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## Vol 49, No 1 (2003)

### Full Issue

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 (Russian)
Clinical endocrinology
Magnetic resonance imaging of the hypothalamopituitary area in the diagnosis of neurogenic diabetes insipidus
Dedov I.I., Vorontsov A.V., Vaks V.V., Marova Y.I., Averkiyeva Y.V., Vladimirova V.P.
###### Abstract

Fifty two patients (23 males and 29 females) with neurogenic diabetes insipidus (DI) were examined to define the role and specific features of brain magnetic resonance imaging (MRI) in the diagnosis of hypothalamopituitary abnormalities in this disease retrospectively after establishing the diagnosis. The group understudy included patients who had symptoms of only DI (they had no signs of chiasm or hypopituitarism). The following abnormalities were detected: no hyperintensive signal of the neurohypophysis body on T1-weighed images in 42.3%) of patients, bulky chiasmocellar masses in 13.6%), the empty ephippium in 15.3%), its structural abnormalities in 1.9%). Abnormal MRI scans were absent in 26.9% of patients with neurogenic DI. Since DI may be the first and only symptom of hypothalamopituitary tumor lesion, a sensitive topical technicjue, such as MRI, should be applied. It was found that brain MRI revealed no hyperintensive signal from the neurohypophysis body on T1-weighed images in 61%) of the patients with idiopathic DI, which may be a pathognomonic sign of idiopathic DI.

Problems of Endocrinology. 2003;49(1):3-7
Chronic gastroduodenal ulcers in patients with diabetes mellitus
Fedorchenko Y.L.
###### Abstract

The clinical picture, course, and treatment of gastroduodenal ulcers (GDU) in diabetics were studied. A total of 395 diabetics were examined; GDU were detected in 36. The incidence of gastric and duodenal ulcers was similar in patients with insulin-dependent diabetes mellitus, while in patients with non-insulin-de- pendent diabetes gastric ulcers predominated. The clinical picture of the disease, gastric acid production, Helicobacter pylori infection, and blood gastrin levels were studied in all patients with ulcers. The efficiency of GDU treatment with quamatel, ranitidine, and antacids was evaluated. The clinical course of GDU in diabetics was asymptomatic. The highest incidence of H. pylori, infection was observed in patients with type 1 diabetes with concomitant peptic ulcers. Serum gastrin levels were more frequently increased in patients with type 1 diabetes and duodenal ulcers and normal in patients with type 2 diabetes. Quamatel therapy was highly effective in diabetics with GDU. Ulcers healed in 85% patients and blood gastrin level significantly decreased after 3- week therapy. GDU in diabetics are characterized by specific location, clinical course, laboratory and instrumental features, which allows a differentiated approach to therapy of these patients.

Problems of Endocrinology. 2003;49(1):7-12
The left ventricular myocardium in children at early stages of diabetes mellitus: Functional and morphometric parameters
Kuznetsova I.G., Filaretova О.V., Nastausheva T.L., Stolnikova T.G., Zazulina О.V.
###### Abstract

Its morphometric parameters (myocardial mass according to the procedure described by R. Devereuxe, end-diastolic volume and their ratio). Sixteen healthy children of the same age formed the control group. A gradual redistribution of transmitral flow towards the atrial systolic phase was found in children with a more than 5-year history of diabetes, which seems to be associated impaired L V myocardial relaxation due to myocardial metabolic changes and to the occurrence of specific microangiopathies. The found changes progressed with time and correlated with the appearance ofperipheral distal neuropathy. Within the first year of diabetes, there was a transient reduction in LV contractility, which appears to be associated with myocardial metabolic imbalance during the decompensation of the disease. There is a trend to LV myocardial hypertrophy 10years after the onset of the disease.

Problems of Endocrinology. 2003;49(1):12-16
Paraoxonase activity and lipid peroxidation in female patients with type 2 diabetes mellitus and without coronary heart disease
Gorshunskaya. M.Y.
###### Abstract

The rate of lipid peroxidation and the parameters of antioxida- tive defense, including the activity of paraoxonase that is essential for the prevention of low-density lipoprotein oxidation, was studied in 229 female patients with type 2 diabetes mellitus with and without coronary heart disease (CHD) under varying glycemic control. Carbohydrate and lipid metabolisms were explored by unified biochemical studies, blood insulin levels were measured by radioimmunological assay. The activity of paraoxonase associated with high-density lipoproteins of ester hydrolase was spectrophotometrically determined by using paraoxan as a substrate. Along with dyslipoproteinemia and insulin resistance, there was a drastically reduced paraoxonase activity that was associated with the high-density lipoproteins of the antioxidant enzyme and more pronounced in diabetics with CHD. A highly significant inverse correlation of the activity of the enzyme with the rate of lipid peroxidation and a less close relationship to basal glycemia have been verified, which substantiates the polygenic nature of decreased paraoxonase activity in diabetes mellitus.

Problems of Endocrinology. 2003;49(1):16-19
Clinical and morphological features of reparative processes in patients with the diabetic foot syndrome
Udovichenko. О.V., Tokmakova A.Y., Antsiferov M.B., Yushkov P.V., Dedov I.I.
###### Abstract

The purpose of the study was to examine structural abnormalities in the skin and granulation tissue, which explain long-term healing and recurrent trophic ulcers in patients with the neuropathic diabetic foot syndrome who received adequate treatment. Biopsy specimens were taken from the granulation tissue and skin of the edge of a trophic ulcer in 12 patients (mean age 48.4$±$6.6 years) with types 1 and 2 diabetes mellitus (its duration was 14.0$±$4. 7 years). All the patients were treated for the neuropathic diabetic foot syndrome according to the generally accepted recommendations. According to the duration of trophic ulcer, the patients were divided into 2 groups: 1) 4 patients with a history of under 3 months and 2) 8 patients with a history of above 3 months. According to the presence or absence of trophic ulcers in the history, all the patients were again divided into Groups A (recurrent ulcers, n = 5) and В (primary ulcers, n = 7). Patients with slowly healing ulcers were found to have an abnormal foot pressure more frequently, as evidenced by computerized pedography (the peak load in the ulcer area being in 63% of them versus 25% in Group 1) and some specific features of granulation tissue (excess of active fibroblasts, immaturity of the extracellular matrix, atrophy and sclerosis of nerve fibers). Patients with recurrent ulcers were older, demonstrated lower vibration perception scores, morphological features, such as abundant active polymorphonuclear leukocytes, mast cells with signs of degranulation and T-helper cells, immature extracellular matrix, fibrinoid necrosis and young capillary vessels with microthrombi. Thus, the detection of the above granulation tissue and wound edge changes (especially with an abnormal foot pressure pattern, significantly low vibration perception scale, and old age) makes it possible to suggest slow healing or recurrence of ulcer.

Problems of Endocrinology. 2003;49(1):19-24
Impact of hypolipidemic therapy on carbohydrate and lipid metabolism and their hormonal regulators in patients with type 2 diabetes mellitus
Starkova N.T., Dolgov V.V., Davydov A.L., Roitman A.P., Baranova L.Y., Trubnikova T.Y., Sazhin N.L.
###### Abstract

The effects of lipostat and maninil on carbohydrate and lipid metabolisms and their hormonal regulators were studied in 36 obese patients aged 50- 70 years who had type 2 diabetes mellitus concurrent with dyslipidemia. A course of therapy with lipostat in a daily dose of 20 mg for 3 months was shown to lead to normalization of lipid metabolism, to diminished glycemia and hyperinsulinemia, and to an increase in fasting plasma somatotropic hormone levels to normal values, these were not observed in the control group.

Problems of Endocrinology. 2003;49(1):24-26
Problems in the diagnosis of the adrenals with incidental
Kuznetsov N.S., Belchenko L.V., Yushkov P.V., Manovitskaya A.V., Opalenov К.V., Mamayeva V.G.
###### Abstract

The paper analyzes the case records of 123 patients with a primary diagnosis ofadrenal incidentaloma who were treated at the Endocrinology Research Center, Russian Academy of Medical Sciences, in the past 5 years. Ofthem, 87patients were operated on at the surgical department of the Center. Thirty  six patients with a clinically verified diagnosis ofhormonally inactive adrenal tumor without signs ofmalignant growth werefollowed up. All the patients underwent comprehensive clinical and instrumental studies (ultrasonography, magnetic resonance imaging, computed tomography, hormonalstudies). Theirresults were comparedwith the morphological data on distal adrenal changes. The hormonal activity ofadrenal tumors was morphologically determined by histochemical techniques. The study revealed true hormonally inactive tumors only in 12 cases. It is concluded that the concepts of adrenal incidentaloma and of true hormonally tumor are not equivalent and that it is necessary to make a comprehensive clinical and morphobiochemical analysis in adrenal tumors to establish a final nosological diagnosis.

Problems of Endocrinology. 2003;49(1):26-31
Magnetic resonance imaging in the diagnosis of hypothalamopituitary diseases in patients with true premature sexual development
Vorontsov A.V., Semicheva T.V., Peterkova V.A.
###### Abstract

То evaluate the incidence of different types of organic abnormalities occurring in premature sexual development (PSD) of central genesis and to study a role of magnetic resonance imaging (MRI) in the diagnosis of PSD, MRI was used in 33 patients with a clinically established diagnosis of true (gonadotropin-dependent) PSD of central genesis. Among the examined children with PSD, hypothalamic hamartoma was detected in 27.3%) of the patients; there were no MRI data suggesting hypothalamopituitary abnormalities in 72.7%o. On the MRI images, a hypothalamic hamartoma appeared as a round or oval mass having a well -defined outlines, a homogeneous structure, 3 to 15 mm in size, which is located in the projection of mamillary bodies or retrochiasmally. Patients with PSD were found to have an increased volume of the adenohypophysis mainly due to the vertical size as compared with a group of healthy children of the same age.

Problems of Endocrinology. 2003;49(1):32-36
Complete elimination of iodine deficiency-induced diseases in the Republic of Serbia through overall salt iodination
Simic M., Banisevic M., Andjelkovic Z., Zivic G., Zikic L., Beloglav D., Simic D., Gerasimov G.A., Mishchenko B.
###### Abstract

The territory of Serbia has been long known to be an iodine deficiency area. The surveys conducted in the past 50 years have shown that iodine deficiency disorders ranged from small goiters to endemic cretinism. The iodine deficiency control program has been implemented in Serbia since 1951, in so doing, the salt concentration of iodine (as potassium iodide) has been gradually increased and at present it is 20 mg/kg. Just in the mid- 1960s, iodine prophylaxis led to cessation of new cases of cretinism and large goiters.This survey was undertaken to determine the iodine provision in Serbia. The survey was conducted in accordance with the WHO, UNICEF, and ICCIDD guidelines and it covered 4598 schoolchildren aged 7 to 15 years from the urban and rural areas of 44 municipalities of Central Serbia and Vojvodina. Among all the examinees, the detection rate of the enlarged thyroid was 2.35% (as evidenced by ultrasound study) and the median urinary iodine concentration was 158 $\mu$g/l. The findings suggest that iodine deficiency has been completely eliminated in Serbia due to the effective program of the overall dietary salt iodination.

Problems of Endocrinology. 2003;49(1):37-40
Chorionic gonadotropin regulation of intracellular potassium levels in human blood immunocompetent cells; a role of phases of a menstrual cycle
Shirshev S.V., Lyalina O.G.
###### Abstract

The effect of chorionic gonadotropin (CG) on intracellular potassium ion( [K${}^{+}$]${}_{i}$) levels in human blood immunocompetent cells was studied by taking into account the phases of a menstrual cycle. CG was used in doses of 10, 50, and 100 IU/ml. Plasma photometry was used to measure the level of [K${}^{+}$]${}_{i}$ in the fractionated peripheral monocytes and lymphocytes from males, as well as from females in the late follicular and luteal phases of a menstrual cycle. CG used in a dose of 100 IU/ml was found to lower flCf in the monocytes and lymphocytes of males and in a dose of 50 IU in the lymphocytes of females in the luteal phase of a menstrual cycle. The hormone used in large doses stabilized the level of [K+]${}_{i}$ in both types of cell of female in the follicular phase. It is concluded that CG can modulate the level of [K+]${}_{i}$ in human peripheral blood immunocompetent cells, and the directionality of its effects depends on a type of hormone-accepting cells and it is determined by a menstrual phase.

Problems of Endocrinology. 2003;49(1):41-44
For practitioners
Forty-year experience in diagnosing and treating chromaffine tissue tumors
Dedov I.I., Kuznetsov N.S., Beltsevich D.G., Kuratev L.V.
###### Abstract

The article devoted to the forty-year experience in diagnosing and treating chromaffine tissue tumors

Problems of Endocrinology. 2003;49(1):44-50
Experimental endocrinology
Neurophysiological effects of corticotropin-releasing factor in the undergoing sections of the olfactory area of the rat cerebral cortex
Mokrushin A.A., Shalyapina V.G.
###### Abstract

Application of corticotrophin-releasing factor (CPF) in concentrations of ${10}^{-9}$-${10}^{-8}$ M on the rat olfactory cortex slices induced activation of the pre- and postsynaptic excitatory components of the focal potentials recorded in the slices. The amplitude and duration of α-amino-3-hydroxy-5-methyl-isoxazole-4-propionic and N-methyl-D-aspartate components of postsynaptic evoking potentials increased upon exposure to CRF, while the amplitude of GABAB-mediated inhibitory postsynaptic potentials was suppressed. At higher concentrations of CRF (${10}^{-8}$ M) epileptiform charges were recorded in the cells. CRF effects were reversible and eliminated after washing. A long (90 min) exposure induced the phenomena similar to long-term posttetanic potentiation. The findings suggest that CRF has pronounced activating properties and ajfects the glutamatergic and GABAergic systems.

Problems of Endocrinology. 2003;49(1):51-53
Endocrine and messenger systems during adaptation to high-altitude conditions
Vishnevsky A.A., Zakirov D.Z., Yakovlev V.M., Zholdubayeva L.Y., Zakharov G.A.
###### Abstract

Of endocrine cascades and messenger systems, the authors conducted this study involving two series of experiments. The pattern of interaction of hormones and secondary messengers while adapting to high-altitude was clarified in the first series. The response of phosphoinositide to exogenous epinephrine was examined in the second series. The experimental data are represented as primary mean values ± standard deviations. Two phases of activity of the hormonal system are shown. These are the excitation of the hypothalamopituitary cascade and the enhanced production of glucocorticoids and catecholamines during early adaptation and the return of activity of the hormonal system to the baseline values by day 14 of adaptation. By that time the rearrangements in the phosphoinositide messenger system consists in increasing the proportion of polyphosphoinositides in the total membranous fraction of inositol-containing phospholipids, in adrenalectomized animals to a greater extent. The above events give rise to new coordination relations between hormones and secondary messengers. It is concluded that that the phosphoinositide messenger system is involved in the formation of defense mechanisms the reduce the destructive effects of large amounts of epinephrine.

Problems of Endocrinology. 2003;49(1):53-56
Obituary
In memory of A. M. Granovsky-Tsvetkova

###### Abstract

The necrology devoted to the memory of A. M. Granovsky-Tsvetkova

Problems of Endocrinology. 2003;49(1):56-56


text:
mersenneforum.org Is it OK to pick up expired exponents?
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 2014-04-03, 18:20 #1 patrik     "Patrik Johansson" Aug 2002 Uppsala, Sweden 52×17 Posts Is it OK to pick up expired exponents? I just want to ask if it is okay to pick up expired exponents (by the "usual" method of putting them into the worktodo.txt and communicating). The reason I ask is that they no longer just disappear from the re-designed exponent status pages, but show as assigned but expired a few days ago. For over three years (IIRC) I have been working in the 10M digit DC range when the DC wavefront hit this range in late January. The reservations made in January are now beginning to expire. (If I reserve them, I'll be putting them only on fast hardware.) Last fiddled with by patrik on 2014-04-03 at 18:21
2014-04-03, 19:12   #2
chalsall
If I May

"Chris Halsall"
Sep 2002

235048 Posts

Quote:
 Originally Posted by patrik I just want to ask if it is okay to pick up expired exponents (by the "usual" method of putting them into the worktodo.txt and communicating).
I would argue yes, so long as Primenet gives you an AID using the "usual method" of putting the candidates into the worktodo.txt without an AID, and having Prime95/mprime communicate with Primenet. Any which don't get an AID should be removed from the worktodo.txt file.

George (of course) can speak in more depth with regards to how Primenet handles this situation.

 2014-04-03, 21:35 #3 patrik     "Patrik Johansson" Aug 2002 Uppsala, Sweden 52×17 Posts As a test, I tried to reserve two exponents. Code: 33340063 Assigned 2014-01-30 ANONYMOUS D expired on 2014-03-31 33343897 Assigned 2014-01-28 ANONYMOUS D expired on 2014-03-29 The first gave me: Code: PrimeNet error 40: No assignment ra: already assigned, exponent 33340063, A: 1, b: 2, c: -1 while the second registered as it should. However, the exponent I picked up now shows as Code: 33343897 Assigned 2014-01-28 Patrik Johansson D expired on 2014-03-29 Maybe someone else already picked up the first one? I discovered, just before posting, that none of the exponents show up on my assignments page (this was expected for the first one, but not for the second). So I think I'd better stick to exponents without the "assigned but expired" attribute.
2014-04-03, 22:37   #4
chalsall
If I May

"Chris Halsall"
Sep 2002

22·7·359 Posts

Quote:
 Originally Posted by patrik I discovered, just before posting, that none of the exponents show up on my assignments page (this was expected for the first one, but not for the second). So I think I'd better stick to exponents without the "assigned but expired" attribute.
Four axioms, if I may...

1. Code is hard.

2. Humans are error prone.

3. Deterministic systems do *exactly* what you tell them to.

4. Testing is good.

2014-04-09, 05:04   #5
snme2pm1

"Graham uses ISO 8601"
Mar 2014
AU, Sydney

11·23 Posts
TF expired

Quote:
 Originally Posted by patrik is okay to pick up expired exponents
I recognise that your thrust is for Lucas–Lehmer or related double tests.
More generally, in accord with the thread title, I might venture that I have no hesitation to recognise an expired Trial Factor assignment within my interest range and simply provide a quick resolution.
By quick, I mean that upon recognising a particularly obstinate delayed job, and prior to PrimeNet making the exercise publicly available, I might prosecute the work but immediately prior to submitting a result, check that nobody has acquired such assignment.
Only once in the past year did I have such reason to abstain from submitting a result due to somebody acquiring the assignment.
(For that exception, it took the acquirer a couple of weeks to resolve even though I knew what the factor gained result would be, but that's ok).

2014-04-09, 18:20   #6
TheMawn

May 2013
East. Always East.

172710 Posts

Quote:
 Originally Posted by snme2pm1 I recognise that your thrust is for Lucas–Lehmer or related double tests. More generally, in accord with the thread title, I might venture that I have no hesitation to recognise an expired Trial Factor assignment within my interest range and simply provide a quick resolution. By quick, I mean that upon recognising a particularly obstinate delayed job, and prior to PrimeNet making the exercise publicly available, I might prosecute the work but immediately prior to submitting a result, check that nobody has acquired such assignment. Only once in the past year did I have such reason to abstain from submitting a result due to somebody acquiring the assignment. (For that exception, it took the acquirer a couple of weeks to resolve even though I knew what the factor gained result would be, but that's ok).
This seems fair, though I might suggest taking work in smaller batches if your assignments expire frequently.

2014-04-09, 21:54   #7
snme2pm1

"Graham uses ISO 8601"
Mar 2014
AU, Sydney

11×23 Posts

Quote:
 Originally Posted by TheMawn I might suggest taking work in smaller batches if your assignments expire frequently.
To clarify, I was talking about languishing TF jobs that other members have abandoned until expiry.
I believe that jobs assigned to me have never been left long enough to expire.

2014-04-09, 22:59   #8
chalsall
If I May

"Chris Halsall"
Sep 2002

22·7·359 Posts

Quote:
 Originally Posted by snme2pm1 To clarify, I was talking about languishing TF jobs that other members have abandoned until expiry.
To clarify, could you give a single example of what you're talking about? Leave off the last three or so digits and replace them with "x"'s if you so wish.

2014-04-09, 23:20   #9
snme2pm1

"Graham uses ISO 8601"
Mar 2014
AU, Sydney

11·23 Posts

Quote:
 Originally Posted by chalsall single example.
The most recent example of an expired exponent that I wanted to clear out was 78086249.
Assigned 2013-01-07;<SomeoneElse>;TF;expired on 2014-04-05
I nudged it from 70 bits up to 72 bits soon after spotting it, to save PrimeNet from reassigning it, then further to 74 bits.

Last fiddled with by snme2pm1 on 2014-04-09 at 23:21

2014-04-09, 23:30   #10
chalsall
If I May

"Chris Halsall"
Sep 2002

274416 Posts

Quote:
 Originally Posted by snme2pm1 The most recent example of an expired exponent that I wanted to clear out was 78086249.
OK. Not terribly valuable work in the short term, but worthwhile in about a year or so. And honorable that you didn't do this work while it was assigned.

http://www.mersenne.org/report_expon...exp_hi=&full=1

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# Unbounded subset of $\mathbb{R}$ with positive Lebesgue outer measure

The set of rational numbers $\mathbb{Q}$ is an unbounded subset of $\mathbb{R}$ with Lebesgue outer measure zero. In addition, $\mathbb{R}$ is an unbounded subset of itself with Lebesgue outer measure $+\infty$. Therefore the following question came to my mind: is there an unbounded subset of $\mathbb{R}$ with positive Lebesgue outer measure?

If there is, can you give me an example?

-
$[0,1]\cup\mathbb Z$ –  Mariano Suárez-Alvarez Jan 17 '11 at 21:53
Thank you, Mariano! I see, the idea is to "add" an unbounded countable set to a set with positive Lebesgue outer measure. –  ragrigg Jan 17 '11 at 21:57

I guess you mean with positive and finite outer measure. An easy example would be something like $[0,1]\cup\mathbb{Q}$. But perhaps you also want to have nonzero measure outside of each bounded interval? In that case, consider $[0,1/2]\cup[1,1+1/4]\cup[2,2+1/8]\cup[3,3+1/16]\cup\cdots$. If you want the set to have positive measure in each subinterval of $\mathbb{R}$, you could let $x_1,x_2,x_3,\ldots$ be a dense sequence (like the rationals) and take a union of open intervals $I_n$ such that $I_n$ contains $x_n$ and has length $1/2^n$.
On the other hand, it is often useful to keep in mind that every set of finite measure is "nearly bounded". That is, if $m(E)<\infty$ and $\epsilon>0$, then there is an $M\gt0$ such that $m(E\setminus[-M,M])<\epsilon$. One way to see this is by proving that the monotone sequence $(m(E\cap[-n,n]))_{n=1}^\infty$ converges to $m(E)$.


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0

They are one and the same, and the opposing model is the Cold Big Bang. Cold big bang is a designation used in cosmology to denote an absolute zero temperature at the beginning of the Universe, instead of a (hot) Big Bang. Also see the Big Bang page of wiki, it also has clues to why it may be called hot. Cold Big Bang

1

We have, at this time, no tools capable of detecting neutrinos at the very low energies of the cosmic neutrino background, and if such tools existed they would have to contend with numerous backgrounds making the experiment ferociously difficult.

1

There is a very nice paper exactly on this topic, where the expressions describing the fit curves are derived: http://arxiv.org/abs/hep-ph/9906447v1 If you have a look at the final expression (formula (6.5) on page 8), you will agree, that the relation between $\Omega_\Lambda$ and the luminosity is hard to describe by words. However, you can try to think ...

1

First of all, we still do not know if the 4 fundamentals forces can be unified. The best we have came up to are GUT theories, which join three of the fundamental forces except gravity. The fact that gravitational waves have been detected, may point towards GUT since they address energies of the GUT scale. Nonetheless, assuming that String theory, which is ...

1

Nope, gravitational waves were predicted by Einstein's General Relativity theory which is a very successful theory of gravity. We still though do not have a working theory of Quantum Gravity and cannot therefore unify these forces as of yet.

0

Leptons annihilate to photons. Hadrons (quarks, 99.97 mass-% of matter) emit about 50% of ${mc^2}$ annihilation energy as neutrinos. If the Big Bang plus cosmic inflation had exactly obeyed all current conservation laws, everything would have exactly cancelled to photons and neutrinos. Pair formation must be antisymmetric re conservation laws, offering no ...

2

The key bit of your question is: why isn't it possible all original matter and anti-matter annihilated each other leaving all pure energy, which in turn and in time could form into matter? since other parts of it are covered by the question Parth Vader linked. Matter and anti-matter don't annihilate to produce pure energy, they annihilate to produce ...

-3

Energy equals mass multiplied by the speed of light squared, and mass equals energy divided by the speed of light squared. What goes up, must come down. Nothing can escape a black hole. Every action has an equal and opposite reaction. Energy cannot be created or destroyed, it only changes form. All things in an enclosed system gain entropy, slow down, ...

-1

The simplest idea is that the collapse of the previous universe caused the big bang. I imagine it as black holes smashing into each other at great speed, immediately causing a nuclear reaction. There are some unproven theories that need to be adjusted, such as the idea that dark energy density remains constant as space expands. Black hole radiation has ...

3

There are several scenarios that point out cyclic cosmologies. The first one is called Matter bounce cosmology which is based on the idea that that the universe originated from a cosmological bounce in which quantum fluctuations develop into a scale-invariant spectrum of curvature perturbations. The bounce is realized beyond Einstein's General Relativity and ...

1

I'm not sure exactly what you're referring to as the three different theories, but recollapsing, accelerating and anything in between universes are not different theories. They are all examples of a solution to the equations of general relativity called the FLRW metric. If we make a few physically plausible assumptions about the distribution of matter in ...

3

The dividing line meets at $(\Omega_m,\Omega_\Lambda)=(0,1)$. From the Friedmann equations, it follows that the scale factor $a(t)$ satisfies the relation $$\frac{\dot{a}^2}{H_0^2} = \Omega_m a^{-1} + (1 - \Omega_m - \Omega_\Lambda) + \Omega_\Lambda a^2.$$ The universe has no big bang singularity if the above expression is negative (or zero) for some ...

0

There is a central point, the point that is the opposite to the expansion of the sphere. We can't see it, but I would assume that in a similar way to how when large stars go supernova they leave behind a black hole, and how there are black holes in the middle of galaxies, there would probably be a black hole in the centre. We would never see it, because ...

0

There are a few other reasons why we think the universe is expanding. Wikipedia lists and explains them far better than I will ever be able to. One thing though, your ruler is not expanding. The intermolecular forces are making sure the distance between molecules stays the same. Empty space is expanding, expanding essentially the distances between weakly ...

-1

Remember, the Big Bang theory is just that- a theory. I predict astrophysicists will soon discover galaxies 15,20,25, billion years old. Then what?

6

There exists a huge gap in the strength of the four forces that we have observed in nature between gravity and the other three: In the following image we see that the radiation decouples from the "soup" at energy densities of 0.25eV. That is the snapshot of CMB, Cosmic Microwave Background radiation . CBR in this plot is the CMB, when we have a ...

1

Carbon has to be produced by the triple-alpha process because there is no stable nucleus with 8 or 5 nucleons. The probability of this is very low, because it requires three different particles to be in the same place at the same time. You'll note that the Wikipedia article says: One consequence of this is that no significant amount of carbon was ...

5

Well, the argument is something like this. Matter and radiation are tightly coupled until the time of last scattering. The photons are in thermal equilibrium with the plasma surrounding them. This means the state of the Universe at the time of last scattering is very precisely imprinted on the photons, and the photons travel from last scattering to us more ...

3

You've overlooked gravitational entropy. The entropy of a black hole horizon is given by: $$S = \frac{kA}{4 \ell_p^2}$$ where $A$ is the area of the horizon, $k$ is Boltzmann's constant and $\ell_p$ is Planck's constant. This entropy is absolutely huge, so if you take a uniformly distributed gas in thermal equilibrium and concentrate it into a black ...

5

Besides the comments, stars have formed well after the Big Bang. Depending on where they are in the universe, it will take a while for us to see them.

6

The time used in describing the evolution of the universe is comoving time. This is the time that would be measured by a freely moving observer on their wristwatch (assuming the high temperatures didn't melt both the observer and the wristwatch :-). Time is not a simple thing to define in general relativity, however we can always unambiguously define proper ...

19

This is a common point of confusion, not only with regards to inflation, but any time an expanding universe comes up... The "cosmic speed limit" as you call it says that no particle or signal can move through spacetime faster than the speed of light. Spacetime is a very specifically defined thing, described with a coordinate system. There is no restriction, ...

-3

I think that dark matter disproves the big bang. The energy excerpted by dark matter on all things would have prevented the universe from coming into existence through the big bag.

4

In curved spacetime, you can no longer compare velocities at different points in the straight-forward manner we use in flat spacetime. Thus the claim that recession velocities should not be considerer 'real' (as in relative) velocities, but rather rates of expansion of space. If you want to get at the former, you need to parallel transport the source's ...

0

After the Universe became transparent to electromagnetic waves, the first wave produced from whole mass lump was Cosmic Microwave Background (CMB) radiation which we can still receive from farthest point of the observable universe (CMB causes your radio to hiss on blank stations). Universe has expanded since then, so CMB is still running trapped inside ...

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# Difference between Larson's Calculus

1. May 9, 2010

### MushroomPirat

I'm looking to buy an undergraduate Calculus textbook, mostly for self-studying, since I'm currently in high school.
I've been looking on here and on amazon for some texts, and the books by Larson seem to be pretty good. However, there's two books which seem to cover almost the exact same thing: the book simply called "Calculus" and the one called "Calculus: Early Transcendental Functions". I was wondering if anyone knew what the difference was between these. I've seen some other calculus textbooks have this distinction as well.

If anyone's interested, here are the links to the google books version of these texts:

Calculus:

Calculus: Early Transcendental Functions:

Just by looking at the table of contents, they seem to have the exact same material, and even looking at the first few pages, they seem to be the same word for word.

EDIT: Oh, also, I was wondering if there are generally significant changes to the textbook in different editions. The older version is a lot cheaper, so I think I much rather get that one, rather than spend over a hundred on a new one if theres no new material. Sorry if these questions seem really basic; this is my first time buying a university level textbook. Thanks :)

Last edited: May 9, 2010
2. May 10, 2010

### Reedeegi

If I were you, I actually would not opt to purchase Larson's text if you wish to truly learn mathematics. Larson's book (as most high school and college texts do) focus on computation and application of mathematical concepts without devoting time to proofs and rigor, which is vitally essential to understanding mathematics. Instead, I'd recommend two rather inexpensive books to learn calculus from.

Elementary Real and Complex Analysis by Georgi Shilov (http://store.doverpublications.com/0486689220.html)

Introduction to Analysis by Maxwell Rosenlicht
http://store.doverpublications.com/0486650383.html

The former is more computational (while still devoting time to proof) whereas the latter is far more proof-based. Given that the two cost roughly $30 when bought together (a far better deal than buying the$110 books) and that they cover far more material than the run-of-the-mill high school calculus text, I recommend that you purchase these texts.

3. May 10, 2010

### MushroomPirat

Hey Reedeegi,

Thanks for your reply! I looked at the books you mentioned quickly on google books and I actually find them quite interesting. It really is a different approach than what I'm used to.

I'm not trying to learn calculus from scratch; I've done calculus up to about the AP Calculus AB level (if you're familiar with that), with more integration techniques like trig sub, partial fractions, and t-sub.

However, I think it would be interesting to re-learn what I do know with those two books you mentioned, with the theory with it. There IS one thing I was thinking about though. I'm planning to go into engineering once I get into university (mechanical or electrical). You mentioned that the "regular" calculus textbooks devote time to the applications of the mathematical concepts. Wouldn't that be important to have when going into engineering? Would learning math from this approach benefit as well?

Sorry if these questions sound really dumb. I'm really not sure how different approaches to math are used by different people.

EDIT: Oh, just thought I should mention. I started looking at calculus textbooks with the intent that I could brush up on Differential and Integral Calculus principles that I haven't covered at school, and then possibly using the same book to self-teach multivariable calculus. So, I'm not sure which direction to go here... To do some analysis and take a more theoretical (I think) approach, or keep going down the 'regular' path. Thanks.

Last edited: May 10, 2010
4. May 11, 2010

### jasonRF

I would actually recommend a more traditional approach of learning basic calculus before advancing to analysis. I am not familiar with Larson (but it is likely reasonable), but I am with Anton ("Calculus") and Thomas and Finney (now called "Thomas's calculus" I think). Both are good. Used copies of old editions are certainly the way to go. A 15 or 20 year old calculus book would be perfectly fine, and can probably be purchased cheap. For example, at Amazon a 1995 edition of Anton's book can be found for less than $5 if you include shipping; https://www.amazon.com/Calculus-Ana...r_1_11?ie=UTF8&s=books&qid=1273579853&sr=8-11 If it doesn't work for you, you are only out$5.

While the google books previews of Rosenlicht and Shilov lead me to guess that they are pretty good books, most of us mortals couldn't jump into a self-study of books at that level without even knowning any calculus ahead of time. The Rosenlicht book is still used at some universities for upper division (undergraduate) analysis courses, all of which require calculus as a prerequisite. They both discuss something called metric spaces before even addressing continuity and derivatives, which is quite the tough road to go. Of course, you can pick up one of those cheap as well and see how it goes, and again you won't be out much.

I am not a mathematician, but I have self-taught myself elementary analysis after I had already taken a fair amount of "easy" undergraduate math classes. In my personal experience, having an intuitive understanding of basic calculus and having experience with a broad variety of functions that a calculus background provides, makes the more abstract definitions and theorems in analysis possible to understand because I have the background to make up my own examples.

jason

Last edited by a moderator: Apr 25, 2017
5. May 24, 2010

### Pinu7

DO NOT LISTEN TO Reedeegi. Those are books on analysis, not Calculus. You should familiarize with the computational techniques in Calculus before touching Analysis

Most calculus books are the exact same, so authors like Larson, Antony, Stewart, or Thomas give more or less the same thing.

Somewhat slightly more advanced texts on Calculus includes Apostol, but that is currently out of print. I suggest using an "easy" book such as Kline for single variable calculus. After that, I would recommend a harder book for multivariable calculus(eg Advanced Calculus of Several Variables- Edwards).

6. Aug 28, 2010

### hitmeoff

May be a few months late here, but I agree with Pinu and jasonRF.

I learned from Larson's Calc text and have seen the other popular ones (Stewart's, etc). Yes they are all pretty similar and you would do well with any, but because I learned from Larson, I obviously am biased towards it. I just think its a super clear text.

If you are self studying and need a cheap calc book, go with Larson and get an edition thats one or two editions old for next to nothing.

If you try to jump right into an analysis text without having a real solid idea of HOW TO DO calculus then you are asking for a lot.

Chances are you have yet to touch a real, serious theoretic-axiomatic approach to math. Though beginning calc is not "axiom axiom theorem theorem proof" its also not just plug-n-chug like most of the math youve taken before. It does force you to think a little more critically about how to set up problems, break big problems down into manageable pieces, etc which are skills needed if you want to succeed at upper level math.

If you jump into analysis right away, you will be trying to learn calculus and learn how do proofs all without first developing your analytical and critical thinking skills. Possible? Sure. Hard? Definitely? My feelings are that this approach will not allow you to learn much "usable" calculus and most of the "analysis" will go over your head.

Get through Larson and maybe pick up an older edition of Anton's Elementary Linear Algebra (i suggest this because you will eventually have to take elementary linear algebra AND because Anton's book does a very good job at the "Elementary" part of Lin Alg, plus you will get introduced to writing proofs as well). From there go on to a solid Analysis text and maybe a more rigorous Linear Algebra book (Friedburg, Axler, etc).

As far as learning analysis for the sake of physics...you have to realize that most analysis, algebra and even linear algebra and ODE/PDE books are written for math majors. Which means not only are they heavy on proofs and light on computation, they also rarely touch on how to apply this more advanced math to other sciences.

I cant tell you how many times ive taken an upper dic math class or picked up a math book because the words "with applications" was part of the description, only to find out that by "applications" they mean applications to other problems in abstract math!

Certainly analysis, algebra, linear algebra, topology, ODE/PDE, differential geometry and so on have their applications in science, especially theoretical physics and chem. But usually you are taught the math you need in the science courses you need it in. I also dont like this approach and most "Math for Physicists" courses only cover that basics of higher level math. So finding a book or taking a course that teaches SERIOUS ADVANCED MATH for SCIENTIST (and not mathematicians) is something Ive been trying to find.


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filtering rRNA from fastq files or mapping results
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6.9 years ago

Hi everybody,

I am new to RNA-Seq. I am working on a RNA-seq data set from the mouse genome. I have ran some quality testing using the fastqc and fastx software. I have found out that the data set is full of rRNA reads.

I would like to filter there reads out, but I am not sure how to go forward with this. I have also read this post, but it didn't help much.

As I have done the mapping against the ensembl GRCM38 version, I guess it is better to work with the ensembl file. But how to do the filtering?

Is there a way to filter the rRNA reads before the mapping?

I know I can convert the gtf files into GenomicRanges object. BUT will that help me to remove the reads mapped to this coordinates after the mapping. if so how?

The last option, as reads in the above mentioned post is to take the list of rRNA ensembl IDs and just filter the list of results from featureCounts or HTseq-count to remove those rRNAs from it.

Is this option better than the other two?

thanks in advance for any information

Tomas

rna-seq gtf rRNA • 15k views
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6.9 years ago

Ensembl puts rRNA in column 2 of its GTF files if the line describes an rRNA, so just grep -wv rRNA Mus_musculus.GRCm38.78.gtf to get rid of them. You'll map them, but not count them. Mapping is fast, you don't need to bother removing them ahead of time.

BTW, you actually don't need to bother removing these at all in most cases. Any decent normalization method (e.g., that used be DESeq2 or edgeR) won't be messed up by differences in rRNA amount.

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Yeah. Filtering some reads in advance of sophisticated normalization could skew the experiment. You want to get rid of rRNA before sequencing, then hand all read data to the analysis software unmolested.

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Yes I know that. The sequencing facility did run rRNA depletion twice, but there still were a lot of rRNA residues in the fastq files I have. So now I need to figure out, what to do with it.

Thanks Devon, I will try it without removing them, but than just filtering them from the list of read counts.

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I think in ribosomal profiling, it might be useful to remove the rRNA before mapping. Ribo-seq data might still contain about 20-30% of total RNA as rRNA. Mapping this together with rest of non rRNA reads might screw the downstream quantification.

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My zebrafish data also has ribosome RNA contamination and my boss mandatorily asked me to use cufflinks...As the cufflinks manual says""Tells Cufflinks to ignore all reads that could have come from transcripts in this GTF file. We recommend including any annotated rRNA, mitochondrial transcripts other abundant transcripts you wish to ignore in your analysis in this file. Due to variable efficiency of mRNA enrichment methods and rRNA depletion kits, masking these transcripts often improves the overall robustness of transcript abundance estimates." I think -M is the right way? Then I tried to get the complete rRNA/tRNA GTF files, at first I though rmsk from UCSC table browser is enough but I found this post in UCSC google groups: " the rmsk table contains coordinates for various repeat families. Some of these repeat families are derived from specific RNA families, such as rRNA or tRNA. These differ from the actual rRNA and tRNAs, the coordinates for which are found in different tracks. " So I am confused, how should I get the right rRNA/tRNA GTF files to use in cufflinks?

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For cufflinks you'll want the repeatmasker track from UCSC (just extract the rRNA and tRNA sites).

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Dear Devon, Karl et al.

I came across an issue with rRNA contamination in my RNA-seq data (poly-A sequencing), and I found several posts about this issue. I observed contradictory or not very straightforward answers regarding what I can do next. For example, here you're recommending to not remove ribosomal RNAs from the mapping as this would introduce a bias in the analysis, whereas I am under the impression that others recommend removing it from the analysis (please see here). I tried removing rRNA from my reads using SortMERNA (using all eukaryotic and prokaryotic databases in this program), but the GC distribution per sample did not change. But in one sample, instead of 3 overrepresented sequences mapping to a ribosomal protein (rpS29, each appearing 10-15% of the reads), I get only one mapping to the same protein with similar level of representation in the RNA-seq data (~10-15% of the reads)). The GC distribution plot did not change.

Is there an objective way to assess if this rRNA contamination will be a problem for my differential expression analysis?

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6.9 years ago
Charles Plessy ★ 2.8k

You can remove rRNA reads with the TagDust 2 program and the human ribosomal DNA complete repeating unit (U13369.1). It might be good to also remove the mitochondrial rRNA sequences (MT-RNR1 and MT-RNR2).

Edited: I got the genome wrong, see below for the mouse reference sequences.

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why do I need to remove the human rRNA ( or for that reason the mitochondrial sequences) if I'm working with the mouse genome? Do I expect to find the human rRNA in my fastq files?

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You don't.

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Oops sorry, for mouse the equivalent sequences are BK000964, mt-Rnr1, and mt-Rnr2.

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Are these the complete rRNA from the mouse?

Which means, if I first map my fastq files to this sequence, I will be able to remove most of my rRNA reads by taking only the unmapped reads in the next step?

Do I need to allow multiple mapping in this step? I guess there might be some multiplications of rRNA reads in the fastq files, if I already know from the fastqc report, that there reads are over-represented.

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The mouse genome version mm10 does not seem to contain the rDNA regions where the rRNA is produced by the RNA polymerase I (assembling these regions this is actually challenging). But these regions or the rRNAs themselves have been the template for repeated elements that are scattered on the chromosomes (the "RNA" part in the RepeatMasker track in the UCSC genome browser). Therefore, when aliging transcriptome reads on a genome, the reads truely originating from rRNA molecules will be matched in a wrong location, regardless of multimapping. As noted in other answers, this is not a problem with some tools for RNA-seq analysis. However, if you look at your data in a less annotation-centric way, it may become confusing and it can be better to remove these reads before alignment.

The removal step with TagDust 2 that I suggested above matches the reads against a short list of user-defined reference sequences, not against a genome, therefore multimapping is not an issue in that step. As a first aproximate (since there are multiple copies in the genome and one can not exclude that some copies have some slight differences), you can assume that the BK000964 sequence represents all rRNAs.


text:
# Working with amplitudes

Don’t worry: I am not going to introduce the Hamiltonian matrix—not yet, that is. But this post is going a step further than my previous ones, in the sense that it will be more abstract. At the same time, I do want to stick to real physical examples so as to illustrate what we’re doing when working with those amplitudes. The example that I am going to use involves spin. So let’s talk about that first.

#### Spin, angular momentum and the magnetic moment

You know spin: it allows experienced pool players to do the most amazing tricks with billiard balls, making a joke of what a so-called elastic collision is actually supposed to look like. So it should not come as a surprise that spin complicates the analysis in quantum mechanics too. We dedicated several posts to that (see, for example, my post on spin and angular momentum in quantum physics) and I won’t repeat these here. Let me just repeat the basics:

1. Classical and quantum-mechanical spin do share similarities: the basic idea driving the quantum-mechanical spin model is that of a electric charge – positive or negative – spinning about its own axis (this is often referred to as intrinsic spin) as well as having some orbital motion (presumably around some other charge, like an electron in orbit with a nucleus at the center). This intrinsic spin, and the orbital motion, give our charge some angular momentum (J) and, because it’s an electric charge in motion, there is a magnetic moment (μ). To put things simply: the classical and quantum-mechanical view of things converge in their analysis of atoms or elementary particles as tiny little magnets. Hence, when placed in an external magnetic field, there is some interaction – a force – and their potential and/or kinetic energy changes. The whole system, in fact, acquires extra energy when placed in an external magnetic field.

Note: The formula for that magnetic energy is quite straightforward, both in classical as well as in quantum physics, so I’ll quickly jot it down: U = −μB = −|μ|·|B|·cosθ = −μ·B·cosθ. So it’s just the scalar product of the magnetic moment and the magnetic field vector, with a minus sign in front so as to get the direction right. [θ is the angle between the μ and B vectors and determines whether U as a whole is positive or negative.

2. The classical and quantum-mechanical view also diverge, however. They diverge, first, because of the quantum nature of spin in quantum mechanics. Indeed, while the angular momentum can take on any value in classical mechanics, that’s not the case in quantum mechanics: in whatever direction we measure, we get a discrete set of values only. For example, the angular momentum of a proton or an electron is either −ħ/2 or +ħ/2, in whatever direction we measure it. Therefore, they are referred to as spin-1/2 particles. All elementary fermions, i.e. the particles that constitute matter (as opposed to force-carrying particles, like photons), have spin 1/2.

Note: Spin-1/2 particles include, remarkably enough, neutrons too, which has the same kind of magnetic moment that a rotating negative charge would have. The neutron, in other words, is not exactly ‘neutral’ in the magnetic sense. One can explain this by noting that a neutron is not ‘elementary’, really: it consists of three quarks, just like a proton, and, therefore, it may help you to imagine that the electric charges inside are, somehow, distributed unevenly—although physicists hate such simplifications. I am noting this because the famous Stern-Gerlach experiment, which established the quantum nature of particle spin, used silver atoms, rather than protons or electrons. More in general, we’ll tend to forget about the electric charge of the particles we’re describing, assuming, most of the time, or tacitly, that they’re neutral—which helps us to sort of forget about classical theory when doing quantum-mechanical calculations!

3. The quantum nature of spin is related to another crucial difference between the classical and quantum-mechanical view of the angular momentum and the magnetic moment of a particle. Classically, the angular momentum and the magnetic moment can have any direction.

Note: I should probably briefly remind you that J is a so-called axial vector, i.e. a vector product (as opposed to a scalar product) of the radius vector r and the (linear) momentum vector p = m·v, with v the velocity vector, which points in the direction of motion. So we write: J = r×p = r×m·v = |r|·|p|·sinθ·n. The n vector is the unit vector perpendicular to the plane containing r and (and, hence, v, of course) given by the right-hand rule. I am saying this to remind you that the direction of the magnetic moment and the direction of motion are not the same: the simple illustration below may help to see what I am talking about.]

Back to quantum mechanics: the image above doesn’t work in quantum mechanics. We do not have an unambiguous direction of the angular momentum and, hence, of the magnetic moment. That’s where all of the weirdness of the quantum-mechanical concept of spin comes out, really. I’ll talk about that when discussing Feynman’s ‘filters’ – which I’ll do in a moment – but here I just want to remind you of the mathematical argument that I presented in the above-mentioned post. Just like in classical mechanics, we’ll have a maximum (and, hence, also a minimum) value for J, like +ħ, 0 and +ħ for a Lithium-6 nucleus. [I am just giving this rather special example of a spin-1 article so you’re reminded we can have particles with an integer spin number too!] So, when we measure its angular momentum in any direction really, it will take on one of these three values: +ħ, 0 or +ħ. So it’s either/or—nothing in-between. Now that leads to a funny mathematical situation: one would usually equate the maximum value of a quantity like this to the magnitude of the vector, which is equal to the (positive) square root of J2 = J= Jx2 + Jy2 + Jz2, with Jx, Jy and Jz the components of J in the x-, y- and z-direction respectively. But we don’t have continuity in quantum mechanics, and so the concept of a component of a vector needs to be carefully interpreted. There’s nothing definite there, like in classical mechanics: all we have is amplitudes, and all we can do is calculate probabilities, or expected values based on those amplitudes.

Huh? Yes. In fact, the concept of the magnitude of a vector itself becomes rather fuzzy: all we can do really is calculate its expected value. Think of it: in the classical world, we have a J2 = Jproduct that’s independent of the direction of J. For example, if J is all in the x-direction, then Jand Jwill be zero, and J2 = Jx2. If it’s all in the y-direction, then Jand Jwill be zero and all of the magnitude of J will be in the y-direction only, so we write: J2 = Jy2. Likewise, if J does not have any z-component, then our JJ product will only include the x- and y-components: JJ = Jx2 + Jy2. You get the idea: the J2 = Jproduct is independent of the direction of J exactly because, in classical mechanics, J actually has a precise and unambiguous magnitude and direction and, therefore, actually has a precise and unambiguous component in each direction. So we’d measure Jx, Jy, and Jand, regardless of the actual direction of J, we’d find its magnitude |J| = J = +√J2 = +(Jx2 + Jy2 + Jz2)1/2.

In quantum mechanics, we just don’t have quantities like that. We say that Jx, Jand Jhave an amplitude to take on a value that’s equal to +ħ, 0 or +ħ (or whatever other value is allowed by the spin number of the system). Now that we’re talking spin numbers, please note that this characteristic number is usually denoted by j, which is a bit confusing, but so be it. So can be 0, 1/2, 1, 3/2, etcetera, and the number of ‘permitted values’ is 2j + 1 values, with each value being separated by an amount equal to ħ. So we have 1, 2, 3, 4, 5 etcetera possible values for Jx, Jand Jrespectively. But let me get back to the lesson. We just can’t do the same thing in quantum mechanics. For starters, we can’t measure Jx, Jy, and Jsimultaneously: our Stern-Gerlach apparatus has a certain orientation and, hence, measures one component of J only. So what can we do?

Frankly, we can only do some math here. The wave-mechanical approach does allow to think of the expected value of J2 = J= Jx2 + Jy2 + Jz2 value, so we write:

E[J2] = E[JJ] = E[Jx2 + Jy2 + Jz2] = ?

[Feynman’s use of the 〈 and 〉 brackets to denote an expected value is hugely confusing, because these brackets are also used to denote an amplitude. So I’d rather use the more commonly used E[X] notation.] Now, it is a rather remarkable property, but the expected value of the sum of two or more random variables is equal to the sum of the expected values of the variables, even if those variables may not be independent. So we can confidently use the linearity property of the expected value operator and write:

E[Jx+ Jy2 + Jz2] = E[Jx2] + E[Jx2] + E[Jx2]

Now we need something else. It’s also just part of the quantum-mechanical approach to things and so you’ll just have to accept it. It sounds rather obvious but it’s actually quite deep: if we measure the x-, y- or z-component of the angular momentum of a random particle, then each of the possible values is equally likely to occur. So that means, in our case, that the +ħ, 0 or +ħ values are equally likely, so their likelihood is one into three, i.e. 1/3. Again, that sounds obvious but it’s not. Indeed, please note, once again, that we can’t measure Jx, Jy, and Jsimultaneously, so the ‘or’ in x-, y- or z-component is an exclusive ‘or’. Of course, I must add this equipartition of likelihoods is valid only because we do not have a preferred direction for J: the particles in our beam have random ‘orientations’. Let me give you the lingo for this: we’re looking at an unpolarized beam. You’ll say: so what? Well… Again, think about what we’re doing here: we may of may not assume that the Jx, Jy, and Jvariables are related. In fact, in classical mechanics, they surely are: they’re determined by the magnitude and direction of J. Hence, they are not random at all ! But let me continue, so you see what comes out.

Because the +ħ, 0 and +ħ values are equally, we can write: E[Jx2] = ħ2/3 + 0/3 + (−ħ)2/3 = [ħ2 + 0 + (−ħ)2]/3 = 2ħ2/3. In case you wonder, that’s just the definition of the expected value operator: E[X] = p1x+ p2x+ … = ∑pixi, with pi the likelihood of the possible value x. So we take a weighted average with the respective probabilities as the weights. However, in this case, with an unpolarized beam, the weighted average becomes a simple average.

Now, E[Jy2] and E[Jz2] are – rather unsurprisingly – also equal to 2ħ2/3, so we find that E[J2] = E[Jx2] + E[Jx2] + E[Jx2] = 3·(2ħ2/3) = 2ħand, therefore, we’d say that the magnitude of the angular momentum is equal to |J| = J = +√2·ħ ≈ 1.414·ħ. Now that value is not equal to the maximum value of our x-, y-, z-component of J, or the component of J in whatever direction we’d want to measure it. That maximum value is ħ, without the √2 factor, so that’s some 40% less than the magnitude we’ve just calculated!

Now, you’ve probably fallen asleep by now but, what this actually says, is that the angular momentum, in quantum mechanics, is never completely in any direction. We can state this in another way: it implies that, in quantum mechanics, there’s no such thing really as a ‘definite’ direction of the angular momentum.

[…]

OK. Enough on this. Let’s move on to a more ‘real’ example. Before I continue though, let me generalize the results above:

[I] A particle, or a system, will have a characteristic spin number: j. That number is always an integer or a half-integer, and it determines a discrete set of possible values for the component of the angular momentum J in any direction.

[II] The number of values is equal to 2j + 1, and these values are separated by ħ, which is why they are usually measured in units of ħ, i.e. Planck’s reduced constant: ħ ≈ 1×10−34 J·s, so that’s tiny but real. 🙂 [It’s always good to remind oneself that we’re actually trying to describe reality.] For example, the permitted values for a spin-3/2 particle are +3ħ/2, +ħ/2, −ħ/2 and −3ħ/2 or, measured in units of ħ, +3/2, +1/2, −1/2 and −3/2. When discussing spin-1/2 particles, we’ll often refer to the two possible states as the ‘up’ and the ‘down’ state respectively. For example, we may write the amplitude for an electron or a proton to have a angular momentum in the x-direction equal to +ħ/2 or −ħ/2 as 〈+x〉 and 〈−x〉 respectively. [Don’t worry too much about it right now: you’ll get used to the notation quickly.]

[III] The classical concepts of angular momentum, and the related magnetic moment, have their limits in quantum mechanics. The magnitude of a vector quantity like angular momentum is generally not equal to the maximum value of the component of that quantity in any direction. The general rule is:

J= j·(j+1)ħ2 > j2·ħ2

So the maximum value of any component of J in whatever direction (i.e. j·ħ) is smaller than the magnitude of J (i.e. √[ j·(j+1)]·ħ). This implies we cannot associate any precise and unambiguous direction with quantities like the angular momentum J or the magnetic moment μ. As Feynman puts it:

“That the energy of an atom [or a particle] in a magnetic field can have only certain discrete energies is really not more surprising than the fact that atoms in general have only certain discrete energy levels—something we mentioned often in Volume I. Why should the same thing not hold for atoms in a magnetic field? It does. But it is the attempt to correlate this with the idea of an oriented magnetic moment that brings out some of the strange implications of quantum mechanics.”

#### A real example: the disintegration of a muon in a magnetic field

I talked about muon integration before, when writing a much more philosophical piece on symmetries in Nature and time reversal in particular. I used the illustration below. We’ve got an incoming muon that’s being brought to rest in a block of material, and then, as muons do, it disintegrates, emitting an electron and two neutrinos. As you can see, the decay direction is (mostly) in the direction of the axial vector that’s associated with the spin direction, i.e. the direction of the grey dashed line. However, there’s some angular distribution of the decay direction, as illustrated by the blue arrows, that are supposed to visualize the decay products, i.e. the electron and the neutrinos.

This disintegration process is very interesting from a more philosophical side. The axial vector isn’t ‘real’: it’s a mathematical concept—a pseudovector. A pseudo- or axial vector is the product of two so-called true vectors, aka as polar vectors. Just look back at what I wrote about the angular momentum: the J in the J = r×p = r×m·v formula is such vector, and its direction depends on the spin direction, which is clockwise or counter-clockwise, depending from what side you’re looking at it. Having said that, who’s to judge if the product of two ‘true’ vectors is any less ‘true’ than the vectors themselves? 🙂

The point is: the disintegration process does not respect what is referred to as P-symmetry. That’s because our mathematical conventions (like all of these right-hand rules that we’ve introduced) are unambiguous, and they tell us that the pseudovector in the mirror image of what’s going on, has the opposite direction. It has to, as per our definition of a vector product. Hence, our fictitious muon in the mirror should send its decay products in the opposite direction too! So… Well… The mirror image of our muon decay process is actually something that’s not going to happen: it’s physically impossible. So we’ve got a process in Nature here that doesn’t respect ‘mirror’ symmetry. Physicists prefer to call it ‘P-symmetry’, for parity symmetry, because it involves a flip of sign of all space coordinates, so there’s a parity inversion indeed. So there’s processes in Nature that don’t respect it but, while that’s all very interesting, it’s not what I want to write about. [Just check that post of mine if you’d want to read more.] Let me, therefore, use another illustration—one that’s more to the point in terms of what we do want to talk about here:

So we’ve got the same muon here – well… A different one, of course! 🙂 – entering that block (A) and coming to a grinding halt somewhere in the middle, and then it disintegrates in a few micro-seconds, which is an eternity at the atomic or sub-atomic scale. It disintegrates into an electron and two neutrinos, as mentioned above, with some spread in the decay direction. [In case you wonder where we can find muons… Well… I’ll let you look it up yourself.] So we have:

Now it turns out that the presence of a magnetic field (represented by the B arrows in the illustration above) can drastically change the angular distribution of decay directions. That shouldn’t surprise us, of course, but how does it work, exactly? Well… To simplify the analysis, we’ve got a polarized beam here: the spin direction of all muons before they enter the block and/or the magnetic field, i.e. at time t = 0, is in the +x-direction. So we filtered them just, before they entered the block. [I will come back to this ‘filtering’ process.] Now, if the muon’s spin would stay that way, then the decay products – and the electron in particular – would just go straight, because all of the angular momentum is in that direction. However, we’re in the quantum-mechanical world here, and so things don’t stay the same. In fact, as we explained, there’s no such things as a definite angular momentum: there’s just an amplitude to be in the +x state, and that amplitude changes in time and in space.

How exactly? Well… We don’t know, but we can apply some clever tricks here. The first thing to note is that our magnetic field will add to the energy of our muon. So, as I explained in my previous post, the magnetic field adds to the E in the exponent of our complex-valued wavefunction a·e(i/ħ)(E·t − px). In our example, we’ve got a magnetic field in the z-direction only, so that U = −μB reduces to U = −μz·B, and we can re-write our wavefunction as:

a·e(i/ħ)[(E+U)·t − px] = a·e(i/ħ)(E·t − px)·e(i/ħ)(μz·B·t)

Of course, the magnetic field only acts from t = 0 to when the muon disintegrates, which we’ll denote by the point t = τ. So what we get is that the probability amplitude of a particle that’s been in a uniform magnetic field changes by a factor e(i/ħ)(μz·B·τ). Note that it’s a factor indeed: we use it to multiply. You should also note that this is a complex exponential, so it’s a periodic function, with its real and imaginary part oscillating between zero and one. Finally, we know that μz can take on only certain values: for a spin-1/2 particle, they are plus or minus some number, which we’ll simply denote as μ, so that’s without the subscript, so our factor becomes:

e(i/ħ)(±μ·B·t)

[The plus or minus sign needs to be explained here, so let’s do that quickly: we have two possible states for a spin-1/2 particle, one ‘up’, and the other ‘down’. But then we also know that the phase of our complex-valued wave function turns clockwise, which is why we have a minus sign in the exponent of our eiθ expression. In short, for the ‘up’ state, we should take the positive value, i.e. +μ, but the minus sign in the exponent of our eiθ function makes it negative again, so our factor is e−(i/ħ)(μ·B·t) for the ‘up’ state, and e+(i/ħ)(μ·B·t) for the ‘down’ state.]

OK. We get that, but that doesn’t get us anywhere—yet. We need another trick first. One of the most fundamental rules in quantum-mechanics is that we can always calculate the amplitude to go from one state, say φ (read: ‘phi’), to another, say χ (read: ‘khi’), if we have a complete set of so-called base states, which we’ll denote by the index i or j (which you shouldn’t confuse with the imaginary unit, of course), using the following formula:

〈 χ | φ 〉 = ∑ 〈 χ | i 〉〈 i | φ 〉

I know this is a lot to swallow, so let me start with the notation. You should read 〈 χ | φ 〉 from right to left: it’s the amplitude to go from state φ to state χ. This notation is referred to as the bra-ket notation, or the Dirac notation. [Dirac notation sounds more scientific, doesn’t it?] The right part, i.e. | φ 〉, is the bra, and the left part, i.e. 〈 χ | is the ket. In our example, we wonder what the amplitude is for our muon staying in the +x state. Because that amplitude is time-dependent, we can write it as A+(τ) = 〈 +at time t = τ | +at time t = 0 〉 = 〈 +at t = τ | +at t = 0 〉or, using a very misleading shorthand, 〈 +x | +x 〉. [The shorthand is misleading because the +in the ket obviously means something else than the +in the bra.]

But let’s apply the rule. We’ve got two states with respect to each coordinate axis only here. For example, in respect to the z-axis, the spin values are +z and −z respectively. [As mentioned above, we actually mean that the angular momentum in this direction is either +ħ/2 or −ħ/2, aka as ‘up’ or ‘down’ respectively, but then quantum theorists seem to like all kinds of symbols better, so we’ll use the +z and −z notations for these two base states here. So now we can use our rule and write:

A+(t) = 〈 +x | +x 〉 = 〈 +x | +z 〉〈 +z | +x 〉 + 〈 +x | −z 〉〈 −z | +x 〉

You’ll say this doesn’t help us any further, but it does, because there is another set of rules, which are referred to as transformation rules, which gives us those 〈 +z | +x 〉 and 〈 −z | +x 〉 amplitudes. They’re real numbers, and it’s the same number for both amplitudes.

〈 +z | +x 〉 = 〈 −z | +x 〉 = 1/√2

This shouldn’t surprise you too much: the square root disappears when squaring, so we get two equal probabilities – 1/2, to be precise – that add up to one which – you guess it – they have to add up to because of the normalization rule: the sum of all probabilities has to add up to one, always. [I can feel your impatience, but just hang in here for a while, as I guide you through what is likely to be your very first quantum-mechanical calculation.] Now, the 〈 +z | +x 〉 = 〈 −z | +x 〉 = 1/√2 amplitudes are the amplitudes at time t = 0, so let’s be somewhat less sloppy with our notation and write 〈 +z | +x 〉 as C+(0) and 〈 −z | +x 〉 as C(0), so we write:

〈 +z | +x 〉 = C+(0) = 1/√2

〈 −z | +x 〉 = C(0) = 1/√2

Now we know what happens with those amplitudes over time: that e(i/ħ)(±μ·B·t) factor kicks in, and so we have:

C+(t) = C+(0)·e−(i/ħ)(μ·B·t) = e−(i/ħ)(μ·B·t)/√2

C(t) = C(0)·e+(i/ħ)(μ·B·t) = e+(i/ħ)(μ·B·t)/√2

As for the plus and minus signs, see my remark on the tricky ± business in regard to μ. To make a long story somewhat shorter :-), our expression for A+(t) = 〈 +x at t | +x 〉 now becomes:

A+(t) = 〈 +x | +z 〉·C+(t) + 〈 +x | −z 〉·C(t)

Now, you wouldn’t be too surprised if I’d just tell you that the 〈 +x | +z 〉 and 〈 +x | −z 〉 amplitudes are also real-valued and equal to 1/√2, but you can actually use yet another rule we’ll generalize shortly: the amplitude to go from state φ to state χ is the complex conjugate of the amplitude to to go from state χ to state φ, so we write 〈 χ | φ 〉 = 〈 φ | χ 〉*, and therefore:

〈 +x | +z 〉 = 〈 +z | +x 〉* = (1/√2)* = (1/√2)

〈 +x | −z 〉 = 〈 −z | +x 〉* = (1/√2)* = (1/√2)

So our expression for A+(t) = 〈 +x at t | +x 〉 now becomes:

A+(t) = e−(i/ħ)(μ·B·t)/2 + e(i/ħ)(μ·B·t)/2

That’s the sum of a complex-valued function and its complex conjugate, and we’ve shown more than once (see my page on the essentials, for example) that such sum reduces to the sum of the real parts of the complex exponentials. [You should not expect any explanation of Euler’s eiθ = cosθ + i·sinθ rule at this level of understanding.] In short, we get the following grand result:

The big question, of course: what does this actually mean? 🙂 Well… Just square this thing and you get the probabilities shown below. [Note that the period of a squared cosine function is π, instead of 2π, which you can easily verify using an online graphing tool.]

Because you’re tired of this business, you probably don’t realize what we’ve just done. It’s spectacular and mundane at the same time. Let me quote Feynman to summarize the results:

“We find that the chance of catching the decay electron in the electron counter varies periodically with the length of time the muon has been sitting in the magnetic field. The frequency depends on the magnetic moment μ. The magnetic moment of the muon has, in fact, been measured in just this way.”

As far as I am concerned, the key result is that we’ve learned how to work with those mysterious amplitudes, and the wavefunction, in a practical way, thereby using all of the theoretical rules of the quantum-mechanical approach to real-life physical situations. I think that’s a great leap forward, and we’ll re-visit those rules in a more theoretical and philosophical démarche in the next post. As for the example itself, Feynman takes it much further, but I’ll just copy the Grand Master here:

Huh? Well… I am afraid I have to leave it at this, as I discussed the precession of ‘atomic’ magnets elsewhere (see my post on precession and diamagnetism), which gives you the same formula: ω= μ·B/J (just substitute J for ±ħ/2). However, the derivation above approaches it from an entirely different angle, which is interesting. Of course, all fits. 🙂 However, I’ll let you do your own homework now. I hope to see you tomorrow for the mentioned theoretical discussion. Have a nice evening, or weekend – or whatever ! 🙂


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# SBPL lattice planner's motion primtive

I am trying to understand the motion primitive file genmprim_unicycle.m used by the SBPL_lattice_planner.

I understand the second part of the file where they use the basemprimendpts* arrays to generate the intermediate poses to generate the path. But I don't understand how they ended up with the array for basemprimendpts*.

I read through all the tutorials in sbpl.net, in particular this tutorial explains how the calculation is done. But the equations which are supposed to be images are not showing up properly since the images are broken due to which I can not understand the equations. (Let me know if it shows up properly for you).

Back to the question. Lets take for example, from this tutorial I see that for moving 8 units at 45 degree, the calculation should be like

x = 8 * cos (45) = 5.66

y = 8 * sin (45) = 5.66

Rounding up the values I can say that (x, y) is (6, 6) which is relatable with

basemprimendpts45_c(2,:) = [6 6 0 forwardcostmult]

But I cannot understand how they calculated the values for forward turn which is represented by

basemprimendpts45_c(3,:) = [5 7 1 forwardandturncostmult]; as seen from the above image. Also I have one more doubt. What does it mean 1/16 theta change. Is it 1/16th change of 45 degree or 1/16th of 360 degree ?.


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# Is the ring of all algebraic integers coherent?

Is the ring of all algebraic integers coherent? Here is the definition of a coherent ring.

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## 1 Answer

Yes. In fact every finitely generated ideal is principal.

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en.wikipedia.org/wiki/B%C3%A9zout_domain – user29743 Nov 20 '12 at 22:16
Dear Countinghaus, How do you prove that the ring of all algebraic integers is a Bezout domain? – Makoto Kato Nov 20 '12 at 22:18
This is a little gross and there's probably an easier argument, but given a finitely generated ideal there is an ideal of an honest number field generated by the same generators; this ideal becomes principal after a finite extension (e.g. by capitulation of ideals in the Hilbert class field) and the generator of that new ideal is also a generator for your original ideal. – user29743 Nov 20 '12 at 22:24


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14
Q:

# The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

 A) 74 B) 94 C) 184 D) 364

Answer:   D) 364

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

$\inline \therefore$Required number = (90 x 4) + 4   = 364.

Q:

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to ?

 A) 13/125 B) 14/57 C) 11/120 D) 16/41

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

Report Error

1 54
Q:

The largest measuring cylinder that can accurately fill 3 tanks of capacity 98, 182 and 266 litres each, is of capacity ?

 A) 7 lts B) 14 lts C) 98 lts D) 42 lts

Answer & Explanation Answer: B) 14 lts

Explanation:

To know the the measuring cylinder that can fill all the given capacities , they must be divisible by the required number.

98,182,266 all are divisible by 14
So  14 litres  is the largest cylinder that can fill all the given cylinders.

(or)

The other method is take HCF of all given capacities i.e 98, 182 and 266.

Report Error

2 71
Q:

If N is the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. What is the sum of the digits of N ?

 A) 6 B) 8 C) 4 D) 9

Answer & Explanation Answer: C) 4

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Report Error

4 39
Q:

Find the greatest number that will divide 964, 1238 and 1400 leaving remainder of 41,31 and 51 respectively ?

 A) 64 B) 69 C) 71 D) 58

Answer & Explanation Answer: C) 71

Explanation:

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.

Report Error

2 130
Q:

A student can divide her books into groups of 5, 9 and 13. what is the smallest possible number of the books ?

 A) 487 B) 585 C) 635 D) 705

Answer & Explanation Answer: B) 585

Explanation:

The smallest possible number of books = L.C.M of 5, 9 and 13.
Therefore, L.C.M of 5,9 and 13 is = 5 x 9 x 13 = 585.

Report Error

4 85


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# Exercise 10 - Chapter 4

edited June 2018

Is it true that a $$\textbf{Bool}$$-profunctor, as in Definition 4.8 is exactly the same as a feasibility relation, as in Definition 4.2, once you peel back all the jargon? Or is there some subtle difference?

• Options
1.
edited June 2018

$$\tag{V-profunctor} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \mathcal{V}$$ $$\tag{Bool-profunctor} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \textbf{Bool}$$ $$\tag{feasibilty relation} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \textbf{Bool}$$ They sure look the same.

Comment Source:$\tag{V-profunctor} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \mathcal{V}$ $\tag{Bool-profunctor} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \textbf{Bool}$ $\tag{feasibilty relation} \Phi : \mathcal{X}^{op} \times \mathcal{Y} \rightarrow \textbf{Bool}$ They sure look the same.


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# Math Is Fun Worksheet

In this multiplication with fractions learning exercise, students problem solve twenty five multiplication equations with fractions. Students show each step of their work.


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Cheap and Secure Web Hosting Provider : See Now

# [Solved]: Computing a Sequence of People Entering and Leaving a Room

, ,
Problem Detail:

I've been working on a problem for my Algorithms class, but I've found myself stuck. The prompt is as follows.

You start with an empty room and a group of n people waiting outside. At each step, you may either admit one person into the room, or let one out. Please design an algorithm which generates a sequence of $2^n − 1$ steps, so that every possible combination of people is achieved exactly once. For example, for $n = 3$, one possible sequence would be

$\qquad\displaystyle e_1, e_2, x_1, e_3, e_1, x_2, x_1$.

where $e_i$ means person $i$ enters the room and $x_j$ means person $j$ exits the room

Hint: study the Hanoi Tower algorithm.

In my research on the prompt, I've found many people referencing Gray Code, but I don't understand how to implement it in an algorithm to compute this combination. I also found a very similar problem, but there wasn't an actual implementation of it.

It would be greatly appreciated if you could point me in the right direction on where to go with this algorithm.

#### Answered By : Rick Decker

Base case It's easy to generate the combinations when $n=1$: just use the operation $e_1$.

Recursion Now assume that you have a sequence $ops_n$ that will generate all the combinations of $n$ people. Define $rops_n$ to be the sequence of operations that "undoes" $ops_n$, namely first form the reverse of $ops_n$ and then replace all "enter" operations with "leave" operations and vice versa. For example, if $ops_2 = \langle\, e_1, e_2, x_1\,\rangle$, then $rops_n =\langle\, e_1, x_2, x_1\,\rangle$. It's not difficult to see that $ops_n, rops_n$ will generate all the combinations of $n$ people and then produce the same combinations in reverse order.

Finally, show that if $ops_n$ generates all the $n$-combinations, then the sequence $ops_n, e_{n+1}, rops_n$ will generate all the $(n+1)$-combinations: first all the ones without person $n+1$ and then all the ones with person $n+1$. From this construction it's clear inductively that you can generate all $n$ combinations with $2^n-1$ operations.

As per the hint you were given, this is very similar in spirit to the Towers of Hanoi problem: To move $n+1$ disks from peg 1 to peg 2, you first move $n$ disks from peg 1 to peg 3, then move the largest disk to peg 2, and finally move the $n$ disks on peg 3 to peg 2.


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# How do you find the zeros of  y = -x^2 - 1/11x + 1/7  using the quadratic formula?

Aug 3, 2016

Zeros are $\frac{7}{22} - \frac{7}{2} \sqrt{\frac{491}{847}}$ and $\frac{7}{22} + \frac{7}{2} \sqrt{\frac{491}{847}}$

#### Explanation:

Quadratic formula gives the zeros of $y = a {x}^{2} + b x + c$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence zeros of $y = - {x}^{2} - \frac{1}{11} x + \frac{1}{7}$ are

$x = \frac{- \left(- \frac{1}{11}\right) \pm \sqrt{{\left(- \frac{1}{11}\right)}^{2} - 4 \left(- 1\right) \left(\frac{1}{7}\right)}}{2 \left(\frac{1}{7}\right)}$

= $\frac{\left(\frac{1}{11}\right) \pm \sqrt{\frac{1}{121} + \frac{4}{7}}}{\frac{2}{7}}$

= $\left(\frac{1}{11} \pm \sqrt{\frac{7 + 484}{847}}\right) \times \frac{7}{2}$

= $\frac{7}{22} \pm \frac{7}{2} \sqrt{\frac{491}{847}}$

Hence zeros are $\frac{7}{22} - \frac{7}{2} \sqrt{\frac{491}{847}}$ and $\frac{7}{22} + \frac{7}{2} \sqrt{\frac{491}{847}}$


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# Commuting smooth maps

Suppose $f:A\to B$ and $g:C\to D$ are smooth embeddings, $h:B\to D$ is a smooth map, and $i:A\to C$ is a continuous map such that $g(i(x))=h(f(x))$.

Then, how to show that $i$ is smooth?

An embedding f is a smooth map such that the inclusion $X\to f(X)$ is a diffeomorphism, and f(X) is a submanifold.

A n-submanifold is a subset S of M, such that for every p in S, there is a chart $x:U\to V$, with p in U, such that $x(U\cap S)=V\cap R^n\times 0$

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Hints, comments, ideas? –  user1708 Mar 27 '12 at 4:03
When a map is an embedding, you can, as a corollary to the local inversion theorem, find suitable charts in which the map has an easy to understand expression. Say $f:M\rightarrow N$ is smooth and an immersion at $p\in M$, then there are (centered) charts $X$ of $M$ in the neighbourhood of $p$ and $Y$ chart of $N$ in the neighborhood of $f(p)$ such that locally $Y\circ f\circ X^{-1}(x_1,\dots,x_m)=(x_1,\dots,x_m,0,\dots,0)$. Work in such a chart and you'll almost be done I think. –  Olivier Bégassat Mar 27 '12 at 4:27