text:
1. Sep 6, 2009

### noblegas

Hope this question is not too personal; I just like to get a good idea on what my financial status is relative to the financial status of other college graduates or soon to be graduates like myself

2. Sep 6, 2009

### negitron

Broke and in debt. The good news is that the higher your degree, the broker and more in debt you're likely to be. Congratulations.

3. Sep 6, 2009

### Topher925

When I got my BS I had zero debt and about 10 grand in the bank.

4. Sep 6, 2009

Same here.

5. Sep 6, 2009

### Pengwuino

Ditto. I'll probably end up with my MS and even more in the bank.

6. Sep 6, 2009

### lisab

Staff Emeritus
I had a modest amount of debt and assets in the neighborhood of $2000. I've always been extremely wary of debt...I chose to go to college part-time and work part-time so that I would not incur excessive debts. Don't know how helpful my answer is since I'm probably atypical. 7. Sep 6, 2009 ### Pengwuino Thank God for state schools :). I'm amazed how one of my friends is making SUCH a stink about tuition increases when we still pay 10-20% of most private universities. Plus his parents pay for his tuition. 8. Sep 6, 2009 ### Astronuc Staff Emeritus I left grad school with about$5-6K. While in grad school, my wife and I bought a new car (Honda Civic Wagon for about 10K), and we paid off her undergrad loans, and saved for a down payment on a house.

I had paid my way through my undergrad program, and I helped my parents support my siblings in their university programs.

My parents subsequently gave me a few $K toward a downpayment on a house. 9. Sep 6, 2009 ### Jimmy Snyder Although I had been able to pay my way through undergraduate studies, I was not able to pay for my first semester as a grad student and had to borrow from my mother. It was humiliating because I thought I had been weaned. For my second semester in grad school, I got a fellowship and was able to pay my way again and within a few more semesters paid my mother back. 10. Sep 6, 2009 ### SticksandStones I'll probably be around$60,000 in debt after I get my B.Sc, and then tack on another $100,000 or so if I end up going to law or pharmacy school. 11. Sep 7, 2009 ### Vanadium 50 Staff Emeritus I graduated college about$9K in debt, and grad school about $7k in debt,$7k in the bank, and a paid-off car.

12. Sep 7, 2009

### Cyrus

That is disgusting amount of debt for a BS. Did you go to Harvard? I hope so for that price.

13. Sep 7, 2009

### Ivan Seeking

Staff Emeritus
I went full time and part time while working part time of full time. Since I went back to school a bit late in life, my situation was different than a typical college student. For one, we moved to Oregon and bought a house right in the middle of the program.

Some of my tuition went on credit cards but we didn't bury ourselves; not by today's standards. However, the cost of tuition was climbing quickly by the time I finished.

For me the far more substantial number was lost income, rather than debt. I had a successful career before returning to college. All in all, going to college probably cost me well over 200K, plus tuition and books. Moving to Oregon probably set us back another 100-200K, but that depends on how you look at it. We bought a five acre lot in the woods that would probably be worth millions anywhere within commuting distance of L.A.

P.S. No regrets. I wouldn't want to do it all again, but it was worth it. I would have liked to have gone on to a graduate program, but that would have been too much to handle financially. By the time I finished, we needed to concentrate on making money.

Last edited: Sep 7, 2009
14. Sep 7, 2009

### Sorry!

Agreed. If I paid that amount of money I wouldn't even want to attend the school. I would just be expecting my degree to come in the mail. :D

I haven't gone to university yet but I'm planning to next year after 4 years I PLAN to have no debt. Universities in Ontario have a lot of excellent scholarship opportunities and the provincial government pays for you to go while you're in school. You only need to pay them once your done. (normally its not the full amount though. I think I don't have to pay back 1 grand every year)

15. Sep 7, 2009

### Dr Transport

about $3K in debt after my Bachelors and 1st Masters. I paid everything off over the next couple of years and when I got my PhD i was in about$10K. I was deployed for the war, so I paid everything back and was completely out of debt except for my mortgage within 18 months.

16. Sep 7, 2009

I'll be graduating from undergrad with about $60k in debt as well. Even though I am a U.S. citizen, I was not eligible for in-state tuition in any state since I moved so often, and also my parents made too much money for me to get any assistance. So, I have financed my education primarily through loans. 17. Sep 7, 2009 ### noblegas Unless you receive any scholarship or your parents paid for all of your college amenities , I cannot imagine someone graduating from harvard with only$60,000 of debt seeing that it cost a student about $60000/year to attend that school 18. Sep 7, 2009 ### turbo I left college before graduating, and then went back a couple of years later to take some courses related to my work as a soils scientist in construction. Anyway, for the first three full years, I ended each year with a little more money in the bank than the previous year. I worked full-time each summer (construction or mill-work) with all the overtime I could get. Every school year, I played guitar in a band and made spending money from frat parties, mixers, etc. I also bought and sold guitars and amps on the side, and repaired and refurbished them. Lots of college kids sell off nice instruments cheap when they get hard-up for cash, so I did quite well with my little side-line. The trick is to be liquid (always have several hundred bucks in your pocket) for when opportunities present themselves, and understand the market well enough to estimate your profits accurately. I never borrowed a dime for college - state universities are a pretty good deal. 19. Sep 7, 2009 ### SticksandStones Nope, in fact I had to go to a lower tier state school because they offered me more scholarships than higher tier schools. It's also not like I did bad in high school. In fact, I was literally the top student in high school. The ironic thing is, since my parents make so little money if I did go to Harvard I'd probably have less debt. 20. Sep 7, 2009 ### Danger College? What's that? I never graduated high-school, but at least I was debt-free when I didn't. 21. Sep 7, 2009 ### Cyrus You're right, I forgot how disgustingly expensive it is to attend that place. 22. Sep 7, 2009 ### LBloom I'm on track to be$22,000 in debt by the time i get my B.S. (unless i get more grants or scholarships. Keeping my fingers crossed). Not sure how I'll pay for grad school.

I don't want to hijack this thread or anything, but does anybody think freshman year is too early for a job or should I wait a year? Don't like the idea of being in so much debt and then worrying about my PhD

23. Sep 7, 2009


text:
Calculates the distribution overlap between a vertical profile ('vp') and the vertical radiation profile of a set of emitted radar beams at various elevation angles as given by beam_profile.

beam_profile_overlap(
vp,
elev,
distance,
antenna,
zlim = c(0, 4000),
noise_floor = -Inf,
noise_floor_ref_range = 1,
steps = 500,
quantity = "dens",
normalize = T,
beam_angle = 1,
k = 4/3,
lat,
re = 6378,
rp = 6357
)

## Arguments

vp

a vertical profile of class vp

elev

numeric vector. Beam elevation(s) in degrees.

distance

the distance(s) from the radar along sea level (down range) for which to calculate the overlap in m.

antenna

radar antenna height. If missing taken from vp

zlim

altitude range in meter, given as a numeric vector of length two.

noise_floor

The system noise floor in dBZ. The total system noise expressed as the reflectivity factor it would represent at a distance noise_floor_ref_range from the radar. NOT YET IMPLEMENTED

noise_floor_ref_range

the reference distance from the radar at which noise_floor is expressed. NOT YET IMPLEMENTED

steps

number of integration steps over altitude range zlim, defining altitude grid size used for numeric integration

quantity

profile quantity to use for the altitude distribution, one of 'dens' or 'eta'.

normalize

Whether to normalize the radiation coverage pattern over the altitude range specified by zlim

beam_angle

numeric. Beam opening angle in degrees, typically the angle between the half-power (-3 dB) points of the main lobe

k

Standard refraction coefficient.

lat

radar latitude. If missing taken from vp

re

rp

## Value

A data.frame with columns distance and overlap.

## Details

This function also calculates the overlap quantity in the output of integrate_to_ppi.

Overlap is calculated as the Bhattacharyya coefficient (i.e. distribution overlap) between the (normalized) vertical profile vp and the (normalized) radiation coverage pattern as calculated by beam_profile. In the calculation of this overlap metric, NA and NaN values in the profile quantity specified by quantity are replaced with zeros.

The current implementation does not (yet) take into account the system noise floor when calculating the overlap.

In the ODIM data model the attribute /how/NEZ or /how/NEZH specifies the system noise floor (the Noise Equivalent Z or noise equivalent reflectivity factor. the H refers to the horizontal channel of a dual-polarization radar). In addition, the attribute /how/LOG gives "security distance above mean noise level (dB) threshold value". This is equivalent to the log receiver signal-to-noise ratio, i.e. the dB above the noise floor for the signal processor to report a valid reflectivity value. We recommend using NEZH+LOG for noise_floor, as this is the effective noise floor of the system below which no data will be reported by the radar signal processor.

Typical values are NEZH = -45 to -50 dBZ at 1 km from the radar. LOG is typically around 1 dB.

Need to evaluate beam by beam the returned signal relative to a uniform beam filling of at least NEZH + LOG If returned signal is lower, the gate is below noise level.

## Examples

# locate example volume file:
pvolfile <- system.file("extdata", "volume.h5", package = "bioRad")

# load the example polar volume file:

# let us use this example vertical profile:
data(example_vp)
example_vp
#>                Vertical profile (class vp)
#>

format: plain


text:
# Finite Abelian Group is Solvable

## Theorem

Let $G$ be a finite abelian group.

Then $G$ is solvable.

## Proof

Proof by strong induction on the order of $G$:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

All abelian groups of order $n$ and below are solvable.

### Basis for the Induction

$\map P 1$ is true, as the trivial group is trivially solvable.

From Prime Group is Simple, both $\map P 2$ and $\map P 3$ hold.

This is the basis for the induction.

### Induction Hypothesis

It remains to be shown that, if $\map P j$ holds, for all $j$ such that $1 \le j \le k$, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ holds.

So this is our induction hypothesis:

All abelian groups of order $k$ and below are solvable.

Then we need to show:

All abelian groups of order $k+1$ and below are solvable.

### Induction Step

This is our induction step:

Let $G$ be an abelian group such that:

$\order G = k + 1$

where $\order G$ denotes the order of $G$.

Then from Positive Integer Greater than 1 has Prime Divisor, there exists some prime $p$ which divides $k + 1$.

From Cauchy's Lemma, $G$ has an element of order $p$.

Since $G$ is abelian, this element generates a subgroup of $G$ which is normal by Subgroup of Abelian Group is Normal.

If $p = n$ and $H = G$ the proof is finished, from Prime Power Group is Solvable.

If $p < n$, then $H$ and $G / H$ (which has order $n / p$) are both solvable from the induction hypothesis.

It follows from Group is Solvable iff Normal Subgroup and Quotient are Solvable that $G$ is solvable.

The result follows by the strong principle of mathematical induction.

$\blacksquare$


text:
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Q.1 Draw a line, say AB, take a point C outside it. Through C draw a line parallel to AB using ruler and compasses only.

(iii) Join the given point C to R and mark $$\angle$$CRA.
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text:
Article provided by Wikipedia

""
""
An illustration of Cantor's diagonal argument (in base 2) for the existence of "uncountable sets. The sequence at the bottom cannot occur anywhere in the enumeration of sequences above.
""
""
An "infinite set may have the same "cardinality as a proper "subset of itself, as the depicted "bijection f(x)=2x from the "natural to the "even numbers demonstrates. Nevertheless, infinite sets of different cardinalities exist, as Cantor's diagonal argument shows.

In "set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument or the diagonal method, was published in 1891 by "Georg Cantor as a "mathematical proof that there are "infinite sets which cannot be put into "one-to-one correspondence with the infinite set of "natural numbers.[1][2][3] Such sets are now known as "uncountable sets, and the size of infinite sets is now treated by the theory of "cardinal numbers which Cantor began.

The diagonal argument was not Cantor's "first proof of the uncountability of the "real numbers, which appeared in 1874.[4][5] However, it demonstrates a powerful and general technique that has since been used in a wide range of proofs,[6] including the first of "Gödel's incompleteness theorems[2] and Turing's answer to the "Entscheidungsproblem. Diagonalization arguments are often also the source of contradictions like "Russell's paradox[7][8] and "Richard's paradox.[9]

## Uncountable set

In his 1891 article, Cantor considered the set T of all infinite "sequences of "binary digits (i.e. each digit is zero or one). He begins with a "constructive proof of the following theorem:

If s1, s2, … , sn, … is any enumeration of elements from T, then there is always an element s of T which corresponds to no sn in the enumeration.

The proof starts with an enumeration of elements from T, for example:

 s1 = (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0, 1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0, 1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...) s7 = (1, 0, 0, 0, 1, 0, 0, ...) ...

Next, a sequence s is constructed by choosing the 1st digit as "complementary to the 1st digit of s1 (swapping 0s for 1s and vice versa), the 2nd digit as complementary to the 2nd digit of s2, the 3rd digit as complementary to the 3rd digit of s3, and generally for every n, the nth digit as complementary to the nth digit of sn. For the example above, this yields:

 s1 = (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0, 1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0, 1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...) s7 = (1, 0, 0, 0, 1, 0, 0, ...) ... s = (1, 0, 1, 1, 1, 0, 1, ...)

By construction, s differs from each sn, since their nth digits differ (highlighted in the example). Hence, s cannot occur in the enumeration.

Based on this theorem, Cantor then uses a "proof by contradiction to show that:

The set T is uncountable.

The proof starts by assuming that T is "countable. Then all its elements can be written as an enumeration s1, s2, … , sn, … . Applying the previous theorem to this enumeration produces a sequence s not belonging to the enumeration. However, this contradicts s being an element of T and therefore belonging to the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable.

### Interpretation

The interpretation of Cantor's result will depend upon one's view of mathematics. To "constructivists, the argument shows no more than that there is no "bijection between the natural numbers and T. It does not rule out the possibility that the latter are "subcountable.["citation needed] In the context of "classical mathematics, this is impossible, and the diagonal argument establishes that, although both sets are infinite, there are actually more infinite sequences of ones and zeros than there are natural numbers.["citation needed]

### Real numbers

The uncountability of the "real numbers was already established by "Cantor's first uncountability proof, but it also follows from the above result. To prove this, an "injection will be constructed from the set T of infinite binary strings to the set R of real numbers. Since T is uncountable, the "image of this function, which is a subset of R, is uncountable. Therefore, R is uncountable. Also, by using a method of construction devised by Cantor, a "bijection will be constructed between T and R. Therefore, T and R have the same cardinality, which is called the ""cardinality of the continuum" and is usually denoted by ${\displaystyle {\mathfrak {c}}}$ or ${\displaystyle 2^{\aleph _{0}}}$.

An injection from T to R is given by mapping strings in T to decimals, such as mapping t = 0111… to the decimal 0.0111…. This function, defined by f(t) = 0.t, is an injection because it maps different strings to different numbers.

Instead of mapping 0111… to the decimal 0.0111…, it can be mapped to the "base b number: 0.0111…b. This leads to the family of functions: fb(t) = 0.tb. The functions fb(t) are injections, except for f2(t). This function will be modified to produce a bijection between T and R.

## General sets

""
""
Illustration of the generalized diagonal argument: The set T = {n∈ℕ: nf(n)} at the bottom cannot occur anywhere in the range of f:"→"P(ℕ). The example mapping f happens to correspond to the example enumeration s in the above picture.

A generalized form of the diagonal argument was used by Cantor to prove "Cantor's theorem: for every "set S, the "power set of S—that is, the set of all "subsets of S (here written as P(S))—has a larger cardinality than S itself. This proof proceeds as follows:

Let f be any "function from S to P(S). It suffices to prove f cannot be "surjective. That means that some member T of P(S), i.e. some subset of S, is not in the "image of f. As a candidate consider the set:

T = { sS: sf(s) }.

For every s in S, either s is in T or not. If s is in T, then by definition of T, s is not in f(s), so T is not equal to f(s). On the other hand, if s is not in T, then by definition of T, s is in f(s), so again T is not equal to f(s); cf. picture. For a more complete account of this proof, see "Cantor's theorem.

### Consequences

This result implies that the notion of the "set of all sets is an inconsistent notion. If S were the set of all sets then P(S) would at the same time be bigger than S and a subset of S.

"Russell's Paradox has shown us that "naive set theory, based on an "unrestricted comprehension scheme, is contradictory. Note that there is a similarity between the construction of T and the set in Russell's paradox. Therefore, depending on how we modify the axiom scheme of comprehension in order to avoid Russell's paradox, arguments such as the non-existence of a set of all sets may or may not remain valid.

The diagonal argument shows that the set of real numbers is "bigger" than the set of natural numbers (and therefore, the integers and rationals as well). Therefore, we can ask if there is a set whose "cardinality is "between" that of the integers and that of the reals. This question leads to the famous "continuum hypothesis. Similarly, the question of whether there exists a set whose cardinality is between |S| and |P(S)| for some infinite S leads to the "generalized continuum hypothesis.

Analogues of the diagonal argument are widely used in mathematics to prove the existence or nonexistence of certain objects. For example, the conventional proof of the unsolvability of the "halting problem is essentially a diagonal argument. Also, diagonalization was originally used to show the existence of arbitrarily hard "complexity classes and played a key role in early attempts to prove "P does not equal NP.

### Version for Quine's New Foundations

The above proof fails for "W. V. Quine's ""New Foundations" set theory (NF). In NF, the "naive axiom scheme of comprehension is modified to avoid the paradoxes by introducing a kind of "local" "type theory. In this axiom scheme,

{ sS: sf(s) }

is not a set — i.e., does not satisfy the axiom scheme. On the other hand, we might try to create a modified diagonal argument by noticing that

{ sS: sf({s}) }

is a set in NF. In which case, if P1(S) is the set of one-element subsets of S and f is a proposed bijection from P1(S) to P(S), one is able to use "proof by contradiction to prove that |P1(S)| < |P(S)|.

The proof follows by the fact that if f were indeed a map onto P(S), then we could find r in S, such that f({r}) coincides with the modified diagonal set, above. We would conclude that if r is not in f({r}), then r is in f({r}) and vice versa.

It is not possible to put P1(S) in a one-to-one relation with S, as the two have different types, and so any function so defined would violate the typing rules for the comprehension scheme.

## References

1. ^ Georg Cantor (1891). "Ueber eine elementare Frage der Mannigfaltigkeitslehre". "Jahresbericht der Deutschen Mathematiker-Vereinigung 1890–1891. 1: 75–78 (84–87 in pdf file). English translation: Ewald, William B. (ed.) (1996). From Immanuel Kant to David Hilbert: A Source Book in the Foundations of Mathematics, Volume 2. Oxford University Press. pp. 920–922. "ISBN "0-19-850536-1.
2. ^ a b Keith Simmons (30 July 1993). Universality and the Liar: An Essay on Truth and the Diagonal Argument. Cambridge University Press. pp. 20–. "ISBN "978-0-521-43069-2.
3. ^ Rudin, Walter (1976). Principles of Mathematical Analysis (3rd ed.). New York: McGraw-Hill. p. 30. "ISBN "0070856133.
4. ^ Gray, Robert (1994), "Georg Cantor and Transcendental Numbers" (PDF), "American Mathematical Monthly, 101 (9): 819–832, "doi:10.2307/2975129, "JSTOR 2975129
5. ^ Bloch, Ethan D. (2011). The Real Numbers and Real Analysis. New York: Springer. p. 429. "ISBN "978-0-387-72176-7.
6. ^ Sheppard, Barnaby (2014). The Logic of Infinity (illustrated ed.). Cambridge University Press. p. 73. "ISBN "978-1-107-05831-6. Extract of page 73
7. ^ "Russell's paradox". Stanford encyclopedia of philosophy.
8. ^ Bertrand Russell (1931). Principles of mathematics. Norton. pp. 363–366.
9. ^ Keith Simmons (30 July 1993). Universality and the Liar: An Essay on Truth and the Diagonal Argument. Cambridge University Press. p. 27. "ISBN "978-0-521-43069-2.
10. ^ See page 254 of Georg Cantor (1878), "Ein Beitrag zur Mannigfaltigkeitslehre", Journal für die Reine und Angewandte Mathematik, 84: 242–258. This proof is discussed in Joseph Dauben (1979), Georg Cantor: His Mathematics and Philosophy of the Infinite, Harvard University Press, "ISBN "0-674-34871-0, pp. 61–62, 65. On page 65, Dauben proves a result that is stronger than Cantor's. He lets "φν denote any sequence of rationals in [0, 1]." Cantor lets φν denote a sequence "enumerating the rationals in [0, 1], which is the kind of sequence needed for his construction of a bijection between [0, 1] and the irrationals in (0, 1).
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Drake
Drake C++ Documentation
SpatialVelocity< T > Class Template Reference

## Detailed Description

### template<typename T> class drake::multibody::SpatialVelocity< T >

This class represents a spatial velocity V (also called a twist) and has 6 elements with an angular (rotational) velocity ω (3-element vector) on top of a translational (linear) velocity v (3-element vector).

Spatial velocity represents the rotational and translational motion of a frame B with respect to a measured-in frame M. This class assumes that both the angular velocity ω and translational velocity v are expressed in the same expressed-in frame E. This class only stores 6 elements (namely ω and v) and does not store the underlying frames B, M, E. The user is responsible for explicitly tracking the underlying frames with monogram notation. For example, V_MB_E denotes frame B's spatial velocity measured in frame M, expressed in frame E and contains ω_MB_E (B's angular velocity measured in M, expressed in E) and v_MBo_E (Bo's translational velocity measured in M, expressed in E), where Bo is frame B's origin point. For an offset frame Bp, the monogram notation V_MBp_E denotes the spatial velocity of frame Bp measured in M, expressed in E. Details on spatial vectors and monogram notation are in sections Spatial Vectors and Multibody Quantities.

Template Parameters
 T The scalar type, which must be one of the default scalars.

#include <drake/multibody/math/spatial_velocity.h>

## Public Member Functions

SpatialVelocity ()
Default constructor. More...

SpatialVelocity (const Eigen::Ref< const Vector3< T >> &w, const Eigen::Ref< const Vector3< T >> &v)
Constructs a spatial velocity V from an angular velocity ω and a translational velocity v. More...

template<typename Derived >
SpatialVelocity (const Eigen::MatrixBase< Derived > &V)
Constructs a spatial velocity V from an Eigen expression that represents a 6-element vector, i.e., two 3-element vectors, namely an angular velocity ω and a translational velocity v. More...

SpatialVelocity< T > & ShiftInPlace (const Vector3< T > &offset)
In-place shift of a SpatialVelocity from a frame B to a frame C, where both B and C are fixed to the same frame or rigid body. More...

SpatialVelocity< T > Shift (const Vector3< T > &offset) const
Shifts a SpatialVelocity from a frame B to a frame C, where both B and C are fixed to the same frame or rigid body. More...

SpatialVelocity< T > ComposeWithMovingFrameVelocity (const Vector3< T > &position_of_moving_frame, const SpatialVelocity< T > &velocity_of_moving_frame) const
Given a frame C's velocity relative to a frame B, and frame B's velocity measured in a frame M, returns C's velocity in M. More...

dot (const SpatialForce< T > &force) const
Calculates the power generated by a spatial force. More...

dot (const SpatialMomentum< T > &momentum) const
Calculates twice (2x) a body B's kinetic energy measured in a frame M. More...

Implements CopyConstructible, CopyAssignable, MoveConstructible, MoveAssignable
SpatialVelocity (const SpatialVelocity &)=default

SpatialVelocityoperator= (const SpatialVelocity &)=default

SpatialVelocity (SpatialVelocity &&)=default

SpatialVelocityoperator= (SpatialVelocity &&)=default

Public Member Functions inherited from SpatialVector< SpatialVelocity, T >
SpatialVector ()
Default constructor. More...

SpatialVector (const Eigen::Ref< const Vector3< T >> &w, const Eigen::Ref< const Vector3< T >> &v)
Constructs a spatial vector from a rotational component w and a translational component v. More...

SpatialVector (const Eigen::MatrixBase< OtherDerived > &V)
Constructs a spatial vector V from an Eigen expression that represents a 6-element vector (3-element rotational vector on top of a 3-element translational vector). More...

int size () const
For 3D (three-dimensional) analysis, the total size of the concatenated rotational vector (3 elements) and translational vector (3 elements) is six (6), which is known at compile time. More...

const T & operator[] (int i) const

T & operator[] (int i)

const Vector3< T > & rotational () const

Vector3< T > & rotational ()

const Vector3< T > & translational () const

Vector3< T > & translational ()

const T * data () const
Returns a (const) bare pointer to the underlying data. More...

T * mutable_data ()
Returns a (mutable) bare pointer to the underlying data. More...

std::tuple< const T, const T > GetMaximumAbsoluteDifferences (const SpatialQuantity &other) const
Returns the maximum absolute values of the differences in the rotational and translational components of this and other (i.e., the infinity norms of the difference in rotational and translational components). More...

decltype(T()< T()) IsNearlyEqualWithinAbsoluteTolerance (const SpatialQuantity &other, double rotational_tolerance, double translational_tolerance) const
Compares the rotational and translational parts of this and other to check if they are the same to within specified absolute differences. More...

decltype(T()< T()) IsApprox (const SpatialQuantity &other, double tolerance=std::numeric_limits< double >::epsilon()) const
Determines whether all six corresponding elements of two spatial vectors are equal to each other to within a specified tolerance epsilon. More...

void SetNaN ()
Sets all the elements in this SpatialVector to NaN. More...

SpatialQuantitySetZero ()
Sets both the rotational and translational components of this SpatialVector to zero. More...

CoeffsEigenTypeget_coeffs ()
Returns a mutable reference to the underlying storage. More...

const CoeffsEigenTypeget_coeffs () const
Returns a constant reference to the underlying storage. More...

SpatialQuantity operator- () const
Unary minus operator. More...

SpatialQuantityoperator+= (const SpatialQuantity &V)

SpatialQuantityoperator-= (const SpatialQuantity &V)
Subtraction assignment operator. More...

SpatialQuantityoperator *= (const T &s)
Multiplication assignment operator. More...

SpatialVector (const SpatialVector &)=default

SpatialVector (SpatialVector &&)=default

SpatialVectoroperator= (const SpatialVector &)=default

SpatialVectoroperator= (SpatialVector &&)=default

## Related Functions

(Note that these are not member functions.)

template<typename T >
SpatialVelocity< T > operator+ (const SpatialVelocity< T > &V1_E, const SpatialVelocity< T > &V2_E)
Adds two spatial velocities by simply adding their 6 underlying elements. More...

template<typename T >
SpatialVelocity< T > operator- (const SpatialVelocity< T > &V1_E, const SpatialVelocity< T > &V2_E)
Subtracts spatial velocities by simply subtracting their 6 underlying elements. More...

Public Types inherited from SpatialVector< SpatialVelocity, T >
enum
Sizes for spatial quantities and its components in 3D (three dimensions). More...

using SpatialQuantity = SpatialVelocity< T >
The more specialized spatial vector class templated on the scalar type T. More...

using CoeffsEigenType = Vector6< T >
The type of the underlying in-memory representation using an Eigen vector. More...

Static Public Member Functions inherited from SpatialVector< SpatialVelocity, T >
static SpatialQuantity Zero ()
Factory to create a zero spatial vector, i.e., a SpatialVector whose rotational and translational components are both zero. More...

## ◆ SpatialVelocity() [1/5]

 SpatialVelocity ( const SpatialVelocity< T > & )
default

## ◆ SpatialVelocity() [2/5]

 SpatialVelocity ( SpatialVelocity< T > && )
default

## ◆ SpatialVelocity() [3/5]

 SpatialVelocity ( )

Default constructor.

In Release builds, all 6 elements of a newly constructed spatial velocity are uninitialized (for speed). In Debug builds, the 6 elements are set to NaN so that invalid operations on an uninitialized spatial velocity fail fast (fast bug detection).

## ◆ SpatialVelocity() [4/5]

 SpatialVelocity ( const Eigen::Ref< const Vector3< T >> & w, const Eigen::Ref< const Vector3< T >> & v )

Constructs a spatial velocity V from an angular velocity ω and a translational velocity v.

## ◆ SpatialVelocity() [5/5]

 SpatialVelocity ( const Eigen::MatrixBase< Derived > & V )
explicit

Constructs a spatial velocity V from an Eigen expression that represents a 6-element vector, i.e., two 3-element vectors, namely an angular velocity ω and a translational velocity v.

This constructor will assert the size of V is six (6) either at compile-time for fixed sized Eigen expressions or at run-time for dynamic sized Eigen expressions.

## ◆ ComposeWithMovingFrameVelocity()

 SpatialVelocity ComposeWithMovingFrameVelocity ( const Vector3< T > & position_of_moving_frame, const SpatialVelocity< T > & velocity_of_moving_frame ) const

Given a frame C's velocity relative to a frame B, and frame B's velocity measured in a frame M, returns C's velocity in M.

Parameters
 [in] position_of_moving_frame which is the position vector p_BoCo_E (from frame B's origin Bo to frame C's origin Co), expressed in frame E. p_BoCo_E must have the same expressed-in frame E as this = V_MB_E. [in] velocity_of_moving_frame which is V_BC_E, frame C's spatial velocity measured in frame B, expressed in the same frame E as this = V_MB_E.
Return values
 V_MC_E frame C's spatial velocity measured in frame M, expressed in frame E.
Note
The returned spatial velocity V_MC_E contains an angular velocity ω_MC_E and translational velocity v_MCo_E that are calculated as:
 ω_MC_E  = ω_MB_E + ω_BC_E
v_MCo_E = v_MBo_E + ω_MB_E x p_BoCo_E + v_BCo_E

If frame C is rigidly fixed to frame B, V_BC_E = 0 and this method produces a Shift() operation (albeit inefficiently). In other words, use Shift() if velocity_of_moving_frame = 0.
SpatialAcceleration::ComposeWithMovingFrameAcceleration().

## ◆ dot() [1/2]

 T dot ( const SpatialForce< T > & force ) const

Calculates the power generated by a spatial force.

For an arbitrary frame B, calculates the dot-product of this = V_MB_E (frame B's spatial velocity measured in a frame M, expressed in a frame E) with F_B_E (frame B's spatial force, expressed in frame E).

Parameters
 [in] force which is F_B_E, frame B's spatial force, expressed in the same frame E as this = V_MB_E.
Returns
Power of spatial force F_B_E in frame M, i.e., F_B_E ⋅ V_MB_E.
Note
Just as equating force 𝐅 to mass * acceleration as 𝐅 = m𝐚 relies on acceleration 𝐚 being measured in a world frame (also called an inertial or Newtonian frame), equating power = dK/dt (where K is kinetic energy) relies on K being measured in a world frame. Hence, it is unusual to use this method unless frame M is the world frame W.
Although the spatial vectors F_B_E and V_MB_E must have the same expressed-in frame E, the returned scalar is independent of frame E.

## ◆ dot() [2/2]

 T dot ( const SpatialMomentum< T > & momentum ) const

Calculates twice (2x) a body B's kinetic energy measured in a frame M.

For any frame (e.g., an offset frame) Bp that is fixed to a rigid body B, calculates the dot-product of this = V_MBp_E (frame Bp's spatial velocity measured in frame M, expressed in frame E) with L_MBp_E (body B's spatial momentum measured in frame M, about Bp's origin, expressed in frame E).

Parameters
 [in] momentum which is L_MBp_E, body B's spatial momentum measured in frame M, about frame Bp's origin, expressed in the same frame E as this = V_MBp_E.
Returns
2*K_MB, twice (2x) body B's kinetic energy measured in frame M.
Note
In general, kinetic energy calculations are only useful when frame M is a world frame (also called a Newtonian or inertial frame). Hence, it is unusual to use this method unless frame M is the world frame W.
Although the spatial vectors V_MBp_E and L_MBp_E must have the same expressed-in frame E, the resulting scalar K_MB is independent of frame E.
As shown below, K_MB can be calculated from any frame Bp fixed on B, including body B's center of mass frame Bcm. This is due to how spatial momentum and spatial velocity shift from Bcm to Bp. For more information, see SpatialMomentum::Shift() and SpatialVelocity::Shift().
  K_MB = 1/2 (L_MBp · V_MBp) = 1/2 (L_MBcm · V_MBcm)

## ◆ operator=() [1/2]

 SpatialVelocity& operator= ( SpatialVelocity< T > && )
default

## ◆ operator=() [2/2]

 SpatialVelocity& operator= ( const SpatialVelocity< T > & )
default

## ◆ Shift()

 SpatialVelocity Shift ( const Vector3< T > & offset ) const

Shifts a SpatialVelocity from a frame B to a frame C, where both B and C are fixed to the same frame or rigid body.

Parameters
 [in] offset which is the position vector p_BoCo_E from frame B's origin to frame C's origin, expressed in frame E. p_BoCo_E must have the same expressed-in frame E as this spatial velocity (this = V_MB_E).
Return values
 V_MC_E which is frame C's spatial velocity measured in frame M, expressed in frame E.
Note
Shift() differs from ShiftInPlace() in that Shift() does not modify this whereas ShiftInPlace() does modify this.

## ◆ ShiftInPlace()

 SpatialVelocity& ShiftInPlace ( const Vector3< T > & offset )

In-place shift of a SpatialVelocity from a frame B to a frame C, where both B and C are fixed to the same frame or rigid body.

On entry, this is V_MB_E (frame B's spatial velocity measured in a frame M and expressed in a frame E). On return this is modified to V_MC_E (frame C's spatial velocity measured in frame M and expressed in frame E).

Parameters
 [in] offset which is the position vector p_BoCo_E from frame B's origin to frame C's origin, expressed in frame E. p_BoCo_E must have the same expressed-in frame E as this spatial velocity.
Return values
 V_MC_E reference to this spatial velocity which has been modified to be frame C's spatial velocity measured in frame M and expressed in frame E. The components of V_MC_E are calculated as:  ω_MC_E = ω_MB_E (angular velocity of this is unchanged). v_MC_E = v_MB_E + ω_MB_E x p_BoCo_E (translational velocity changes).
Shift() to shift spatial velocity without modifying this.

## ◆ operator+()

 SpatialVelocity< T > operator+ ( const SpatialVelocity< T > & V1_E, const SpatialVelocity< T > & V2_E )
related

Parameters
 [in] V1_E spatial velocity expressed in the same frame E as V2_E. [in] V2_E spatial velocity expressed in the same frame E as V1_E.
Note
The general utility of this operator+() function is questionable and it should only be used if you are sure it makes sense. One use case is for calculating the spatial velocity V_MC of a frame C measured in a frame M when frame C is moving relative to a frame B and one has pre-calculated V_MBc (frame Bc's spatial velocity measured in frame M, where frame Bc is instantaneously coincident with frame C). For this use case, the operator+ function returns V_MC_E = V_MBc_E + V_BC_E, where the precalculated V_MBc_E is equal to V_MBo_E.Shift(p_BoCo_E).
Shift(), ShiftInPlace(), and ComposeWithMovingFrameVelocity().

## ◆ operator-()

 SpatialVelocity< T > operator- ( const SpatialVelocity< T > & V1_E, const SpatialVelocity< T > & V2_E )
related

Subtracts spatial velocities by simply subtracting their 6 underlying elements.

Parameters
 [in] V1_E spatial velocity expressed in the same frame E as V2_E. [in] V2_E spatial velocity expressed in the same frame E as V1_E.
Note
The general utility of this operator-() function is questionable and it should only be used if you are sure it makes sense. One use case is for calculating relative spatial velocity, e.g., a frame C's spatial velocity relative to a frame B, measure in a frame M. This use case calculates V_M_BC_E = V_MC_E - V_MB_E, which contains ω_BC (C's angular velocity measured in B) and v_M_BoCo (Co's velocity relative to Bo, measured in M), where both ω_BC and v_M_BoCo are expressed in frame E.
 ω_BC  = ω_MC - ω_MB
v_M_BoCo = v_MCo - v_MBo = DtM(p_BoCo)

where DtM(p_BoCo) is the time-derivative in frame M of p_BoCo (position vector from Bo to Co). A second use case has to do with a frame C that is moving on a frame B and calculates frame C's spatial velocity measured in frame B. It assumes you have pre-calculated V_MBc (frame Bc's spatial velocity measured in frame M, where frame Bc is fixed to B and instantaneously coincident with frame C. This use case returns V_BC_E = V_MC_E - V_MBc_E, where the precalculated V_MBc_E is equal to V_MBo_E.Shift(p_BoBc_E).


text:
# Koethe conjecture

First of all, has the Koethe conjecture been solved ?! Does a proof of the following statement really imply the proof of the conjecture ?

• The sum of two right nil ideals in any ring is nil.

Are there other equivalent statements that imply the conjecture ?

Thanks, SB

-
This seems to be a nice survey: math.bas.bg/serdica/2001/2001-159-170.pdf –  Julian Kuelshammer Oct 17 '11 at 17:58
Thanks for the link ! As a matrix fan, I also found the following statement books.google.co.uk/… interesting : - If $I$ is a nil ideal in $R$ then $M_n( I)$ is a nil ideal in $M_n(R)$ ! –  Serifo Blade Oct 17 '11 at 23:10
I'd never yet met a matrix fan :) –  Mariano Suárez-Alvarez Dec 18 '11 at 1:37
Oh yes I am a big fan of matrix rings !:) –  Serifo Blade Dec 18 '11 at 17:21


text:
# Laplacian of vector proof

## Homework Statement

With ##\vec{r}## the position vector and ##r## its norm, we define
$$\vec{f} = \frac{\vec{r}}{r^n}.$$
Show that
$$\nabla^2\vec{f} = n(n-3)\frac{\vec{r}}{r^{n+2}}.$$

## Homework Equations

Basic rules of calculus.

## The Attempt at a Solution

From the definition of the Laplacian
$$\nabla^2 \vec{f} = \nabla \cdot \nabla \vec{f}\\ = \left( \partial_i \partial_j \delta_{ij} \right) \frac{\vec{r}}{r^n}\\ = \vec{r}\left( \partial_i \partial_i \right) r^{-n}\\ = \vec{r}(-n)(-n-1) r^{-n-2}\\ = n(n+1) \frac{\vec{r}}{r^{n+2}}.$$

Clearly this does not agree with the proposed solution. I think my error is assuming ##vec{r}## is constant, but we know this is not always true in other coordinate systems. Any ideas?

Dr Transport
Gold Member
Laplacian in spherical coordinates is not simply the second derivative wrt $r$ and you have ignored the $\vec{r}$.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

With ##\vec{r}## the position vector and ##r## its norm, we define
$$\vec{f} = \frac{\vec{r}}{r^n}.$$
Show that
$$\nabla^2\vec{f} = n(n-3)\frac{\vec{r}}{r^{n+2}}.$$

## Homework Equations

Basic rules of calculus.

## The Attempt at a Solution

From the definition of the Laplacian
$$\nabla^2 \vec{f} = \nabla \cdot \nabla \vec{f}\\ = \left( \partial_i \partial_j \delta_{ij} \right) \frac{\vec{r}}{r^n}\\ = \vec{r}\left( \partial_i \partial_i \right) r^{-n}\\ = \vec{r}(-n)(-n-1) r^{-n-2}\\ = n(n+1) \frac{\vec{r}}{r^{n+2}}.$$

Clearly this does not agree with the proposed solution. I think my error is assuming ##vec{r}## is constant, but we know this is not always true in other coordinate systems. Any ideas?

Your ##\vec{f}## is a vector whose ##i##th component is ##f_i = x_i/r^n##. What is ##\nabla^2 f_i##?

Your ##\vec{f}## is a vector whose ##i##th component is ##f_i = x_i/r^n##. What is ##\nabla^2 f_i##?

##\nabla^2 f_i = \partial_j \partial_j f_i = \partial_j \partial_j (x_i/r^n)## Does this look correct?

Ray Vickson
Homework Helper
Dearly Missed
##\nabla^2 f_i = \partial_j \partial_j f_i = \partial_j \partial_j (x_i/r^n)## Does this look correct?

Why not finish the computations? When you have done that, if your final result agrees with what you were asked to show, then you know you have done it correctly---at least if the dimensionality is 3 (that is, vectors are ##(x,y,z)## or ##(x_1,x_2,x_3)##). For dimensions other than 3 the results ought to be a bit different.

Why not finish the computations?

Ok, so I think we continue by saying ##r^{-n}\partial_j \partial_j(x_i) + x_i\partial_j \partial_j(r^{-n})## by product rule, right? However, I really don't know how to proceed from here. Partial derivatives are with respect to x,y,z and ##x_i## is technically not the same thing. Any help in the right direction here?

Ray Vickson
Homework Helper
Dearly Missed
Ok, so I think we continue by saying ##r^{-n}\partial_j \partial_j(x_i) + x_i\partial_j \partial_j(r^{-n})## by product rule, right? However, I really don't know how to proceed from here. Partial derivatives are with respect to x,y,z and ##x_i## is technically not the same thing. Any help in the right direction here?

No, not right. First compute ##f_{ij} = \partial_j (x_i/r^n)##, which will certainly involve using the product rule. Then, compute ##\partial_j f_{ij}##, first without summing over ##j##, then doing ##j##-sum; those derivatives will involve the product rule again. So, when you differentiate twice, you cannot just use the product rule on ##\partial^2##; it is a bit more complicated than that:
$$\partial^2 (fg) =(\partial^2 f) g + 2 (\partial f) (\partial g) + f (\partial^2 g)$$
However, using this formula is not much quicker than just going ahead and differentiating twice, re-using the product rule each time.

Anyway, I do not understand the rest of your difficulties: of course ##x_i## is the same thing as one of the ##x##, ##y## or ##z##; in fact, when YOU resorted to the use of the Einstein summation convention in post #1 you must have been assuming that your coordinates are ##x_1, x_2, x_3##. These are just names; instead of using ##x_i, i=1,2,3## or ##x,y,z## I could call the coordinates red, white and blue, or Tom, Dick and Harry. The only thing about them that matters is how they appear in formulas and how they can be manipulated algebraically.

No, not right. First compute ##f_{ij} = \partial_j (x_i/r^n)##, which will certainly involve using the product rule. Then, compute ##\partial_j f_{ij}##, first without summing over ##j##, then doing ##j##-sum; those derivatives will involve the product rule again. So, when you differentiate twice, you cannot just use the product rule on ##\partial^2##; it is a bit more complicated than that:
$$\partial^2 (fg) =(\partial^2 f) g + 2 (\partial f) (\partial g) + f (\partial^2 g)$$
However, using this formula is not much quicker than just going ahead and differentiating twice, re-using the product rule each time.
Yea, I spaced it big time! So just to be clear, how about we call ##\vec{r} = r_i \vec{e_i}##. Then we have $$\partial_j \partial_j (r_i \vec{e_i} r^{-n}) = \partial_j (\partial_j (r_i \vec{e_i} r^{-n}))\\ = \vec{e_i} \partial_j(\partial_j (r_i r^{-n})\\ = \vec{e_i} \partial_j(r^{-n} \partial_j r_i + r_i \partial_j r^{-n})$$
But how do you simplify the term ##r^{-n} \partial_j r_i + r_i \partial_j r^{-n}##?

Anyway, I do not understand the rest of your difficulties: of course ##x_i## is the same thing as one of the ##x##, ##y## or ##z##; in fact, when YOU resorted to the use of the Einstein summation convention in post #1 you must have been assuming that your coordinates are ##x_1, x_2, x_3##. These are just names; instead of using ##x_i, i=1,2,3## or ##x,y,z## I could call the coordinates red, white and blue, or Tom, Dick and Harry. The only thing about them that matters is how they appear in formulas and how they can be manipulated algebraically.
I was referring to the fact that if we define the components of ##\vec{r}## to be ##x_i## then ##x_i## is not the same as x,y,z since we don't know that ##\vec{r} = <x,y,z>##.

Ray Vickson
Homework Helper
Dearly Missed
No, not right. First compute ##f_{ij} = \partial_j (x_i/r^n)##, which will certainly involve using the product rule. Then, compute ##\partial_j f_{ij}##, first without summing over ##j##, then doing ##j##-sum; those derivatives will involve the product rule again. So, when you differentiate twice, you cannot just use the product rule on ##\partial^2##; it is a bit more complicated than that:
$$\partial^2 (fg) =(\partial^2 f) g + 2 (\partial f) (\partial g) + f (\partial^2 g)$$
However, using this formula is not really much easier than just going ahead and differentiating twice, re-using the product rule each time.

Anyway, I do not understand the rest of your difficulties: of course ##x_i## is the same thing as one of the ##x##, ##y## or ##z##; in fact, when YOU resorted to the use of the Einstein summation convention in post #1 you must have been assuming that your coordinates are ##x_1, x_2, x_3##. These are just names; instead of using ##x_i, i=1,2,3## or ##x,y,z## I could call the coordinates red, white and blue, or Tom, Dick and Harry. The only thing about them that matters is how they appear in formulas and how they can be manipulated algebraically.
Yea, I spaced it big time! So just to be clear, how about we call ##\vec{r} = r_i \vec{e_i}##. Then we have $$\partial_j \partial_j (r_i \vec{e_i} r^{-n}) = \partial_j (\partial_j (r_i \vec{e_i} r^{-n}))\\ = \vec{e_i} \partial_j(\partial_j (r_i r^{-n})\\ = \vec{e_i} \partial_j(r^{-n} \partial_j r_i + r_i \partial_j r^{-n})$$
But how do you simplify the term ##r^{-n} \partial_j r_i + r_i \partial_j r^{-n}##?

I was referring to the fact that if we define the components of ##\vec{r}## to be ##x_i## then ##x_i## is not the same as x,y,z since we don't know that ##\vec{r} = <x,y,z>##.

Call the components whatever you want. If you like ##\vec{r}=(x_1, x_2, x_3)## then go ahead and use those; if you prefer ##\vec{r}=(x,y, z)## then go with those; if you prefer ##\vec{r}=(\psi, \lambda, \sigma)## then use those instead. All that matters is how you use them.

However, if you want to use summation notation (and, possibly, the Einstein summation convention) then you had better use something like ##\vec{r}=(x_1, x_2, x_3)## or ##\vec{r}=(\xi_1,\xi_2, \xi_3)## or ##\vec{r}=(p_1, p_2, p_3)##, or even ##\vec{r} = (r_1,r_2,r_3)## as you seem to like best. It really and truly does not matter, but books and papers tend to use ##x,y,x## or ##x_1,x_2,x_3## most commonly, and the latter is what I was referring to.

As to the rest of your question: you seem to be saying that you do not know how to find derivatives like ##\partial x_i /\partial x_j## (of course, for ##i = j## and for ##i \neq j##), or derivatives like ##(\partial /\partial x_j) (x_1^2+x_2^2+x_3^2)^{-n/2}##, and I find that hard to believe.

All of your post makes sense. I think what I'm having trouble with is thinking about ##\vec{r}##. Let's say ##\vec{r} = r_i \vec{e_i} = <r_1,r_2,r_3>##. Now lets get concrete and define ##<r_1,r_2,r_3> \equiv <2x,z,y+2>##. Then ##\partial_1 r_1 = \partial (r_1) / \partial x_1 = 2 \neq 1##. However, if we say ##\vec{r} = x_i \vec{e_i} = <x_1,x_2,x_3>\equiv <2x,z,y+2>## then ##\partial_1 x_1 = \partial (x_1) / \partial x_1 = 1 \neq 2##. Can you see what I'm confused about?

Ray Vickson
Homework Helper
Dearly Missed
All of your post makes sense. I think what I'm having trouble with is thinking about ##\vec{r}##. Let's say ##\vec{r} = r_i \vec{e_i} = <r_1,r_2,r_3>##. Now lets get concrete and define ##<r_1,r_2,r_3> \equiv <2x,z,y+2>##. Then ##\partial_1 r_1 = \partial (r_1) / \partial x_1 = 2 \neq 1##. However, if we say ##\vec{r} = x_i \vec{e_i} = <x_1,x_2,x_3>\equiv <2x,z,y+2>## then ##\partial_1 x_1 = \partial (x_1) / \partial x_1 = 1 \neq 2##. Can you see what I'm confused about?

No. The ##x,y,z## (or ##r_1, r_2, r_3## if you prefer) are not functions of anything in this problem. Something else is a function of THEM, though, and that is what you have. That is, you are given a vector function
$$\vec{f} = \vec{i} \frac{x}{(x^2+y^2+z^2)^{n/2}}+ \vec{j} \frac{y}{(x^2+y^2+z^2)^{n/2}}+ \vec{k} \frac{z}{(x^2+y^2+z^2)^{n/2}},$$
where ##\vec{i}##, ##\vec{j}## and ##\vec{k}## are the unit vectors along the x-, y- and z-axes. You want to compute
$$\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \vec{f},$$
and this can be done by doing it separately for the x-, y- and z-components of ##\vec{f}## (because the unit coordinate vectors ##\vec{i}##, ##\vec{j}## and ##\vec{k}## are constants).

Just to simplify the work---not for any other reason at all---I prefer to use the symbols ##x_1, x_2, x_3## instead of the symbols ##x,y,z## and to speak of components 1,2, and 3 of ##\vec{f}## instead of the x-, y- and z-components. That's all, nothing mysterious about it. Anyway, that is what YOU did in post #1 when you, yourself used the notation ##\partial_i##, etc. I am just following YOUR lead!

joshmccraney
No. The ##x,y,z## (or ##r_1, r_2, r_3## if you prefer) are not functions of anything in this problem. Something else is a function of THEM, though, and that is what you have.
This is exactly what I was confused about. So "position vector" implies the vector ##<x,y,z>##, or yea, any other name you want. It never (unless other wise stated) would imply something that I wrote, like ##<x^2,z,4>##, right? This makes sense since that must be what your position is. Sorry for all the other stuff; at the end of the day this was my real problem. Thanks!

In fact, that's why I seemingly didn't know derivatives or understand the dummy-variables behind the symbols ##x_i## and ##r_i##, etc. To me, it seemed like taking ##\partial r## was as arbitrary as taking ##\partial f##.


text:
# Find the mean of first ten even natural numbers.

Question:

Find the mean of first ten even natural numbers.

Solution:

The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20

$\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$

$=\frac{2+4+6+8+10+12+14+16+18+20}{10}=11$

Mean = 11


text:
1. ## Antiderivative

Which is the easiest method to find the antiderivative of (2t-3).

I know that this may be an easy one, I just need to see ways of finding out how to find an antiderivative.

2. Originally Posted by katchat64
Which is the easiest method to find the antiderivative of (2t-3).

I know that this may be an easy one, I just need to see ways of finding out how to find an antiderivative.
Integrate it with respect to t!

$\int 2t-3 dt = \frac{2t^2}{2} - 3t + C$

$= t^2 - 3t + C$

3. Ok.... I already have to do that, so then it just makes the problem a lot easier right?

4. Say I need to substitute 2 & 0, how would the problem look?

5. If you're a beginner, it helps to write every term (if possible) in the form

$ax^n$.

Then you can remember that the antiderivative is

$\frac{a}{n + 1}x^{n + 1} + C$.

So in this case, you'd have $2t^1 + 3t^0$.

Can you find this antiderivative using the formula?

If you're finding a definite integral, once you have the antiderivative, substitute the upper limit and see what you have ( $F(b)$, substitute the lower limit ( $F(a)$) and see what you have, then do $F(b) - F(a)$.

6. ## Evaluate the Integral

$\int_0^2{(2t-3)\,dx}$

What I got for the antiderivative of (2t-3) is (t^2-3t)

I substituted 2 & 0 and the answer I got is (-2)

7. Originally Posted by katchat64
$\int_0^2{(2t-3)\,dx}$

What I got for the antiderivative of (2t-3) is (t^2-3t)

I substituted 2 & 0 and the answer I got is (-2)

Your antiderivative $F(x) = t^2 - 3t$.

$F(2) = 2^2 - 3(2) = 4 - 6 = -2$.

$F(0) = 0^2 - 3(0) = 0$

$F(2) - F(0) = -2 - 0 = -2$.

So yes it's correct

8. Thanks so much!


text:
# A necessary and sufficient condition for a measure to be continuous.

If $(X,\mathcal{M})$ is a measurable space such that $\{x\}\in\mathcal{M}$ for all $x\in$$X, a finite measure \mu is called continuous if \mu(\{x\})=0 for all x\in$$X$.

Now let $X=[0,\infty]$, $\mathcal{M}$ be the collection of the Lebesgue measurable subsets of $X$. Show that $\mu$ is continuous if and only if the function $x\to\mu([0,x])$ is continuous.

One direction is easy: if the function is continuous, I can get that $\mu$ is continuous. But the other direction confuses me. I want to show the function is continuous, so I need to show for any $\epsilon>0$, there is a $\delta>0$ such that $|\mu([x,y])|<\epsilon$ whenever $|x-y|<\delta$.But I can't figure out how to apply the condition that $\mu$ is continuous to get this conclusion.

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By the way, you mean $\{x \} \in \mathcal M$. The set $\mathcal M$ contains subsets of $X$. Same for $\mu(\{x\})$, not $\mu(x)$. And you want to show that for any $\varepsilon > 0$, there exists a $\delta > 0$ such that $|\mu(]x,y])| < \varepsilon$ whenever $0 \le y-x < \delta$ if $y - x > \delta$, otherwise you have $\mu(]y,x])$ if $-\delta < y-x \le 0$. –  Patrick Da Silva Dec 4 '12 at 2:02
@PatrickDaSilva: The braces were there, but the OP wrote {x} instead of \{x\}. I fixed it. –  Nate Eldredge Dec 4 '12 at 2:09
@Nate : Oh, then it's a good thing you did, and even more good thing that you mentioned it. But there was still an issue (an important one) with the $|\mu([x,y])| < \varepsilon$. –  Patrick Da Silva Dec 4 '12 at 2:12

It seems like the contrapositive is a good way to go. Suppose that $x\mapsto\mu([0,x])$ is not continuous, say at the point $x_0$. Then there exists an $\epsilon>0$ such that for all $\delta>0$ there is a $y$ such that $\vert x_0-y\vert<\delta$ but $\vert\mu([x_0,y])\vert\geq\epsilon$. Thus we can construct a sequence $(y_n)$ which converges to $x_0$, but such that $\vert\mu([x_0,y_n])\vert\geq\epsilon$ for all $n$. Hence we can conclude that $\mu(x_0)\geq\epsilon>0$. Hopefully you can fill in the details from there!

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To show that $f$ is continuous at $x \in [0,\infty]$, it suffices to see that the left/right limits exist and are equal to the value of the function. Take any sequence $x_n \searrow x$. Then $$\mu([0,x_n]) = \mu([0,x]) + \mu([0,x_n]) - \mu([0,x]) = \mu([0,x]) + \mu(]x,x_n]) \longrightarrow \mu([0,x]) + \mu(\varnothing) = \mu([0,x])$$ because measures have the following property (which is usually called "continuity"... let's not use that word for now) : when $\mu(A_1) < \infty$ and $A_1 \supset A_2 \supset \dots \supset A_n \supset \dots,$ then $$\lim_{n \to \infty} \mu(A_n) = \mu \left( \bigcap_{n=1}^{\infty} A_n \right).$$ In our case, $]x, x_n]$ is a decreasing sequence of subsets because $x_n \searrow x$, so the property applies. Note that I need the fact that $\mu$ is a finite measure to do so, because I need to ensure $\mu(]x,x_n]) < \infty$.

The case where $x_n \nearrow x$ is treated similarly.

Hope that helps,

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Are you around? –  Pedro Tamaroff Dec 4 '12 at 2:51
@Peter : Yes, what is it? By the way, that's one thing I always wanted to do with this website ; how does one leave a message without posting on a question??? –  Patrick Da Silva Dec 4 '12 at 3:36
You can ping someone in the Mathematics room. Can you drop by? –  Pedro Tamaroff Dec 4 '12 at 3:56


text:
# Do in our atmosphere at least two positions exist where there is no wind?

If we look at a two-dimensional sphere and draw a continuous vector field on it, you can compare this with a tennis ball whose hairs you comb in a continuous way. Two points on the ball will have an undefined direction of the hair. If you comb them from up-down, or from the right to the left, or whatever way you try to comb it, two points arise where there is no direction of the balls hair.

The wind field in the atmosphere is a continuous vector field (which can be compared with the directions of the hair on the ball). The difference is that the atmosphere in which the wind field is situated is three-dimensional. But I can't imagine that this one extra degree of freedom makes any difference for the same two points.

So my question is: are there in the atmosphere of the earth (at least?) two points in which the wind field is not defined, and does this mean the same as saying that there are two positions in the atmosphere where there is no wind?

• Related and possibly answered, but I am not capable of judgement Cyclones on Earth – user154420 Jun 20 '17 at 12:09

You are right that the degree of freedom in the altitude / height in the atmosphere does not change things greatly for the implications of the hairy ball theorem for meteorology. All you need to do is apply the theorem to every constant altitude surface or any continuous deformation thereof contained in the atmosphere. The theorem then dictates that there must be at least one point somewhere on that surface where the wind vector vanishes - i.e. a cyclonic / anticyclonic eye - or a source / sink (where air flows upwards / downwards and leaves a sink / source in each constant altitude layer). The hairy ball theorem applied to the 2-sphere $\mathbb{S}^2$ simply says that there must be at least one point where the wind vector field, assuming that the field is continuous, vanishes.
You don't need two singular points as you seem to think: consider the flow lines resulting from the projection of the lines $\mathrm{Re}(z)=const$ of constant real part (or any family of parallel straight lines) on the complex plane onto the Riemann sphere through stereographic projection. There is one singular point at the North pole.


text:
New Question

# Revision history [back]

### openstack launch instance hangs on 'Getting metadata...'

Hi,

I've prepped a windows install image locally on KVM with the last step being install of cloudbase-init. After importing this image into glance and then launching the instance, cloudbase-init.exe shell hangs on the following message:

It seems fetching metadata is somehow failing. I can close this window to continue, but not surprisingly, hostname renaming, password etc are not setup.

It's probably worth noting that I have no problems with the metadata service in my openstack environment when launching linux instances: they pickup the SSH key injection and hostname just fine, so I'm assuming it's a configuration problem on the windows instance side.

I'm assuming that this may be a setting in the cloudbase-init.conf file that I need to get right, but I can't find any documentation on how.

The github readme explains what plugins are available, but again, I don't see any documentation anywhere explaining how to use them.

any help is appreciated!

### openstack launch instance hangs on 'Getting metadata...'

Hi,

I've prepped a windows install image locally on KVM with the last step being install of cloudbase-init. After importing this image into glance and then launching the instance, cloudbase-init.exe shell hangs on the following message:

2014-12-19 09:36:41.155 1516 DEBUG cloudbaseinit.metadata.services.httpservice [-] Getting metadata from: http://169.254.169.254/openstack/latest/metadata.json _getdata http://169.254.169.254/openstack/latest/meta_data.json _get_data c:\program files (x86)\Cloudbase Solutions\Cloudbase-Init\Python27\lib\site-packages\cloudbaseinit\metadata\services\httpservice.py:72Solutions\Cloudbase-Init\Python27\lib\site-packages\cloudbaseinit\metadata\services\httpservice.py:72

(please pardon any minor typos above - typed by hand)

It seems fetching metadata is somehow failing. I can close this window to continue, but not surprisingly, hostname renaming, password etc are not setup.

It's probably worth noting that I have no problems with the metadata service in my openstack environment when launching linux instances: they pickup the SSH key injection and hostname just fine, so I'm assuming it's a configuration problem on the windows instance side.

I'm assuming that this may be a setting in the cloudbase-init.conf file that I need to get right, but I can't find any documentation on how.

The github readme explains what plugins are available, but again, I don't see any documentation anywhere explaining how to use them.

any help is appreciated!

### openstack launch instance hangs on 'Getting metadata...'

Hi,

I've prepped a windows install image locally on KVM with the last step being install of cloudbase-init. After importing this image into glance and then launching the instance, cloudbase-init.exe shell hangs on the following message:

2014-12-19 09:36:41.155 1516 DEBUG cloudbaseinit.metadata.services.httpservice [-] Getting metadata from: http://169.254.169.254/openstack/latest/meta_data.json _get_data c:\program files (x86)\Cloudbase Solutions\Cloudbase-Init\Python27\lib\site-packages\cloudbaseinit\metadata\services\httpservice.py:72

(please pardon any minor typos above - typed by hand)

It seems fetching metadata is somehow failing. I can close this window to continue, but not surprisingly, hostname renaming, password etc are not setup.

It's probably worth noting that I have no problems with the metadata service in my openstack environment when launching linux instances: they pickup the SSH key injection and hostname just fine, so I'm assuming it's a configuration problem on the windows instance side.

I'm assuming that this may be a setting in the cloudbase-init.conf file that I need to get right, but I can't find any documentation on how.

The github readme explains what plugins are available, but again, I don't see any documentation anywhere explaining how to use them.

any help is appreciated!

edit: Here are the logs:

http://paste.openstack.org/show/159132/

Note that the 'Getting metadata' line is the last entry in the logs since it hung there indefinitely, so I closed the shell window

### openstack launch instance hangs on 'Getting metadata...'

Hi,

I've prepped a windows install image locally on KVM with the last step being install of cloudbase-init. After importing this image into glance and then launching the instance, cloudbase-init.exe shell hangs on the following message:

2014-12-19 09:36:41.155 1516 DEBUG cloudbaseinit.metadata.services.httpservice [-] Getting metadata from: http://169.254.169.254/openstack/latest/meta_data.json _get_data c:\program files (x86)\Cloudbase Solutions\Cloudbase-Init\Python27\lib\site-packages\cloudbaseinit\metadata\services\httpservice.py:72

(please pardon any minor typos above - typed by hand)

It seems fetching metadata is somehow failing. I can close this window to continue, but not surprisingly, hostname renaming, password etc are not setup.

It's probably worth noting that I have no problems with the metadata service in my openstack environment when launching linux instances: they pickup the SSH key injection and hostname just fine, so I'm assuming it's a configuration problem on the windows instance side.

I'm assuming that this may be a setting in the cloudbase-init.conf file that I need to get right, but I can't find any documentation on how.

The github readme explains what plugins are available, but again, I don't see any documentation anywhere explaining how to use them.

any help is appreciated!

edit: Here are the logs:

http://paste.openstack.org/show/159132/

Note that the 'Getting metadata' line is the last entry in the logs since it hung there indefinitely, so I closed the shell window

edit2: I've also configured DHCP in Neutron to tell clients to use an MTU of 1438. Basically following instructions found here:


text:
# Number of ways to represent a number from a given set of numbers

I want to know in how many ways can we represent a number $x$ as a sum of numbers from a given set of numbers $\{a_1.a_2,a_3,...\}$. Each number can be taken more than once.

For example, if $x=4$ and $a_1=1, a_2=2$, then the ways of representing $x=4$ are:

• $1+1+1+1$
• $1+1+2$
• $1+2+1$
• $2+1+1$
• $2+2$

Thus the number of ways $=5$.

I want to know if there exists a formula or some other fast method to do so. I can't brute force through it. I want to write code for it.

Note: $x$ can be as large as $10^{18}$. The number of terms $a_1,a_2,a_3,\dots$ can be up to $15$, and each of $a_1,a_2,a_3,\dots$ can also be only up to $15$.

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It looks to me that you want to count the number of restricted compositions, based on the fact that you consider $1+1+2$ and $1+2+1$ to be distinct. – J. M. Nov 6 '11 at 6:08
What do you mean by product sum? If $x=6$, $a_1=2$, and $a_2=3$, do you allow the combination $2,3$ as well as $2,2,2$ and $3,3$? With the same $a_1$ and $a_2$ and $x=8$ do you allow $2,2,3$ (since $2\cdot 3+2=8$)? – Brian M. Scott Nov 6 '11 at 6:29
I'd say in general this is going to be difficult - but have a look at the Wikipedia page about partitions – Juan S Nov 6 '11 at 6:57
@Brian M.Scott Sorry for the confusion. That term 'product sum' was misleading. I've edited the question. – Andrew Nov 6 '11 at 6:58
@J.M. But please note that there's no exact closed range for the values of a1,a2,a3... Like, for example, it can be that x=18 and { a1=5, a2=3, a3=7, a4=13 }. – Andrew Nov 6 '11 at 7:43

## 2 Answers

Let $c(x)$ be the number of sequences that sum to $x$. How precisely do you need to know $c(x)$? These things tend to grow exponentially, so we should expect $c(10^{18})$ to have about a billion billion digits, up to some smallish factor. An exact answer will be practically impossible to compute. But you can possibly get at least the exact number of digits and the first few significant digits by this method:

Since the order of the term matters to you, the ${1,2}$ case is easy to solve: $c(x)$ is simply the $x+1$th Fibonacci number. You can see this because the last term in a sum totalling $x$ must be either $1$ or $2$, so therefore $c(x)=c(x-1)+c(x-2)$, which is the recurrence formula for the Fibonacci numbers. This agrees with your brute-force solution for $c(4)$, because $F_5=5$.

This suggests a general plan of writing down the linear recurrence that arises from your set of terms. In general the solutions of such a recurrence are linear combinations of exponential functions, whose bases are the roots of the characteristic polynomial. For large $x$ such as $10^{18}$, the term with the largest base will dominate all others completely, so we can just ignore all of the others. The trick is then just to find its coefficient.

One way to do this would be to set $c(x)=0$ for $-15\le x<0$ and $c(0)=0$, and then solve 16 linear equations in 16 unknowns.

Another way is simply to run the recurrence up to $c(N)$ for some $N$ between, say, 100 and 1000, and assume that by that time all of the smaller bases have faded into insignificance. Then we can estimate the coefficient as $\frac{c(N)}{r^N}$, and our approximation for large $x$ is $$\log c(x) \approx \log c(N) +(x-N)\log r$$ It is necessary to work with logarithms in this step, because even $c(1000)$ is likely to overflow the hardware floating-point representation in your computer. In fact, one could choose $N$ adaptively such that $c(N)$ is around $2^{500}$.

You can use $\frac{c(N)}{c(N-1)}$ as a first approximation of $r$ and refine it using Newton-Raphson iteration on the characteristic polynomial.

Special cases: If the numerically largest root of the characteristic polynomial is complex, the recurrence can have oscillating solutions that are never dwarfed by the exponential term. I think the only way this can happen is if the $a_i$s share a divisor $\ne 1$. So start by dividing out the greatest common divisor from the $a_i$s as well as from $x$.

If the largest root of the characteristic polynomial is a multiple root, the growth rates from the master theorem look slightly different. I think this cannot happen when the recurrence is constructed from your problem, but it is probably worth checking for explicitly when you have found $r$.

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This would be very difficult to code. But thanks. – Andrew Nov 6 '11 at 18:52
Answer now rephrased so it would be more practical to code. – Henning Makholm Nov 12 '11 at 17:25
Henning, thanks a lot. But I was able to solve the problem by "Solving Linear Recurrences in Logarithmic time". Here's a link- fusharblog.com/… – Andrew Nov 16 '11 at 5:32
And yes, it would be impossible to get the exact answer for large values, I only needed to get the answer modulo a large prime no (10^9 + 7). – Andrew Nov 16 '11 at 5:36
It would have helped if you had told that before everyone tried to help you without this crucial information. If you need only the last digits, it's simple to work modulo $10^9+7$ (a smallish prime) and use exponentiation by squaring on the transition matrix of the linear recurrence. – Henning Makholm Nov 16 '11 at 11:16

Edited in the light of Byron's correction to the original:

The answer is the coefficient of $u^x$ in the expansion of $$\sum_{y=0}^x(u^{a_1}+u^{a_2}+\cdots+u^{a_r})^y$$ which can be helpful if there is something special about $a_1,a_2,\dots,a_r$ enabling you to simplify this expression; otherwise, it's just a rewording of your problem. But this means your problem is no easier than that of finding a particular coefficient in this expansion, which suggests that most of the time there is no method significantly better than brute force; it's just a question of how cleverly you can arrange for the force to be applied.

As for making it "solvable in a few seconds," you may be asking for too much. Let's look in some detail at a very small example of what you want to do: how many ways to get 100 with 2s, 3s, and 5s?

Well, there's one way using just 5s. You could use 17 5s and 5 3s; there are 22-choose-5 ways to do that. Or you could use 14 5s and 10 3s; 24-choose-10 ways. Carrying on, we see that if you use just 3s and 5s, there are $$1+{22\choose5}+{24\choose10}+\cdots+{32\choose30}$$ ways. Now there's a similar formula for using just 2s and 5s, and another one for just 2s and 3s. Now what if we use 2s, 3s, and 5s? We could use 19 5s, a 3, and a 2; that gets you a multinomial coefficient, 21-choose-19,1,1. Then there's a 22-choose-18,2,2, and there's a 23-choose-17,3,3 and a 24-choose-17,1,6 and so on and so forth. A simple formula? Maybe, but I don't see it. Although I'd be interested to see what Henning's method does with this example.

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He treats zero differently from other numbers, so $2+2+0+0$ is the same as $0+0+2+2$. What he wants is the coefficient of $u^x$ in $\sum_{y=0}^x(u^{a_1}+\cdots +u^{a_r})^y$. – Byron Schmuland Nov 6 '11 at 14:16
Can you give some computational insight into it. I need to write a code for the same. And I can't afford even a linear complexity since x can be as large as 10^18. So I think there must be some formula or something which makes it solvable in a few seconds. – Andrew Nov 6 '11 at 18:50
@Gerry You have to remove the "1+" under the power of $y$. Alternatively, summing over all $0\leq y<\infty$, the generating function is $1/(1-u^{a_1}-\cdots -u^{a_r})$. – Byron Schmuland Nov 7 '11 at 3:05


text:
# algebra problem

• June 24th 2009, 05:27 AM
the undertaker
algebra problem
Let a and b be positive real numbers. Prove that

(a+2)(b+2)(a+b) > 16ab
• June 24th 2009, 05:40 AM
TheAbstractionist
By AM–GM,

$a+2\ \ge\ 2\sqrt{2a}$

$b+2\ \ge\ 2\sqrt{2b}$

$a+b\ \ge\ 2\sqrt{ab}$

The result follows.