 2 and ∠ 3 are adjacent to each other, Solution: ∠ APD and ∠ DPC are adjacent to each other∠ DPC and ∠ CPB are adjacent to each other. If Two Angles Form A Linear Pair, The Angles Are Supplementary. In the figure above, the two angles ∠BAC and ∠CAD share a common side (the blue line segment AC). In the diagram below, ∠ABC and ∠DBE are supplementary since 30°+150°=180°, but they do not form a linear pair since they are not adjacent. Common side. Two Obtuse Angles Can Be Adjacent Angles. v. Angles which are neither complementary nor adjacent. At times, in geometry, the pair of angles are used. The two angles will change so that they always add to 180°, In the figure above, the two angles ∠PQR and ∠JKL are supplementary because they always add to 180°. If the two supplementary are adjacent to each other then they are called linear pair. A. D. The angles are congruent and are right angles. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. Draw a linear pair of angles. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add … Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Non-common side makes a straight line or Sum of angles is 180°. The endpoints of the ray from the side of an angle are called the vertex of an angle. The angles in a linear pair are supplementary. Yes. O. Additionally, the ray OB is the common arm between these two angles. Adjacent Angles Are Two Angles That Share A Common Vertex, A Common Side, And No Common Interior Points. Vertically Opposite Angles: When two lines intersect, then the angles that are opposite one another at the intersection are called Vertically Opposite Angles. Two angles that overlap, one inside the other sharing a side and vertex in the figure on the right, the two angles ∠PSQ and ∠PSR overlap. Both sets (top and bottom) are supplementary but only the top ones are linear pairs because these ones are also adjacent. Example : Logical Equivalence Lateral Area . Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pair. iv. See the first picture below. Related Questions to study. Another way of defining them is: “two angles that share a side and a vertex, but do not share any interior points”. Smenevacuundacy and 26 more users found this answer helpful Adjacent angles can be a complementary angle or supplementary angle when they share the common vertex and side. Example: If following angles make a linear pair, find the value of q. The measure of rotation of a ray, when it is rotated about its endpoint is known as the angle formed by the ray between its initial and final position. Two Adjacent Angle Can Be Complementary Too If They Add Up To 90°. In this picture there is a flatscreen Toshiba television. They share a common side, but not a common vertex. On the other hand, two right angles will always make a linear pair as their sum is equal to 180°. Two angles form a linear pair. Before you know all these pairs of angles there is another important concept which is called ‘angles on a straight line’. A Linear Pair Forms A Straight Angle Which Contains 180º, So You Have 2 Angles Whose Measures Add To 180, Which Means They Are Supplementary. Linear pair forms two supplementary angles. This television has a pair of supplementary angles that are not a linear pair (one green, one blue). Two adjacent angles forming a linear pair are in the ratio 7 : 5 find the angles. Solution: 130° + 50° = 180°Since the sum of these angles is equal to two right angles, so they can make a linear pair. This statement is correct. It can also be said that angles of the linear pair are always supplementary to each other. A: Angles that are next to each other are known as adjacent angles, i.e., two angles with one common arm. Two angles are said to be supplementary if the sum of both the angles is 180 degrees. Adjacent angles which are not in linear pair. Adjacent angles, Linear pair, Vertically opposite angles. In A Right Triangle, The Altitude From A Right-angled Vertex Will Split The Right Angle Into Two Adjacent Angle; 30°+60°, 40°+50°, Etc. The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees. The two angles are said to be adjacent angles when they share the common vertex and side. Draw the pairs of angles as described below. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. | Definition & Examples - Tutors.com In the figure, ∠1 and ∠3 are non-adjacent angles. If that is not possible, say why. linear pair. Let us take example of the angles, shown in following figure: ∠ COB and ∠ BOA have a common vertex, i.e. … A linear pair of angles has two defining characteristics: 1) the angles must be supplmentary; 2) The angles must be adjacent ; In the picture below, you can see two sets of angles. Although they share a common side (PS) and a common vertex (S), they are not considered adjacent angles when they overlap like this. 1 answer. Linear Pair of Angles. Two Adjacent Angle Can Be Supplementary Too, If They Add Upto 180°. (i v) When two lines intersect opposite angles are supplementary. Adjacent Angles Are Angles That Come Out Of The Same Vertex. Adjacent angles are angles that are side by side. Supplementary angles a and b do not form linear pair. Vertical Angles Are Always Congruent, Which Means That They Are Equal. Solution: 110° + 70° = 180°Since the sum of these angles is equal to two right angles, so they can make a linear pair. Can Two Obtuse Angles Be Adjacent? If two angles form a linear pair, the angles are supplementary. Note: ∠ APD and ∠ CPB are not adjacent to each other, because they don’t have a common arm in spite of having a common vertex. A pair of adjacent angles formed by intersecting lines is called a Linear Pair of Angles. 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# Question #2538f

Dec 27, 2017

My hunch is to use the Ratio Test to prove the $\Leftarrow$ direction and the Divergence Test to prove the contrapositive of the $\Rightarrow$ direction (which also proves the $\Rightarrow$ implication). Not sure if it will work, but here goes.

#### Explanation:

($\Leftarrow$) Suppose that $\left({u}_{n}\right)$ is a sequence that converges to a negative number $u$ and that $a > 1$. Let ${b}_{n} = {a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{n}} = {a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{n}}$.

Then $| {b}_{n + 1} \frac{|}{|} {b}_{n} | = \frac{{a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{n}} \cdot {a}^{{u}_{n + 1}}}{{a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{n}}} = {a}^{{u}_{n + 1}} \to {a}^{u}$ as $n \to \infty$.

Since $a > 1$ and $u < 0$, $| {a}^{u} | = {a}^{u} < 1$.

Therefore, by the Ratio Test, the series ${\sum}_{n = 1}^{\infty} \left({a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{n}}\right)$ converges.

Therefore, the $\Leftarrow$ direction is proven.

($\Rightarrow$) Suppose that $\left({u}_{n}\right)$ is a sequence that converges to a negative number $u$ and that $a \le 1$. Choose $N \in \mathbb{N}$ so that ${u}_{n} < 0$ for all $n > N$.

Let $A = {a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{N}} = {a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{N}}$ and note that $A \ne 0$.

Next, suppose $n > N$. Then ${a}^{{u}_{1} + {u}_{2} + \cdots {u}_{n}} = {a}^{{u}_{1}} \cdot {a}^{{u}_{2}} \cdots {a}^{{u}_{N}} \cdot {a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}} = A \cdot {a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}}$. Since $a \le 1$ and ${u}_{N + 1} < 0$, ${u}_{N + 2} < 0$, ..., ${u}_{n} < 0$, it follows that ${a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}}$ is the product of a bunch of numbers that are all $\ge 1$.

Since $A \ne 0$, it follows that $A \cdot {a}^{{u}_{N + 1}} \cdots {a}^{{u}_{n}}$ does not converge to zero as $n \to \infty$. But this means that ${a}^{{u}_{1} + {u}_{2} + \cdots {u}_{n}}$ does not converge to zero as $n \to \infty$.

By the Divergence Test, the series ${\sum}_{n = 1}^{\infty} \left({a}^{{u}_{1} + {u}_{2} + \cdots + {u}_{n}}\right)$ diverges.

Thus, the contrapositive of the $\Rightarrow$ implication is proved, meaning the original implication itself is also proved.


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Is hierarchical clustering of significant genes 'supervised' or 'unsupervised' clustering?
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5.5 years ago
steve ★ 3.1k

I have differential gene expression data from an RNA-Seq experiment. After filtering to include only genes from a previously developed Gene Ontology list, and filtering for significant genes based on adjusted p value, I am using the R pheatmap package to make clustering heatmaps.

Does this count as 'supervised' or 'unsupervised' clustering?

The descriptions here and here seem to suggest that hierarchical clustering is 'unsupervised'.

But does the pre-filtering of significant genes of interest make this 'supervised'? Is there more to the designation?

RNA-Seq heatmap clustering • 7.7k views
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Since you are using outside knowledge (differences between 2 known groups of samples) this would fall under supervised or semi-supervised clustering... However in a paper you could describe it as unsupervised clustering of differentially expressed genes and everyone would understand that it was semi supervised.

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Sorry for re-upping this post (it is always better than creating a new thread, I guess). So, if I got it right:

1. when you extract RNA from samples (let's say treated with something or even a specific phenotype) and,
2. when you have your genes annotated, and

you're asking how these genes cluster together then you are doing an unsupervised hierarchical clustering, correct?

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Clustering is typically an unsupervised approach. Unsupervised means you don't use external information to group your data points/items, i.e. grouping is based only on the data. In supervised learning, you make use of external information to form the groups, typically category labels to train a classifier. There are also intermediate situations called semi-supervised learning in which clustering for example is constrained using some external information.
So if you apply hierarchical clustering to genes represented by their expression levels, you're doing unsupervised learning.

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Thanks so much, Jean-Karim. Very helpful.

regards,

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5.5 years ago
Steven Lakin ★ 1.8k

This distinction has more to do with machine learning algorithm categories. While clustering is considered a subcategory of "machine learning," in your case what you're doing is mostly considered linear algebra.

Pre-filtering does not affect the category: the algorithm sees only the data, which in this case is an N-dimensional geometric space from which some sort of sample-wise distance is calculated. You can influence the way that clustering happens within pheatmap by using a different distance metric (e.g. "euclidean", "maximum", "manhattan", "canberra", "binary" or "minkowski") or by changing algorithm parameters (pheatmap uses k-means, so changing k).

You can also read more about different hierarchical joining methods by reading up on hclust, which is the function underlying pheatmap:

Ward's minimum variance method aims at finding compact, spherical clusters. The complete linkage method finds similar clusters. The single linkage method (which is closely related to the minimal spanning tree) adopts a ‘friends of friends’ clustering strategy. The other methods can be regarded as aiming for clusters with characteristics somewhere between the single and complete link methods. Note however, that methods "median" and "centroid" are not leading to a monotone distance measure, or equivalently the resulting dendrograms can have so called inversions or reversals which are hard to interpret, but note the trichotomies in Legendre and Legendre (2012).

Supervised in most machine learning contexts means using prior information (prior data) in order to inform a decision about new data, given some category of algorithm.

Unsupervised means using only the data itself to make some decisions about the data, again given some category of algorithm.

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Thanks for the clarification. I had gone looking for the documention on pheatmap to figure this out but couldn't find a vignette anywhere to describe the algorithms used. Agreed that it doesn't seem like much of a practical distinction, the motivation for finding this out was based more on the literal semantics of 'supervised' vs 'unsupervised' in regards to the experiment.

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# Math Help - Area of region.

1. ## Area of region.

Sketch the region enclosed by y = e^4x, y = e^9x, and x= 1. Decide whether to integrate with respect to x or y. The find the area of the region.

2. You learned how to sketch back in algebra, and also how to solve systems of equations. So you should have a nice picture with labelled intersection points.

Where are you stuck after that?


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94-310 Jorba A., Simo C.
On Quasiperiodic Perturbations of Elliptic Equilibrium Points (107K, LaTeX) Oct 10, 94
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers

Abstract. In this work we study quasiperiodic time-dependent perturbations of ordinary differential equations near elliptic equilibrium points, that is, equations like $$\dot{x}=(A+\varepsilon Q(t,\varepsilon))x+\varepsilon g(t,\varepsilon)+ h(x,t,\varepsilon),$$ where $A$ is elliptic and $h$ is ${\cal O}(x^2)$. Our results show that, under suitable hypothesis of analyticity, nonresonance and nondegeneracy with respect to $\varepsilon$, there exists a Cantorian set ${\cal E}$ such that for all $\varepsilon\in{\cal E}$ there exists a quasiperiodic solution such that it goes to zero when $\varepsilon$ does. This quasiperiodic solution has the same set of basic frequencies as the perturbation. Moreover, the relative measure of the set $[0,\,\varepsilon_0]\setminus{\cal E}$ in $[0,\,\varepsilon_0]$ is exponentially small in $\varepsilon_0$. The case $g\equiv 0$, $h\equiv 0$ (quasiperiodic Floquet theorem) is also considered. Finally, the Hamiltonian case is studied. In this situation, most of the invariant tori that are near the equilibrium point are not destroyed, but only slightly deformed and shaken" in a quasiperiodic way. This quasiperiodic shaking" has the same basic frequencies as the perturbation.

Files: 94-310.tex


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- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnline Submission ㆍMy Manuscript - For Reviewers - For Editors
 Eventual shadowing for chain transitive sets of $C^1$ generic dynamical systems J. Korean Math. Soc. 2021 Vol. 58, No. 5, 1059-1079 https://doi.org/10.4134/JKMS.j190083Published online July 27, 2021Printed September 1, 2021 Manseob Lee Mokwon University Abstract : We show that given any chain transitive set of a $C^1$ generic diffeomorphism $f$, if a diffeomorphism $f$ has the eventual shadowing property on the locally maximal chain transitive set, then it is hyperbolic. Moreover, given any chain transitive set of a $C^1$ generic vector field $X$, if a vector field $X$ has the eventual shadowing property on the locally maximal chain transitive set, then the chain transitive set does not contain a singular point and it is hyperbolic. We apply our results to conservative systems (volume-preserving diffeomorphisms and divergence-free vector fields). Keywords : Shadowing, eventual shadowing, chain transitive, locally maximal, generic, hyperbolic MSC numbers : Primary 37C50; Secondary 37D20 Downloads: Full-text PDF   Full-text HTML