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# Quaternions and spatial rotation

Quaternions and spatial rotation

Unit quaternions provide a convenient mathematical notation for representing orientations and rotations of objects in three dimensions. Compared to Euler angles they are simpler to compose and avoid the problem of gimbal lock. Compared to rotation matrices they are more efficient and more numerically stable. Quaternions have found their way into applications in computer graphics, robotics, navigation, and orbital mechanics of satellites. ["Quaternions and rotation Sequences: a Primer with Applications to Orbits, Aerospace, and Virtual Reality." Kuipers, Jack B., Princeton University Press copyright 1999.]

Quaternion rotation operations

A very strict explanation of the properties used in this section is given by Altmann. ["Rotations, Quaternions, and Double Groups." Altmann, Simon L., Dover Publications, 1986.]

Visualizing the space of rotations

Unit quaternions represent in a very straightforward way the mathematical space of rotations in three dimensions. The correspondence between rotations and quaternions can be understood by first visualizing the space of rotations itself.

Any rotation in three dimensions is a rotation by some angle about some axis. When the angle is zero the axis does not matter, so rotation by zero degrees is a single point in the space of rotations (the identity rotation). For a tiny but nonzero angle, the set of possible rotations is like a small sphere surrounding the identity rotation, where each point on the sphere represents an axis pointing in a particular direction (compare the celestial sphere). Rotations through increasingly large angles are increasingly far from the identity rotation, and we can think of them as concentric spheres of increasing radius. Thus, near the identity rotation, the abstract space of rotations looks similar to ordinary three-dimensional space (which can also be seen as a central point surrounded by concentric spheres of every radius). However, as the rotation angle increases toward 360°, rotations about different axes stop diverging and become more similar to each other, becoming identical (and equal to the identity rotation) when the angle reaches 360°.

We can see similar behavior on the surface of a sphere. If we start at the north pole and draw straight lines (that is, lines of longitude) in many directions, they will diverge but eventually converge again at the south pole. Concentric circles of increasing radius drawn around the north pole (lines of latitude) will eventually collapse to a point at the south pole once the radius reaches the distance between the poles. If we think of different directions from the pole (that is, different longitudes) as different rotation "axes", and different distances from the pole (that is, different latitudes) as different rotation "angles", we have an analogy to the space of rotations. Since the sphere's surface is two dimensional while the space of rotations is three dimensional, we must actually model the space of rotations as a hypersphere; however, we can think of the ordinary sphere as a slice through the full hypersphere (just as a circle is a slice through a sphere). We can take the slice to represent, for example, just the rotations about axes in the "xy" plane. Note that the angle of rotation is "twice" the latitude difference from the north pole: points on the equator represent rotations of 180°, not 90°, and the south pole represents a rotation of 360°, not 180°.

The north pole and the south pole represent the same rotation, and in fact this is true of any two antipodal points: if one is a rotation by "α" around the axis v, the other is a rotation by 360° − "α" around the axis −v. In fact, then, the space of rotations is not the (hyper)sphere itself but the (hyper)sphere with antipodal points identified. But for many purposes we can think of rotations as points on the sphere, even though they are twofold redundant (a so-called double cover).

Parameterizing the space of rotations

We can parameterize the surface of a sphere with two coordinates, such as latitude and longitude. But latitude and longitude are ill-behaved (degenerate) at the north and south poles, though the poles are not intrinsically different from any other points on the sphere. It can be shown that no two-parameter coordinate system can avoid such degeneracy (the hairy ball theorem). We can avoid such problems by embedding the sphere in three-dimensional space and parameterizing it with three Cartesian coordinates (here "w","x","y"), placing the north pole at ("w","x","y") = (1,0,0), the south pole at ("w","x","y") = (−1,0,0), and the equator at "w" = 0, "x"2 + "y"2 = 1. Points on the sphere satisfy the constraint "w"2 + "x"2 + "y"2 = 1, so we still have just two degrees of freedom though there are three coordinates. A point ("w","x","y") on the sphere represents a rotation around the ("x","y",0) axis by an angle $alpha= 2 cos^\left\{-1\right\} w = 2 sin^\left\{-1\right\} sqrt\left\{x^2+y^2\right\}$.

In the same way the hyperspherical space of 3D rotations can be parameterized by three angles (Euler angles), but any such parameterization is degenerate at some points on the hypersphere, leading to the problem of gimbal lock. We can avoid this by using four Euclidean coordinates "w","x","y","z", with "w"2 + "x"2 + "y"2 + "z"2 = 1. The point ("w","x","y","z") represents a rotation around the ("x","y","z") axis by an angle $alpha = 2 cos^\left\{-1\right\} w = 2 sin^\left\{-1\right\} sqrt\left\{x^2+y^2+z^2\right\}.$

Quaternions briefly

The complex numbers can be defined by introducing an abstract symbol i which satisfies the usual rules of algebra and additionally the rule i2 = −1. This is sufficient to reproduce all of the rules of complex number arithmetic: for example, $\left(a+bmathbf\left\{i\right\}\right)\left(c+dmathbf\left\{i\right\}\right) = ac + admathbf\left\{i\right\} + bmathbf\left\{i\right\}c + bmathbf\left\{i\right\}dmathbf\left\{i\right\} = ac + admathbf\left\{i\right\} + bcmathbf\left\{i\right\} + bdmathbf\left\{i\right\}^2 = \left(ac - bd\right) + \left(bc + ad\right) mathbf\left\{i\right\}$.

In the same way the quaternions can be defined by introducing abstract symbols i, j, k which satisfy the rules i2 = j2 = k2 = ijk = −1 and the usual algebraic rules "except" the commutative law of multiplication (a familiar example of such a noncommutative algebra is matrix algebra). From this all of the rules of quaternion arithmetic follow: for example, one can show that $\left(a + bmathbf\left\{i\right\} + cmathbf\left\{j\right\} + dmathbf\left\{k\right\}\right) \left(e + fmathbf\left\{i\right\} + gmathbf\left\{j\right\} + hmathbf\left\{k\right\}\right) = \left(ae - bf - cg - dh\right) + \left(af + eb + ch - dg\right) mathbf\left\{i\right\} + \left(ag + ce + df - bh\right) mathbf\left\{j\right\} + \left(ah + de + bg - cf\right) mathbf\left\{k\right\}$.

The imaginary part $bmathbf\left\{i\right\} + cmathbf\left\{j\right\} + dmathbf\left\{k\right\}$ of a quaternion behaves like a vector $\left(b, c, d\right) in mathbb\left\{R\right\}^3$, and the real part "a" behaves like a scalar in $mathbb\left\{R\right\}$. When quaternions are used in geometry, it is more convenient to define them as "a scalar plus a vector" and use the modern language of vector cross and dot products (which were originally inspired by the quaternions). In place of the rules i2 = j2 = k2 = ijk = −1 we have the vector multiplication rule $mathbf\left\{vw\right\} = mathbf\left\{v\right\} imes mathbf\left\{w\right\} - mathbf\left\{v\right\} cdot mathbf\left\{w\right\}$. Vector multiplication is noncommutative (because of the cross product), while scalar-scalar and scalar-vector multiplications commute. From these rules it follows immediately that $\left(s + mathbf\left\{v\right\}\right) \left(t + mathbf\left\{w\right\}\right) = \left(st - mathbf\left\{v\right\} cdot mathbf\left\{w\right\}\right) + \left(smathbf\left\{w\right\} + tmathbf\left\{v\right\} + mathbf\left\{v\right\} imes mathbf\left\{w\right\}\right)$.

The (left and right) multiplicative inverse of a nonzero quaternion is $\left(s + mathbf\left\{v\right\}\right)^\left\{-1\right\} = frac\left\{s - mathbf\left\{v\left\{s^2 + |mathbf\left\{v\right\}|^2\right\}$, as can be verified by direct calculation.

Parameterizing the space of rotations with quaternions

Let ("w","x","y","z") be the coordinates of a rotation as previously described. Define the quaternion:$q = w + xmathbf\left\{i\right\} + ymathbf\left\{j\right\} + zmathbf\left\{k\right\} = w + \left(x,y,z\right) = cos \left(alpha/2\right) + mathbf\left\{u\right\} sin \left(alpha/2\right)$where u is a unit vector. Then it can be shown (see next section) that the quaternion product:$q mathbf\left\{v\right\} q^\left\{-1\right\}$yields the vector v rotated by an angle "α" around the u axis. The rotation is clockwise if our line of sight points in the direction u. This operation is known as conjugation by "q".

It follows that quaternion multiplication is composition of rotations, for if "p" and "q" are quaternions representing rotations, then rotation (conjugation) by "pq" is:$p q mathbf\left\{v\right\} \left(p q\right)^\left\{-1\right\} = p q mathbf\left\{v\right\} q^\left\{-1\right\} p^\left\{-1\right\}$,which is the same as rotating (conjugating) by "q" and then by "p".

The quaternion inverse of a rotation is the opposite rotation, since $q^\left\{-1\right\} \left(q mathbf\left\{v\right\} q^\left\{-1\right\}\right) q = mathbf\left\{v\right\}$. The square of a quaternion rotation is a rotation by twice the angle around the same axis. More generally "q""n" is a rotation by "n" times the angle around the same axis as "q". This can be extended to arbitrary real "n", allowing for smooth interpolation between spatial orientations; see Slerp.

Proof of the quaternion rotation identity

Let u be a unit vector (the rotation axis) and let $q = cos frac\left\{alpha\right\}\left\{2\right\} + mathbf\left\{u\right\} sin frac\left\{alpha\right\}\left\{2\right\}$. Our goal is to show that:$mathbf\left\{v\right\}\text{'} = q mathbf\left\{v\right\} q^\left\{-1\right\} = left\left( cos frac\left\{alpha\right\}\left\{2\right\} + mathbf\left\{u\right\} sin frac\left\{alpha\right\}\left\{2\right\} ight\right) , mathbf\left\{v\right\} , left\left( cos frac\left\{alpha\right\}\left\{2\right\} - mathbf\left\{u\right\} sin frac\left\{alpha\right\}\left\{2\right\} ight\right)$yields the vector v rotated by an angle "α" around the axis u. Expanding out, we have:where and $mathbf\left\{v\right\}_\left\{$ are the components of v perpendicular and parallel to u respectively. This is the formula of a rotation by "α" around the u axis.

Example

Consider the rotation "f" around the axis u = "i" + "j" + "k", with a rotation angle of 120°, or ⁄3 radians.

:$alpha = frac\left\{2 pi\right\}\left\{3\right\} = 120^circ$

The length of u is √3, the half angle is π⁄3 (60°) with cosine ½, (cos 60° = 0.5) and sine √3⁄2, (sin 60° = 0.866...). We are therefore dealing with a conjugation by the unit quaternion

:$z = cosfrac\left\{alpha\right\}\left\{2\right\} + sinfrac\left\{alpha\right\}\left\{2\right\},hatmathbf\left\{u\right\}$

:$z = cos 60^circ + sin 60^circ,hatmathbf\left\{u\right\}$

:$z = frac\left\{1\right\}\left\{2\right\} + frac\left\{sqrt\left\{3\left\{2\right\}cdot,hatmathbf\left\{u\right\}$

:$z = frac\left\{1\right\}\left\{2\right\} + frac\left\{sqrt\left\{3\left\{2\right\}cdot frac\left\{\left(i+j+k\right)\right\}\left\{sqrt\left\{3$

:$z = frac\left\{1 + i + j + k\right\}\left\{2\right\}$.

Concretely,:"f"("ai" + "bj" + "ck") = "z" ("ai" + "bj" + "ck") "z"-1 .

Note that "z"-1 = 1/"z", as "z" has unit modulus; here"z"-1 = "z"* = (1−"i"−"j"−"k")/2. This can be simplified, using the ordinary rules for quaternion arithmetic, to:"f"("ai" + "bj" + "ck") = "ci" + "aj" + "bk",

as expected: the rotation corresponds to keeping a cube held fixed at one point, and rotating it 120° about the long diagonal through the fixed point (observe how the three axes are permuted cyclically).

Quaternion non-commutativity and rotations

The multiplication of quaternions is non-commutative. Since this operation corresponds to a three dimensional rotation, this property can be easily demonstrated by showing that three dimensional rotations are not commutative in general. A simple exercise of applying two rotations to an asymmetrical object (e.g., a book) can explain it. First, rotate a book 90 degrees clockwise around the z axis. Next rotate it 180 degrees clockwise around the x axis and memorize the result. Then restore the original orientation, so that the book title is again readable, and apply those rotations in opposite order. Compare the outcome to the earlier result. This shows that, in general, the composition of two different rotations around two distinct spatial axes will not commute.

Quaternions versus other representations of rotations

The representation of a rotation as a quaternion (4 numbers) is more compact than the representation as an orthogonal matrix (9 numbers). Furthermore, for a given axis and angle, one can easily construct the corresponding quaternion, and conversely, for a given quaternion one can easily read off the axis and the angle. Both of these are much harder with matrices or Euler angles.

In computer games and other applications, one is often interested in “smooth rotations”, meaning that the scene should slowly rotate and not in a single step. This can be accomplished by choosing a curve such as the spherical linear interpolation in the quaternions, with one endpoint being the identity transformation 1 (or some other initial rotation) and the other being the intended final rotation. This is more problematic with other representations of rotations.

When composing several rotations on a computer, rounding errors necessarily accumulate. A quaternion that’s slightly off still represents a rotation after being normalised—a matrix that’s slightly off need not be orthogonal anymore and therefore is harder to convert back to a proper orthogonal matrix.

Quaternions also avoid a phenomenon called gimbal lock which can result when, for example in pitch/yaw/roll rotational systems, the pitch is rotated 90° up or down, so that yaw and roll then correspond to the same motion, and a degree of freedom of rotation is lost. In a gimbal-based aerospace inertial navigation system, for instance, this could have disastrous results if the aircraft is in a steep dive or ascent.

The orthogonal matrix corresponding to a rotation by the unit quaternion "z" = "a" + "bi" + "cj" + "dk" (with |z| = 1) is given by:

Beware the vector convention: There are two conventions for rotation matrices: one assumes row vectors on the left; the other assumes column vectors on the right; the two conventions generate matrices that are the transpose of each other. The above matrix assumes row vectors on the left. In general, a matrix for vertex transpose is ambiguous unless the vector convention is also mentioned. Historically, the column-on-the-right convention comes from math and classical mechanics, whereas row-vector-on-the-left comes from computer graphics, where typesetting row vectors was easier back in the early days.

(Compare the equivalent general formula for a 3 × 3 rotation matrix in terms of the axis and the angle.)

Note that quaternions, like any other rotation or linear transform, are not "handed" (as in left-handed vs right-handed). Handedness of a coordinate system comes from the interpretation of the numbers in physical space. No matter what the handedness convention, rotating the X vector 90 degrees around the Z vector will yield the Y vector -- the math and numbers are the same.

Performance comparisons with other rotation methods

This section discusses the performance implications of using quaternions versus other methods (axis/angle or rotation matrices) to perform rotations in 3D. A brief summary:

* Note: angle-axis can be stored as 3 elements by multiplying the unit rotation axis by the rotation angle; however, before using it the unit axis and angle must be recovered by normalizing, costing additional math operations.

There are three basic approaches to rotating a vector "v":
# Compute the matrix-vector product of a 3x3 rotation matrix R and the original 3x1 vector "v", giving $v_\left\{new\right\} = mathbf\left\{R\right\}v$. This requires 3*(3 multiplications + 2 additions) = 9 multiplications and 6 additions, the most efficient method for rotating a vector.
# Use the quaternion rotation formula derived above of $v_\left\{new\right\}= z mathbf\left\{v\right\} z^\left\{-1\right\}$. Computing this result is equivalent to transforming the quaternion to a rotation matrix R using the formula above then multiplying with a vector. Performing some common subexpression elimination yields an algorithm that costs 21 multiplies and 18 adds. As a second approach, the quaternion could first be converted to its equivalent angle/axis representation then the angle/axis representation used to rotate the vector. However, this is both less efficient and less numerically stable when the quaternion nears the no rotation point.
# Use the angle-axis formula to convert an angle/axis to a rotation matrix R then multiplying with a vector. Converting the angle/axis to R using common subexpression elimination costs 14 multiplies, 2 function calls (sin, cos), and 10 add/subtracts; from item 1, rotating using R adds an additional 9 multiplications and 6 additions for a total of 23 multiplies, 16 add/subtracts, and 2 function calls (sin, cos).

Pseudo-code for rotating using a quaternion: given a quaternion "z" = "a" + "bi" + "cj" + "dk" (with |z| = 1) and a vector "v" with elements "v1", "v2", and "v3", the following code performs a rotation. Note the use of temporary variables "t"xx. Also note one optimization of the diagonal entries of the R matrix: since $a^2 + b^2 + c^2 + d^2 = 1$, rewrite the top-left entry as $a^2 + b^2 + c^2 + d^2 - 2c^2 - 2d^2 = 1 - 2c^2 - 2d^2$; the other two diagonal entries can be similarly rewritten. t2 = a*b t3 = a*c t4 = a*d t5 = -b*b t6 = b*c t7 = b*d t8 = -c*c t9 = c*d t10 = -d*d v1new = 2*( (t8 + t10)*v1 + (t6 - t4)*v2 + (t3 + t7)*v3 ) + v1 v2new = 2*( (t4 + t6)*v1 + (t5 + t10)*v2 + (t9 - t2)*v3 ) + v2 v3new = 2*( (t7 - t3)*v1 + (t2 + t9)*v2 + (t5 + t8)*v3 ) + v3

Pseudo-code for creating an angle/axis matrix where the unit axis is ("v1", "v2", "v3") and the angle is "θ": t1 = cos(theta) t2 = 1 - t1 t3 = v1*v1 t6 = t2*v1 t7 = t6*v2 t8 = sin(theta) t9 = t8*v3 t11 = t6*v3 t12 = t8*v2 t15 = v2*v2 t19 = t2*v2*v3 t20 = t8*v1 t24 = v3*v3 R [1, 1] = t1 + t2*t3 R [1, 2] = t7 - t9 R [1, 3] = t11 + t12 R [2, 1] = t7 + t9 R [2, 2] = t1 + t2*t15 R [2, 3] = t19 - t20 R [3, 1] = t11 - t12 R [3, 2] = t19 + t20 R [3, 3] = t1 + t2*t24

Pairs of unit quaternions as rotations in 4D space

A pair of unit quaternions "z"l and "z"r can represent any rotation in 4D space. Given a four dimensional vector v, and pretending that it is a quaternion, we can rotate the vector v like this:

It is straightforward to check that for each matrix "M" "M"T = "I", that is, that each matrix (and hence both matrices together) represents a rotation. Note that since ("z""l" "v") "z""r" = "z""l" ("v" "z""r"), the two matrices must commute. Therefore, there are two commuting subgroups of the set of four dimensional rotations. Arbitrary four dimensional rotations have 6 degrees of freedom, each matrix represents 3 of those 6 degrees of freedom.

Since an infinitesimal four-dimensional rotation can be represented by a pair of quaternions (as follows), all (non-infinitesimal) four-dimensional rotations can also be represented.

$z_l=left\left(1+\left\{dt_\left\{ab\right\}+dt_\left\{cd\right\}over 2\right\}i+\left\{dt_\left\{ac\right\}-dt_\left\{bd\right\}over 2\right\}j+\left\{dt_\left\{ad\right\}+dt_\left\{bc\right\}over 2\right\}k ight\right)$

$z_r=left\left(1+\left\{dt_\left\{ab\right\}-dt_\left\{cd\right\}over 2\right\}i+\left\{dt_\left\{ac\right\}+dt_\left\{bd\right\}over 2\right\}j+\left\{dt_\left\{ad\right\}-dt_\left\{bc\right\}over 2\right\}k ight\right)$

*Slerp — spherical linear interpolation
*conversion between quaternions and Euler angles
*rotation group
*coordinate rotations
*Clifford algebras
*spinor group
*covering map
*3-sphere
*SO(4)

References

* [ftp://ftp.cis.upenn.edu/pub/graphics/shoemake/quatut.ps.Z Shoemake, Ken. Quaternion tutorial]
* [http://graphics.stanford.edu/courses/cs348c-95-fall/software/quatdemo/ Hart, Francis, Kauffman. Quaternion demo]
* [http://www.diku.dk/publikationer/tekniske.rapporter/1998/98-5.ps.gz Dam, Koch, Lillholm. Quaternions, Interpolation and Animation]
* [http://home.ewha.ac.kr/~bulee/quaternion.pdf Byung-Uk Lee, Unit Quaternion Representation of Rotation]
* [http://www.itk.org/CourseWare/Training/QuaternionsI.pdf Ibanez, Luis, Quaternion Tutorial I]
* [http://www.itk.org/CourseWare/Training/QuaternionsII.pdf Ibanez, Luis, Quaternion Tutorial II]
* [http://citeseer.ist.psu.edu/vicci01quaternions.html Leandra Vicci, Quaternions and Rotations in 3-Space: The Algebra and its Geometric Interpretation]

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# The Orbital Index

Issue No. 173 | Jun 22, 2022

🚀 🌍 🛰

 ¶SpaceX in the news.  SpaceX conducted three perfect Falcon 9 launches and first stage recoveries in just 36 hours—the fastest three launch sequence ever by a commercial launch company. The first delivered 53 Starlink v1.5 sats to LEO from KSC—the constellation is now >2,400 strong in orbit. This was followed by a Vandenberg launch of SARah-1, a SAR surveillance satellite for the German military. Finally, a backup Globalstar-15 FM15 communications satellite with undisclosed secondary payloads from a defense customer was launched back in Florida.The booster flying the weekend’s Starlink mission was their fleet’s life-leader, completing its 13th launch and landing. This flight also hit a 100 orbital-class booster reflight milestone. After the two following launches, SpaceX is now at 102 reflights, 126 landings, and 164 total launches.A group of SpaceX employees published an open letter asking that the company address Musk’s increasing Twitter activity and questionable content, more clearly define its “no asshole” policy, and hold leadership more accountable. Company president and COO Gwynne Shotwell responded with a company-wide email stating that five employees had been fired for the “overreaching activism” of the letter and for the way it was sent to all employees at the company. (We stand by our previously stated option that the corporate culture at SpaceX needs to change, while also acknowledging that this may not have been an acceptable use of company resources, and that this may not have been the best response from SpaceX leadership.)With the FAA environmental review out of the way, Musk tweeted that Starship will be ready to conduct an orbital test next month. Inputting that prediction into the Elon time converter gives a more reasonable sounding estimate of August-November.And to make all this happen, the company recently closed $1.68 billion in additional funding—marginally less than it was planning—at a ~$125 billion valuation.
 The Falcon 9 carrying a Globalstar-15 FM15 communications satellite... and undisclosed secondary defense payloads launched from Cape Canaveral. Photo credit: Julia Bergeron.
 The Orbital Index is made possible through generous sponsorship by:
 ¶Type I Civilization by 2371? In 1964, Russian astronomer Nikolai Kardashev categorized civilizations into three types based on their ability to capture and use energy (paper). Type I civilizations can capture the energy received by an entire planet (~1017 Watts for Earth), Type II civilizations are able to capture the energy output of a single star (Sol generates ~4 × 1026 W), and Type III civilizations capture energy equivalent to their local galaxy (ours is ~5 × 1036 W). Carl Sagan adapted the Kardashev scale to be continuous and calculated that we were a Type 0.7 civilization in 1973—more recently that’s been updated to 0.73. You can, of course, use other scales for measuring human development, such as the UN Human Development Index (education, life expectancy, and per capita income), Sagan’s own Information Mastery (although Earth’s data production appears to be growing significantly faster than Sagan hypothesized), the Global Peace Index, and sadly, historic global atmospheric carbon levels. A new analysis takes the Kardashev scale and estimates when our planet might actually reach the Type I milestone (paper), while controlling for continued avoidance of the great filter (given current trends, perhaps this is a species’s ability to drastically alter the climate of its home planet). Taking into account fossil, nuclear, and renewable energy production, along with limitations from the United Nations Framework Convention on Climate Change (as opposed to the largely unsustainable energy consumption growth of the past 200 years) the authors estimate that we’ll consume an Earth-worth of energy sometime around 2371. What about Type II? It’s all incredibly speculative, but we might harvest energy using Dyson spheres, and later black hole accretion discs and jets, and/or matter-antimatter annihilation. “As an approximation to the more familiar, it can be estimated that to obtain 1×1026 W [Type II], the water mass contained in approximately 14,700 typical backyard swimming pools, typically 20,000 gallons each, would need to be annihilated every second, assuming 100% efficiency in converting mass to captured energy.” 💧💥
 ¶(Short) Papers. New Horizon data appear to show cryovolcanic slushy-ice-water volcanoes on Pluto which resurface large areas over time, leaving regions younger than their surroundings (paper). Picture a slow moving “slushy icy-water mix or even possibly a solid flow like glaciers.”Meanwhile, lava flows interacting with sulfur dioxide frost on Jupiter’s moon Io may create dunes (paper).Titan has similar dunes, but made of frozen hydrocarbons, left by an almost-Earth-like seasonal liquid transport cycle (paper).Not only is Europa’s subsurface ocean an intriguing site to look for extraterrestrial life, but based on its ice shell it “may be less of a barrier and more of a dynamic system – and site of potential habitability in its own right.” This data comes from ice-penetrating radar scans of Europa-like features in Greenland, finding abundant water pockets in the ice itself (paper). Hubble confirmed that comet C/2014 “Bernardinelli-Bernstein” has the largest cometary nucleus ever seen, at 130 km across (paper). The object that killed the dinosaurs was probably only ~10 km across. Worry not, C/2014’s closest approach to the Sun is 1.6 billion km.Speaking of which, shards of the dinosaur-killing asteroid may have been found at the fascinating Tanis site in North Dakota.Meanwhile, the 200-meter-wide comet 323P/SOHO circles the Sun every four years, getting as close as 5 million km from the surface. On its 2021 approach, the comet had multiple house-sized chunks blown off of it, probably due to superheating and its likely-YORP-induced half-hour rotation period (the fastest of any known comet).On Earth, life may have evolved as early as 300 million years after the planet formed (paper). It’s kind of hard to imagine that life could show up on Earth basically as soon as it could possibly exist, and yet wouldn’t have evolved somewhere else too.However, humanity will need to survive 300,000 years before we should expect to detect a signal from another communicating extraterrestrial intelligent civilization (CETI). This is the best case scenario according to a recent paper. The pessimistic case is 50 million years. We’ll keep looking, though… unless China has already detected aliens.
 XKCD #2569
 ¶News in brief. With the ISS back on track for Russian support through 2024, Rogozin is now directing his threats towards forcibly taking over control of eROSITA, the German x-ray telescope aboard Spektr-RG ● ESA broke ground on a new deep-space antenna to support ongoing exploration missions—their fourth in Australia ● With OFT-2 out of the way, new crewmembers have been assigned to the upcoming Starliner CFT-1 ● Momentus’ first Vigoride tug mission, launched on Transporter 5, is struggling, with solar array and comms failures, and only two of the nine customer satellites onboard having been deployed ● ESA and NASA are collaborating on Lunar Pathfinder, with a NASA CLPS mission in ~2025 scheduled to launch the lunar farside communications satellite for ESA—the beginning of its Moonlight lunar communications network ● Artemis I’s fourth WDR attempt aborted at T-29 seconds on Monday due to a LH2 leak in the boosters quick disconnect—second stage testing was completed on Tuesday ● Now dependent on SpaceX for launch and likely mutually fearful of Project Kuiper, OneWeb and Starlink have stopped their bickering over spectrum ● Cygnus aborted it first ISS reboost test this week, shutting down engines just 5 seconds into a 5+ minute burn—a next attempt is planned for the 25th ● Before that, the docked Progress 81 capsule had to boost the station to avoid a piece of debris generated by recent Russian ASAT testing that could have come within 0.8 km ● South Korea successfully launched its Nuri rocket with several satellites ● Stratolaunch’s enormous Roc hypersonic-launch-test-platform climbed to its highest altitude yet, 8,200 m, on its seventh test flight from the Mojave Air and Space Port (landing video). 🛫🛫🛬🛬
 Stratolaunch’s Roc on its record breaking flight.
 ¶Jobs. Neuralink is hiring a Senior Mechanical Engineer to work on hardware solutions for brain-computer interface devices (including surgical robotics, automation, microfabrication, and more) in Fremont, CA.
 ¶Etc. Cats always land with their feet down… but what if there is no down?A lovely thread from Dr. Phil (a comm/nav engineer on the Shuttle before he became a physicist) on the exceptionally complex process to keep Florida rain from soaking into the thermal protection tiles on the Space Shuttle, which would have added many tons to its takeoff weight.Percy appears to have picked up a pet rock. Space Flight Laboratory (SFL) announced that their 3.5 kg, 3U CanX-7 cubesat which launched in 2016 has successfully deorbited via deployable drag sail. The four part, 4-square-meter sail allowed the satellite, initially in a 700 km orbit, to deborbit in 5 years, when it would have taken about 175 years without the additional drag. The CHIME radio telescope sifts through close to a petabyte of data per day as it searches for extragalactic fast radio bursts (FRBs). When detected, it now releases a notification in the Virtual Observatory Event (VOEvent) format which anyone can subscribe to in order to quickly point their own telescopes. You can also subscribe to the SuperNova Early Warning System (detecting the neutrinos that herald a supernova), Chirp for gravitational wave event alerts, and of course space weather alerts from NOAA.
 Detailed in a recent paper, Perseverance caught some of the most intense dust activity ever witnessed on Mars—hundreds of dust devils and the first video of wind gusts lifting a Martian dust cloud. Below is a video of dust devils in Jezero Crater taken on July 20, 2021. The unusual quantity of dust devils at Jezero may be due to the roughness of the crater’s surface, making it easier for wind to lift dust.

© 2020 The Orbital Index. All rights reserved.

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text:
## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 37

#### Answer

The equation of the tangent line: $$(l):y=2x+2$$ The equation of the normal line: $$(m):y=-\frac{1}{2}x+2$$

#### Work Step by Step

$$y=f(x)=x^4+2e^x$$ 1) Find the derivative of $y$$y'=\frac{d}{dx}(x^4)+2\frac{d}{dx}(e^x)$$ $$y'=4x^3+2e^x$$ 2) Find$y'(0)$$$y'(0)=4\times0^3+2e^0$$$$y'(0)=0+2\times1=2$$ 3)$y'(0)$is the slope of the tangent line$(l)$to the curve$f(x)$at point$(0,2)$Knowing the slope of the tangent line$(l)y'(0)$and 1 point of (l)$(0,2)$, the equation of the tangent line$(l)$would be: $$(l):(y-2)=y'(0)(x-0)$$ $$(l):y-2=2x$$ $$(l):y=2x+2$$ 4) The normal line$(m)$at$(0,2)$is the line perpendicular to the tangent line$(l)$at$(0,2)$. So, the slope$a$of$(m)$is: $$a=\frac{-1}{y'(0)}=\frac{-1}{2}$$ (the product of the slopes of 2 perpendicular lines equals$-1$) Knowing the slope of$(m)$and 1 point of$(m)(0,2)$, the equation of the normal line$(m)\$ would be $$(m):(y-2)=a(x-0)$$ $$(m):y-2=-\frac{1}{2}x$$ $$(m):y=-\frac{1}{2}x+2$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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Posted by8 years ago
Archived

[Ask Math] I want to learn financial stochastic differential equations. What are the best pdf's for this job out there ?

I am interested in something easy to understand maybe somewhat applied textbook on Stochastic differential equations. Something that has exercises and someone with an engineer background can keep his interest on. The thrugh however is that any help would be greatly admired. Thanks in advance guys!

EDIT: I asked for some responses and so many people volunteered to help, this is WAY too good. Thank you reddit for being so cool!

82% Upvoted
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level 1
2 points · 8 years ago

I've read a couple of books on this and I thinks Shreve's books are really good introductory texts:

They focus heavily on applications in finance, so I don't know if that's your thing. Also they've been critizised for a their use of handwaving arguments, which is fair, but not bad when you're just getting started.

Neftci's book is also pretty well written and only assume basic calculus.

It also really depends what maths background you have exactly. To really get into it you'd need a solid background in measure theory and probability theory as well as calculus.

level 2
Original Poster3 points · 8 years ago

Thank you thank you you are my star!

level 3
1 point · 8 years ago

No worries, man. Let me know if you need any more pointers. Have fun with those SDEs :)

Oh and one more tip: I found the trick to understanding stochastic calculus is to really understand the notation. Try to really get what dW means and why dWdW != 0 before moving on to the "real" stuff. That's what makes it different from ordinary calculus, basically.

level 4
Original Poster2 points · 8 years ago

Thanks I will keep this in mind!

level 4
[deleted]
1 point · 8 years ago

Thank you so much!

level 1
1 point · 8 years ago · edited 8 years ago

Ima gonna do PhD in Financial mathematics starting next month. My general take on this topic besides the aforementioned Shreve and Neftci book If you want to have a better understanding of the working and maths: (the previous two books focus more on the products themselves, especially the Neftci (who used to taught in the school where I got my master before he passed btw)) Like someone has said, it will be beneficial if u have some idea of probability theory (Richard Durrett has a classic text on it, but u'd probably not need it if you are not doing hardcore stochastic analysis), and perhaps real analysis background (measure theory derived from real analysis, also lesbegue integrals, the idea extended to construction of some of the stochastic theory that are applied) I recommend Tom Apostol's mathematical analysis for this

for me, the two best book on SDE and application of finance are the Bernt Oksandel and the Fima Klebaner

Klebaner is easier than Oksandel, but the latter has more application in finance.

Four of the most important theory/idea in Financial math that you will need a good grasp on: Ito's Lemma (pretty much like construction of differentiation in normal calculus), Girsanov theorem, Feymann-Kac (which is just Kolmogorov equations tbh) and martingale representation theorem.

Good luck, this field is huge, have fun.

Edit: BTW, used to be a EE graduated from Purdue, so I know what u meant by 'engineering approach'.

level 2
Original Poster1 point · 8 years ago

I bow to your superior stochastic-fu arts!

level 3
1 point · 8 years ago

o I almost forgot, the Klebaner/Oksandel books are very comprehensive, but if you want a REALLY introductory textbook on Stochastic calculus, check this one out: Thomas Mikosh that's my very first stochastic calculus book, it's only about 200 pages and a lot of graph, easy math. Good intro book

I would probably do: Mikosh -> Klebaner/Oksandel (depends which book u like better) alongside with the Neftci to know more about pricing of various products. Shreve is self-containing, but sometimes it lacks explanation of finer detail, good reference text though. (his another book Brownian motion and stochastic calculus co-authored with Ioannis Karatzas is a classic text that every graduate student must read though) If you want to know more about the 'financial insight' (not just math) of financial engineering, there is no better book than Mark Joshi's concept of mathematical finance. Okay, I should stop before I overwhelmed you with all these books. So to conclude, my idea to study FE would be split into two routes:

Mathematics: Mikosh -> Klebaner/Oksandel

Finance/Product-pricing: Neftci -> Mark Joshi -> Shreve (maybe)

level 1
1 point · 8 years ago

level 1
Probability1 point · 8 years ago

I second CorenliusJack's recommendation with slight modification: I would mostly use Øksendal and Shreve (although Shreve is a little on the simplistic side and doesn't cover SDEs very extensively). Most of the research now is done in stochastic processes without the assumption of continuous sample paths. I'm not sure what your experience level is, but at a bare minimum you should have a good grasp of measure theory and basic stochastic processes in order to tackle SDEs. Full disclosure, I am a PhD student in mathematical finance in my 5th year.

level 1
-1 points · 8 years ago

I've been taught the subject twice, and I can say that the time I actually understood it we were using the course notes from the Institute and Faculty of Actuaries' CT8 subject - produced by ActEd. Unfortunately you won't find a (legal) copy of them online.

level 2
Original Poster1 point · 8 years ago

Damn it would be so good to get our hands on it. Thanks a lot for the tip!

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text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Existence of positive solutions of a nonlinear fourth-order boundary value problem. (English) Zbl 1195.34037

Summary: We study the existence of positive solutions of fourth-order boundary value problem

${u}^{\left(4\right)}\left(t\right)=f\left(t,u\left(t\right),{u}^{\text{'}\text{'}}\left(t\right)\right),t\in \left(0,1\right)·$
$u\left(0\right)=u\left(1\right)={u}^{\text{'}\text{'}}\left(0\right)={u}^{\text{'}\text{'}}\left(1\right)=0,$

where $f:\left[0,1\right]×\left[0,\infty \right)×\left(-\infty ,0\right]\to \left[0,\infty \right)$ is continuous. The proof of our main result is based upon the Krein-Rutman theorem and the global bifurcation techniques.

##### MSC:
 34B18 Positive solutions of nonlinear boundary value problems for ODE 47N20 Applications of operator theory to differential and integral equations 47J15 Abstract bifurcation theory
##### References:
 [1] Gupta, C. P.: Existence and uniqueness theorems for a bending of an elastic beam equation, Appl. anal. 26, 289-304 (1988) · Zbl 0611.34015 · doi:10.1080/00036818808839715 [2] Gupta, C. P.: Existence and uniqueness results for the bending of an elastic beam equation at resonance, J. math. Anal. appl. 135, 208-225 (1988) · Zbl 0655.73001 · doi:10.1016/0022-247X(88)90149-7 [3] Agarwal, R. P.: On fourth order boundary value problems arising in beam analysis, Differential integral equations 2, No. 1, 91-110 (1989) · Zbl 0715.34032 [4] Del Pino, M. A.; Mansevich, R. F.: Existence for a fourth-order boundary value problem under a two-parameter nonresonance condition, Proc. amer. Math. soc. 112, No. 1, 81-86 (1991) · Zbl 0725.34020 · doi:10.2307/2048482 [5] Ma, Ruyun; Wang, Haiyan: On the existence of positive solutions of fourth-order ordinary differential equations, Appl. anal. 59, 225-231 (1995) · Zbl 0841.34019 · doi:10.1080/00036819508840401 [6] Bai, Zhanbing; Wang, Haiyan: On positive solutions of some nonlinear fourth-order beam equations, J. math. Anal. appl. 270, No. 2, 357-368 (2002) · Zbl 1006.34023 · doi:10.1016/S0022-247X(02)00071-9 [7] Li, Yongxiang: Positive solutions of fourth-order boundary value problems with two parameters, J. math. Anal. appl. 281, 477-484 (2003) · Zbl 1030.34016 · doi:10.1016/S0022-247X(03)00131-8 [8] Chai, Guoqing: Existence of positive solutions for fourth-order boundary value problem with variable parameters, Nonlinear anal. 26, 289-304 (2007) [9] Ma, Ruyun: Existence of positive solutions of a four-order boundary value problem, Appl. math. Comput. 168, 1219-1231 (2005) [10] Ma, Ruyun: Nodal solutions of boundary value problems of fourth-order ordinary differential equations, J. math. Anal. appl. 319, 424-434 (2006) · Zbl 1098.34012 · doi:10.1016/j.jmaa.2005.06.045 [11] Ma, Ruyun: Nodal solutions for a fourth-order two-point boundary value problem, J. math. Anal. appl. 314, 254-265 (2006) · Zbl 1085.34015 · doi:10.1016/j.jmaa.2005.03.078 [12] Dancer, E. N.: Global solution branches for positive mappings, Arch. ration. Mech. anal. 52, 181-192 (1973) · Zbl 0275.47043 · doi:10.1007/BF00282326 [13] Zeidler, E.: Nonlinear functional analysis and its applications, I. Fixed-point theorems, (1985) [14] Deimling, K.: Nonlinear functional analysis, (1985) · Zbl 0559.47040


text:
mantel_test: Mantel test \loadmathjax In metan: Multi Environment Trials Analysis

Description

Performs a Mantel test between two correlation/distance matrices. The function calculates the correlation between two matrices, the Z-score that is is the sum of the products of the corresponding elements of the matrices and a two-tailed p-value (null hypothesis: \mjsdeqnr = 0).

Usage

 1 mantel_test(mat1, mat2, nboot = 1000, plot = FALSE)

Arguments

 mat1, mat2 A correlation matrix or an object of class dist. nboot The number of permutations to be used. Defaults to 1000. plot if plot = TRUE, plots the density estimate of the permutation distribution along with the observed Z-score as a vertical line.

Value

• mantel_r The correlation between the two matrices.

• z_score The Z-score.

• p-value The quantile of the observed Z-score. in the permutation distribution.

Author(s)

Tiago Olivoto tiagoolivoto@gmail.com

pairs_mantel()
 1 2 3 4 5 6 library(metan) # Test if the correlation of traits (data_ge2 dataset) # changes between A1 and A2 levels of factor ENV A1 <- corr_coef(data_ge2 %>% subset(ENV == "A1"))[["cor"]] A2 <- corr_coef(data_ge2 %>% subset(ENV == "A2"))[["cor"]] mantel_test(A1, A2, plot = TRUE)


text:
DDP Recipes Forum

Ground bison with hominy and wild rice

Ground bison with hominy and wild rice

Here is a recipe (with some additions).

 RECIPE- Bison with Hominy and Wild Rice Locale for ingredients Price Ground Bison 1 lbWild rice - 3/4 cupHominy - four cupsKosher salt to taste Northern Meats in Ishpeming (across from movie theatre)Nett LakeWal-martWal-mart $8.95 per pound (but had shopping show certificates from great lakes radio)$12/lbhuge can was \$2.98. (total of 12 cups)unsure

Okay, next time I might not use as much hominy in the future but I felt like I had to cook it. I have a ton of it. Anyway, I browned the ground bison. Cooked wild rice -- boil on high for ten minutes, then cover on low until the rice breaks open and starts to curl. Cook hominy - basically I just let it boil for a while, then put it on low. Keep stirring both rice and hominy to make sure nothing sticks to the bottom. Mix all together when everything is cooked. Add salt to taste (and it will need something). Didn't have leeks, but will add minced leeks next time.


text:
# Articles by category: tools

#### Finding the Most Popular SciHub Articles with Approximate Heavy Hitters

01 Oct 2021

I’ve added heavy hitters functionality to the dsrs crate (in addition to a variant of Count-Min). It’s another streaming algorithm which helps us find the most popular repeated lines in a stream. In this blog post, we’ll see how this approximate algorithm saves memory over an exact approach.For instance, maybe we have access logs which contain IP addresses like so:1.1.1.13.1.2.31.1.1.14.2.1.21.1.1.1where there could be millions of unique IP addresses accessing our server, but we’d only be int...

#### Amazon Reviewers With Sketches

29 Jun 2021

To show off a recent command line tool for sketching, dsrs, let’s plot the rolling 28-day average daily count of active reviewers on Amazon.The raw data here is item,user,rating,timestamp so this would map to a sophisticated GROUP BY with a COUNT DISTINCT over 28-day windows in SQL. But since the data’s only available as CSV, how can we get to the same answer? If we’re just interested in an approximate solution, can we do this without using a bunch of memory or custom (shuffle-inducing…) slid...

#### Fast SVMlight to CSR in Python with Rust

17 Jan 2021

Lots of sparse datasets are kept around in a convenient text format called SVMlight. It’s easy to manipulate with unix tools and very easily compressed so it’s perfect to distribute.However, the main way that’s available to access this format in python is dreadfully slow due to a natural lack of parallelism. svm2csr is a quick python package I wrote with a rust extension that parallelizes SVMlight parsing for faster loads in python. Check it out!P.S., here’s what this format looks like:-1 2:1...

#### L-BFGS vs GD

01 Nov 2020

Curiously, the original L-BFGS convergence proof essentially reduces the L-BFGS iteration to GD. This establishes L-BFGS converges globally for sufficiently regular functions and also that it has local linear convergence, just like GD, for smooth and strongly convex functions.But if you look carefully at the proof, the construction is very strange: the more memory L-BFGS uses the less it looks like GD, the worse the smoothness constants are for the actual local rate of convergence. I go to in...

#### Linear Degeneracy and Standardization

25 Oct 2020

This notebook comes from my Linear Regression Analysis notes.In the ordinary least squares setting, we model our outputs $\mathbf{y}=X\boldsymbol\beta+\boldsymbol\varepsilon$ where $\boldsymbol\varepsilon\sim N(\mathbf{0}, \sigma^2 I)$, with $\boldsymbol\beta,\sigma^2$ unknown.As a result the OLS fit $\hat{\boldsymbol\beta}\sim N\left((X^\top X)^{-1}X^\top \mathbf{y},\sigma^2 (X^\top X)^{-1}\right)$ (an distribution with an unknown variance scaling factor which we must still S...

#### RCT2 Solution

12 Aug 2020

Last time, we discussed the RCT2 problem, which we won’t delve into in great detail, but at a high level, we have an inductively defined Markov chain, parameterized by $n$, with special start and end states and the following outgoing arrows, such that for $k\in[n]$, we have the following transition dynamics:from IPython.display import ImageImage(filename='transitions.png') We already went over how to solve the expected hitting time for the end state for a given, known $n$. We now ...

#### Roller Coaster Tycoon 2 (RCT2) Problem

11 Aug 2020

Roller Coaster Tycoon 2 (RCT2) ProblemIn Roller Coaster Tycoon 2, you play as the owner of an amusement park, building rides to attract guests. One ride you can build is a maze. It turns out that, for visual appeal, guests traverse mazes using a stochastic algorithm.Marcel Vos made an entertaining video describing the algorithm, which has in turn spawned an HN thread and Math SE post.I really enjoyed solving the problem and wanted to share the answer as well as some further thoughts.Mazes in ...

#### Discrete Residues

05 Jul 2020

In the previous post, we discussed a particular quantity, $\mathbb{P}\{X=1\}$, where $X$ follows a Poisson Binomial distribution, parameterized by $\{p_j\}_{j=1}^n$. This means that $X=\sum_jX_j$, where $X_j$ are Bernoulli- $p_j$ independent random variables.We came up with an $O(1)$ memory and $O(n)$ time approach to computing the desired probability, and gave an example where even the best approximations can be poor.What about a more general question? Let’s ta...

#### Complex Hash Collisions

04 Jul 2020

This article investigates a fun thought experiment about the Poisson-Binomial distribution.Let’s imagine we’re designing a large hash table. We know up-front we’re going to get $n$ distinct keys, so let’s number them $j\in [n]$.Ahead of time, we’re allowed to see the set of hashes the $j$-th key will belong to. It is allowed to take on one of $c_j$ distinct hashes, $S_j=\{h_1, \cdots, h_{c_j}\}\subset\mathbb{N}$, where at run time the real hash is sampled uniformly (and in...

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#### The Triple Staple

30 Nov 2019

When reading, I prefer paper to electronic media. Unfortunately, a lot of my reading involves manuscripts from 8 to 100 pages in length, with the original document being an electronic PDF.Double-sided printing works really well to resolve this issue partway. It lets me convert PDFs into paper documents, which I can focus on. This works great up to 15 pages. I print the page out and staple it. I’ve tried not-stapling the printed pages before, but then the individual papers frequently get out o...

#### Numpy Gems, Part 2

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Numpy Gems 1: Approximate Dictionary Encoding and Fast Python MappingWelcome to the first installment of Numpy Gems, a deep dive into a library that probably shaped python itself into the language it is today, numpy.I’ve spoken extensively on numpy (HN discussion), but I think the library is full of delightful little gems that enable perfect instances of API-context fit, the situation where interfaces and algorithmic problem contexts fall in line oh-so-nicely and the resulting code is clean, ...

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Beating TensorFlow Training in-VRAMIn this post, I’d like to introduce a technique that I’ve found helps accelerate mini-batch SGD training in my use case. I suppose this post could also be read as a public grievance directed towards the TensorFlow Dataset API optimizing for the large vision deep learning use-case, but maybe I’m just not hitting the right incantation to get tf.Dataset working (in which case, drop me a line). The solution is to TensorFlow harder anyway, so this shouldn’t reall...

#### Jupyter Tricks

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Jupyter TricksHere’s a list of my top-used Juypter tricks, and what they do.UII find the UI to be intuitive, Help > User Interface Tour describes more. There are command (enter by pressing the escape button or clicking outside of a cell) and edit (enter by typing in a cell) modes. You can tell you’re in edit mode if the “pencil” corner indicator is present:It’s also faster to use the commands as listed in Help > Keyboard Shortcuts; with those you can also remove the toolbar with View &g...


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# Question #40e06

Nov 16, 2016

54 words/minute

#### Explanation:

A “unit rate” for anything is a basic unit per a basic time unit. In this case we have “words” and “minutes”. The rate is a ratio of the two:

$\frac{297}{5.5}$

Performing the indicated division gives us the desired unit rate of 54 words/minute


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# Braid relations $n_\alpha n_\beta n_\alpha \ldots = n_\beta n_\alpha n_\beta \ldots$ in arbitrary reductive groups

I'm currently trying to prove or disprove the following claim. First let me set up some notation.

Let $G$ be a connected reductive group over a field $K$, let $S \leq Z \leq N \leq G$ be respectively a maximal split torus, its centralizer and its normalizer. Let $\Phi$ be the root system of the pair $(G,S)$, and for $\alpha \in \Phi$ let $U_\alpha \leq G$ be the associated root subgroup (possibly noncommutative, when $G$ is not split). Given $u \in U_\alpha(K)-\{1\}$, there exist unique elements $m(u) \in N(K)$ and $v,v' \in U_{-\alpha}(K)$ such that $vuv' = m(u)$. The image of the element $m(u)$ in the Weyl group $W_0 = N(K)/Z(K)$ is the reflection $s_\alpha$ associated to the root $\alpha$, where $W_0$ is naturally identified with the Weyl group of $\Phi$ via the action of $N(K)$ on $X^\ast(S)$.

Now for the claim: For any two roots $\alpha, \beta \in \Phi$ with $\alpha,\beta$ not parallel (an obvious necessary condition), and any $u \in U_\alpha(K)-\{1\}$ and $v \in U_\beta(K)-\{1\}$ the braid relation

$n_\alpha n_\beta n_\alpha \ldots = n_\beta n_\alpha n_\beta \ldots$

in $N(K)$ holds true, where the number of factors on both sides equal the order of $s_\alpha s_\beta$, and $n_\alpha = m(u)$ and $n_\beta = m(v)$.

I think I have a proof of this in the case when $\Phi$ is reduced. So a counterexample should involve non-reduced root systems. I tried to disprove the claim using the example $G = SU(h)$ of a quasi-split unitary group (example 1.15 in Tits Corvallis article), but unfortunately (or fortunately) the claim seems to hold in this case. (edit: Even for unitary groups over division algebras the claim seems to hold true.)

So if you either have a counterexample to the above claim or a proof (with or without reference), it would help me a lot.

Despite what I wrote earlier the claim is not true, despite Proposition 6.1.8 of Bruhat-Tits (first volume). Said proposition states that

Proposition 6.1.8: Let $\Phi$ be a root system of rank $2$ and let $(T, (U_a,M_a)_{a \in \Phi})$ be a generative root datum. Order the set $\Phi^{\text{red}} = \{ a \in \Phi : \frac{1}{2}a \not\in \Phi\}$ of reduced roots in a circular order, i.e. write $\Phi^{\text{red}} = \{a_1,\ldots{},a_{2k}\}$ with $\Phi \cap (\mathbb{R}_{> 0} a_{i-1} + \mathbb{R}_{> 0} a_{i+1}) = \{a_{i-1},a_i,a_{i+1}\}$. Then $\Delta = \{a_1, a_k\}$ is a basis of $\Phi^{\text{red}}$ with $\{a_1,\ldots,a_k\}$ being the set of positive roots. Let $a = a_1$ and $b = a_k$.

For any $u \in U_a$, $u' \in U_b$ with $u,u' \neq 1$ the relation $mm'm\ldots = m'mm'\ldots$ holds true, where the number of factors on both sides is $k$, and $m = m(u)$, $m' = m(u')$.

I thought that my claim would follow from this by constructing for two given non-parallel roots $\alpha,\beta$ a rank two root subsystem $\Psi \subseteq \Phi$ for which $\{\alpha, \beta\}$ forms a basis. Although it is possible to do so, one cannot restrict the root datum $(T,(U_a,M_a)_{a \in \Phi})$ defined by the reductive group in question to a root datum $(T, (U_a,M_a)_{a \in \Psi})$ in general, since the condition, that for $a,b \in \Psi$ the commutator subgroup $(U_a, U_b)$ lies in the subgroup generated by the $U_{pa+qb}$ with $p,q \in \mathbb{N}_{\geq 1}$, might be violated, unless $\Psi = \Phi \cap \mathbb{Z}\Psi$. Hence the proposition implies the claim only for roots $\alpha, \beta$ which are (up to a multiple) part of a root basis.

To see that the claim is false in general, consider the following example of a special unitary group over a division algebra (which I had already considered, but the roots $\alpha, \beta$ I chose formed part of a basis). Let $D$ be a central division algebra over a field $K$ which is equipped with an involution $\sigma: D \rightarrow D$ (i.e. $\sigma(ab) = \sigma(b)\sigma(a)$ and $\sigma^2 = \text{id}$) which is the identity on $K$. Let $n \geq 1$, $I = \{\pm 1,\ldots{},\pm n\}$ and let $V$ be the free right-$D$-module with basis $e_{-n},\ldots{},e_n$. Consider the hermitian form $h$ on $V$ given in terms of coordinates by $h(x,y) = \sum_{i \in I} \sigma(x_i)y_{-i} + \sigma(x_0)y_0$. Let $G$ be the algebraic group $\mathbf{SU}(V,h)$ considered naturally as a subgroup of $\mathbf{GL}_D(V)$. Let $S$ be the 'obvious' maximal split subtorus of $G$, i.e. on rational points we have

$S(K) = \{ \text{diag}(d_{-n},\ldots{},d_n) : d_i \in K^\times \text{ for }i \in I \cup \{0\}, d_{-i}d_i = d_0 = 1\text{ for } i \in I \}$

The root system $\Phi = \Phi(G,S)$ is then given as follows. Let $\chi_1,\ldots,\chi_n$ be the 'obvious' $\mathbb{Z}$-basis of $X^\ast(S)$, i.e. $\chi_i(\text{diag}(d_{-n},\ldots{},d_n)) = d_i$. Then we have

$\Phi = \{ \chi_i - \chi_j : i,j \in I,\ i \neq \pm j\} \cup \{ \chi_i, 2\chi_i : i \in I \}$

(type $\text{BC}_n$) or

$\Phi = \{ \chi_i - \chi_j : i,j \in I,\ i \neq \pm j\} \cup \{ \chi : i \in I \}$

(type $\text{B}_n$), according to whether or not $D$ contains an element $x$ with $\sigma(x) = -x$. In either case, the root subgroups are given as follows. For $i,j \in I$, $i \neq \pm j$ we have

$U_{\chi_i - \chi_j}(K) = \{ u_{i,j}(d) : d \in D \}$

Here $u_{i,j}(d) = 1 + d e_{i,j} - \sigma(d)e_{-j,-i}$ as a sum in the matrix ring $\text{End}_D(V)$ and $e_{i,j}$ denotes the matrix which has a one in the $i$-th row and $j$-th column, and zeroes everywhere else. For $i \in I$

$U_{\chi_i}(K) = \{ u_i(c,d) : c,d \in D,\ \sigma(c)c + d + \sigma(d) = 0 \}$

where $u_i(c,d) = 1 + ce_{0,-i} - \sigma(c)e_{i,0} + de_{i,-i}$. The elements $m(u)$ are then given as follows.

$m(u_{i,j}(d)) = \tau \text{diag}(d_{-n},\ldots{},d_n)$

Here $\tau$ is the permutation $(i\ j)(-i\ -j)$ of $I\cup \{0\}$ considered as a monomial matrix in $\text{GL}_D(V)$, and $d_i = d$, $d_{-i} = \sigma(d)^{-1}$, $d_j = -d^{-1}$, $d_{-j} = -\sigma(d)$ and $d_k = 1$ for all other $k$. For the other roots we have

$m(u_i(c,d)) = \tau \text{diag}(d_{-n},\ldots{},d_n)$

where $\tau = (i\ -i)$ and $d_i = d$, $d_{-i} = \sigma(d)^{-1}$, $d_0 = cd^{-1}\sigma(c) + 1$ and $d_k = 1$ for all other $k$.

Now let $i \neq \pm j$, $\alpha = \chi_i$, $\beta = \chi_j$ and $n_\alpha = m(u_i(c,d))$, $n_\beta = m(u_j(\tilde{c},\tilde{d}))$. Since $s_\alpha s_\beta$ is of order two, the braid relations reduces to the commutativity $n_\alpha n_\beta = n_\beta n_\alpha$. But an easy computation shows that $n_\alpha, n_\beta$ commute if and only if $cd^{-1}\sigma(c)+1$ and $\tilde{c}\tilde{d}^{-1}\sigma(\tilde{c}) + 1$ commute, which is easily seen to be false in general. For instance take $K = \mathbb{R}$, $D$ the quaternions and let $c = \tilde{c} = 1$, $d = -\frac{1}{2} + \frac{1}{2}i$, $\tilde{d} = -\frac{1}{2} + \frac{1}{2}j$. Then $cd^{-1}\sigma(c) + 1 = -\frac{1}{2}i$ and $\tilde{c}\tilde{d}^{-1}\sigma(\tilde{c}) + 1 = -\frac{1}{2}j$.


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## Algebra 1: Common Core (15th Edition)

We will put x=2 and y=2 in the inequality. $2\leq-2(2)+1$ By simplifying- $2\leq-4+1$ $2\leq-3$ No


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## Basic College Mathematics (10th Edition)

Multiples of 12 : $12\times1, 12\times2, 12\times3, 12\times4, 12\times5,$ = 12, 24, 36, 48, 60 Multiples of 18 : $18\times1, 18\times2, 18\times3, 18\times4, 18\times5,$ = 18, 36, 54, 72, 90 Now 36 is the smallest common number in both the lists. so 36 is LCM of 12 & 18.


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# A compound has 2.3% H, 32.6% P and remaining O. What is emprical formula? Then write...

###### Question:

A compound has 2.3% H, 32.6% P and remaining O. What is emprical formula? Then write balanced equation showing rxn between tetraphosphorus decoxide and water.

Using the eqn, determine theorertical yield if we have 51.6g of tetraphosphorus and 39.2g water (find limiting and excess regant first)

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Curvy Equation

Age 16 to 18Challenge Level

Siddhant from Singapore International School in India and Wiktor and Jiali from LAE Tottenham in the UK sketched the function correctly.

Jiali found the $x$-intercept:

Siddhant wrote:
At $x=0,$ the $\dfrac{\ln{x}}{x}$ is undefined, but $0$ is not in the domain of $\text{h}(x)$

To investigate what happens as $x\rightarrow0$, Jiali substituted in some values of $x.$ Jiali also substituted in larger values of $x$ to investigate the shape of the curve:

Wiktor described what happens for large values of $x,$ using the graphs of $y=x$ and $y=\ln{x}$ to help:

Siddhant, Wiktor and Jiali all differntiated $\text{h}(x)$ using the quotient rule and Siddhant used $\text{h}'(x)$ to understand the shape of the graph. Siddhant wrote:
\begin{align}\text{h}'(x) &= \frac{\frac{\text{d}}{\text{d}x}\left[\ln{x}\right].x - \ln{x}.\frac{\text{d}}{\text{d}x}\left[x\right]}{x^2}\\ &=\frac{1-\ln{x}}{x^2}\end{align} Stationary point:
$\dfrac{1-\ln{x}}{x^2} = 0\\ 1-\ln{x}=0\\ \ln{x}=1\\ x=e$

Therefore $\text{h}(x)$ is stationary at $x=e$

$\rightarrow$ Therefore as $x$ approaches $e,$ $\text{h}(x)$ increases
$\rightarrow$ As $x$ increases from $e,$ $\text{h}(x)$ decreases because $\text{h}'(x)$ becomes negative (and so must have a horizontal asymptote since Jiali showed it doesn't cross the $x$ axis again)

Siddhant, Wiktor and Jiali used the graph to solve the equation $n^m=m^n$. Jiali began:

Siddhant showed how the equation can be related to the graph:
$n^m=m^n$
Take $\ln$ on both sides \begin{align} \ln{n^m} &= \ln{m^n}\\ m\ln{n} &= n\ln{m} \\ \frac{\ln{n}}n &=\frac{\ln{m}}m\end{align} We know:
$\text{h}(m) = \dfrac{\ln{m}}{m}\\ \text{h}(n) = \dfrac{\ln{n}}{n}$

Therefore, $n^m = m^n$ can be written as $\text{h}(m) = \text{h}(n)$

Using the graph, $\text{h}(x)$ is many-one after $x=1$ and is stationary at $x=e$.

Jiali and Wiktor showed this graphically. This is Wiktor's work:

Wiktor then substituted $2$ and $4$ into the equation to check that they work. Siddhant found $4$ by trial.

Well done!


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# Questions in Equilibrium [closed]

I have two inquiries concerning equilibrium:

1. Does $$K_c = 5.0$$ indicate that the equilibrium mixture contains both products and reactants at an approximately equal amount? I know that $$K_c \approx 1.0$$ is said to be so, however I am unsure of if $$5.0$$ would be considered to be approximately $$1.0$$ in this case.

2. Is a reaction at equilibrium if the forward reaction does not change? I believe it would as an indication of equilibrium is that both the forward and reverse reactions occur at the same rate.

## closed as off-topic by A.K., Mithoron, Mathew Mahindaratne, Todd Minehardt, airhuffApr 20 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• I suppose the OP wants confirmation the equilibrium means a dynamic equilibrium, with the equal rate of the opposite reactions. Yes, it is . By "reaction does not change" he probably means the concentrations are constant. – Poutnik Apr 19 at 17:44
• You won't get quicker or better answers if you spam Chemistry.SE with the same questions. This is not a paid service, there is no staff constantly sitting on-line and answering all the questions, and you cannot demand a fast and accurate response (plus you already got one answer in less than an hour, and in addition to that I already voted to reopen your first question). Please visit Help Center to familiarize yourself with the basic principles of this site. – andselisk Apr 20 at 3:11
• If were familiar with the rules, you would've not re-posted the same question just because you were dissatisfied with the first answer given and want a quicker response. I feel pity for you if you see this as aggression or fight, I'm just informing that what you've done wasn't correct; also, as I see it, the answer was as decent as the question itself. – andselisk Apr 20 at 3:27
• @James It is very unusual the rate of reaction being constant. Do you have a particular equilibrium in mind ? – Poutnik Apr 20 at 3:35
• @James correct me if I'm wrong but I choose to disagree. A specific example of an equilibrium refers to an equilibrium example with reactants and products for example the equilibrium formed by iron thiocyanate ions in solution. Without a specific example it is difficult to tell if an equilibrium constant 5.0 corresponds to roughly similar amounts of reactants and products – sab hoque Apr 20 at 4:25

A reaction is in equilibrium,
if the reaction quotient $$Q=K_{\rm c}$$,

For reaction $$\ce{A <=> B}$$ $$K_{\rm c}=5$$ means concentration ratio 1:5. It does not seem to me to be approximately the same concentrations.

The equilibrium constant is the thermodynamical quantity, determining the position of equilibrium. It's value does not say if the reaction is in equilibrium.

About the same concentrations, it means rather the same product of concentrations of reagents and products.

If there is a reaction $$\ce{A + B <=> C + D}$$ with $$K_{\rm c}=1$$ and if concentrations are $$c_{\rm A}=1000c_{\rm B}=100c_{\rm C}=10c_{\rm D}$$ we can hardly speak about equal concentrations at equilibrium. as $$K_c=1=\frac{10\cdot 100}{1000\cdot 1}$$

Additionally, if a reaction is not symmetric in counts of reagents of both sides, equilibrium concentration ratios change with concentrations.

If there is a reaction $$\ce{A <=> 2 B}$$ The same concentrations during equilibrium with $$K_{\rm c}=1$$ occur only for $$c_{\rm A}=1$$. If we consider concentrations of $$\ce{A}$$ and $$\ce{B}$$, the following concentrations are in equilibrium:

• A B
• 0.0001 0.01
• 0.01 0.1
• 1 1
• 100 10
• 10000 100

Yes, you think right, at equilibrium, the rates of the forward and reverse reactions are equal.

If \begin{align} rate_{\rm forward}&=const\\ rate_{\rm backward}&=f([products])\\ \end{align} ("Forward reaction does not change"=has a constant rate, as was explained)

..then as reaction is progressing, the backward reaction rate increases, until it matches the forward rate and reaction is at equilibrium ( regardless of the $$K_{\rm c}$$ value.)

• If you are familiar with the concept of the equilibrium constant, which I expect you are, you would know that if Kc, which is the denotation for the equilibrium constant, is approximately 1.0, the reaction is said to be at equilibrium. The issue arises to what is meant by "approximately". This is in fact what my question is asking. – James Apr 19 at 23:21
• @James The Kc value and reaction being at equilibrium are 2 independent things. Kc value is matter of reaction thermodynamics, describing the position of the equilibrium state with minimal value of Gibbs energy. Being at equilibrium is matter of reaction kinetics. The Kc value does not say if the reaction already reached equilibrium or not. – Poutnik Apr 20 at 3:28
• @James A reaction can be in equilibrium for any value of Kc, e.g $10^{52}$ – Poutnik Apr 20 at 5:12


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## Differential and Integral Equations

### Nonlocal interaction equations: Stationary states and stability analysis

Gaël Raoul

#### Abstract

In this paper, we are interested in the long-time behavior of solutions to a nonlocal interaction equation in dimension $1$. We show that, up to an extraction, the solution converges to a steady-state. Then, we study the structure of stable steady-states.

#### Article information

Source
Differential Integral Equations, Volume 25, Number 5/6 (2012), 417-440.

Dates
First available in Project Euclid: 20 December 2012


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# $KO_*$ groups of $\mathbb{R}P^\infty$, “Snaiths” theorem for $KO$

I wonder whether anyone has taken the time to compute $$KO_*(\mathbb{R}P^\infty)$$?

The standard tools to compute these Groups in the complex case rest on the requirement for the cohomology theories $$E$$ to be complex orientable. Naturally, I looked up whether something like real orientable cohomology theories exist in the literature but found out that $$KO$$ is not real-oriented. Anyway, there is a way to "circumvent" Snaiths theorem for the spectrum $$K$$ if one is only interestd in the algebra of cooperations, in the sense that one can show that $$K_*(\mathbb{C}P^\infty) \xrightarrow{i_*} K_*K$$ is an injection of rings, where $$i$$ is induced from the inclusion $$\mathbb{C}P^\infty \simeq BU(1) \hookrightarrow BU$$. In fact, one only needs to invert the Bott element $$\beta$$ to turn it into an isomorphism, so it is a localization. This can be concluded from

Robert M. Switzer. Algebraic topology—homotopy and homology. Classics in Mathematics. Springer-Verlag, Berlin, 2002. Reprint of the 1975 original [Springer, New York; MR0385836 (52 #6695)].

17.33, which states

$$K_*K$$ is generated over $$\mathbb{Z}[u,u^{-1},v^{-1}]$$ by the polynomials $$\{p_1,p_2,\ldots\}$$.

By a process reminiscent of

J. F. Adams. Stable homotopy and generalised homology. University of Chicago Press, Chicago, Ill.-London, 1974. Chicago Lectures in Mathematics.

p. 44 we can describe the relations of the generators of $$\beta_i$$ of $$K_*(\mathbb{C}P^\infty) = K_* \{\beta_0 , \beta_1 , \ldots \}$$ such that

$$\beta_1\beta_n = n \beta_n +(n+1)\beta_{n+1}$$

and by setting

$$\binom{x}{i} = \frac{x(x-1)\cdots (x-(i-1))}{i!} \in \mathbb{Q}[x]$$

with $$x:=\beta_1$$ one can see that

$$K_*(\mathbb{C}P^\infty)\otimes \mathbb{Q}$$ is the polynomial algebra $$K_* \otimes \mathbb{Q}[x]$$ over $$K_*\otimes \mathbb{Q} = \mathbb{Q} [t,t^{-1}]$$ and $$K_*(\mathbb{C}P^\infty)$$ can be identified with the subalgebra of $$K_* \otimes \mathbb{Q}[x]$$ generated by $$\binom{x}{i}$$ for $$i=0,1,2, \ldots$$,

where we set $$\binom{x}{0}=1$$.

While snaiths theorem works on the spectrum level and the aforementioned result follows, I wonder whether a similar result holds in the real case, i.e. $$KO_*(\mathbb{R}P^\infty)[\alpha^{-1}] \cong KO_*KO$$ for some element $$\alpha \in KO_*(\mathbb{R}P^\infty)$$?


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Transcribed image text: Find the slope of the tangent line to the curve y = x/1-x at the point (1/2, 1). Find the equation of that tangent line. On your calculator, graph both the curve and the tangent line on the same set of axes with viewing window x: [-1, 1, y: [-2, 2. [, except for anything concerning the graphs you made on. Wolfram|Alpha Widgets Sign In. Slope of the tangent line to a curve. Slope of the tangent line to a curve. y=. (X,Y) Submit. Added Feb 16, 2015 by razer65 in Mathematics. Find the slope of the tangent line to a curve y=f (x) at a point (X, Y). This calculator helps in finding the normal line and eases the process of finding this line. This calculator is user friendly with its simple instructions and steps that can be easily understood. This calculator provides the functions result step by step. The use of this calculator is free of cost. It does not require any financial requirement. How To Find The Equation Of A Tangent Line Math Wonderhowto. Slope Of Tangent Line Geogebra. Calc Iii Parametric Equations Of A Tangent Line You. Secant Slope Calculator Geogebra. Finding The Tangent Line Equation With Derivatives Calculus Problems You. How To Find The Equation Of A Tangent Line 8 Steps. Sage Calculus Tutorial Tangent Lines. The slope of a line is a measure of how steep it is. Slope Calculator Solutions Input two points using numbers, fractions, mixed numbers or decimals. The slope calculator shows the work. What is the slope of the tangent line that touches the graph of Iny + e2x-1 = 3y– 4x at the point .1) 5 A) 2 в з 1 D m/2 Question Transcribed Image Text: What is the slope of the tangent line that. When using the slope of tangent line calculator, the slope-intercept formula for a line is found by the formula below: y = mx + b. Where. m stands for the slope of the line. b is the y-intercept. For instance, when you enter the curve, y= 4x^2-4x+1 at x=1, in our tangent line finder, the result will be as follows: y= 4x2-4x+1 at x=1. Sketch the tangent line going through the given point. (Remember, the tangent line runs through that point and has the same slope as the graph at that point.) Example 1: Sketch. cheap rv living youtube 2022 • To use this online calculator for Slope of Tangent Line given Slope of Normal Line, enter Slope of Line 1 (m 1) and hit the calculate button. Here is how the Slope of Tangent Line given Slope of Normal Line calculation can be explained with given input values -> -0.333333 = -1/3 . • Related calculator: Tangent Line Calculator Let's see how to derive equation of tangent line when we are given equation of curve $${r}={f{{\left(\theta\right)}}}$$$ in polar coordinates. We will proceed in the same fashion as with tangent lines to parametric curves , because polar coordinates in some sense similar to parametric curves.


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# All Questions

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I've heard that light can't escape from a black hole. Can it? If not, why?
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### Why do sunspots appear dark?

Sunspots, such as this one, appear dark: Why?
110 views

### Amateur observing targets for binary star systems?

Are there any easily resolvable, binary star, observing targets visible from ~N40°? I'd like to be able to show, in one observing session, Mizar and Alcor (naked eye resolvable), another with ...


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Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 1.3 - Problem 1
Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 1.3 - Problem 1

×

# ?In Exercises 1–10, evaluate the function at the given value(s) of the independent variable. Simplify the results. 1. f(x) = 7x - 4 (

ISBN: 9781285774770 141

## Solution for problem 1 Chapter 1.3

Calculus: Early Transcendental Functions | 6th Edition

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Problem 1

In Exercises 1–10, evaluate the function at the given value(s) of the independent variable. Simplify the results.

1. f(x) = 7x - 4

(a) f(0)

(b) f(-3)

(c) f(b)

(d) f(x-1)

Step-by-Step Solution:

Step 1 of 5) In Exercises 1–10, evaluate the function at the given value(s) of the independent variable. Simplify the results.

Step 2 of 2

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Calculus: Early Transcendental Functions : Integration Techniques, LHpitals Rule, and Improper Integrals
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Statistics: Informed Decisions Using Data : Applications of the Normal Distribution
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# Basic open sets in the Zariski topology are also compact.

Let $A$ be a commutative ring and $X = \text{Spec}(A)$. The closed sets are those of the form $V(E) = \{$ prime ideals $\hat{p} \subset A$ containing $E \}$. And the open sets are the complements of these.

I've already shown that $X$ is compact by showing that $1$ is in a certain ideal and so must equal a finite sum.

How do you show that in general $X_f = X \setminus V(f)$ is compact in the Zariski topology?

I tried generalizing: that it's true of any basis element in a compact topological space. But that never left the runway.

Your generalization is false, since for example open arcs in $S^1$ form a basis for its topology, but are not compact.

You could use there's a homeomorphism ${\rm Spec}(A_f)\simeq X_f$, and use what you've already proven.

You can also give a direct proof, in the same lines as the one for $X$. Suppose we cover $X_f$ with the basic open sets $X_{f_i}$. Since $X_f\cap X_{f_i}= X_{ff_i}$, we may suppose that $X_f=\bigcup_i X_{g_i}$, and $g_i\in (f)$. Translating into open sets, we get that $$V((g_i:i\in I))=V(f)$$

This means there is an equation of the form $f^n=a_1g_1+\cdots+a_ng_n$. This means that $f$ is in the radical of $(g_1,\ldots,g_n)$ so that $V(f)=V(g_1,\ldots,g_n)$ (since we already had $g_i\in (f)$), so that indeed $X_f=\bigcup\limits_{i=1}^n X_{g_i}$.

• What's $A_f$ ? Thanks. I'm not opposed to more advanced methods. I know what a homeomorphism is, so this should be interesting. – BananaCats Category Theory App Mar 4 '15 at 18:44
• @EnjoysMath $A_f$ is $A$ localized at $\{f,f,^2,f^3,\ldots\}$, or $A[f^{-1}]$. – Pedro Tamaroff Mar 4 '15 at 18:47
• okay, I haven't gotten that far. What do you recommend I read up on? Localization of rings? I'm going by Atiyah, and localization hasn't been presented yet. Is $A[f^{-1}]$ a ring extension in which the inverse of $f$ is included or defined to be included? – BananaCats Category Theory App Mar 4 '15 at 18:50
• @EnjoysMath That book should have all you need for you to understand the above. Look at the exercises, and the chapter on localization. I think it is the third. – Pedro Tamaroff Mar 4 '15 at 18:53
• Thanks dude! I'm still on exercises of chapter 1. But I'll skip ahead and back. – BananaCats Category Theory App Mar 4 '15 at 18:54


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# Preparation of a standard solution by dissolving

- Calculate the mass of solid which must be dissolved in the given volume. For example, to prepare of $250\; ml$ of a $0.2$ molar solution of sodium chloride, we need $n$ $=$ $[NaCl]\cdot V$ $=$ $0.2\cdot 0.25$ $=$ $0.05\; mol$, ie $m$ $=$ $n\cdot M$ $=$ $0.5\cdot 58.44$ $=$ $2.922\; g$ $NaCl$ - Weigh accurately that mass: - Introduce in a volumetric flask: - Rinse with distilled water into the volumetric flask: - Introduce distilled water: - Stir to dissolve the solid: - Fill distilled water to the mark: - Stopper the flask and stir: - The standard solution (the molarity is known) is ready to use !

Pictures taken from: →   Lycée Clemenceau de Reims


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# nLab Guifre Vidal

Selected writings

## Selected writings

On tree tensor networks related to entanglement entropy:

• Sukhwinder Singh, Robert N. C. Pfeifer, Guifre Vidal, Tensor network decompositions in the presence of a global symmetry, Phys. Rev. A82:050301, 2010 (arXiv:0907.2994)

• Sukhwinder Singh, Robert N. C. Pfeifer, Guifre Vidal, Tensor network states and algorithms in the presence of a global $U(1)$ symmetry, Phys. Rev. B 83, 115125 (2011) (arXix:1008.4774)

• Sukhwinder Singh, Guifre Vidal, Tensor network states and algorithms in the presence of a global $SU(2)$ symmetry, Phys. Rev. B 86, 195114 (2012) (arXiv:1208.3919)

category: people

Last revised on February 9, 2020 at 05:31:52. See the history of this page for a list of all contributions to it.


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## Tuesday, March 31, 2015

### LeetCode OJ - Excel Sheet Column Number

Problem:

Solution:

As you will be able to read from the code - the algorithm is pretty simple. The key idea is to understand how did I get there (i.e. the thought process).

If it is possible at all, I would start any problem from its simplest case. In this particular example, a single letter problem is quite simple, isn't it? So all we need to do is to have s[0] - 'A' + 1 and we are done.

Now we start with the second letter. I knew if I know how to do "AA", it is simply the next of "Z", which is 27. "AZ" is 52, therefore "BA" is 53. I conjectured that it is just the second digit times 26 + the first digit, and it is indeed the case, because for the second character to advance, the first character must have gone through 'A' - 'Z', which is 26 characters.

The rest follows.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/excel-sheet-column-number/

#include "LEET_EXCEL_SHEET_COLUMN_NUMBER.h"
#include <map>
#include <iostream>
#include <vector>

using namespace std;

namespace _LEET_EXCEL_SHEET_COLUMN_NUMBER
{
class Solution
{
public:
int titleToNumber(string s)
{
int result = 0;
int digitFactor = 1;
for (int i = s.length() - 1; i >= 0; i--)
{
result += (s[i] - 'A' + 1) * digitFactor;
digitFactor *= 26;
}

return result;
}
};
};

using namespace _LEET_EXCEL_SHEET_COLUMN_NUMBER;

int LEET_EXCEL_SHEET_COLUMN_NUMBER()
{
Solution solution;
cout << solution.titleToNumber("A") << endl;
cout << solution.titleToNumber("Z") << endl;
cout << solution.titleToNumber("AA") << endl;
cout << solution.titleToNumber("AZ") << endl;
cout << solution.titleToNumber("BA") << endl;
cout << solution.titleToNumber("ZZ") << endl;
cout << solution.titleToNumber("AAA") << endl;
cout << solution.titleToNumber("AAZ") << endl;
cout << solution.titleToNumber("ABA") << endl;
cout << solution.titleToNumber("AZZ") << endl;
cout << solution.titleToNumber("BAA") << endl;
return 0;
}


text:
# Research-aptitude questions set 10

By | January 5, 2017

# JOIN FREE COACHING

## Important Research aptitude questions

 Question 1
WATS lines provided by long distance telephone service at fixed rates. In this regard, WATS is the abbreviation of
 A West Africa Theological Seminary B Washtenaw Area Transportation Study C Wide Area Telecommunications Service D World Air Transport Statistics
 Question 2
To obtain the freest opinion of the respondent, when we ask general question before a specific question then this procedure is called as the
 A Research technique B Qualitative technique C Funnel technique D Quantitative technique
 Question 3
A small scale trial run of a particular component is known as
 A Pilot testing B Pre-testing C Lab experiments D Both A & B
 Question 4
Which of the following is not a characteristic of a good theory or explanation?
 A It is parsimonious B It is testable C It is general enough to apply to more than one place, situation, or person D All of the above are characteristics of good theories
 Question 5
Which of the following is not a basic assumption of science?
 A Science cannot provide answers to all questions B It is possible to distinguish between more and less plausible claims C Researchers should follow certain agreed upon norms and practices D Science is best at solving value conflicts, such as whether abortion is immoral
 Question 6
Behaviouristic School was established by
 A William James B W. Kohler C J.B. Watson D K. Koffka
 Question 7
The most effective method of studying psychology is
 A Experimental Method B Observation Method C Introspection Method D Survey Method
 Question 8
Anything which evokes a response in the Organism is called
 A Stimulus B Thing C Situation D Incidence
 Question 9
A systematic study of facts according to a reliable and correct method of study is called a
 A Biological Study B Social Technique C Scientific Study D Methodology
 Question 10
That, which cannot be observed by another person, is called
 A Experience B Activity C Action D Exercise
 Question 11
The first psychological laboratory was established in Leipzig by Wilhelm Wundt in the year
 A 1789 B 1879 C 1899 D 1897
 Question 12
Woodworth's approach to define Psychology is
 A Constant B Fluctuating C Dynamic D Complex
 Question 13
The study of the interaction of heredity and environment is made by
 A Physiological Psychologists B Child Psychologists C Comparative Psychologists D Educational Psychologists
 Question 14
Field Observation is always
 A Subjective B Objective C Neutral D Complex
 Question 15
Regulating the Relevant Variable'' is called
 A Hypothesis B Constant Error C Experimental Control D Ant hypothesis
 Question 16
When more than one independent variable works in an experimental situation, it is called
 A Situational Crisis B Interaction C Multisituational Effect D Variable Crisis
 Question 17
It does not possible or even desirable to rule out the differential effects of all relevant variables by means of experimental control. In such instances psychologists employ controlled conditions by mathematical methods. This is called
 A Statistical Control B Mathematical Control C Psychological Control D Physiological Control
 Question 18
Gestalt School was founded by
 A Max Wertheimer (1880-1943) B Wolfgang Kohler (1887-1937) C Kurt Koffka (1886-1941) D Kurt Lewin (1890-1947)
 Question 19
The theory of evolution was propounded by
 A William James (1842-1910) B Kurt Lewin (1890-1947) C Sir Charles Darwin (1809-1882) D John Deway (1859-1952)
 Question 20
The central theme of Structural Psychology'' was borrowed from
 A Physics and Biology B Biology and Chemistry C Sociology and Neurology D Physics and Chemistry
 Question 21
Who has constructed the first intelligence test as a psychologist and made important contributions to our understanding of the thought process?
 A William James B Alfred Binet C William Mc Dougall D J.B. Watson
 Question 22
A provisional theory to explain observed facts is known as
 A Construct B Theory C Hypothesis D Event
 Question 23
An experiment has been aptly described as a
 A Question put to self B Question put to C Question put to psychologists D Question put to nature stimuli
 Question 24
An experiment is set up to confirm or refute some
 A Hypotheses B Questions C Enquiry D Events
 Question 25
Psychology has come into existence as a separate discipline a little over
 A A hundred years ago B Two hundred years ago C Three hundred years ago D Four hundred years ago
There are 25 questions to complete.