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http://comments.gmane.org/gmane.comp.mathematics.maxima.general/34883 | 1,369,285,253,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702810651/warc/CC-MAIN-20130516111330-00037-ip-10-60-113-184.ec2.internal.warc.gz | 53,763,160 | 3,448 | 2 Jul 2011 02:12
## How can I add simplification rule depending relation between two factors?
```Hi,
I am trying to make simplification rules for divisor function divsum.
1. divsum(p^n) = (p^(n+1)-1)/(p-1) when p is prime.
defrule was great to implement this rule though I was happier if there is a feature of "prime".
(example)
matchdeclare(pp, primep, nn, lambda([x], (integerp(x) or featurep(x, integer)) and x > 0))\$
defrule(sigma_simp1, divsum(pp^nn, 1), (pp^(nn + 1) - 1)/(pp - 1))\$
declare(n, integer)\$
assume(n > 0)\$
sigma_simp1(divsum(3^n));
=> (3^(n+1)-1)/2
2. divsum(a*b) = divsum(a)*divsum(b) if a and b are relatively prime.
I lost my way for this simplification.
Are there any manners to let Maxima check if two variables are relatively prime?
Thanks,
-
Yuji
```
2 Jul 2011 02:18
### Re: How can I add simplification rule depending relation between two factors?
```On 7/1/2011 5:12 PM, ICHIKAWA, Yuji wrote:
> Hi,
>
> I am trying to make simplification rules for divisor function divsum.
>
> 1. divsum(p^n) = (p^(n+1)-1)/(p-1) when p is prime.
> defrule was great to implement this rule though I was happier if there is a feature of "prime".
>
> (example)
> matchdeclare(pp, primep, nn, lambda([x], (integerp(x) or featurep(x, integer)) and x> 0))\$
> defrule(sigma_simp1, divsum(pp^nn, 1), (pp^(nn + 1) - 1)/(pp - 1))\$
>
> declare(n, integer)\$
> assume(n> 0)\$
> sigma_simp1(divsum(3^n));
> => (3^(n+1)-1)/2
>
> 2. divsum(a*b) = divsum(a)*divsum(b) if a and b are relatively prime.
>
> I lost my way for this simplification.
> Are there any manners to let Maxima check if two variables are relatively prime?
>
> Thanks,
> -
> Yuji
> _______________________________________________
> Maxima mailing list
> Maxima <at> math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
a. gcd(R,S) will be 1 if R and S are relatively prime.
```
2 Jul 2011 03:54
### Re: How can I add simplification rule depending relation between two factors?
```Fateman-san,
I recognized that there is no reliable way to attach a condition for matching that involves 2 parameters in
the match.
By the way, you can see a^b even if a, b are numbers, when you use a returned value of "factor".
It works well in the first case, though you need to change divsum to noun in order to see the result.
I think that the case of a*b is same.
Thanks,
-
Yuji
On 2011/07/02, at 9:18, Richard Fateman wrote:
> On 7/1/2011 5:12 PM, ICHIKAWA, Yuji wrote:
>> Hi,
>>
>> I am trying to make simplification rules for divisor function divsum.
>>
>> 1. divsum(p^n) = (p^(n+1)-1)/(p-1) when p is prime.
>> defrule was great to implement this rule though I was happier if there is a feature of "prime".
>>
>> (example)
>> matchdeclare(pp, primep, nn, lambda([x], (integerp(x) or featurep(x, integer)) and x> 0))\$
>> defrule(sigma_simp1, divsum(pp^nn, 1), (pp^(nn + 1) - 1)/(pp - 1))\$
>>
>> declare(n, integer)\$
>> assume(n> 0)\$
```
5 Jul 2011 02:21
### simplification function for divsum
```Hi,
I made a simplication function for divsum.
It works when second argument of divsum is 1.
I hope that someone will enjoy it.
Thanks,
-
Yuji
```
```
On 2011/07/02, at 10:54, ICHIKAWA, Yuji wrote:
> Fateman-san,
>
> I recognized that there is no reliable way to attach a condition for matching that involves 2 parameters in
the match.
>
> By the way, you can see a^b even if a, b are numbers, when you use a returned value of "factor".
> It works well in the first case, though you need to change divsum to noun in order to see the result.
> I think that the case of a*b is same.
>
> Thanks,
> -
> Yuji
``` | 1,142 | 3,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2013-20 | latest | en | 0.769759 |
https://www.thefreelibrary.com/A+compactness+property+for+solutions+of+the+Ricci+flow-a017250308 | 1,571,212,761,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00370.warc.gz | 1,119,692,942 | 28,060 | # A compactness property for solutions of the Ricci flow.
1. In this paper we prove that, given a sequence of complete solutions to the Ricci flow with their curvatures uniformly bounded above and their injectivity radii uniformly bounded below, we can find a convergent subsequence. To make this precise we need a few definitions.
To begin with, we fix a time interval A [less than or equal to] t [less than or equal to] [Omega] with -[infinity] [less than or equal to] A [less than] 0 and 0 [less than] [Omega] [less than or equal to] [infinity]. If M is a manifold, we look for a complete time-dependent metric on M with bounded curvature, measuring a smooth one-parameter family of metrics G = G(t) for A [less than] t [less than] [Omega] where each G(t) = {[g.sub.ij](x, t)[dx.sup.i][dx.sup.j]} is a complete metric with Riemannian curvature bounded by a constant B independent of t. A marking on M is a choice of a point Q [element of] M which we call the origin, and an orthonormal frame q at Q at t = 0 with respect to the metric G(0). For simplicity we shall refer to such a collection M = {M, G, Q, q} as an evolving complete marked Riemannian manifold. We say that M is a solution to the Ricci flow if it solves the equation
[Delta]/[Delta]t [g.sub.ij] = 2[R.sub.ij]
for the evolution of the metric by its Ricci tensor.
Definition 1.1. We say that a sequence [M.sub.k] = {[M.sub.k], [G.sub.k], [Q.sub.k], [q.sub.k]} of evolving complete marked Riemannian manifolds converges to the evolving complete marked Riemannian manifold M = {M, G, Q, q} if there exists a sequence of open sets [U.sub.k] in M containing Q and a sequence of diffeomorphisms [F.sub.k] of the sets [U.sub.k] in M to open sets [V.sub.k] in [M.sub.k] mapping Q to [Q.sub.k] and q to [q.sub.k], such that any compact set in M eventually lies in all [U.sub.k] and the pull-backs [G.sub.k] of the metrics [G.sub.k] by the maps [F.sub.k] converge to G on every compact subset of M x (A, [Omega]) uniformly together with all their derivatives.
Because of the choice of an origin and a frame at the origin, it is clear that if the limit exists it is unique up to a unique isometry. If we did not fix the origin, the limit metric might not be unique; for example a sequence of paraboloids with the same shape but the origin moving away can converge to the flat plane. It is also clear that a limit of solutions to the Ricci flow is again a solution to the Ricci flow.
We can now state our precise result.
MAIN THEOREM 1.2. Let [M.sub.k] = {[M.sub.k], [G.sub.k], [Q.sub.k], [q.sub.k]} be a sequence of evolving complete marked Riemannian manifolds which are solutions to the Ricci flow. Suppose that (a) the absolute value of the Riemannian sectional curvatures of the [M.sub.k] at all times A [less than] t [less than] [Omega] are uniformly bounded above by a constant B [less than] [infinity] independent of k, and (b) the injectivity radii of the [M.sub.k] at the origin [Q.sub.k] at time t = 0 are uniformly bounded below by a constant [Delta] [greater than] 0. Then there exists a subsequence which converges to an evolving complete marked Riemannian manifold M = { M, G, Q, q} which is also a solution of the Ricci flow, with its Riemannian sectional curvatures bounded above by B and its injectivity radius at the origin Q at time t = 0 bounded below by [Delta].
2. Here we sketch the main line of argument in the proof. The reader is no doubt aware of the work of Gromov [G], Peters [P1] and [P2], and Green and Wu [G&W], on limits of compact manifolds with bounded curvature, diameter and injectivity radius, and it will suffice to adapt these ideas to our present situation. One thing at least is much easier. The work of W. X. Shi [S] shows that for solutions of the Ricci flow bounds on the curvature automatically produce at subsequent times bounds on all the derivatives of the curvature. We use these in the following way. First it suffices to prove the theorem in the case A [greater than] - [infinity] and [Omega] [less than] [infinity]; for this being done, of A = -[infinity] we take a sequence [A.sub.j] [approaches] - [infinity] and if [Omega] = [infinity] we take a sequence [[Omega].sub.j] [approaches] [infinity], apply the theorem on each finite interval, and by the usual diagonalization trick extract a subsequence converging on the union of the finite interval. Similarly, if we can prove the theorem with the extra assumption of derivative bounds on the curvature, then given a solution on a finite interval A [less than] t [less than] [Omega] for each [[Epsilon].sub.j] [greater than] 0 we get a solution with derivative bounds on the curvature for A + [[Epsilon].sub.j] [less than] t [less than] [Omega] and now letting [[Epsilon].sub.j] [approaches] 0 and finding a subsequence converging on each smaller interval, we find by diagonalization a subsequence converging on all of A [less than] t [less than] [Omega]. Hence we may add to the hypotheses of Theorem 1.1 the following.
ASSUMPTION 2.1. The limits A and [Omega] are finite. Also for every p we can find a bound [B.sub.p] such that the [p.sup.th] covariant derivatives of the curvature tensors [Rm.sub.k] of the metrics [G.sub.k] satisfy.
[absolute value of [D.sup.p][Rm.sub.k]] [less than or equal to] [B.sub.p],
the length being measured in [G.sub.k].
The previous authors have exerted great ingenuity to obtain the best possible regularity of the limit without the use of bounds on the derivatives of the curvature. Here all this cleverness is wasted. Thus the reader who only wishes to understand this case can spare himself all the machinery of harmonic coordinates and Sobolev estimates.
There are however other new problems to face. First off there is the injectivity radius. Clearly we need to keep some control over it, or the limit may collapse to a lower dimensional manifold or worse. Actually it would be very nice if we could let this happen and still get something in the limit which we could recognize as a solution to the Ricci flow. Perhaps some ambitious reader will wish to peruse the papers of Fukawa [F1] and [F2] with this in mind. Still the present result will be usable in some cases. In particular we want to be able to consider the limit of blow-ups of singularities such as a neck pinch. In this case, although each [M.sub.k] is compact, the limit M is a cylinder, which is not. Thus we studiously avoid any diameter bound. This raises the interesting question of what happens to the injectivity radius at a point P as the distance of P from the origin Q goes to infinity. Fortunately there is a very nice answer in the paper of Cheeger, Gromov and Taylor [GG&T], where they show that the injectivity radius [Rho](P) at P falls off at worst exponentially; in particular
[Rho](P) [greater than or equal to] c/[square root of B] [([Delta][square root of B]).sup.n] [e.sup.-C[square root of Bd(P,Q)]]
where B is an upper bound on the absolute value of the sectional curvatures, [Delta] a lower bound on the injectivity radius at Q with [Delta] [less than or equal to] c/[square root of B], and c [greater than] 0 and C [less than] [infinity] are constants depending only on the dimension while d(P, Q) is the distance from P to Q and n is the dimension. (Their statement and proof are quite elegant, involving a comparison with the volume of hyperbolic space. The form given here follows from knowing that the volume of the ball of radius r in hyperbolic space grows exponentially in r.) Considering a hyperbolic tube shrinking down at infinity shows that the exponential decay of the injectivity radius is possible. Fortunately such an exponential decay rate is not so fast as to seriously annoy any of the usual constructions involved in coveting a Riemannian manifold with suitable coordinate balls; it is only necessary to let the radii of the balls shrink exponentially as their centers go to infinity. This technique avoids having to assume a uniform lower bound for the injectivity radius over the whole manifold, a hypothesis which would be much harder to satisfy in applications.
The next step is to extract a subsequence of the metrics at t = 0 which will converge. This is a purely Riemannian problem and involves nothing from the Ricci flow, except that we allow ourselves the use of derivative bounds on the curvature. We state the result as a definition and a theorem.
Definition 2.2. A sequence [M.sub.k] = ([M.sub.k], [G.sub.k], [Q.sub.k], [q.sub.k]) of complete marked Riemannian manifolds converges to a complete marked Riemannian manifold M = (M, G, Q, q) if we can find a sequence of open sets [U.sub.k] in M containing Q and a sequence of diffeomorphisms [F.sub.k] of [U.sub.k] in M to open sets [V.sub.k] in [M.sub.k] taking Q to [Q.sub.k] and q to [q.sub.k], such that every compact set in M is eventually in all the [U.sub.k] and the pull-backs [Mathematical Expression Omitted] of the metrics [G.sub.k] by the maps [F.sub.k] converge on every compact subset of M uniformly together with all their derivatives.
THEOREM 2.3. Given any sequence [M.sub.k] = ([M.sub.k], [G.sub.k], [Q.sub.k], [q.sub.k]) as before such that we have uniform bounds on the Riemannian curvature tensors [RM.sub.k] and their [p.sup.th] covariant derivatives
[absolute value of [D.sup.p] [Rm.sub.k]] [less than or equal to] [B.sub.p]
for all p [greater than or equal to] 0 with constants [B.sub.p] independent of k, and such that we have uniform lower bounds on the injectivity radii [Rho]([Q.sub.k]) of the manifolds [M.sub.k] at the origins [Q.sub.k]
[Rho]([Q.sub.k]) [greater than or equal to] [Delta]
for some [Delta] [greater than] 0 independent of k, we can find a convergent subsequence.
We will prove this result in the next section. Here we show how to use it on solutions to the Ricci flow. Given a sequence [G.sub.k] of complete marked solutions to the Ricci flow on [M.sub.k] x (A, [Omega]), with curvature and injectivity radius bounds we can by the previous theorem find a subsequence which converges at t = 0 to a metric G on a manifold M. Then we get a sequence of open sets [U.sub.k] in M containing Q and a sequence of maps [F.sub.k] of [U.sub.k] in M to open sets [V.sub.k] in [M.sub.k] taking Q to [Q.sub.k] and q to [q.sub.k], such that the pull-backs [Mathematical Expression Omitted] of the metrics [G.sub.k] at time t = 0 by the maps [F.sub.k] converge to G uniformly together with all derivatives. However, the pull-backs [Mathematical Expression Omitted] are defined at all times A [less than] t [less than] [Omega] (although G is not, yet). We also have uniform bounds on the curvatures of the [Mathematical Expression Omitted] and their derivatives, independent of k. What we claim next is that we can find uniform bounds on all the covariant derivatives of the [Mathematical Expression Omitted] taken with respect to the fixed metric G.
LEMMA 2.4. Let M be a Riemannian manifold with metric G, K a compact subset of M, and [G.sub.k] a collection of solutions to the Ricci flow defined on neighborhoods of K x [[Beta], [Psi]] with the time interval [[Beta], [Psi]] containing t = 0. Let D denote the covariant derivative with respect to G and [Mathematical Expression Omitted] the length of a tensor with respect to G, while [D.sub.k] and [Mathematical Expression Omitted] are the same for [G.sub.k]. Suppose that
(a) the metrics [G.sub.k] are all uniformly equivalent to G at t = 0 on K, so that
cG(V, V) [less than or equal to] [G.sub.k](V, V) [less than or equal to] CG(V, V)
for some constants c [greater than] 0 and C [less than] [infinity] independent of k; and
(b) the covariant derivatives of the metrics [G.sub.k] with respect to the metric G are all uniformly bounded at t = 0 on K, so that
[absolute value of [D.sup.p][G.sub.k]] [less than or equal to] [C.sub.p]
for some constants [C.sub.p] [less than] [infinity] independent of k for all p [greater than or equal to] 1; and
(c) the covariant derivatives of the curvature tensors [Rm.sub.k] of the metrics [G.sub.k] are uniformly bounded with respect to the [G.sub.k] on K x [[Beta], [Psi]], so that
[Mathematical Expression Omitted]
for some constants [C[prime].sub.p] independent of k for all p [greater than or equal to] 0.
Then the metrics [G.sub.k] are uniformly bounded with respect to G on K x [[Beta], [Psi]], so that
[Mathematical Expression Omitted]
for some constants [Mathematical Expression Omitted] and [Mathematical Expression Omitted] independent of k, and the covariant derivatives of the metrics [G.sub.k] with respect to the metric G are uniformly bounded on K x [[Beta], [Psi]], so that
[Mathematical Expression Omitted]
for some constants [Mathematical Expression Omitted] independent of k with [Mathematical Expression Omitted], [Mathematical Expression Omitted] and [Mathematical Expression Omitted] depending only on c, C, [C.sub.p] and [C[prime].sub.p] and the dimension.
Proof. First we use the fact that the metrics [G.sub.k] are solutions to the Ricci flow to get bounds
[Mathematical Expression Omitted]
on K x [[Beta], [Psi]] for some constants [Mathematical Expression Omitted] and [Mathematical Expression Omitted]. Since
[Delta]/[Delta]t [G.sub.k](V, V) = -2R[c.sub.k](V, V)
and we have
[absolute value of R[c.sub.k](V, V)] [less than or equal to] [C[prime].sub.0] [where] [G.sub.k] (V, V)
which gives
[absolute value of [Delta]/[Delta]t ln [G.sub.k](V, V)] [less than or equal to] 2 [C[prime].sub.0]
Now the bounds on ln [G.sub.k](V, V) at t = 0 easily extend to [Beta] [less than or equal to] t [less than or equal to][Psi].
Next we observe that the evolution of the connection is given by the first covariant derivative of the Ricci curvature, so that there is a formula of the form
[Delta]/[Delta]t [[Gamma].sub.k] = [DR[c.sub.k].
The connection [[Gamma].sub.k] of the metric [G.sub.k] is not a tensor, but if [Gamma] is the connection of G then the difference [[Gamma].sub.k] - [Gamma] is a tensor. Taking [Gamma] to be fixed in time, we get
[Delta]/[Delta]t ([[Gamma].sub.k] - [Gamma]) = DR[c.sub.k].
[Mathematical Expression Omitted]
for a constant [C[double prime].sub.1] depending only on [C[prime].sub.1] and the dimension. But the tensor [[Gamma].sub.k] - [Gamma] is equivalent to the tensor D[G.sub.k], and this tensor is bounded at t = 0 by
[absolute value of D[G.sub.k]] [less than or equal to] [C.sub.1].
Integrating over time we get bounds
[Mathematical Expression Omitted]
for all of [Beta] [less than or equal to] t [less than or equal to] [Psi], using the fact that we already know [Mathematical Expression Omitted] and [Mathematical Expression Omitted] are equivalent.
Next we want to bound [D.sup.2][G.sub.k]. Regarding D as fixed in time, we see that [Delta]/[Delta]t commutes with D. Thus
[Delta]/[Delta]t [D.sup.2][G.sub.k] = -2[D.sup.2][G.sub.k] = 2[D.sup.2]R[c.sub.k].
Now we can write
[Mathematical Expression Omitted].
We already have bounds on [Mathematical Expression Omitted] in [Mathematical Expression Omitted] and hence in [Mathematical Expression Omitted]. We also know that there is a formula
(D - [D.sub.k])R[c.sub.k] = D[G.sub.k] * R[c.sub.k]
where * denotes some inner tensor product, since [[Gamma].sub.k] - [Gamma] is equivalent to D[G.sub.k].
Then there is some formula
D(D - [D.sub.k])R[c.sub.k] = [D.sup.2][G.sub.k] * R[c.sub.k] + D[G.sub.k] * DR[c.sub.k].
We already have bounds on R[c.sub.k], DR[c.sub.k] and D[G.sub.k]. Finally there is a formula
(D - [D.sub.k])DR[c.sub.k] = D[G.sub.k] * DR[c.sub.k]
also. Combining we get an estimate
[absolute value of [Delta]/[Delta]t[D.sup.2][G.sub.k]] [less than or equal to] [C.sup.#] [absolute value of [D.sup.2][G.sub.k]] + [C.sup.#]
for some constant [C.sup.#]. Then our bounds on [absolute value of [D.sup.2][G.sub.k]] at t = 0 integrate to bounds on all of [Beta] [less than or equal to] t [less than or equal to] [Psi]. This gives
[Mathematical Expression Omitted]
for some constant [Mathematical Expression Omitted] independent of k. The bounds on the higher derivatives are the same. This proves the lemma.
We apply the lemma to the pull-back metrics [Mathematical Expression Omitted]. Since all of the [Mathematical Expression Omitted] are uniformly bounded with respect to the fixed metric G, we can find a subsequence which converges uniformly together with all its derivatives on every compact subset of M x (A, [Omega]). The limit metric will agree with that obtained previously at t = 0, where we knew it converged already. The limit G is now clearly itself a solution of the Ricci flow. This completes the proof of the Main Theorem, except for the proof of Theorem 2.3.
3. We turn now to the proof of Theorem 2.3. We are given a sequence [M.sub.k] = ([M.sub.k], [G.sub.k], [Q.sub.k], [q.sub.k]) of complete marked Riemannian manifolds with uniform bounds on the curvature and its covariant derivatives, and uniform lower bounds on the injectivity radii at the origins [Q.sub.k]. By our previous remark we can find a function [Rho](r) = [ce.sup.-Cr] such that if P is any point at distance r from the origin then the injectivity radius at P is large compared to [Rho](r); in fact let us make it bigger by a factor 500[[Gamma].sup.2] where [Gamma] = [e.sup.10cC].
In each [M.sub.k] we choose inductively a sequence of points [Mathematical Expression Omitted] for [Alpha] = 0, 1, 2, ... in the following way. First we let [Mathematical Expression Omitted]. The rest we choose so that if [Mathematical Expression Omitted] is the distance from [Mathematical Expression Omitted] and [Mathematical Expression Omitted], then [Mathematical Expression Omitted] is as small as possible, subject to the requirement that the open balls [Mathematical Expression Omitted] around [Mathematical Expression Omitted] of radius [Mathematical Expression Omitted] are all disjoint.
LEMMA 3.1. The balls [Mathematical Expression Omitted] cover [M.sub.k]. In fact for any r we can find [Lambda](r) independent of k such that the given balls for [Alpha] [less than or equal to] [Lambda](r) cover the ball B([Q.sub.k], r) of radius r around [Q.sub.k].
Proof. Let P [element of] [M.sub.k] and let r be the distance from P to the origin [Q.sub.k] and let [Rho] = [Rho](r). Consider those [Alpha] with [Mathematical Expression Omitted]. Then [Mathematical Expression Omitted]. Using the curvature bound and the injectivity radius bound, each ball [Mathematical Expression Omitted] has volume at least [Epsilon][[Rho].sup.n] where [Epsilon] [greater than] 0 is some constant and n is the dimension. These balls are all disjoint and contained in the ball B([Q.sub.k], r + c) since each [Mathematical Expression Omitted]. But we can estimate the volume of this ball from above by a function of r (exponential in fact). This gives us a bound [Lambda](r) on the number of [Alpha] with [Mathematical Expression Omitted]. Now I claim that the given point P must lie in one of the balls [Mathematical Expression Omitted]. If not, we could choose the next point in the sequence of [Mathematical Expression Omitted] to be P instead, for since [Mathematical Expression Omitted] the [Mathematical Expression Omitted] with [r.sub.k] [less than or equal to] r. But this is a contradiction. This proves the lemma.
Now it is desirable from the beginning to work with balls of a standard size. To this end observe that [Mathematical Expression Omitted] because with all [Mathematical Expression Omitted] this is as far out as we would have to go to find [Mathematical Expression Omitted]. Hence by passing to a subsequence using a diagonalization argument we may assume that [Mathematical Expression Omitted] converges to some [r.sup.[Alpha]] for each [Alpha]. Then [Mathematical Expression Omitted] converges to [[Rho].sup.[Alpha]] = [Rho]([r.sup.[Alpha]]). Then for all [Alpha] we can find k([Alpha]) such that if k [greater than or equal to] k([Alpha]) then [Mathematical Expression Omitted] and [Mathematical Expression Omitted].
LEMMA 3.2. For every r there exists a number k(r) such that if k [greater than or equal to] k(r) then the balls [Mathematical Expression Omitted] for [Alpha] [less than or equal to] [Lambda](r) cover the ball B([Q.sub.k], r) of radius r around the origin.
Proof. Let k(r) = max{k([Alpha]): [Alpha] [less than or equal to] [Lambda](r)}. Then [Mathematical Expression Omitted] and
[Mathematical Expression Omitted]
so the balls [Mathematical Expression Omitted] for [Alpha] [less than or equal to] [Lambda](r) cover B([Q.sub.k], r) by Lemma 3.1.
Suppose that [Mathematical Expression Omitted] and [Mathematical Expression Omitted] meet for k [greater than or equal to] k([Alpha]) and k [greater than or equal to] k([Beta]). Then we must have
[Mathematical Expression Omitted]
and since [[Rho].sup.[Alpha]] [less than or equal to] c and [[Rho].sup.[Beta]] [less than or equal to] c we get
[Mathematical Expression Omitted].
This then makes
[Mathematical Expression Omitted]
with [Gamma] = [e.sup.8Cc], since [Rho](r) = [ce.sup.-Cr]. It follows that
[[Rho].sup.[Beta]] [less than or equal to] 4 [Gamma][[Rho].sup.[Alpha]].
Hence if we let [Mathematical Expression Omitted] then [Mathematical Expression Omitted] whenever [Mathematical Expression Omitted] and [Mathematical Expression Omitted] meet and k [greater than or equal to] k([Alpha]) and k [greater than or equal to] k([Beta]).
Next we define balls [Mathematical Expression Omitted] and [Mathematical Expression Omitted] and [Mathematical Expression Omitted] and [Mathematical Expression Omitted]. Since we have
[Mathematical Expression Omitted]
the [Mathematical Expression Omitted] disjoint. Since [Mathematical Expression Omitted] the [Mathematical Expression Omitted] cover B([Q.sub.k], r) for [Alpha] [less than or equal to] [Lambda](r) as before. If [Mathematical Expression Omitted] meets [Mathematical Expression Omitted] with k, l [greater than or equal to] k([Alpha]) then
[Mathematical Expression Omitted]
and since [[Rho].sup.[Alpha]] [[Rho].sup.[Beta]] [less than or equal to] c we get
[Mathematical Expression Omitted]
and this makes
[Mathematical Expression Omitted]
since [Mathematical Expression Omitted] and [Rho](r) = [c.sup.-Cr. This makes
[[Rho].sup.[Beta]] [less than or equal to] 4[Gamma][[Rho].sup.[Alpha]].
Now any point in [Mathematical Expression Omitted] has distance at most
5[[Rho].sup.[Alpha]] + 5[[Rho].sup.[Beta]] + 5[[Rho].sup.[Beta]] [less than or equal to] 45[Gamma][[Rho].sup.[Alpha]]
for [Mathematical Expression Omitted] and hence [Mathematical Expression Omitted]. Likewise any point in [Mathematical Expression Omitted] has distance at most
5[[Rho].sup.[Alpha]] + 5[[Rho].sup.[Beta]] + 45[Gamma][[Rho].sup.[Beta]] [less than or equal to] 205[[Gamma].sup.2][[Rho].sup.[Alpha]]
and hence [Mathematical Expression Omitted]. Since the injectivity radius at [Mathematical Expression Omitted] is larger than [Mathematical Expression Omitted] and [Mathematical Expression Omitted] we see that [Mathematical Expression Omitted] is still a nice embedded ball.
LEMMA 3.3. There exists a number N such that for any given [Alpha] there are at most N choices of [Beta] for which [Mathematical Expression Omitted] meets [Mathematical Expression Omitted] when k [greater than or equal to] max{k([Alpha]), k([Beta])}.
Proof. If [Mathematical Expression Omitted] meets [Mathematical Expression Omitted] then [Mathematical Expression Omitted] and the balls [Mathematical Expression Omitted] are all disjoint. Moreover [[Rho].sup.[Beta]] is at least a fraction of [[Rho].sup.[Alpha]] as before. Thus (using the curvature bound and using [[Rho].sup.[Alpha]] [less than or equal to] c) each [Mathematical Expression Omitted] has volume at least a fraction of [([[Rho].sup.[Alpha]]).sup.n] where n is the dimension, while [Mathematical Expression Omitted] has volume at most a multiple of [([[Rho].sup.[Alpha]]).sup.n]. The ratio of these numbers gives a bound N. This proves the lemma.
Next observe that by passing to a subsequence we can guarantee that for any pair [Alpha] and [Beta] we can find a number k([Alpha], [Beta]) such that if k [greater than or equal to] k([Alpha], [Beta]) then either [Mathematical Expression Omitted] always meets [Mathematical Expression Omitted] or it never does. Let us increase k(r) if necessary so that k(r) [greater than or equal to] k([Alpha], [Beta]) if [Alpha], [Beta] [less than or equal to] [Lambda](r). Then we have achieved the following situation.
LEMMA 3.4. For every r we can find [Lambda](r) and k(r) such that if k [greater than or equal to] k(r) then the ball B([Q.sub.k], r) of radius r around the origin in [M.sub.k] is covered by the balls [Mathematical Expression Omitted] for [Alpha] [less than or equal to] [Lambda](r). Moreover no more than N balls ever meet, and for any [Alpha], [Beta] [less than or equal to][Lambda](r) either [Mathematical Expression Omitted] meets [Mathematical Expression Omitted] for all k [greater than or equal to] k(r) or for none.
Now we let [Mathematical Expression Omitted], [E.sup.Alpha], [Mathematical Expression Omitted] and [Mathematical Expression Omitted] be the balls of radii 4[[Rho].sup.[Alpha]], 5[[Rho].sup.[Alpha]], 45[Gamma][[Rho].sup.[Alpha]], and 205[[Gamma].sup.2][[Rho].sup.[Alpha]] around the origin in Euclidean space. At each point [Mathematical Expression Omitted] we choose an orthonormal frame [Mathematical Expression Omitted], taking care to pick [Mathematical Expression Omitted] at [Mathematical Expression Omitted], but choosing the others as we please. Then we define coordinate charts [Mathematical Expression Omitted] as the composition of the linear isometry of Euclidean space to the tangent space at [Mathematical Expression Omitted] taking the standard frame to the frame [Mathematical Expression Omitted] with the exponential map at [Mathematical Expression Omitted] into [M.sub.k]. We also get maps [Mathematical Expression Omitted] and [Mathematical Expression Omitted] in the same way. We let [Mathematical Expression Omitted] (and [Mathematical Expression Omitted] and [Mathematical Expression Omitted]) be the pull-backs of the metrics [G.sub.k] on [M.sub.k] by [Mathematical Expression Omitted]. We also consider the coordinate transition functions [Mathematical Expression Omitted] and [Mathematical Expression Omitted] defined by
[Mathematical Expression Omitted].
Clearly [Mathematical Expression Omitted]. Moreover [Mathematical Expression Omitted] is an isometry from [Mathematical Expression Omitted] to [Mathematical Expression Omitted] and [Mathematical Expression Omitted] from [Mathematical Expression Omitted] to [Mathematical Expression Omitted]. Since the metrics [Mathematical Expression Omitted] are in geodesic coordinates and have their curvatures and their covariant derivatives uniformly bounded, it follows by Corollary 4.10 in the next section that by passing to a subsequence we can guarantee that for each [Alpha] (and indeed for all [Alpha] by diagonalization) the metrics [Mathematical Expression Omitted] converge uniformly with all their derivatives to a smooth metric [G.sup.[Alpha]] on [E.sup.[Alpha]] (or [Mathematical Expression Omitted] or [Mathematical Expression Omitted]) which is also in geodesic coordinates.
Look now at any pair [Alpha], [Beta] for which the balls [Mathematical Expression Omitted] and [Mathematical Expression Omitted] always meet for large k, and thus the maps [Mathematical Expression Omitted] (and [Mathematical Expression Omitted] and [Mathematical Expression Omitted] and [Mathematical Expression Omitted]) are always defined for large k. By Theorem 5.1 it follows that the isometries [Mathematical Expression Omitted] (and [Mathematical Expression Omitted] and [Mathematical Expression Omitted] and [Mathematical Expression Omitted]) always have a convergent subsequence, so we may assume [Mathematical Expression Omitted] (and [Mathematical Expression Omitted] and [Mathematical Expression Omitted] and [Mathematical Expression Omitted]). The limit maps [Mathematical Expression Omitted] and [Mathematical Expression Omitted] are isometries in the limit metrics [G.sup.[Beta]] and [G.sup.[Alpha]]. Moreover
[Mathematical Expression Omitted].
We are now done picking subsequences, and everything is prepared to converge.
Definition 3.5. We say that a diffeomorphism F between Riemannian manifolds is an ([[Epsilon].sub.1], [[Epsilon].sub.2],..., [[Epsilon].sub.p]) approximate isometry if
[absolute value of tDF [center dot] DF - I] [less than] [[Epsilon].sub.1]
and
[absolute value of [D.sup.2]F] [less than] [[Epsilon].sub.2],..., [absolute value of [D.sup.p]F] [less than] [[Epsilon].sub.p]
where [D.sup.p]F is the [p.sup.th] covariant derivative of F with respect to the given metrics (regarded as a p-multilinear map of the tangent bundle of the domain into the pull-back by F of the tangent bundle of the range).
THEOREM 3.6. Take the subsequence [M.sub.k] = ([M.sub.k], [G.sub.k], [Q.sub.k], [q.sub.k]) chosen above. Then for every r and every ([[Epsilon].sub.1], [[Epsilon].sub.2],..., [[Epsilon].sub.p]), and for all k and l sufficiently large in comparison, we can find a diffeomorphism [F.sub.kl] of a neighborhood of the ball [B.sub.k](r) of radius r around [Q.sub.k] in [M.sub.k] into an open set in [M.sub.l] which is an ([[Epsilon].sub.1], [[Epsilon].sub.2],...,[[Epsilon].sub.p]) approximate isometry.
Proof. The idea (following Peters [P1] or Greene and Wu [G& W]) is to define the map [Mathematical Expression Omitted] of [Mathematical Expression Omitted] to [Mathematical Expression Omitted] for k and l large compared to [Alpha] so as to be the identity map on [E.sup.[Alpha]] in the coordinate charts [Mathematical Expression Omitted] and [Mathematical Expression Omitted], and then to define [F.sub.kl] on a neighborhood of B([Q.sub.k], r) for k, l [greater than or equal to] k(r) by averaging the maps [Mathematical Expression Omitted] for [Alpha] [less than or equal to] [Lambda](r). To describe the averaging process on [Mathematical Expression Omitted] we only need to consider those [Mathematical Expression Omitted] with [Beta] [less than or equal to] [Lambda](r) which meet [Mathematical Expression Omitted]; there are never more than N of them, and they are the same for k and l when k, l [greater than or equal to] k(r). The averaging process is defined by taking [F.sub.kl](X) to be the center of mass of the [Mathematical Expression Omitted] for [Mathematical Expression Omitted] averaging over those [Beta] where [Mathematical Expression Omitted] meets [Mathematical Expression Omitted] using weights [Mathematical Expression Omitted] defined by a partition of unity. The center of mass of the points [Y.sup.[Beta]] with weights [[Mu].sup.[Beta]] is defined to be the point Y such that
[exp.sub.Y] [V.sup.[Beta]] = [Y.sup.[Beta]] and [summation of] [[Mu].sup.[Beta]][V.sup.[Beta]] = 0.
When the points [Y.sup.[Beta]] are all close and the weights [[Mu].sup.[Beta]] satisfy 0 [less than or equal to] [[Mu].sup.[Beta]] [less than or equal to] 1 then there will be a unique solution Y close to [Y.sup.[Beta]] which depends smoothly on the [Y.sup.[Beta]] and the [[Mu].sup.[Beta]] (see [G&K] for the details). The point Y is found by the inverse function theorem, which also provides bounds on all the derivatives of Y as a function of the [Y.sup.[Beta]] and the [[Mu].sup.[Beta]].
Since [Mathematical Expression Omitted] and [Mathematical Expression Omitted], the map [Mathematical Expression Omitted] can be represented in local coordinates by the map
[Mathematical Expression Omitted]
defined by
[Mathematical Expression Omitted].
Since [Mathematical Expression Omitted] as k [approaches] [infinity] and [Mathematical Expression Omitted] as l [approaches] [infinity] and [Mathematical Expression Omitted], we see that the maps [Mathematical Expression Omitted] as k, l [approaches] [infinity] for each choice of [Alpha] and [Beta]. The weights [Mathematical Expression Omitted] are defined in the following way. We pick for each [Beta] a smooth function [[Psi].sup.[Beta]] which equals 1 on [Mathematical Expression Omitted] and equals 0 outside [E.sup.[Beta]]. We then transfer [[Psi].sup.[Beta]] to a function [Mathematical Expression Omitted] on [M.sub.k] by the coordinate map [Mathematical Expression Omitted]. Then let
[Mathematical Expression Omitted]
as usual. In the coordinate chart [E.sup.[Alpha]] the function [Mathematical Expression Omitted] looks like the composition of [Mathematical Expression Omitted] with [[Psi].sup.[Beta]]. Call this function
[Mathematical Expression Omitted].
Then as k [approaches] [infinity], [Mathematical Expression Omitted] where
[[Psi].sup.[Alpha][Beta]] = [[Psi].sup.[Beta]] [convolution] [J.sup.[Beta][Alpha]].
In the coordinate chart [E.sup.[Alpha]] the function [Mathematical Expression Omitted] looks like
[Mathematical Expression Omitted]
and [Mathematical Expression Omitted] as k [approaches] [infinity] where
[[Mu].sup.[Alpha][Beta]] = [[Psi].sup.[Alpha][Beta]] / [summation over [Gamma]] [[Psi].sup.[Alpha][Gamma]].
Since the sets [Mathematical Expression Omitted] cover B([Q.sub.k], r), it follows that [Mathematical Expression Omitted] on this set and hence [Mathematical Expression Omitted] there is no problem bounding all these functions and their derivatives. There is a small problem in that we want to guarantee that the averaged map still takes [Q.sub.k] to [Q.sub.l] and [q.sub.k] to [q.sub.l]. This is true at least for the map [Mathematical Expression Omitted]. Therefore it will suffice to guarantee that [Mathematical Expression Omitted] in a neighborhood of [Q.sub.k] if [Alpha] [not equal to] 0. This happens if the same is true for [Mathematical Expression Omitted]. If not, we can always replace [Mathematical Expression Omitted] by [Mathematical Expression Omitted] which still leaves [Mathematical Expression Omitted] or [Mathematical Expression Omitted] everywhere, and this is sufficient to make [Mathematical Expression Omitted] everywhere. Now in the local coordinates [E.sup.[Alpha]] we are averaging maps [Mathematical Expression Omitted] which converge to the identity with respect to weights [Mathematical Expression Omitted] which converge. It follows that the averaged map converges to the identity in these coordinates. Thus [F.sub.kl] can be made to be an ([[Epsilon].sub.1], [[Epsilon].sub.2], ..., [[Epsilon].sub.p]) approximate isometry on B([Q.sub.k], r) when k and l are suitably large. At least the estimates
[absolute value of tD[F.sub.kl] [multiplied by] D[F.sub.kl] - I] [less than] [[Epsilon].sub.1]
and [absolute value of [D.sup.2][F.sub.kl]] [less than] [[Epsilon].sub.2], ..., [absolute value of [D.sup.p][F.sub.kl]] [less than] [[Epsilon].sub.p] on B([Q.sub.k], r) follow from the local coordinates. We still need to check that [F.sub.kl] is a diffeomorphism on a neighborhood of B([Q.sub.k], r).
This, however, follows quickly enough from the fact that we also get a map [F.sub.lk] on a slightly larger ball B([Q.sub.l], r[prime]) which contains the image of [F.sub.lk] on B([Q.sub.k], r) if we take r[prime] = (1 + [[Epsilon].sub.1])r, and [F.sub.lk] also satisfies the above estimates. Also [F.sub.kl] and [F.sub.lk] fix the markings, so the composition [F.sub.lk] [convolution] [F.sub.kl] satisfies the same sort of estimates and fixes the origin [Q.sub.k] and the frame [q.sub.k]. Then [F.sub.lk] [convolution] [F.sub.kl] must be very close to the identity on B([Q.sub.k], r), and it follows that [F.sub.kl] is invertible. For if [F.sub.kl](X) = [F.sub.kl](Y), then we conclude that X is very close to Y. But on each ball [E.sub.[Alpha]] in local coordinates [F.sub.kl] is almost the identity, so once X and Y fall in the same ball [Mathematical Expression Omitted] they must agree. Now X lies in some [Mathematical Expression Omitted], and if the distance from X to Y is no more than [[Rho].sup.[Alpha]] then X and Y both lie in [Mathematical Expression Omitted]. If we need [Mathematical Expression Omitted] to help cover B([Q.sub.k], r) then [Mathematical Expression Omitted] and since [[Rho].sup.[Alpha]] [less than or equal to] c we get [Mathematical Expression Omitted] and [Mathematical Expression Omitted] with [Rho] = [Rho](r). Then [Mathematical Expression Omitted], and this provides a lower bound on [[Rho].sup.[Alpha]] in terms of r only. So choosing ([[Epsilon].sub.1], [[Epsilon].sub.2], ..., [[Epsilon].sub.p]) small compared to r will make X and Y close enough to lie in some [Mathematical Expression Omitted] for [Alpha] [less than or equal to] [Lambda](r). This completes the proof of Theorem 3.2.
We now know the manifolds [M.sub.k] are nearly isometric for large k. We still need to construct the limit manifold. In the case where the diameters are uniformly bounded this is not a problem, since there are only a finite number of diffeomorphism types. In our case, however, life is not so easy. One can imagine a limit of surfaces with [M.sub.k] having k holes as k [right arrow] [infinity] to see the problem.
To handle this, we first note another problem, which is that the approximate isometries we constructed don't compose to give each other, that is
[F.sub.lm] [convolution] [F.sub.kl] [not equal] [F.sub.km].
We can, however, rectify this situation. For each integer r we choose the numbers ([[Epsilon].sub.1](r), [[Epsilon].sub.2](r), ..., [[Epsilon].sub.r](r)) so small that when we choose k(r) large in comparison and find the maps [F.sub.k(r),k(r + 1)] constructed above on neighborhoods of B([Q.sub.k(r)], r) in [M.sub.k(r)] into [M.sub.k(r + 1)] the image always lies in B([Q.sub.k(r + 1)], r + 1) and the composition of [F.sub.k(r),k(r + 1)] with [F.sub.k(r+1),k(r+2)] and ... and [F.sub.k(s - 1),k(s)] for any s [greater than] r is still an ([[Eta].sub.1](r), [[Eta].sub.2](r), ..., [[Eta].sub.r](r)) isometry for any choice of [[Eta].sub.j](r), say [[Eta].sub.j](r) = 1/r for 1 [less than or equal to] j [less than or equal to] r. Basically this is just the idea that by making the terms [A.sub.s] in a series small enough for s [greater than or equal to] r we can always make [summation over] s [greater than or equal to] r [A.sub.s] as small as we like. Now we simplify the notation by writing [M.sub.r] in place of [M.sub.k](r) and [F.sub.r] in place of [F.sub.k(r),k(r + 1)].
Then [F.sub.r] is a diffeomorphism
[F.sub.r] : B([Q.sub.r], r) [right arrow] B([Q.sub.r + 1], r + 1)
and the composition
[F.sub.s - 1] [convolution] ... [convolution] [F.sub.r] : B([Q.sub.r], r) [right arrow] B([Q.sub.s], s)
is always an ([[Eta].sub.1], ..., [[Eta].sub.r]) approximate isometry.
We now construct the limit manifold M as a topological space by identifying the balls B([Q.sub.r], r) with each other using the homeomorphisms [F.sub.r]. Given any two points X and Y in M, we have X [element of] B([Q.sub.r], r) and Y [element of] B([Q.sub.s], s) for some r and s. If r [less than or equal to] s then X [element of] B([Q.sub.s], s) also, by identification. A set in M is open if and only if it intersects each B([Q.sub.r], r) in an open set. Then choosing disjoint neighborhoods of X and Y in B([Q.sub.s], s) gives disjoint neighborhoods of X and Y in M. Thus M is a Hausdorff space.
Any smooth chart on B([Q.sub.r], r) also gives a smooth chart on B([Q.sub.s], s) for all s [greater than or equal to] r. The union of all such charts gives a smooth atlas on M. It is fairly easy to see that the metrics [G.sub.r] on B([Q.sub.r], r) converge to a smooth metric G on M uniformly together with all derivatives on compact sets. For since the [F.sub.r] are very good approximate isometries, the [G.sub.r] are very close to each other, and hence form a Cauchy sequence (together with their derivatives, in the sense that the covariant derivatives of [G.sub.r] with respect to [G.sub.s] are very small when r and s are both large). One checks in the usual way that such a Cauchy sequence converges. One can also see that the limit metric is complete. For if we look at the closed ball of radius r - 1 in [M.sub.r] around the origin [Q.sub.r], which is identified with the origin Q in the limit M, we see this ball is compact and contains the closed ball of radius r - 2 around Q in the limit M when r is large. But a closed subset of a compact set is compact, so all the closed balls around Q in M are compact. Thus M is complete.
The origins [Q.sub.r] are identified with each other, and hence with an origin Q in M. Likewise the frames [q.sub.r] in [M.sub.r] are identified with each other, and hence with a frame q in M. Now it is the inverses of the maps identifying B([Q.sub.r], r) with open subsets of M that provide the diffeomorphisms of open sets in M to the manifolds [M.sub.r] such that the pull-backs of the metrics [G.sub.r] converge to G. This completes the proof of Theorem 2.3, except we still owe the reader proof of Corollary 4.10.
4. In this section we give proofs of Theorems 4.1 and 4.2 quoted earlier. First we look at geodesic coordinates. Recall that a metric [g.sub.ij](x)d[x.sup.i]d[x.sup.j] defined in a ball around the origin is in geodesic coordinates if every line through the origin is a geodesic (parametrized proportional to arc length) and if [g.sub.ij] = [I.sub.ij] at the origin x = 0.
LEMMA 4.1. The metric [g.sub.ij] is in geodesic coordinates if and only if
[g.sub.ij][x.sup.i] = [I.sub.ij][x.sup.i].
Geometrically this says that the metric inner product with the radial vector agrees with the Euclidean inner product.
Proof. A curve [x.sup.i] = [x.sup.i](t) is a geodesic (parametrized proportional to arc length) if it satisfies the equation
[Mathematical Expression Omitted].
Then the line [x.sup.i] = t[v.sup.i] for fixed [v.sub.i] are geodesics if and only if
[Mathematical Expression Omitted]
which is one set of equations for geodesic coordinates. Suppose this is true. Since the Christoffel symbols are given by
[Mathematical Expression Omitted]
we see that
[x.sup.j][x.sup.k] [Delta]/[Delta][x.sup.l] [g.sub.jk] = 2[x.sup.j][x.sup.k] [Delta]/[Delta][x.sup.j] [g.sub.kl].
From this we compute [x.sup.i][x.sup.j][x.sup.k] [Delta]/[Delta][x.sup.i] [g.sub.jk] = 0 and
[x.sup.i] [Delta]/[Delta][x.sup.i] [[g.sub.jk][x.sup.j][x.sup.k]] = 2[g.sub.jk][x.sup.j][x.sup.k]
which can also be written as
[x.sup.i] [Delta]/[Delta][x.sup.i] ln [[g.sub.jk][x.sup.j][x.sup.k]] = 2.
But we also have
[x.sup.i] [Delta]/[Delta][x.sup.i] ln[[I.sub.jk][x.sup.j][x.sup.k]] = 2
(since [I.sub.jk] is also in geodesic coordinates). Therefore the ratio
[g.sub.jk][x.sup.j][x.sup.k]/[I.sub.jk][x.sup.j][x.sup.k]
is constant along radial lines. But [g.sub.jk] [right arrow] [I.sub.jk] as x [right arrow] 0 along any radial line, so the ratio must be identically one. Thus we have
[g.sub.jk][x.sup.j][x.sup.k] = [I.sub.jk][x.sup.j][x.sup.k].
If we differentiate this identity we get
[x.sup.j][x.sup.k] [Delta]/[Delta][x.sup.l] [g.sub.jk] + 2[g.sub.kl][x.sup.k] = 2[I.sup.kl][x.sup.k]
which by our previous equation gives
[x.sup.j][x.sup.k] [Delta]/[Delta][x.sup.j] [g.sub.kl] + [g.sub.kl][x.sup.k] = [I.sub.kl][x.sup.k].
But this implies that
[x.sup.j] [Delta]/[Delta][x.sup.j] [[g.sub.kl][x.sup.k] - [I.sub.kl][x.sup.k]] = 0
and hence the function [g.sub.kl][x.sup.k] - [I.sub.kl][x.sup.k] is constant along radial lines. But [g.sub.kl] [right arrow] [I.sub.kl] as x [right arrow] 0, so the function must vanish everywhere. Hence [G.sub.kl][x.sup.k] = [I.sub.kl][x.sup.k] in geodesic coordinates.
Conversely if [g.sub.kl][x.sup.k] = [I.sub.kl][x.sup.k], then differentiating we get
[x.sup.k] [Delta]/[Delta][x.sup.j] [g.sub.kl] + [g.sub.jl] = [I.sub.jl]
from which we conclude
[x.sup.k][x.sup.l] [Delta]/[Delta][x.sup.j][g.sub.kl] = 0 and [x.sup.j][x.sup.l] [Delta]/[Delta][x.sup.j] [g.sub.kl] = 0.
Then we get [Mathematical Expression Omitted] which shows we are in geodesic coordinates. This proves the lemma.
Note in particular that in geodesic coordinates
[[absolute value of x].sup.2] = [g.sub.ij][x.sup.i][x.sup.j] = [I.sub.ij][x.sup.k][x.sub.j]
is unambiguously defined. Also if we let [x.sup.i] = t[v.sup.i] then for t [not equal to] 0
[Mathematical Expression Omitted]
and then letting t [right arrow] 0 we conclude that [Mathematical Expression Omitted]. Hence from
[Mathematical Expression Omitted]
we see that all the first derivatives of [g.sub.jk] vanish at the origin.
Introduce the symmetric tensor
[A.sub.ij] = 1/2 [x.sup.k] [Delta]/[Delta][x.sup.k] [g.sub.ij].
Since we have [g.sub.jk][x.sup.k] = [I.sub.jk][x.sup.k], we get
[x.sup.k] [Delta]/[Delta][x.sup.i] [g.sub.jk] = [I.sub.ij] - [g.sub.ij] = [x.sup.k] [Delta]/[Delta][x.sup.j] [g.sub.ik]
and hence from the formula for [Mathematical Expression Omitted]
[Mathematical Expression Omitted].
Hence [A.sub.kl][x.sup.k] = 0. Let [D.sub.i] be the covariant derivative with respect to the metric [g.sub.ij]. Then
[Mathematical Expression Omitted].
Introduce the potential function
p = [[absolute value of x].sup.2]/2 = 1/2[g.sub.ij][x.sup.i][x.sup.j].
We can use the formulas above to compute
[D.sub.i]p = [g.sub.ij][x.sup.j].
Also we get
[D.sub.i][D.sub.j]p = [g.sub.ij] + [A.sub.ij].
The defining equation for p gives
[g.sup.ij][D.sub.i]p[D.sub.j]p = 2p.
If we take the covariant derivative of this equation we get
[g.sup.kl][D.sub.j][D.sub.k]p [convolution] [D.sub.l]p = [D.sub.j]p
which is equivalent to [A.sub.jk][x.sup.k] = 0. But if we take the covariant derivative again we get
[g.sup.kl][D.sub.i][D.sub.j][D.sub.k]p [convolution] [D.sub.l]p + [g.sup.kl][D.sub.j][D.sub.k]p [convolution] [D.sub.i][D.sub.l]p = [D.sub.i][D.sub.j]p.
Now switching derivatives
[D.sub.i][D.sub.j][D.sub.k]p = [D.sub.i][D.sub.k][D.sub.j]p = [D.sub.k][D.sub.i][D.sub.j]p + [R.sub.ikjl][g.sup.lm][D.sub.m]p
and if we use this and [D.sub.i][D.sub.j]p = [g.sub.ij] + [A.sub.ij] and [g.sup.kl][D.sub.l]p = [x.sup.k] we find that
[x.sup.k][D.sub.k][A.sub.ij] + [A.sub.ij] + [g.sup.kl][A.sub.ik][A.sub.jl] + [R.sub.ikjl][x.sup.k][x.sup.l] = 0.
Let us now use [absolute value of [T.sub.i...j]] to denote the length of a tensor with respect to the metric [g.sub.ij], so that
[[absolute value of [T.sub.i...j]].sup.2] = [g.sup.ik] ... [g.sup.jl][T.sub.i...j][T.sub.k...l].
From our assumed curvature bounds we can take [absolute value of [R.sub.ijkl]] [less than or equal to] [B.sub.0]. Then we get the following estimate.
LEMMA 4.1. We have
[absolute value of [x.sup.k][D.sub.k][A.sub.ij] + [A.sub.ij]] [less than or equal to] C[[absolute value of [A.sub.ij]].sup.2] + C[B.sub.0][r.sup.2]
on the ball [absolute value of x] [less than or equal to] r for some constant C depending only on the dimension.
We now show how to use the maximum principle on such equations.
THEOREM 4.2. Let f be a function on a ball [absolute value of x] [less than or equal to] r, and let [Lambda] [greater than] 0 be a constant. Then
[lambda] sup [absolute value of f] [less than or equal to] sup [absolute value of [x.sup.k] [Delta]f/[Delta][x.sup.k] + [Lambda]f].
Proof. We can assume the sup is positive by interchanging f with -f if necessary. If it occurs in the interior, we can prove the estimate on a smaller ball. If it occurs on the boundary, then since [x.sup.k][Delta]/[Delta][x.sup.k] is pointing outward we have
[x.sup.k] [Delta]f/[Delta][x.sup.k] [greater than or equal to] 0
at that point. The result follows.
COROLLARY 4.3. For any tensor T = {[T.sub.i...j]} and any constant [Lambda] [greater than] 0 we have
[Lambda] sup [absolute value of T] [less than or equal to] sup [absolute value of [x.sup.k][D.sub.k]T = [Lambda]T].
Proof. Apply the preceding theorem to f = [[absolute value of T].sup.2] and get
[Mathematical Expression Omitted]
and expand the latter as twice the inner product of T with [x.sup.k][D.sub.k]T + [Lambda]T. This gives
2[Lambda] sup [[absolute value of T].sup.2] [less than or equal to] 2 sup [absolute value of T] sup [absolute value of [x.sup.k][D.sub.k]T + T]
and we can divide by 2 sup [absolute value of T] to get the result.
We apply this to the tensor [A.sub.ij].
COROLLARY 4.4. On the ball [absolute value of x] [less than or equal to] r we get
[Mathematical Expression Omitted]
for some constant depending only on the dimension.
THEOREM 4.5. There exist constants c [greater than] 0 and [C.sub.0] [less than] [infinity] such that if the metric [g.sub.ij] is in geodesic coordinates with [absolute value of [R.sub.ijkl]] [less than or equal to] [B.sub.0] in the ball of radius r [less than or equal to]c/[square root of [B.sub.0]] then
[absolute value of [A.sub.ij]] [less than or equal to] [C.sub.0][B.sub.0][r.sup.2].
Proof. We use a barrier argument. Since the derivatives of [g.sub.ij] vanish at the origin, so does [A.sub.ij]. Hence the estimate holds near the origin. But Corollary 4.4 says that [absolute value of [A.sub.ij]] avoids an interval when c is chosen small. In fact the inequality
X [less than or equal to] C[X.sup.x] + D
is equivalent to
[absolute value of 2CX - 1] [greater than or equal to] [square root of 1 - 4CD]
which makes X avoid an interval if 4CD [less than] 1. (Hence in our case we need to choose c with 4[C.sup.2][c.sup.2] [less than] 1.) Then if X is on the side containing 0 we get
X [less than or equal to] 1 - [square root of 1 - 4CD]/2c [less than or equal to] 2D.
This gives [absolute value of [A.sub.ij]] [less than or equal to] [C.sub.0][B.sub.0][r.sup.2] with [C.sub.0] = 2C.
We can also derive bounds on all the covariant derivatives of p in terms of bounds on the covariant derivatives of the curvature. To simplify the notation, we let
[D.sup.q]p = {[D.sub.j1][D.sub.j2] ... [D.sub.jq]p}
denote the [q.sub.th] covariant derivative, and in estimating [D.sup.q]p we will lump all the lower order terms into a general slush term [[Phi].sup.q] which will be a polynomial in [D.sup.1]p, [D.sup.2]p, ..., [D.sup.q - 1]p and Rm, [D.sup.j]Rm, ..., [D.sup.q - 2]Rm. We already have estimates on a ball of radius r
p [less than or equal to] [r.sup.2]/2
[absolute value of [D.sup.1]p] [less than or equal to] r
[absolute value of [A.sub.ij]] [less than or equal to] [C.sub.0][B.sub.0][r.sup.2]
and since [D.sub.i][D.sub.j]p = [g.sub.ij] + [A.sub.ij] and r [less than or equal to] c / [square root of B] if we choose c small we can make
[absolute value of [A.sub.ij] [less than or equal to] 1/2.
and we get
[absolute value of [D.sup.2]p] [less than or equal to] [C.sub.2]
for some constant [C.sub.2] depending only on the dimension.
Start with the equation [g.sup.ij][D.sub.i]p[D.sub.j]p = 2p and apply repeated covariant derivatives. Observe that we get an equation which starts out
[g.sup.ij][D.sub.i]p[D.sup.q][D.sub.j]p + ... = 0
where the omitted terms only contain derivatives [D.sup.q]p and lower. If we switch two derivatives in a term [D.sup.q + 1]p or lower, we get a term which is a product of a covariant derivative of Rm of order at most q - 2 (since the two closest to p commute) and a covariant derivative of p of order at most q - 1; such a term can be lumped in with the slush term [[Phi].sup.q]. Therefore up to terms in [[Phi].sup.q] we can regard the derivatives as commuting. Then paying attention to the derivatives in [D.sup.q]p we get an equation
[g.sub.ij][D.sub.i]p[D.sub.j][D.sub.[k.sub.1]] ... [D.sub.[k.sub.q]]p + [g.sup.ij][D.sub.i][D.sub.[k.sub.1]]p[D.sub.j][D.sub.[k.sub.2]] ... [D.sub.[k.sub.q]]p
+ [g.sup.ij][D.sub.i][D.sub.[k.sub.2]]p[D.sub.j][D.sub.[k.sub.1]][d.sub.[k.sub.3]] ... [D.sub.[k.sub.q]]p
+ ...
+ [g.sup.ij][D.sub.i][D.sub.kq]p[D.sub.j][D.sub.[k.sub.1]] ... [D.sub.[k.sub.q - 1]p
= [D.sub.[k.sub.1]] ... [D.sub.[k.sub.q]]p + [[Phi].sup.q].
Recalling that [D.sub.i][D.sub.j]p = [g.sub.ij] + [A.sub.ij] we can rewrite this as
[g.sup.ij][D.sub.i]p[D.sub.j][D.sub.[k.sub.1]] ... [D.sub.[k.sub.q]]p + (q - 1)[D.sub.[k.sub.1]] ... [D.sub.[k.sub.q]]p
+ [g.sup.ij][A.sub.i[k.sub.1]][D.sub.j][D.sub.[k.sub.2]] ... [D.sub.[k.sub.q]]p + ...
= [g.sup.ij][A.sub.ikq][D.sub.j][D.sub.[k.sub.1]] ... [D.sub.[k.sub.q - 1]]p = [[Phi].sup.q].
Estimating the product of tensors in the usual way gives
[absolute value of [x.sup.i][D.sub.i][D.sup.q]p + (q - 1)[D.sup.q]p] [less than or equal to] q [absolute value of A] [absolute value of [D.sup.q]p] + [absolute value of [[Phi].sup.q]].
Applying Corollary 4.3 with T = [D.sup.q]p gives
(q - 1)[absolute value of [D.sup.q]p] [less than or equal to] q [absolute value of A] [absolute value of [D.sup.q]p] + [absolute value of [[Phi].sup.q]].
Now we can make [absolute value of A] [less than or equal to] 1/2 by making r [less than or equal to] c [square root of B] with c small; it is important here that c is independent of q! Then we get
(q - 2)[absolute value of [D.sup.q]p] [less than or equal to] 2[absolute value of [[Phi].sup.q]]
which is a good estimate for q [greater than or equal to] 3. The term [[Phi].sup.q] is estimated inductively from the terms [D.sup.q - 1]p and [D.sup.q - 2]Rm and lower. This proves the following result.
THEOREM 4.6. There exist constants [C.sub.q] for q [greater than or equal to] 3 depending only on q and the dimension and on [absolute value of [D.sup.j]Rm] for j [less than or equal to] q - 2 such that
[absolute value of [D.sup.q]p] [less than or equal to] [C.sub.q]
on the ball r [less than or equal to] c / [square root of [B.sub.0]].
Now we turn our attention to estimating the Euclidean metric [I.sub.jk] and its covariant derivatives with respect to [g.sub.jk]. In this we need a slightly different estimate on our partial differential equation.
THEOREM 4.7. Suppose that f is a function on a ball [absolute value of x] [less than or equal to] r with f(0) = 0
[absolute value of [x.sup.i][Delta]f/[Delta][x.sup.i]] [less than or equal to] C[[absolute value of x].sup.2]
for some constant C. Then
[absolute value of f] [less than or equal to] C[[absolute value of x].sup.2]
for the same constant C.
Proof. Consider the point where [absolute value of f] is largest on the ball [absolute value of x] [less than or equal to] s for s [less than or equal to] r. By switching f with -f if necessary we can assume f [greater than or equal to] 0 there. Rotate so that the point where this occurs is [x.sup.1] = s, [x.sup.2] = ... = [x.sup.n] = 0. Then by the Mean Value Theorem
f(x, 0, ..., 0) = s [Delta]f/[Delta][x.sup.1] ([x.sup.1], 0, ..., 0)
for some [x.sup.1] with 0 [less than or equal to] [x.sup.1] [less than or equal to] s. But by our hypothesis
[x.sup.1] [Delta]f/[Delta][x.sup.1] ([x.sup.1], 0, ..., 0) [less than or equal to] C[([x.sup.1]).sup.2]
so
f(s, 0, ...,0) [less than or equal to] Cs[x.sup.1] [less than or equal to] C[s.sup.2]
and this is enough to prove the theorem.
COROLLARY 4.8. If T = {[T.sub.j...k]} is a tensor which vanishes at the origin and if
[absolute value of [x.sup.i][D.sub.i]T] [less than or equal to] C[[absolute value of x].sup.2]
on a ball [absolute value of x] [less than or equal to] r then [absolute value of T] [less than or equal to] C[[absolute value of x].sup.2] with the same constant C.
Proof. We would like to apply the above theorem to the function f = [absolute value of T]. In case this is not smooth, we use
f = [square root of [[absolute value of T].sup.2] + [[Epsilon].sup.2]] - [Epsilon]]
and let [Epsilon] [right arrow] 0. Note that if T = 0 at the origin then f = 0 also. Also
[absolute value of T] [less than or equal to] f + [Epsilon].
Now we compute
[x.sup.i] [Delta]f/[Delta][x.sup.i] = 1/f + [Epsilon] <T, [x.sup.i][D.sub.i]T>
and hence
[absolute value of [x.sup.i] [Delta]f/[Delta][x.sup.i]] [less than or equal to] C[[absolute value of x].sup.2]
with the same constant C. Then we get f [less than or equal to] C[[absolute value of x].sup.2] for every [Epsilon], and when [Epsilon][right arrow] 0 we get [absolute value of T] [less than or equal to] C[[absolute value of x].sup.2] as desired.
Our application will be to the tensor [I.sub.jk] which gives the Euclidean metric as a tensor in geodesic coordinates. We have
[Mathematical Expression Omitted]
and since
[Mathematical Expression Omitted]
we get the equation
[x.sup.i][D.sub.i][I.sub.jk] = -[g.sup.pq][A.sub.jp][I.sub.kq] - [g.sub.pq][A.sub.kp][I.sub.jq].
We already have [absolute value of [A.sub.jk]] [less than or equal to] [C.sub.0][B.sub.0][[absolute value of x].sup.2] for [absolute value of x] [less than or equal to] r [less than or equal to] c/[square root of [B.sub.0]]. The tensor [I.sub.jk] doesn't vanish at the origin, but the tensor
[h.sub.jk] = [I.sub.jk] - [g.sub.jk]
surely does. We can then use
[x.sup.i][D.sub.i][h.sub.jk] = -[g.sup.pq][A.sub.jp][h.sub.kq] - [g.sup.pq][A.sub.kp][h.sub.jq] - 2[A.sub.jk].
Suppose [absolute value of [h.sub.jk]] [less than or equal to] M(s) on the ball [absolute value of x] [less than or equal to] s. Then
[absolute value of [x.sup.i][D.sub.i][h.sub.jk]] [less than or equal to] 2[1 + M(s)]][C.sub.0][B.sub.0][[absolute value of x].sup.2]
and we get
[absolute value of [h.sub.jk]] [less than or equal to] 2[1 + M(s)][C.sub.0][B.sub.0][[absolute value of x].sup.2]
on [absolute value of x] [less than or equal to] s from Corollary 4.8. This makes
M(s) [less than or equal to] 2[1 + M(s)][C.sub.0][B.sub.0][s.sup.2].
Then for s [less than or equal to] r [less than or equal to] c/[square root of [B.sub.0]] with c small compared to [C.sub.0] we get 2[C.sub.0][C.sub.0][s.sup.2] [less than or equal to] 1/2 and M(s) [less than or equal to]4[C.sub.0][B.sub.0][s.sup.2]. Thus
[absolute value of [h.sub.jk]] [less than or equal to] 4[C.sub.0][B.sub.0][[absolute value of x].sup.2]
for [absolute value of x] [less than or equal to] r. We state this result.
THEOREM 4.9. if [absolute value of x] [less than or equal to] r [less than or equal to] c/[square root of [B.sub.0]]. then
[absolute value of [I.sub.jk] - [g.sub.jk]] [less than or equal to] 4[C.sub.0][B.sub.0][[absolute value of x].sup.2]
and hence for c small enough
1/2[g.sub.jk] [less than or equal to] [I.sub.jk] [less than or equal to] 2[g.sub.jk].
Thus the metrics are comparable. Note that this estimate only needs r small compared to [B.sub.0] and does not need any bounds on the derivatives of the curvature.
Now to obtain bounds on the covariant derivatives of the Euclidean metric [I.sub.kl] with respect to the Riemannian metric [g.sub.kl] we want to start with the equation
[x.sup.i][D.sub.i][I.sub.kl] + [g.sup.mn][A.sub.km][I.sub.ln] + [g.sup.mn][A.sub.lm][I.sub.kn] = 0
and apply q covariant derivatives [D.sub.[j.sub.1]] ... [D.sub.[j.sub.q]]. Each time we do this we must interchange [D.sub.j] and [x.sup.i][D.sub.i], and since this produces a term which helps we should look at it closely. If we write [R.sub.ji] = [[D.sub.j], [D.sub.i]] for the commutator, this operator on tensors involves the curvature but no derivatives. Since
[Mathematical Expression Omitted]
we can compute
[[D.sub.j], [x.sup.i][D.sub.i]] = [D.sub.j] + [g.sup.im][A.sub.jm][D.sub.i] + [x.sup.i][R.sub.ji]
and the term [D.sub.j] in the commutator helps, while [A.sub.jm] can be kept small and [R.sub.ji] is zero order. It follows that we get an equation of the form
[x.sup.i][D.sub.i][D.sub.[j.sub.1]] ... [D.sub.[j.sub.q]][I.sub.kl] + q[D.sub.[j.sub.1]] ... [D.sub.[j.sub.q]][I.sub.kl]
+ [summation of] [g.sup.im][A.sub.jkm][D.sub.[j.sub.1]] where h = 1 to q ... [D.sub.[j.sub.h - 1]][D.sub.i][D.sub.[j.sub.h + 1]] ... [D.sub.[j.sub.q]][I.sub.kl]
+ [g.sup.mn][A.sub.km][D.sub.[j.sub.1]] ... [D.sub.[j.sub.q]][I.sub.ln]
+ [g.sup.mn][A.sub.lm][D.sub.[j.sub.1]] ... [D.sub.[j.sub.q]][I.sub.ln] + [[Psi].sup.q] = 0
where the slush term [[Psi].sup.q] is a polynomial in derivatives of [I.sub.kl] of degree no more than q - 1 and derivatives of p of degree no more than q + 2 (remember [x.sup.i] = [q.sup.ij][D.sub.j]p and [A.sub.ij] = [D.sub.i][D.sub.j]p - [g.sub.ij]) and derivatives of the curvature Rm of degree no more than q - 1. We now estimate
[D.sup.q][I.sub.kl] = {[D.sub.j1] ... [D.sub.[j.sub.q]][I.sub.kl]}
by induction on q using Corollary 4.3 with [Lambda] = q. Noticing a total of q + 2 terms contracting [A.sub.ij] with a derivative of [I.sub.kl] of degree q, we get the estimate
q sup [absolute value of [D.sup.q][I.sub.kl]] [less than or equal to] (q + 2) sup [absolute value of A] sup [absolute value of [D.sup.q][I.sub.kl]] + sup [absolute value of [[Psi].sup.q]].
Now since [absolute value of A] [less than or equal to] [C.sub.0][B.sub.0][r.sup.2] and r [less than or equal to] c/[square root of [B.sub.0]], if we take c sufficiently small so that sup [absolute value of A] [less than or equal to] 1/4, then for all q [greater than or equal to] 1
(3q - 2) sup [absolute value of [D.sup.q][I.sub.kl]] [less than or equal to] 4 sup [absolute value of [[Psi].sup.q]]
and everything works. This proves the following result.
THEOREM 4.10. There exists a constant c [greater than] 0 depending only on the dimension, and constants [C.sub.q] depending only on the dimension and q and bounds [B.sub.j] on the curvature and its derivatives for j [less than or equal to] q where [absolute value of [D.sub.j]Rm] [less than or equal to] [B.sub.j], so that for any metric [g.sub.kl] in geodesic coordinates in the ball [absolute value of x] [less than or equal to] r [less than or equal to] c/[square root of [B.sub.0]] the Euclidean metric [I.sub.kl] satisfies
1/2[g.sub.kl] [less than or equal to] [I.sub.kl] [less than or equal to] 2[g.sub.kl]
and the covariant derivatives of [I.sub.kl] with respect to [g.sub.kl] satisfy
[absolute value of [D.sub.j1] ... [D.sub.[j.sub.q]][I.sub.kl]] [less than or equal to] [C.sub.q].
COROLLARY 4.11. We also have estimates
1/2[I.sub.kl] [less than or equal to] [g.sub.kl] [less than or equal to] 2[I.sub.kl]
and
[Mathematical Expression Omitted]
for similar constants [Mathematical Expression Omitted].
Proof. The difference between a covariant derivative and an ordinary derivative is given by the connection [Mathematical Expression Omitted], which we can recover from solving
[Mathematical Expression Omitted]
to get
[Mathematical Expression Omitted].
This gives us bounds on [Mathematical Expression Omitted]. We then obtain bounds on the first derivatives of [g.sub.ij] from
[Mathematical Expression Omitted].
Now we get bounds on covariant derivatives of [Mathematical Expression Omitted] from the covariant derivatives of [I.sub.pk] and bounds on the ordinary derivatives of [Mathematical Expression Omitted] by relating them to the covariant derivatives using the [Mathematical Expression Omitted], and bounds on the ordinary derivatives of the [g.sub.jk] from bounds on the ordinary derivatives of the [Mathematical Expression Omitted], always proceeding inductively on the order of the derivative. This proves the corollary.
5. In this section we show how to estimate the derivatives of an isometry.
THEOREM 5.1. Let y = F(x) be an isometry from a ball in Euclidean space with a metric [g.sub.ij]d[x.sup.i]d[x.sup.j] to a ball in Euclidean space with a metric [h.sub.pq]d[y.sup.p]d[y.sup.q]. Then we can bound all of the derivatives of y with respect to x in terms of bounds on the derivatives of [g.sub.ij] with respect to x and bounds on the derivatives of [h.sub.pq] with respect to y.
Proof. Since y = F(x) is an isometry we have the equation
[h.sub.pq] [Delta][y.sup.p]/[Delta][x.sup.j] [Delta][y.sup.q]/[Delta][x.sup.k] = [g.sub.jk].
Using bounds [g.sub.ij] [less than or equal to] C[I.sub.jk] and [h.sub.pq] [greater than or equal to] c[I.sub.pq] comparing to the Euclidean metric, we easily get estimates
[absolute value of [Delta][y.sup.p]/[Delta][x.sup.j]] [less than or equal to] C.
Now if we differentiate the equation with respect to [x.sup.i] we get
[h.sub.pq][[Delta].sup.2][y.sup.p]/[Delta][x.sup.i][Delta][x.sup.j] [Delta][y.sup.q]/[Delta][x.sup.k] + [h.sub.pq][Delta][y.sup.p]/[Delta][x.sup.j] [[Delta].sup.2][y.sup.q]/[Delta][x.sup.i][Delta][x.sup.k]
= [Delta][g.sub.jk]/[Delta][x.sup.i] - [Delta][h.sub.pq]/[Delta][y.sup.r] [Delta][y.sup.r]/[Delta][x.sup.i] [Delta][y.sup.p]/[Delta][x.sup.j] [Delta][y.sup.q]/[Delta][x.sup.k].
Now let
[T.sub.ijk] = [h.sub.pq][Delta][y.sup.p]/[Delta][x.sup.i] [[Delta].sup.2][y.sup.q]/[Delta][x.sup.j][Delta][x.sup.k]
and let
[U.sub.ijk] = [Delta][g.sub.jk]/[Delta][x.sup.i] - [Delta][h.sub.pq][Delta][y.sup.r] [Delta][y.sup.r]/[Delta][x.sup.i] [Delta][y.sup.p]/[Delta][x.sup.j] [Delta][y.sup.q]/[Delta][x.sup.k].
Then the above equation says
[T.sub.kij] + [T.sub.jik] = [U.sub.ijk].
Using the obvious symmetries [T.sub.ijk] = [T.sub.ikj] and [U.sub.ijk] = [U.sub.ikj] we can solve this in the usual way to obtain
[T.sub.ijk] = 1/2([U.sub.jik] + [U.sub.kij] - [U.sub.ijk]).
We can recover the second derivatives of y with respect to x from the formula
[[Delta].sup.2][y.sup.p]/[Delta][x.sup.i][Delta][x.sup.j] = [g.sup.kl][T.sub.kij][Delta][y.sup.p]/[Delta][x.sup.l].
Combining these gives an explicit formula giving [[Delta].sup.2][y.sup.p]/[Delta][x.sup.i][x.sup.j] as a function of [g.sub.ij], [h.sub.pq], [Delta][g.sub.jk]/[Delta][x.sup.i], [Delta][h.sub.pq]/[Delta][y.sup.r], and [Delta][y.sup.p]/[Delta][y.sup.i]. This gives bounds
[absolute value of [[Delta].sup.2][y.sup.p]/[Delta][y.sup.i][Delta][y.sup.j]] [less than or equal to] C
and bounds on all higher derivatives follow by differentiating the formula and using induction.
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CALIFORNIA, SAN DIEGO, LA JOLLA, CA 92093-0112
REFERENCES
[CG&T] J. Cheeger, M. Gromov and M. Taylor, Finite propagation speed, kernel estimates for functions of the Laplace operator, and the geometry of complete Riemannian manifolds, J. Differential Geom. 17 (1982), 15-53.
[F1] K. Fukaya, Collapsing Riemannian manifolds to ones of lower dimension, J. Differential Geom. 25 (1987), 139-156.
[F2] -----. J. Differential Geom. (1989).
[G&W] R. Greene and H. Wu, Lipschitz convergence of Riemannian manifolds, Pacific J. Math. 131 (1988), 119-141.
[G] M. Gromov, Structures Metriques pour les Varietes Riemanniennes, Textes Math., vol. 1, Cedric-Nathan, Paris, 1981.
[G&K] K. Grove and H. Karcher, How to conjugate [C.sup.1] close group actions, Math. Z. 132 (1973), 11-20.
[P1] S. Peters, Cheeger's finiteness theorem for diffeomorphism classes of Riemannian manifolds, J. Reine Angew. Math. 349 (1984), 77-82.
[P2] -----, Convergence of Riemannian manifolds, Comp. Math. 62 (1987), 3-16.
[S] W. X. Shi, Deforming the metric on complete Riemannian manifolds, J. Differential Geom. 30 (1989), 223-301; Ricci deformation of the metric on complete noncompact Riemannian manifolds, J. Differential Geom. 30 (1989), 303-394.
COPYRIGHT 1995 Johns Hopkins University Press
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https://math.stackexchange.com/questions/434787/matsunagas-method-for-solving-x2y2-p | 1,695,867,703,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00627.warc.gz | 406,795,766 | 39,467 | # Matsunaga's Method for solving $x^2+y^2=p$
In his history of number theory, Dickson mentions an 18th century algorithm due to Matsunago [Sic --- he means, presumably, Matsunaga Ryohitsu a.k.a. Matsunaga Yoshisuke] for finding two numbers whose squares sum to a given integer $p$. Dickson's source Mikami can be found describing the Matsunaga method in entry 15 on page 233 here:
http://quod.lib.umich.edu/u/umhistmath/ACD4271.0008.001/244?rgn=full+text;view=pdf
Quoting:
To solve the equation $x^2+y^2=k$, let $\tfrac{1}{2}k$ be $=r^2+R$, where $r^2$ is the greatest square contained in $k$. Form the quantities
$$a_1 = 2r-1 \quad a_2 = a_1 - 2 \quad a_3 = a_2 -2 \ldots$$ $$b_1 = 2r+1 \quad b_2 = b_1 + 2 \quad b_3 = b_2 + 2 \ldots$$
and from $2R=A$ successively subtract $b_1$, $b_2$, $\ldots$. When these subtractions are impossible, successively add $a_1$, $a_2$, $\ldots$ and carry out the subtractions. If we come at last to a remainder that vanishes, let the values of $a$ and $b$ employed in the last place be $a'$ and $b'$. Then $x = \tfrac{1}{2}(a'+1)$ and $y = \tfrac{1}{2}(b'-1)$ is a solution required."
I have a lot of questions about this, but I'll start with: Can you make this work? I've tried several different interpretations of the instructions but have been unable to make it work to produce the representations $5 = 1^2 + 2^2$, $10 = 1^2 + 3^2$, $13 = 2^2 + 3^2$, $17 = 4^2 + 1^2$, $41 = 4^2 + 5^2$, ... in a reliable manner. I desire a systematic way to follow these instructions that produces each of these representations (and others, if possible!). Or perhaps this does not work for all $k$ (i.e., the remainder 0 is not reached). But an exposition of how it works when it does work would be appreciated.
Incidentally, the wording differs slightly from that given by Dickson. I doubt he would include this result as stated if it didn't hold some merit. Perhaps his rewording of the instructions (p.229 of volume 2 of his "Number Theory") was clarifying exactly how the algorithm was to be performed?
• I can't figure out what this bit is referring to at all (let alone much of the rest being rather unclear): "… and carry out the subtractions". Which subtractions? Jul 2, 2013 at 23:06
I was able to reverse-engineer the method of Matsunaga Ryohitsu and find that it does work, modulo understanding the following two points about it, that are not necessarily clear from the wording (either Mikami's or Dickson's):
• When it is impossible to subtract a $b_i$ from the current remainder $A$, what we must add to $A$ is the next unused $a_j$ (after the latest $a_{j-1}$ used), which is not necessarily the "corresponding" $a_i$ (i.e., $j$ may be a smaller index than $i$). Alternatively, instead of changing $a_{j+1} = a_j - 2$ at every step, we change it only when required. That is, the $a_i$'s and $b_i$'s aren't meant to be recalculated simultaneously at each step, but the next $a_i$ is only to be picked when necessary. This is clearer from Mikami's wording than Dickson's.
• Mikami's wording "let the values of $a$ and $b$ employed in the last place be $a'$ and $b'$" (or Dickson's "$a', b'$ are the values last employed") are probably not what the original intended, which seems to be to take $a'$ and $b'$ to be the next ones after the ones last used in an addition or subtraction. This can be sort of coerced to fit the original wording, if we imagine that as soon as we use up an $a_j$ or a $b_i$, we write the next ones immediately, and so $a'$ and $b'$ are taken to be the last ones writen down.
With that in mind, the method can be viewed as follows. We want $x^2 + y^2 = k$, so we want to search over all pairs of positive integers $(x, y)$ until we find a pair satisfying the equation. We will organize the search as follows. Let $\frac{k}2 = r^2 + R$ (where $r^2$ is the largest square not greater than $\frac12k$, and $R$ is the "remainder") so that $k = 2r^2 + 2R$. If we assume wlog that $x \le y$, we see that we must have $\color{red}{y \ge r}$ (because if $x \le y < r$, then $x^2 + y^2 < 2r^2 \le k$) and that $\color{red}{x \le r}$ (because if $r < x \le y$, then $x^2 + y^2 \ge 2(r+1)^2 > k$).
So let's start with $x, y$ both equal to $r$, for which $x^2 + y^2 = 2r^2 \le k$. We will successively increment $y$ by $1$ at a time as long as $x^2 + y^2 \le k$, then when we have overshot $k$ we will decrement $x$ by $1$ and continue, and so on. It is clear that this method will always work and find a solution if one exists, as it's an exhaustive search over all potential solutions $(x,y)$: for every $y \ge r$, we try the largest $x$ such that $x^2 + y^2 \le k$.
The beauty of this method is that accomplishes the exhaustive search without any squaring or multiplications, just additions and subtractions, that too of relatively smaller numbers.
More concretely, for a fixed $x$, suppose we increase $y$ by $1$. When we replace $y$ with $y+1$, the left-hand side (namely $x^2 + y^2$) increases to $x^2 + (y+1)^2$, i.e., it increases by $2y + 1$: twice the current candidate value for $y$, plus $1$. Or, if we consider candidates for $2y$ instead, the left-hand-side increases by the current candidate (for $2y$) plus $2$. This is what the $b$'s are doing in the method. (As the final $y = \frac12(b' - 1)$, this means $b' = 2y + 1$.)
Similarly, when we increase $y$ by too much, the left-hand side $x^2 + y^2$ will overshoot the right-hand side, and to compensate, it is time for try the next lower value for $x$ (or equivalently, for $2x$). When we replace $x$ with $x-1$, the left-hand side changes from $x^2 + y^2$ to $(x-1)^2 + y^2$, decreasing by $2x-1$, or twice the current value minus $1$. Or, if we consider candidate values for $2x$, it decreases by the current value for $2x$ minus $2$. These are the $a_j$'s, shifted by $1$. (As the final $x = \frac12(a' + 1)$, this means that $a' = 2x - 1$.)
A complete description of the method
We maintain, at all times $t$, a pair of integers $(x_t, y_t)$ such that $x_t^2 + y_t^2 \le k$. However, we won't work directly with the numbers $(x_t, y_t, k)$ except to output a solution; instead we work with the numbers \begin{align} a_t &= 2x_t - 1 ,\\ b_t &= 2y_t + 1, \\ A_t &= k - (x_t^2 + y_t^2). \end{align}
1. Initially, at $t = 1$, (with the idea that $x_t = y_t = r$) set $a_t = 2r - 1$, $b_t = 2r + 1$, and $A_t = k - 2r^2$, where $r^2$ is the largest square not greater than $\frac{k}2$.
2. If $A_t = 0$, output the solution $x_t = \frac12(a_t + 1)$, $y_t = \frac12(b_t - 1)$.
3. While $A_t < b_t$, [need to decrease $x_t$ to make room for $y_t + 1$]
• Set $A_{t+1} = A_t + a_t$ and $a_{t+1} = a_t - 2$, keeping $b_{t+1} = b_t$.
• Increase $t$ to $t+1$.
4. [Now that $A_t \ge b_t$, can update $y_{t+1} = y_t + 1$.]
Set $A_{t+1} = A_t - b_t$ and $b_{t+1} = b_t + 2$, keeping $a_{t+1} = a_t$.
Update $t$ to $t+1$ and go back to step 2.
(Note that when we output a solution, we don't have to stop. If we keep going, we'll find all other solutions as well, if they exist. We can stop when $a_t < 0$.)
Some examples of the method (we don't have to restrict ourselves to even $k$), including all the ones in your question and more:
• $k = 5$. In this case, $\frac12k = \frac52 = 1^2 + \frac32$, so $r = 1$ and $A_1 = 2R = 3$. Start with $a_1 = 2r - 1 = 1$, and $b_1 = 2r + 1 = 3$. Subtract, from $A = 3$, the current $b$: you already get $0$, so $a' = a_1 = 1$, and $b' = b_2 = b_1 + 2 = 5$. Now $x = \frac12 (a' + 1) = 1$ and $y = \frac12 (b' - 1) = 2$ is indeed a solution.
• $k = 10$. In this case, $\frac12k = 5 = 2^2 + 1$, so $r = 2$ and $2R = 2$. Start with $a_1 = 2r - 1 = 3$ and $b_1 = 2r + 1 = 5$. It is already impossible to subtract $b_1$ from $A = 2R$, so add $a_1$ and carry out the subtraction: we get the new remainder $A$ to be $2R + a_1 - b_1 = 2 + 3 - 5 = 0$. So $a' = a_2 = a_1 -2 = 1$, and $b' = b_2 = b_1 + 2 = 7$. Now $x = \frac12(a' + 1) = 1$ and $y = \frac12(b' - 1) = 3$ is indeed a solution.
• $k = 13$. We have $\frac12k = 2^2 + \frac52$, so $r = 2$ and $2R = 5$. $a_1 = 2r - 1 = 3$, $b_1 = 5$. Subtracing $b_1$ from $A=2R$ gives remainder $0$, so $a' = a_1 = 3$, and $b' = b_2 = 7$. This gives the solution $x' = \frac12(a' + 1) = 2$ and $y' = \frac12(b' - 1) = 3$.
• $k = 17$. We have $\frac12k = 2^2 + \frac92$, so $r = 2$ and $2R = 9$. $a_1 = 2r - 1 = 3$, $b_1 = 5$. Update $A=2R$ to $A - b_1 = 4$. Next we cannot subtract $b_2 = 7$ from it, so add $a_1 = 3$ and subtract. Remainder is $0$, so $a' = a_2 = 1$, $b' = b_3 = 9$. This gives the solution $x' = \frac12(a' + 1) = 1$ and $y' = \frac12(b' - 1) = 4$.
• $k = 41$. In this case, $\frac12k = 4^2 + \frac92$, so $r = 4$ and $2R = 9$. Start with $a_1 = 2r - 1 = 7$, and $b_1 = 2r + 1 = 9$. Subtracting $b_1$ from $A = 2R$ already gives remainder $0$, so $a' = a_1 = 7$ and $b' = b_2 = b_1 + 2 = 11$. Now $x' = \frac12(a' + 1) = 4$ and $y' = \frac12(b' - 1) = 5$ is indeed a solution.
• For something less trivial, consider $k = 218$. In this case, $\frac12k = 109 = 10^2 + 9$, so $r = 10$ and $2R = 18$. Start with $a_1 = 2r - 1 = 19$, and $b_1 = 2r + 1 = 21$. Subtracting $b_1$ from $A = 2R$ is already impossible, so add $a_1$ and carry out the subtraction: the new value of $A$ is $2R + a_1 - b_1 = 16$, meanwhile we compute new values $a_2 = a_1 - 2 = 17$ and $b_2 = b_1 + 2 = 23$. Again, subtracting $b_2$ from $A$ is impossible, so we add $a_2$, thus setting $A$ now to $16 + a_2 - b_2 = 10$. Next compute $a_3 = a_2 - 2 = 15$, and $b_3 = b_2 + 2 = 25$. This time, subtracting $b_3$ is still not possible, but magically, it works if we add $a_3$, and indeed we get remainder $0$. So $a' = a_4 = a_3 -2 = 13$, and $b' = b_4 = b_3 + 2 = 27$. Now, $x' = \frac12(a' + 1) = 7$ and $y' = \frac12(b' - 1) = 13$ indeed a solution.
• I've enjoyed watching your great answer grow into a fantastic answer! Thanks a lot. The main point I was missing was your second bullet at the top --- I never tried the next unused a and b. I'm not surprised since I was doing it all in my head, and as you said, it helps to interpret the wording as writing down the next a and b the instant you've used the prior ones. Jul 5, 2013 at 13:44
• That helped. I put together a simple Matlab function to demonstrate the algorithm for everyone to view and download. I still found your "A complete description of the method" section not completely straightforward, so let me know if the code is lacking in any way. @BarrySmith, my code does output multiple solutions if found, e.g., k=50 or k=325. Jul 5, 2013 at 22:55
• @horchler: Nice, your code is exactly what I meant. Which part did you find not entirely straightforward, so that I can improve it? (One thing I did genuinely miss is the "&& a >= 0" condition on the inner loop, that we may have to stop the loop when there's no longer hope of termination / when the termination condition of the whole program is reached.) Jul 6, 2013 at 3:19
• @ShreevatsaR: Following your steps 1–4, partly what confused me was your comments in square brackets, which I didn't realize were comments at the time. Also, in your examples you use 2*R (=A), which isn't in your method and which, despite being in the originals is unnecessary. And yes, the && a >= 0 was necessary to not lock up in some cases -I surmised that from your comment about stopping when a < 0. Finally, I'm not sure that it helps to use an iteration index t even though the original formulations use one. I find it superfluous and think that the method could be described with out it. Jul 6, 2013 at 14:38
• Great work!(+1) @horchler Nice code! Jul 10, 2013 at 20:02 | 4,028 | 11,569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-40 | latest | en | 0.913781 |
http://lesswrong.com/r/discussion/lw/bn9/help_me_teach_bayes/ | 1,371,636,536,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708664942/warc/CC-MAIN-20130516125104-00075-ip-10-60-113-184.ec2.internal.warc.gz | 142,837,211 | 14,906 | Less Wrong is a community blog devoted to refining the art of human rationality. Please visit our About page for more information.
# Help me teach Bayes'
2 11 April 2012 05:56PM
Next Monday I am supposed to introduce a bunch of middle school students to Bayes' theorem.
I've scoured the Internet for basic examples where Bayes' theorem is applied. Alas, all explanations I've come cross are, I believe, difficult to grasp for the average middle school student.
So what I am looking for is a straightforward explanation of Bayes' theorem that uses the least amount of Mathematics and words possible. (Also, my presentation has to be under 3 minutes.)
I think that it would be efficient in terms of learning for me to use coins or cards, something tangible to illustrate what I'm talking about.
What do you think? How should I teach 'em Bayes' ways?
PS: I myself am new to Bayesian probability.
## Comments (19)
Sort By: Best
Comment author: 11 April 2012 06:17:12PM 13 points [-]
This is my favorite explanation of the theorem so far:
http://oscarbonilla.com/2009/05/visualizing-bayes-theorem/
But I doubt you can explain it to middle school students in only 3 min. If I were you, I wouldn't discuss the theorem itself, just the cancer patient problem. Have the students try to figure out the answer for themselves, and then surprise them with the real answer (and justify it by talking about a population of 1 million people or whatever; your explanation doesn't have to use probabilities, although the problem statement could).
Comment author: 12 April 2012 12:25:12AM 4 points [-]
Actually, you might want to come up with a different example than the standard one so students who happen to encounter the standard one later on will appreciate it. (I was turned off by Eliezer's Bayesian theorem explanation initially, because it started off by challenging me to solve the standard disease example, which I already knew the trick for.)
Comment author: 13 April 2012 04:42:27AM 0 points [-]
Nice one, I like it!
But there's something I fail to understand: where's the 9.6% rendered?
"9.6% of the area outside of event A." - wait, doesn't that little area outside A represent the women with cancer?
Comment author: 14 April 2012 06:23:00AM 1 point [-]
Pretty sure the 9.6% is the section of the green circle that doesn't overlap with the red circle.
Comment author: 12 April 2012 03:09:17AM 3 points [-]
Middle-school students may be a little hazy on what a theorem is, or have a notion of it that derives only from memorizing the teacher's password. ("Math has theorems, science has theories.")
Comment author: 11 April 2012 08:01:41PM 3 points [-]
Probability is a mathematical object called a measure, which means it obeys exactly the same rules as area or volume. This is why the "visualizing Bayes' theorem" link is exactly true. Probabilities are like circles (or other shapes) with area equal to their probability, and these circles overlap when two things happen together. So I think the Venn diagram explanation might help students remember it.
Comment author: 14 April 2012 06:26:33AM 0 points [-]
I was under the impression that ET Jaynes did not like the circle diagram because it implied an infinitude of outcomes:
http://www-biba.inrialpes.fr/Jaynes/cc02m.pdf
Comment author: 14 April 2012 07:52:18AM * 0 points [-]
Hm, good point. For example, for his statements "It will rain today" and "the roof will leak," the points in the Venn diagram you'd draw to show how these probabilities overlap don't correspond to anything real. On the other hand, it's really useful to picture this stuff, and you can imagine "chunking up" your space into regions corresponding to the different discrete outcomes (like a bar graph), and the exact same rules are followed, except now it seems a bit more meaningful.
Comment author: 11 April 2012 08:15:44PM 2 points [-]
Possibly relevant what specific grade(s) are these students from, and are they in any sort or gifted program or is it a normal middle school population.
Comment author: 12 April 2012 02:14:48PM 1 point [-]
I'm guessing 6th or 7h grade, average flock.
Comment author: 11 April 2012 08:07:06PM * 2 points [-]
This isn't an approach I've seen much before, and so it may not be wise to do this your first time (I also recommend adapting this explanation), but maybe focus on that Bayes' theorem is when you have two competing hypotheses, and you get evidence that is more probable under one hypothesis than the other.
When you get evidence, you keep track of the probability of each hypothesis separately, but what matters is their normalized probability. (I'll use frequencies since those are easier for people to manipulate than probabilities.) You might start off with the knowledge that the chance someone has a rare disease is 100 out of a million, but the chance that someone has a common disease is 9,800 out of a million. Everyone with the rare disease goes to the doctor, but only half of the people with the common disease go to the doctor, and no healthy people visit the doctor- so in a city of one million people, 100 people with the rare disease visit the doctor, and 4,900 people with the common disease visit the doctor, and the probability someone at the doctor's office has the rare disease is 2%.
Then the doctor runs a test- it gives an A result 99 times out of a hundred for people with the rare disease, and a B result 1 time out of a hundred for people with the rare disease. It gives an A result 2 times out of a hundred for people with the common disease, and a B result 98 times out of a hundred. Now we have 99 people with rare and A, 1 person with rare and B, 98 people with common and A, and 4802 people with common and B. Of the people who got an A result, they have about a 50% chance to have either disease- but of people with a B result, almost all of them have the common disease.
The main mental strategy that Bayes' theorem helps people with is "keep multiple hypotheses in your head at once," and you may want to emphasize that to people just hearing about it.
Comment author: 11 April 2012 06:27:10PM 3 points [-]
``````P(A and B) = P(B and A)
P(A and B) = P(A) * P(B|A)
``````
=>
``````P(A and B) = P(A) * P(B|A)
P(B and A) = P(A) * P(B|A)
P(B) * P(A|B) = P(A) * P(B|A)
P(A|B) = P(A) * P(B|A) / P(B)
``````
Comment author: 11 April 2012 07:13:37PM * 0 points [-]
P(B and A) = P(A) * P(B|A)
To make it clearer, shouldn't this step be = P(B) * P(A|B)?
Also, are middle school pupil in the U.S. familiar with the notation? Maybe one should state it in English instead?
The probability of A if B is known to occur is equal to the probability of A times the probability of B if A is known to occur divided by the probability of B.
ETA
P(A and B) = P(A) * P(B|A)
The probability of A and B to occur at the same time is the same as the probability of A to occur alone times the probability of B to occur under the condition that A is known to occur.
Comment author: 11 April 2012 07:06:40PM -1 points [-]
Spot the typo.
Comment author: 11 April 2012 07:28:51PM 0 points [-]
Don't see one. Could you please tell me where?
Comment author: 11 April 2012 07:34:05PM 0 points [-]
I think the second line after the => was supposed to be `P(B and A) = P(B) * P(A|B)`.
Comment author: 11 April 2012 07:52:55PM 1 point [-]
No, it's a substitution on the left hand side. I substituted P(A and B) for P(B and A)
Comment author: 12 April 2012 08:14:02PM 1 point [-]
You test 1,000 people for cancer. Of the people tested, only 10 actually have cancer. If the person has cancer, there's a 90% chance that the test will catch it. It will also claim that 1% of healthy people have cancer.
This means that of the 990 healthy people, 9 will get marked as having cancer. It also means that of the 10 ill people, only 9 of them will be noticed. So your odds of actually having cancer, given a positive test result, are only 50/50.
I mostly teach this to people who can follow the actual math without blinking, but I've found it's the fastest way to give a very basic explanation of what Bayes means. It's a specific, concrete example, and it's also one that feels intuitively useful - you've now learned about false positives and false negatives, and how this affects the meaning of actual results.
From there, you can go in to the math, but I've found most people who are bad at math can still work it out this way, and most people who are good at math can quickly derive the formulas from the example :)
I'd probably go in to some cool examples of what else it's been used for, possibly with worked examples if you have the time - 3 minutes is probably a bit short for anything beyond a single concrete example and a few points of why it's cool (being used to locate nuclear submarines in WW2, breaking the Enigma code, etc. :))
Possibly a handout with a few fun / cool problems in case kids want to practice, and maybe some pointers towards literate on the subject, but I suspect a pre-college audience won't generally be receptive to such a thing. You're probably better off just hooking their interest and trusting that the ones who find it cool will have the sense to look it up on Google :)
Comment author: 11 April 2012 08:27:46PM 1 point [-]
I have a post here including a short primer on Bayes' theorem.
http://lesswrong.com/r/discussion/lw/8lr/logodds_or_logits/ | 2,369 | 9,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2013-20 | latest | en | 0.953978 |
http://www.jiskha.com/display.cgi?id=1338372184 | 1,496,086,175,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612537.91/warc/CC-MAIN-20170529184559-20170529204559-00222.warc.gz | 703,741,630 | 3,830 | # physics
posted by on .
a train starts from rest and accelerates uniformly at 100m/min^2 for 10 min.then it maintains a constant velocity for remaining 20 min.brakes are applied and train uniformly retards and comes to rest in next 5 min.
a)the maximum velocity reached
b)the retardation in last 5 min
c)total distance traveled
d)the average velocity of the train.
• physics - ,
v=a•t1 =100•10 = 1000 m/min.
s1 = a1•t1²/2 =100•100/2 =5000 m.
s2=v•t2 =1000•20 =20000 m.
0 =v +a2•t3,
a2 = -v/t3 = - 1000/5 = - 200 m/min,
s3 = v •t3 – a2•t3²/2 =1000•5 - 200•25/2 =2500 m
s = s1 + s2 +s3 = 5000 + 20000+ 2500 = 27500 m.
t =t1+t2+t3 = 10 +20 =5 = 35 min.
v(ave) =s/t = 27500/35 =785 m/min.
a) 1000 m/min,
b) - 200 m/min,
c) s = 27500 m.
d) v(ave) =785 m/min. | 315 | 761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-22 | latest | en | 0.627335 |
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# Fourier Analysis
### Fourier Cosine Transforms
Please see attached file. Please show all steps in detail. In the two problems below find the Fourier Cosine Transform of the given f(x) and write f(x) as a Fourier integral. 1) -1, -Pi <x<0 f(x) = 1, 0<x<Pi 0, |x| > Pi 2) 2x+2a -a<x<0 f(x) = -2x+2a
### Mathematica : Fourier Series Expansions
Please help me in solving this problem. Please see attached file for full problem description. Using Mathematica and computer facilities express the function f(x) in terms of Fourier series expansion and show that the series converges as the number of terms increases: f(x) = x e-x/4 sin(x/3) -π < x < π
### Fourier Transform
Please see attachment. 1. What is the Fourier Transform for the convolution of sin(2t)*cos(2t). 2. Compute the inverse Fourier transform for X(w)= sin^2*3w 3. A continuous time signal x(t) has the Fourier transform X(w) = 1/jw+b where b is a constant. Determine the Fourier transform for v(t) = x*(5t-4)
### Periodic Functions: Fourier Sine and Cosine Series
Please see attachment. #1 for following periodic functions acting on the given interval Do the following: a) Sketch 4 periods of the given function of period b) Expand the function in a sine - cosine Fourier Series f(x) = 2-x, -2<x<2 #2) Expand the function in a sine-cosine series and
### Fourier Series of Even and Odd Functions
Find the Fourier series expansion of the functions: f(t) = 1 if Pi/3<|t|<2Pi/3 0 everywhere else f(t) = 1 Pi/3 < t < 2Pi/3 -1 -2Pi/3 < t < Pi/3 0 everywhere else In the interval [-Pi , Pi]
### Fourier Transforms and Wave Analysis
The question is the example on page 2 of the attachment (entitled 'Uniform Transducer'). it states that the center of the finger is at z'=L/4. I assume this is an arbitrary position. For Eq (2.4.6), the contribution from the left-hand finger is added. I'm not entirely sure how this equation is arrived at. It does not look like a
### Fourier Transforms : Time-Frequency Duality and Time Differentiation Property
Please see the attached file for the fully formatted problems. You don't have to worry about (iii).
### Solving a Fourier Transform
Suppose f(x) has the Fourier transform F(Ω). If a ≠ 0 show that f(ax) has the Fourier Transform 1/|a| F (Ω/a). Please see the attached file for the fully formatted problems.
### Exponential Fourier Series
Find the exponential Fourier series for x(t), y(t) and z(t). In each of three cases it is not necessary to do any integration. ω=2πf t = n/256 (t goes from 0 to 1 in increments of 1/256) x(t)= cos ω;t frequency= 2Hz y(t)= cos ωt frequency= 16Hz z(t)= the product of x(t) and y(t)
### Fourier Integral and Convergence
Given the function below: Expand the function in an Fourier integral and determine what this integral converges to. f(x) = xe^-|4x| keywords: integration, integrates, integrals, integrating, double, triple, multiple
### Fourier Transforms, Gaussian and Ricker Wavelets and Waveforms and Dirac Delta Function
Please see the attached files for the fully formatted problems.
### Fourier Series : Expansions and Differentiation
Let f (x) = |x| for x greater or equal to -1, less than or equal to +1 a) Write the Fourier series for f (x) on [-1,1]. b) Show that this series can be differentiated term by term to yield the Fourier expansion of f'(x) on [-1,1] c) Determine f'(x) and write it's Fourier series on [-1,1] d) Compare b and c. key
### Examples of Fourier series and sums of numerical series.
We use the Fourier expansions of certain poynomial functions to compute the sum of some useful numerical series. The formulas are quite general and give, at the end, the Fourier expansion of every polynomial function. By the way, these formulas can be also used for a numerical approximation of pi=3.14....
### Set of functions (Fourier Series and Signal Spaces)
Functions, Interval. See attached file for full problem description. Given the set of functions f1(t) = A1*exp(-t) f2(t) = A2*e^(-2t) Defined on the interval (0, infinity). (a) Find A1 such that f1(t) is normalized to unity on (0, infinity). Call this function PHI_1(t). (b) Find B such that PHI(t) and f2
### Fourier Transform Integrals
Using the Fourier transform integral, find Fourier transforms of the following signals. xa(t) = t *exp(-αt) * u(t), α > 0; xb(t) = t2 * u(t) * u(1 – t) xc(t) = exp(-αt) * u(t) * u(1 – t), α > 0;
### Fourier series, Fourier Transform and Partial Differential Equations
Please see the attached file for the fully formatted problems. ODE: 1. Solve ()'sinyxy=+. 2. Find the complete solution of the ODE ()()42212cosyyyx−−=. 3. Find the complete solution of the ODE ()46sinyy−=. 4. Find a second order ODE whose solution is a family of circle with arbitrary radius and center on t
### Proof of sum of a given infinite series of constants (closed form).
The sum of the infinite series, 1/2^2 - 2/3^2 + 3/4^2 - 4/5^2 + ... is given as pi^2/12 - log 2 on pages 64-65 in the book "Summation of Series" by L. B. W. Jolley, 2nd ed., 1961, Dover Pubs. Inc. (the ^ symbol denotes exponentiation in the above series and sum). For most of the series in his book, he lists a source (referen
### Derivation of Fourier Transform of a Gaussian
From equation 6 of the attached, derive equation 7. In the expression for s(v), the natural log does not apply to the term [(v-vo)/(dv/2)]^2 , should just be ln(2). Yes, the limits from minus infinity to plus inifinity are adequate. Thanks for the help!
### Fourier coefficients outputs
Fourier coefficients / b1, b2, b3, b4, b5... b11. -------------------------------------------------------------------------------- I have an output of an electronic device (full wave rectifier) that gives a sine wave with the negative part transposed symmetric to xx so that the function is always positive. I have to find the f
### Matlab plots of the FFT of sequence
(See attached file for full problem description) For sequence x[n]=[1 1 1 1 0 0 0 0] for n=0:7, so N=8 Using above x[n]: a) stem(x); b) Use the shift theorm to plot x delayed by 1, 4, 5, 6, and 8 samples, and plot the result for each. Remember the shift theorem says a delay by t0 seconds is equal to multiplying the spe
### Discrete time Fourier transform of sequence and Matlab plot
Please see the attached file for full description. Calculate by hand the X(omega), DTFT of the sequence x[n]=[1 1 1 1 0 0 0 0] for n=0:7, zero else. Using Matlab, plot the real and imaginary components of your result for X(omega) for omega=0:0.01:2*pi, one plot for the real, one part for the imaginary. On the same plot
### Fourier transform of functions
Using Fourier transforms where possible, derive the Fourier transforms of the following functions using the relationship: a.) f(x) = exp[i2po(x/lamda)sin(theta)] b.) f(x) = exp(- /ax/ ) See the attached file for full description.
### Inverse Fourier transform,.,,
Find the inverse Fourier transform of each of the following Fourier transforms: X(x) = jw The answer I have is x[n] = (-1)^n / n (for n not equal to zero) 0 (for n = 0) I don't know how to get there.
### Fourier Series of Signal
(See attached file for full problem description) Consider a periodic function f(x) with period L. Over one period, f(x) = sin(2*pi*x/L) over the interval -L/4 to L/4, f(x) = 0 over the intervals -L/2 to -L/4, and L/4 to L/2. Derive an expression for the nth Fourier series coefficient, an. In the Fourier series expansion
### Applications of Fourier transform
Define a Fourier Transform. What are its properties and application areas ? Describe its application in signal processing.
### Find the fourier transform of a positive integer
Find the fourier transform of x^k, where k is a positive integer and x is a single real variable.
### Fourier Transform of a Signal: Scaling and Time-Shift Property
Use tables and the scaling and time-shift property to find the Fourier transform of the signal below (which is zero for all other values of t than those shown, that is, it's not periodic) Plot the magnitude of the spectrum of this signal. Hint: Find the transform of the pulse centered around t=0, then use the time-shift p
### Fourier series
Find the Fourier series in trigonometric form for f(t) = |sin(pi*t)|. Graph its power spectrum.
### Fourier series by using MATLAB
(See attached file for full problem description) Find the Fourier Series coefficients for the signal (see the attached file). Use Matlab to plot the truncated Fourier series reconstruction for the signal, using the first 15 terms of the sum. Given an and bn, what would be the complex coefficients if you had instead calcula
### Bessel and Legendre Series : Fourier-Legendre Expansions
8. The first three Legendre polynomials are P0(x) = 1, P1(x) = x, and P2(x) = 1/2(3x2- 1). If x = cosθ , then P0( cosθ ) = 1 and P1( cosθ ) = cos θ . Show that P2( cosθ ) = 1/4( 3cos2θ + 1 ). 9. Use the results of problem 8, to find a Fourier-Legendre expansion ( F (θ) = )of F( | 2,432 | 9,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-34 | latest | en | 0.815443 |
https://www.nagwa.com/en/videos/375185090763/ | 1,611,433,949,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00335.warc.gz | 909,719,904 | 13,076 | # Question Video: Finding the First Derivative of an Exponential Function
Use the integral test to determine whether the series β_(π = 1)^(β) (1/π) converges or diverges.
02:51
### Video Transcript
Use the integral test to determine whether the series, which is the sum from π equals one to β of one over π, converges or diverges.
Letβs start by recalling the integral test. The integral test tells us that if π is a continuous, positive, decreasing function on the interval between π and β and that π of π is equal to π π. Then, if the integral from π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals π to β of π π. And secondly, if the integral from π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π equals π to β of π π. Now, in our case, our series is the sum from π equals one to β of one over π. Therefore, π π is equal to one over π. Using the fact that π of π is equal to π π, we can say that π of π₯ is equal to one over π₯.
Now, we need to check that π of π₯ is a continuous, positive, and decreasing function on the interval between π and β. Noting that since our sum goes from π equals one, π must be equal to one. Letβs start by checking the continuity of our function over this interval. The only discontinuity of π of π₯ occurs when π₯ is equal to zero. However, zero is not in our interval between one and β. Therefore, this function must be continuous between one and β. On this interval, π₯ is always positive. Therefore, one over π₯ must also be positive. And so, weβve satisfied that condition too. Now, if π₯ starts at one and gets larger and larger and larger, then one over π₯ will get smaller and smaller and smaller. Therefore, we can say that this is a decreasing function on our interval.
Now that weβve satisfied these three conditions, weβre able to use the integral test. We need to work out whether the integral from one to β of one over π₯ with respect to π₯ is convergent or divergent. Now, we know that the differential of the natural logarithm of π₯ with respect to π₯ is equal to one over π₯. Therefore, the natural logarithm of π₯ is the antiderivative of one over π₯. Hence, we have that our integral is equal to the natural logarithm of π₯ between one and β. When using the bound of β, we need to take the limit as π₯ tends to β of the natural logarithm of π₯. Then, for our lower bound of one, we simply subtract the natural logarithm of one.
Now, letβs evaluate this limit, the natural logarithm as an increasing function. Therefore, as π₯ gets larger, the natural logarithm of π₯ also gets larger. Hence, we can say that this limit is also equal to β. We also have that the natural logarithm of one is equal to zero. However, regardless of what constant this term is, this will not affect our answer of β. This tells us that our integral must be divergent. Hence, by the integral test, we can say that the sum from π equals one to β of one over π is divergent. | 973 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-04 | latest | en | 0.903801 |
https://www.knowpia.com/knowpedia/Ring_homomorphism | 1,716,568,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00234.warc.gz | 745,842,854 | 19,246 | BREAKING NEWS
Ring homomorphism
## Summary
In mathematics, a ring homomorphism is a structure-preserving function between two rings. More explicitly, if R and S are rings, then a ring homomorphism is a function ${\displaystyle f:R\to S}$ that preserves addition, multiplication and multiplicative identity; that is,[1][2][3][4][5]
{\displaystyle {\begin{aligned}f(a+b)&=f(a)+f(b),\\f(ab)&=f(a)f(b),\\f(1_{R})&=1_{S},\end{aligned}}}
for all ${\displaystyle a,b}$ in ${\displaystyle R.}$
These conditions imply that additive inverses and the additive identity are preserved too.
If in addition f is a bijection, then its inverse f−1 is also a ring homomorphism. In this case, f is called a ring isomorphism, and the rings R and S are called isomorphic. From the standpoint of ring theory, isomorphic rings have exactly the same properties.
If R and S are rngs, then the corresponding notion is that of a rng homomorphism,[a] defined as above except without the third condition f(1R) = 1S. A rng homomorphism between (unital) rings need not be a ring homomorphism.
The composition of two ring homomorphisms is a ring homomorphism. It follows that the rings forms a category with ring homomorphisms as morphisms (see Category of rings). In particular, one obtains the notions of ring endomorphism, ring isomorphism, and ring automorphism.
## Properties
Let f : RS be a ring homomorphism. Then, directly from these definitions, one can deduce:
• f(0R) = 0S.
• f(−a) = −f(a) for all a in R.
• For any unit a in R, f(a) is a unit element such that f(a)−1 = f(a−1) . In particular, f induces a group homomorphism from the (multiplicative) group of units of R to the (multiplicative) group of units of S (or of im(f)).
• The image of f, denoted im(f), is a subring of S.
• The kernel of f, defined as ker(f) = {a in R | f(a) = 0S}, is a two-sided ideal in R. Every two-sided ideal in a ring R is the kernel of some ring homomorphism.
• An homomorphism is injective if and only if kernel is the zero ideal.
• The characteristic of S divides the characteristic of R. This can sometimes be used to show that between certain rings R and S, no ring homomorphism RS exists.
• If Rp is the smallest subring contained in R and Sp is the smallest subring contained in S, then every ring homomorphism f : RS induces a ring homomorphism fp : RpSp.
• If R is a field (or more generally a skew-field) and S is not the zero ring, then f is injective.
• If both R and S are fields, then im(f) is a subfield of S, so S can be viewed as a field extension of R.
• If I is an ideal of S then f−1(I) is an ideal of R.
• If R and S are commutative and P is a prime ideal of S then f−1(P) is a prime ideal of R.
• If R and S are commutative, M is a maximal ideal of S, and f is surjective, then f−1(M) is a maximal ideal of R.
• If R and S are commutative and S is an integral domain, then ker(f) is a prime ideal of R.
• If R and S are commutative, S is a field, and f is surjective, then ker(f) is a maximal ideal of R.
• If f is surjective, P is prime (maximal) ideal in R and ker(f) ⊆ P, then f(P) is prime (maximal) ideal in S.
Moreover,
• The composition of ring homomorphisms ST and RS is a ring homomorphism RT.
• For each ring R, the identity map RR is a ring homomorphism.
• Therefore, the class of all rings together with ring homomorphisms forms a category, the category of rings.
• The zero map RS that sends every element of R to 0 is a ring homomorphism only if S is the zero ring (the ring whose only element is zero).
• For every ring R, there is a unique ring homomorphism ZR. This says that the ring of integers is an initial object in the category of rings.
• For every ring R, there is a unique ring homomorphism from R to the zero ring. This says that the zero ring is a terminal object in the category of rings.
• As the initial object is not isomorphic to the terminal object, there is no zero object in the category of rings; in particular, the zero ring is not a zero object in the category of rings.
## Examples
• The function f : ZZ/nZ, defined by f(a) = [a]n = a mod n is a surjective ring homomorphism with kernel nZ (see modular arithmetic).
• The complex conjugation CC is a ring homomorphism (this is an example of a ring automorphism).
• For a ring R of prime characteristic p, RR, xxp is a ring endomorphism called the Frobenius endomorphism.
• If R and S are rings, the zero function from R to S is a ring homomorphism if and only if S is the zero ring (otherwise it fails to map 1R to 1S). On the other hand, the zero function is always a rng homomorphism.
• If R[X] denotes the ring of all polynomials in the variable X with coefficients in the real numbers R, and C denotes the complex numbers, then the function f : R[X] → C defined by f(p) = p(i) (substitute the imaginary unit i for the variable X in the polynomial p) is a surjective ring homomorphism. The kernel of f consists of all polynomials in R[X] that are divisible by X2 + 1.
• If f : RS is a ring homomorphism between the rings R and S, then f induces a ring homomorphism between the matrix rings Mn(R) → Mn(S).
• Let V be a vector space over a field k. Then the map ρ : k → End(V) given by ρ(a)v = av is a ring homomorphism. More generally, given an abelian group M, a module structure on M over a ring R is equivalent to giving a ring homomorphism R → End(M).
• A unital algebra homomorphism between unital associative algebras over a commutative ring R is a ring homomorphism that is also R-linear.
## Non-examples
• The function f : Z/6ZZ/6Z defined by f([a]6) = [4a]6 is a rng homomorphism (and rng endomorphism), with kernel 3Z/6Z and image 2Z/6Z (which is isomorphic to Z/3Z).
• There is no ring homomorphism Z/nZZ for any n ≥ 1.
• If R and S are rings, the inclusion RR × S that sends each r to (r,0) is a rng homomorphism, but not a ring homomorphism (if S is not the zero ring), since it does not map the multiplicative identity 1 of R to the multiplicative identity (1,1) of R × S.
## Category of rings
### Endomorphisms, isomorphisms, and automorphisms
• A ring endomorphism is a ring homomorphism from a ring to itself.
• A ring isomorphism is a ring homomorphism having a 2-sided inverse that is also a ring homomorphism. One can prove that a ring homomorphism is an isomorphism if and only if it is bijective as a function on the underlying sets. If there exists a ring isomorphism between two rings R and S, then R and S are called isomorphic. Isomorphic rings differ only by a relabeling of elements. Example: Up to isomorphism, there are four rings of order 4. (This means that there are four pairwise non-isomorphic rings of order 4 such that every other ring of order 4 is isomorphic to one of them.) On the other hand, up to isomorphism, there are eleven rngs of order 4.
• A ring automorphism is a ring isomorphism from a ring to itself.
### Monomorphisms and epimorphisms
Injective ring homomorphisms are identical to monomorphisms in the category of rings: If f : RS is a monomorphism that is not injective, then it sends some r1 and r2 to the same element of S. Consider the two maps g1 and g2 from Z[x] to R that map x to r1 and r2, respectively; fg1 and fg2 are identical, but since f is a monomorphism this is impossible.
However, surjective ring homomorphisms are vastly different from epimorphisms in the category of rings. For example, the inclusion ZQ is a ring epimorphism, but not a surjection. However, they are exactly the same as the strong epimorphisms.
## Notes
1. ^ Some authors use the term "ring" to refer to structures that do not require a multiplicative identity; instead of "rng", "ring", and "rng homomorphism", they use the terms "ring", "ring with identity", and "ring homomorphism", respectively. Because of this, some other authors, to avoid ambiguity, explicitly specify that rings are unital and that homomorphisms preserve the identity.
## Citations
1. ^ Artin 1991, p. 353
2. ^ Eisenbud 1995, p. 12
3. ^ Jacobson 1985, p. 103
4. ^ Lang 2002, p. 88
5. ^ Hazewinkel 2004, p. 3
## References
• Artin, Michael (1991). Algebra. Englewood Cliffs, N.J.: Prentice Hall.
• Atiyah, Michael F.; Macdonald, Ian G. (1969), Introduction to commutative algebra, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., MR 0242802
• Bourbaki, N. (1998). Algebra I, Chapters 1–3. Springer.
• Eisenbud, David (1995). Commutative algebra with a view toward algebraic geometry. Graduate Texts in Mathematics. Vol. 150. New York: Springer-Verlag. xvi+785. ISBN 0-387-94268-8. MR 1322960.
• Hazewinkel, Michiel (2004). Algebras, rings and modules. Springer-Verlag. ISBN 1-4020-2690-0.
• Jacobson, Nathan (1985). Basic algebra I (2nd ed.). ISBN 9780486471891.
• Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556 | 2,489 | 8,859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.906569 |
https://mathoverflow.net/questions/265160/density-of-numbers-whose-prime-factors-belong-to-given-arithmetic-progressions?noredirect=1 | 1,606,468,281,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141191511.46/warc/CC-MAIN-20201127073750-20201127103750-00212.warc.gz | 391,704,245 | 29,106 | # Density of numbers whose prime factors belong to given arithmetic progressions
By a theorem of Landau, the number of integers $n\leq x$ whose prime divisors belong to only arithmetic progressions $a_1,\dots,a_r$ mod $q$, with $r\leq\varphi(q)$ and $a_i$ coprime to $q$ for each $i$, is
$$Cx (\log x)^{r/\varphi(q)-1} + O(x (\log x)^{r/\varphi(q)-2}),$$
for some constant $C$ depending on $r$ and $q$.
This is an old result. I am sure much more is known about the distribution of such $n$, but I do not know the literature well enough, and would be happy if someone could provide suggestions regarding the state of the art (or other classical theorems) regarding results of this nature. Thanks!
This result has been generalised a fair bit. The main generalisation is that you can replace congruence conditions by so-called "Frobenian conditions", namely conditions of the type which arise in the Chebotarev density theorem. There have also been some improvements in the error term, but substantial improvements are not really possible without assuming GRH.
Serre has written quite a bit about such topics. See for example Theorem 2.8 of the paper:
Serre - Divisibilité de certaines fonctions arithmétiques.
The associated zeta functions don't admit a meromorphic continuation to all of $\mathbb{C}$ in general; they have natural boundaries along the line $\mathrm{re}(s) = 0$. You can read more about this in the paper:
Hashimoto - Partial zeta functions.
(the author works in an even greater generality than I describe here).
• I'll accept the answer, though I realize now that what I was really looking for was this constant $C$ explicitly. I couldn't find this exact derivation in Landau.. – Tian An Apr 2 '17 at 16:58
• It is quite easy to write down the constant $C$ explicitly. One can write the associated Dirichlet series in terms of Dirichlet $L$-functions. One then obtains the asymptotic formula and thus $C$ via a Tauberian theorem ($C$ will be written in terms of special values of Dirichlet $L$-functions, which can be "calculated" using the class number formula). These methods are all explained very well in the cited paper of Serre; I would very much recommend that you study it. – Daniel Loughran Apr 2 '17 at 18:03
• The constant in the closely related case of representations of integers by binary quadratic forms can be found in: Brink, Moree, Osburn - Principal forms $x^2+ny^2$ representing many integers. – Daniel Loughran Apr 2 '17 at 18:05 | 628 | 2,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-50 | latest | en | 0.923944 |
https://socratic.org/questions/what-is-the-average-speed-of-an-object-that-is-moving-at-3-m-s-at-t-0-and-accele-2 | 1,604,045,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00631.warc.gz | 551,309,353 | 6,324 | # What is the average speed of an object that is moving at 3 m/s at t=0 and accelerates at a rate of a(t) =t^2-1 on t in [0,2]?
May 8, 2017
The average speed is $= 2.67 m {s}^{-} 1$
#### Explanation:
The speed is the integral of the acceleration
$a \left(t\right) = {t}^{2} - 1$
$v \left(t\right) = \int \left({t}^{2} - 1\right) \mathrm{dt}$
$v \left(t\right) = \frac{1}{3} {t}^{3} - t + C$
Plugging in the initial conditions
$v \left(0\right) = 0 - 0 + C = 3$
So,
$C = 3$
and
$v \left(t\right) = \frac{1}{3} {t}^{3} - t + 3$
The average speed is
$2 \overline{v} = {\int}_{0}^{2} \left(\frac{1}{3} {t}^{3} - t + 3\right) \mathrm{dt}$
$= {\left[\frac{1}{12} {t}^{4} - \frac{1}{2} {t}^{2} + 3 t\right]}_{0}^{2}$
$= \left(\frac{16}{12} - \frac{4}{2} + 6\right) - \left(0\right)$
$= \frac{16}{3}$
$\overline{v} = \frac{\frac{16}{3}}{2} = 2.67 m {s}^{-} 1$ | 397 | 870 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-45 | latest | en | 0.536225 |
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1 CS6702 GRAPH THEORY AND APPLICATIONS 2 MARKS QUESTIONS AND ANSWERS 1 UNIT I INTRODUCTION CS6702 GRAPH THEORY AND APPLICATIONS QUESTION BANK 1. Define Graph. 2. Define Simple graph. 3. Write few problems solved by the applications of graph theory. 4. Define incidence, adjacent and degree. 5. What are finite and infinite graphs? 6. Define Isolated and pendent vertex. 7. Define null graph. 8. Define Multigraph 9. Define complete graph 10. Define Regular graph 11. Define Cycles 12. Define Isomorphism. 13. What is Subgraph? 14. Define Walk, Path and Circuit. 15. Define connected graph. What is Connectedness? 16. Define Euler graph. 17. Define Hamiltonian circuits and paths 18. Define Tree 19. List out few Properties of trees. 20. What is Distance in a tree? 21. Define eccentricity and center. 22. Define distance metric. 23. What are the Radius and Diameter in a tree. 24. Define Rooted tree 25. Define Rooted binary tree 1. Explain various applications of graph. 2. Define the following kn, cn, kn,n, dn, trail, walk, path, circuit with an example. 3. Show that a connected graph G is an Euler graph iff all vertices are even degree. 4. Prove that a simple graph with n vertices and k components can have at most (n-k)(n-k+1)/2 edges. 5. Are they isomorphic? 6. Prove that in a complete graph with n vertices there are (n-1)/2 edges-disjoint Hamiltonian circuits, if n is odd number Prove that, there is one and only one path between every pair of vertices in a tree T. 8. Prove the given statement, A tree with n vertices has n-1 edges. 9. Prove that, any connected graph with n vertices has n-1 edges is a tree. 10. Show that a graph is a tree if and only if it is minimally connected. 11. Prove that, a graph G with n vertices has n-1 edges and no circuits are connected.
2 CS6702 GRAPH THEORY AND APPLICATIONS 2 MARKS QUESTIONS AND ANSWERS 2 UNIT II TREES, CONNECTIVITY & PLANARITY 1. Define Spanning trees. 2. Define Branch and chord. 3. Define complement of tree. 4. Define Rank and Nullity. 5. How Fundamental circuits created? 6. Define Spanning trees in a weighted graph. 7. Define degree-constrained shortest spanning tree. 8. Define cut sets and give example. 9. Write the Properties of cut set 10. Define Fundamental circuits 11. Define Fundamental cut sets 12. Define edge Connectivity. 13. Define vertex Connectivity. 14. Define separable and non-separable graph. 15. Define articulation point. 16. What is Network flows. 17. Define max-flow and min-cut theorem (equation). 18. Define component (or block) of graph. 19. Define 1-Isomorphism. 20. Define 2-Isomorphism. 21. Briefly explain Combinational and geometric graphs. 22. Distinguish between Planar and non-planar graphs. 23. Define embedding graph. 24. Define region in graph. 25. Why the graph is embedding on sphere. 1. Find the shortest spanning tree for the following graph. 2. Explain 1 - isomarphism and 2 - isomarphism of graphs with your own example. 3. Prove that a connected graph G with n vertices and e edges has e-n+2 regions. 4. Write all possible spanning tree for K5. 5. Prove that every cut-set in a connected graph G must contain at least one branch of every spanning tree of G. 6. Prove that the every circuit which has even number of edges in common with any cut-set. 7. Show that the ring sum of any two cut-sets in a graph is either a third cut set or en edge disjoint union of cut sets. 8. Explain network flow problem in detail. 9. If G 1 and G 2 are two 1-isomorphic graphs, the rank of G 1 equals the rank of G 2 and the nullity of G 1 equals the nullity of G 2, prove this. 10. Prove that any two graphs are 2-isomorphic if and only if they have circuit correspondence.
3 CS6702 GRAPH THEORY AND APPLICATIONS 2 MARKS QUESTIONS AND ANSWERS 3 UNIT III MATRICES, COLOURING AND DIRECTED GRAPH 1. What is proper coloring? 2. Define Chromatic number 3. Write the properties of chromatic numbers (observations). 4. Define Chromatic partitioning 5. Define independent set and maximal independent set. 6. Define uniquely colorable graph. 7. Define dominating set. 8. Define Chromatic polynomial. 9. Define Matching (Assignment). 10. What is Covering? 11. Define minimal cover. 12. What is dimer covering? 13. Define four color problem / conjecture. 14. State five color theorem 15. Write about vertex coloring and region coloring. 16. What is meant by regularization of a planar graph? 17. Define Directed graphs. 18. Define isomorphic digraph. 19. List out some types of directed graphs. 20. Define Simple Digraphs. 21. Define Asymmetric Digraphs (Anti-symmetric). 22. What is meant by Symmetric Digraphs? 23. Define Simple Symmetric Digraphs. 24. Define Simple Asymmetric Digraphs. 25. Give example for Complete Digraphs. 26. Define Complete Symmetric Digraphs. 27. Define Complete Asymmetric Digraphs (tournament). 28. Define Balance digraph (a pseudo symmetric digraph or an isograph). 29. Define binary relations. 30. What is Directed path? 31. Write the types of connected digraphs. 32. Define Euler graphs. 1. Prove that any simple planar graph can be embedded in a plane such that every edge is drawn as a straight line. 2. Show that a connected planar graph with n vertices and e edges has e-n+2 regions. 3. Define chromatic polynomial. Find the chromatic polynomial for the following graph. 4. Explain matching and bipartite graph in detail. 5. Write the observations of minimal covering of a graph. 6. Prove that the vertices of every planar graph can be properly colored with five colors. 7. Explain matching in detail. 8. Prove that a covering g of graph G is minimal iff g contains no path of length three or more. 9. Illustrate four-color problem. 10. Explain Euler digraphs in detail.
4 CS6702 GRAPH THEORY AND APPLICATIONS 2 MARKS QUESTIONS AND ANSWERS 4 UNIT IV PERMUTATIONS & COMBINATIONS 1. Define Fundamental principles of counting 2. Define rule of sum. 3. Define rule of Product 4. Define Permutations 5. Define combinations 6. State Binomial theorem 7. Define combinations with repetition 8. Define Catalan numbers 9. Write the Principle of inclusion and exclusion formula. 10. Define Derangements 11. What is meant by Arrangements with forbidden (banned) positions. 1. Explain the Fundamental principles of counting. 2. Find the number of ways of ways of arranging the word APPASAMIAP and out of it how many arrangements have all A s together. 3. Discuss the rules of sum and product with example. 4. Determine the number of (staircase) paths in the xy-plane from (2, 1) to (7, 4), where each path is made up of individual steps going 1 unit to the right (R) or one unit upward (U). iv. Find the coefficient of a 5 b 2 in the expansion of (2a - 3b) State and prove binomial theorem. 6. How many times the print statement executed in this program segment? 7. Discuss the Principle of inclusion and exclusion. 8. How many integers between 1 and 300 (inc.) are not divisible by at least one of 5, 6, 8? 9. How 32 bit processors address the content? How many address are possible? 10. Explain the Arrangements with forbidden positions.
5 CS6702 GRAPH THEORY AND APPLICATIONS 2 MARKS QUESTIONS AND ANSWERS 5 UNIT V GENERATING FUNCTIONS 1. Define Generating function. 2. What is Partitions of integer? 3. Define Exponential generating function 4. Define Maclaurin series expansion of e x and e -x. 5. Define Summation operator 6. What is Recurrence relation? 7. Write Fibonacci numbers and relation 8. Define First order linear recurrence relation 9. Define Second order recurrence relation 10. Briefly explain Non-homogeneous recurrence relation. 1. Explain Generating functions 2. Find the convolution of the sequences 1, 1, 1, 1,.. and 1,-1,1,-1,1, Find the number of non negative & positive integer solutions of for x1+x2+x3+x4= Find the coefficient of x5 in(1-2x)7. 5. The number of virus affected files in a system is 1000 and increases 250% every 2 hours. 6. Explain Partitions of integers 7. Use a recurrence relation to find the number of viruses after one day. 8. Explain First order homogeneous recurrence relations. 9. Solve the recurrence relation an+2-4an+1+3an=-200 with a0=3000 and a1= Solve the Fibonacci relation Fn = Fn-1+Fn Find the recurrence relation from the sequence 0, 2, 6, 12, 20, 30, 42,. 12. Determine (1+ 3i) Discuss Method of generating functions. All the Best No substitution for hard work.
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Planar Graph (7A) Copyright (c) 2015 2018 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later
### MATHEMATICAL STRUCTURES FOR COMPUTER SCIENCE
MATHEMATICAL STRUCTURES FOR COMPUTER SCIENCE A Modern Approach to Discrete Mathematics SIXTH EDITION Judith L. Gersting University of Hawaii at Hilo W. H. Freeman and Company New York Preface Note to the | 9,489 | 35,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-18 | latest | en | 0.860812 |
http://complete-markets.com/category/working-papers/ | 1,571,446,233,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00483.warc.gz | 50,585,231 | 8,906 | # Interactive Graphic for new Working Paper: 'Fake News'
My coauthor (William Pratt) and I will post our new working paper 'Fake News' here and to SSRN shortly. It is an analysis of how false information of a takeover of Twitter on July 14 2015 was incorporated into market (stock and option) prices.
To accompany the paper, I created an interactive graphic to help understand the realtionship between Twitter's stock price and options implied volatility. I am posting the graphic HERE, and linking to it in the journal article. Take a look, and feedback is always welcome.
# New Working Paper: Parameter Variation & the Components of Natural Gas Price Volatility
I have posted a new working paper of mine to the 'Working Paper' section. The paper, Parameter Variation & the Components of Natural Gas Price Volatility, hypothesizes that parameters linking natural gas returns to fundamental variables will tend to change as market participants learn. I therefore estimate the parameters using the Kalman filter. I also decompose conditional natural gas volatility into portions attributable to each variable. The abstract is below.
Abstract
Estimating a static coefficient for a deseasoned gas storage or
weather variable implicitly assumes that market participants react
identically throughout the year (and over each year) to that variable.
In this analysis we model natural gas returns as a linear function of
gas storage and weather variables, and we allow the coefficients of this
function to vary continuously over time. This formulation takes into
account that market participants continuously try to improve their
forecasts of market prices, and this likely means they continuously
change the scale of their reaction to changes in underlying variables.
We use this model to also calculate conditional natural gas volatility
and the proportion of volatility attributable to each factor. We find
that return volatility is higher in the winter, and this increase is at-
tributable to increases in the proportion of volatility due to weather
and natural gas storage. We provide time series estimates of the chang-
ing proportion of volatility attributable to each factor, which is useful
for hedging and derivatives trading in natural gas markets.
# State Dependence in the Natural Gas and Rig Count Relationship
I just posted a new working paper, 'State Dependence in the Natural Gas and Rig Count Relationship' to the USAEE working paper series on SSRN. The paper is available for download here: http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2371753
In a nutshell, the paper finds that changes in the north American natural gas rig count does affect future changes in natural gas prices (previous research has not found this result). The relationship is state dependent however. When natural gas prices are above \$6.74/MMBtu then increases in the rig count will drive down natural gas prices. Below this threshold the rig count does not affect gas prices (however gas prices affect the rig count). In sum, the evidence is consistent with gas producers, 'killing the rally' in gas prices by markedly increasing gas production. | 631 | 3,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-43 | latest | en | 0.89637 |
http://blog.csdn.net/qq_18661257/article/details/46808129 | 1,516,576,008,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00753.warc.gz | 42,213,220 | 14,308 | Count the Colors
B - Count the Colors
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Description
Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.
Your task is counting the segments of different colors you can see at last.
Input
The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.
Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:
x1 x2 c
x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.
All the numbers are in the range [0, 8000], and they are all integers.
Input may contain several data set, process to the end of file.
Output
Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.
If some color can't be seen, you shouldn't print it.
Print a blank line after every dataset.
Sample Input
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1
Sample Output
1 1
2 1
3 1
1 1
0 2
1 1
/*
Author: 2486
Memory: 232 KB Time: 110 MS
Language: C++ (g++ 4.7.2) Result: Accepted
*/
//暴力可过，或者是线段树
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn=8000+5;
int n,a,b,c,ans[maxn],col[maxn];
int main() {
while(~scanf("%d",&n)) {
memset(ans,0,sizeof(ans));
memset(col,0,sizeof(col));
int Max=0;
for(int i=0; i<n; i++) {
scanf("%d%d%d",&a,&b,&c);
for(int j=a; j<b; j++) {
col[j]=c+1;
}
Max=max(Max,b);
}
for(int i=0; i<=Max; i++) {
while(i!=0&&col[i]&&col[i]==col[i-1]) {
i++;
}
if(col[i]) {
ans[col[i]-1]++;
}
}
for(int i=0; i<=8001; i++) {
if(ans[i]) {
printf("%d %d\n",i,ans[i]);
}
}
printf("\n");
}
}
/*
Author: 2486
Memory: 356 KB Time: 20 MS
Language: C++ (g++ 4.7.2) Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=8000+5;
int sum[maxn],col[maxn<<2],res[maxn];
int n,x1,x2,c;
void pushup(int rt) {}
void pushdown(int rt,int l,int r) {
if(col[rt]!=-1) {
col[rt<<1]=col[rt<<1|1]=col[rt];//直接往下递归就可以了，不用管其他
col[rt]=-1;
}
}
void build(int rt,int l,int r) {
col[rt]=-1;
if(l==r)return;
int mid=(l+r)>>1;
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
}
void update(int L,int R,int c,int rt,int l,int r) {
if(L<=l&&r<=R) {
col[rt]=c;
return;
}
pushdown(rt,l,r);
int mid=(l+r)>>1;
if(L<=mid)update(L,R,c,rt<<1,l,mid);
if(mid<R)update(L,R,c,rt<<1|1,mid+1,r);
pushup(rt);
}
void query(int L,int R,int rt,int l,int r) {
if(l==r) {
sum[l]=col[rt];//赋值，放回区间的位子颜色
return;
}
pushdown(rt,l,r);
int mid=(l+r)>>1;
if(L<=mid)query(L,R,rt<<1,l,mid);
if(mid<R)query(L,R,rt<<1|1,mid+1,r);
pushup(rt);
}
int main() {
while(~scanf("%d",&n)) {
memset(sum,-1,sizeof(sum));
memset(res,0,sizeof(res));
build(1,0,maxn-1);
for(int i=0; i<n; i++) {
scanf("%d%d%d",&x1,&x2,&c);
if(x1==x2)continue;
update(x1,x2-1,c,1,0,maxn-1);
}
query(0,maxn-1,1,0,maxn-1);//将所有数据都更新到底部
for(int i=0; i<maxn; i++) {
while(i!=0&&sum[i]!=-1&&sum[i]==sum[i-1])//如果没有颜色就加一，如果与之前相等那么也是加一
i++;
res[sum[i]]++;
}
for(int i=0; i<maxn; i++)if(res[i])printf("%d %d\n",i,res[i]);
printf("\n");
}
return 0;
}
• 本文已收录于以下专栏：
举报原因： 您举报文章：Count the Colors 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 1,290 | 3,430 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-05 | latest | en | 0.643315 |
https://www.etutorworld.com/5th-grade-math-worksheets/patterns-including-numbers.html | 1,675,771,048,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00432.warc.gz | 769,559,627 | 46,140 | Select Page
# Patterns Including Numbers
Home >> 5th Grade Math Worksheets >> Patterns including numbers
### Patterns
Patterns are present everywhere! They are predictable arrangements following a certain rule.
In our daily life, we see patterns in animals, arrangement of leaves; numbers etc….These patterns make learning interesting for us and help us unwind our mind.
### Simple Patterns
Are you familiar with these patterns?
3, 6, 9, 12, 15, ____
### Triangular Numbers
Emma was watching a sports channel. She saw the following picture and asked her father what game they are playing?
The father replied that this is a game of bowling pins. The players roll a ball and knock down as many bowling pins as possible. She was surprised at the arrangement of the bowling pins. There were a total of 10 bowling pins. She drew the arrangement using dots in an increasing pattern.
She counted the number of dots in each pattern. They are 1, 3, 6 and 10.
The number of dots that can form a triangle-shaped pattern are called triangular numbers.
Let us observe the pattern and understand the rule to find the next numbers.
### Square Numbers
The next game on the channel was chess. Emma liked the pattern of squares on the chessboard and wanted to count them.
She started drawing the squares using dots in an increasing pattern.
She noted the number of dots in each square as 1, 4, 9 and 16.
Let us observe the pattern and understand the rule to find the next numbers.
### Turn the shape
Mia is printing on a drawing sheet with a block. Look at the beautiful pattern she has made:
Has she used different blocks to make the pattern?
We can create a pattern by simply turning a block in a particular way. We can also turn a block in different ways to create different designs.
Here, Mia has used the same block for printing. She turns the block one- fourth in a clockwise direction each time. Since the block is rotated by a one-fourth turn ever time, it is called a quarter turn.
Another way of creating a pattern is by rotating a block by half a turn each time. Since the block is rotated by turn, it is called half turn.
### Interesting number and alphabet patterns
The number 101 is used to make a pattern by taking a 1/4th turn every time.
After which turn does the number look the same?
The word ZOOZ is used to make a pattern by taking a 1/4th turn every time. After 1/4th turn, the word is read as NOON, but vertically.
After which turn does the number look the same?
### Check Point
1. What is the missing number in this pattern?
1, 4, 27, 256, ______, 46656
1. Which number should be placed in the empty triangle?
3. Which of these shapes will look the same after 1/2 a turn?
4. Study the pattern carefully and complete it:
1 × 1 × 1 = 1 = 1
2 × 2 × 2 = 8 = 3 + 5
3 × 3 × 3 = 27 = 7 + 9 + 11
4 × 4 × 4 = 64 = 13 + ____ + ____ + ____
5 × 5 × 5 = ___ = ____ + ____ + ____ + ____ + ____
5.Complete the pattern:
1. 3125 as 11 = 1, 22 = 4, 33 = 9, 44 = 256, 55 = 3125, 66 = 46656
2. (7 – 3) × 3 = 4 × 3 = 12
(8 – 6) × 4 = 2 × 4 = 8
(8 – 5) × 3 = 3 × 3 = 9
(7 – 6) × 4 = 1 × 4 = 4
3.
4. 4 × 4 × 4 = 64 = 13 + 15 + 17 + 19
5 × 5 × 5 = 125 = 21 + 23 + 25 + 27 + 29
5.
## IN THE NEWS
Our mission is to provide high quality online tutoring services, using state of the art Internet technology, to school students worldwide. | 928 | 3,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5.09375 | 5 | CC-MAIN-2023-06 | longest | en | 0.965094 |
https://www.electronicspoint.com/forums/threads/capacitor-and-inductor-reactance.272939/ | 1,624,140,002,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00162.warc.gz | 674,753,372 | 12,330 | # Capacitor and inductor reactance
Discussion in 'Electronics Homework Help' started by Zeeman2000, Mar 2, 2015.
1. ### Zeeman2000
6
1
Mar 2, 2015
Hello everyone,
I am currently working through a self study book on electronics. I have starting a chapter which is an introduction to Alternating Current. The chapter discusses capacitor and inductor reactance using the following formulas:
XL = 2pifL
XC = 1/2pifC
At the end of the chapter there is a self test which gives a value of a capacitor reactance and the value of a capacitor or inductor and asks to find the frequecy necessary to cause reactance.
The equation for this was not dicussed im the chapter. Can someone explain the equation needed to find the answer?
I tried the equation in reverse which worked for:
100Hz
10H
6.28 * 100 * 10 = 6280 ohms
6280 / 10 / 6.28 = 100Hz
In the self test the values are:
L = 50 μH, XL = 320 ohms, f =
320 / 50 / 6.28 = 1.019108280254777
Not sure how they are getting this answer. Any help much appreciatted.
2. ### Laplace
1,252
184
Apr 4, 2010
Units in the formula are Ohm, Hertz, and Henry. 50 μH = 0.000050 H.
3. ### Zeeman2000
6
1
Mar 2, 2015
Yes, 50 μH = 0.000050 H
Turns out the question in the book was wrong. I have since checked the errata and it should have read 50mH not 50μH
As even using 0.000050 H gave and answer in MHz
4. ### BobK
7,682
1,688
Jan 5, 2010
Yep, at 50mH the answer in the book is right.
Bob | 452 | 1,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-25 | latest | en | 0.905263 |
https://learn-scikit.oneoffcoder.com/pyod.html | 1,679,603,285,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00451.warc.gz | 427,026,350 | 11,214 | 11. PyOD, Anamoly Detection
PyOD is a unified outlier detection API. Let’s see how we can use it to detect outliers.
11.1. Data
We will generate inlier and outlier data according to normal distributions. The inlier distribution is as follows.
• $$\mathcal{N}(0, 1)$$
The outlier distributions are as follows.
• $$\mathcal{N}(2, 1)$$
• $$\mathcal{N}(3, 1)$$
• $$\mathcal{N}(4, 1)$$
• $$\mathcal{N}(10, 1)$$
We will sample 1,000 samples from the inlier distribution and 100 samples from each outlier distribution for a total of 1,400 samples. Here, $$X$$ will be all the samples merged and $$y$$ will be a binary vector where 0 indicates inlier and 1 indicates outlier.
[1]:
import numpy as np
import pandas as pd
import random as rand
np.random.seed(37)
rand.seed(37)
N = 1_000
a = np.random.normal(0, 1, N)
b = np.random.normal(2, 1, 100)
c = np.random.normal(3, 1, 100)
d = np.random.normal(4, 1, 100)
e = np.random.normal(10, 1, 100)
X = pd.DataFrame({'x0': np.concatenate([a, b, c, d, e])})
y = np.concatenate([
np.zeros(N),
np.ones(100),
np.ones(100),
np.ones(100),
np.ones(100)
])
X.shape, y.shape
[1]:
((1400, 1), (1400,))
Let’s visualize the distribution of $$X$$ and $$y$$. We will visualize $$X$$ at different bandwidths for the kernel density estimation (KDE).
[2]:
import matplotlib.pyplot as plt
plt.style.use('ggplot')
fig, ax = plt.subplots(1, 2, figsize=(15, 4))
for bw in [0.1, 0.5, 0.9, 1.0]:
_ = X['x0'].plot(kind='kde', bw_method=bw, ax=ax[0], label=f'{bw:.1f}')
_ = ax[0].legend()
_ = ax[0].set_title('KDE of data at different bandwidths')
s = pd.Series(y).value_counts().sort_index()
s.index = ['inlier', 'outlier']
_ = s.plot(kind='bar', ax=ax[1])
_ = ax[1].set_title('Distribution of inlier and outlier')
plt.tight_layout()
11.2. Isolation Forest
Isolation Forest is an approach to detect outliers. The predictions are -1 for outlier and 1 for inlier.
[3]:
from sklearn.ensemble import IsolationForest
iso = IsolationForest(n_jobs=-1, random_state=37)
y_pred = iso.fit_predict(X)
Here, we visualize the distribution of the predictions.
[4]:
fig, ax = plt.subplots(1, 2, figsize=(10, 4))
s = pd.Series(y_pred).value_counts().sort_index()
s.index = ['outlier', 'inlier']
s.plot(kind='bar', ax=ax[0])
s.plot(kind='pie', ax=ax[1])
_ = ax[1].set_ylabel('')
_ = ax[0].set_title('Distribution of predicted outlier and inlier')
_ = ax[1].set_title('Proportion of predicted outlier and inlier')
plt.tight_layout()
We can benchmark the (non-validated) performance of the Isolation Forest model using
[5]:
from sklearn.metrics import accuracy_score, f1_score, average_precision_score, roc_auc_score
y_p = np.array([0 if v == 1 else 1 for v in y_pred])
s = pd.Series([
accuracy_score(y, y_p),
f1_score(y, y_p),
average_precision_score(y, y_p),
roc_auc_score(y, y_p)
], index=['acc', 'f1', 'aps', 'roc'])
_ = s.plot(kind='bar', title='Performance of Isolation Forest')
Here is the confusion matrix.
• P_N: predicted negative
• P_P: predicted positive
• T_N: true negative
• T_P: true positive
[6]:
from sklearn.metrics import confusion_matrix
pd.DataFrame(confusion_matrix(y, y_p), columns=['P_N', 'P_P'], index=['T_N', 'T_P'])
[6]:
P_N P_P
T_N 909 91
T_P 117 283
Now let’s plot the following distributions.
• true vs predicted inliers
• true vs predicted outliers
It seems Isolation Forest is having problems detecting outliers more than inliers.
[7]:
fig, ax = plt.subplots(1, 2, figsize=(15, 4))
_ = X[y == 0].x0.plot(kind='kde', bw_method=0.5, ax=ax[0], label='true_inlier')
_ = X[y_p == 0].x0.plot(kind='kde', bw_method=0.5, ax=ax[0], label='pred_inlier')
_ = X[y == 1].x0.plot(kind='kde', bw_method=0.5, ax=ax[1], label='true_outlier')
_ = X[y_p == 1].x0.plot(kind='kde', bw_method=0.5, ax=ax[1], label='pred_outlier')
_ = ax[0].legend()
_ = ax[1].legend()
_ = ax[0].set_title('KDE of true vs pred inliers')
_ = ax[1].set_title('KDE of true vs pred outliers')
plt.tight_layout()
Now, let’s plot the distributions of
• TN: true negative
• FP: false positive
• FN: false negative
• TP: true positive
Look at the KDE plots for FP and FN.
• The FP samples belong to the inlier class, but were predicted as outlier; the distribution is to the left of the TN curve.
• The FN samples belong to the outlier class, but were predicted as inlier; the distribution is to the right of the TN cruve.
[8]:
fig, ax = plt.subplots(figsize=(15, 4))
_ = X[(y== 0) & (y_p == 0)].x0.plot(kind='kde', label='TN', ax=ax)
_ = X[(y== 0) & (y_p == 1)].x0.plot(kind='kde', label='FP', ax=ax)
_ = X[(y== 1) & (y_p == 0)].x0.plot(kind='kde', label='FN', ax=ax)
_ = X[(y== 1) & (y_p == 1)].x0.plot(kind='kde', label='TP', ax=ax)
_ = ax.set_title('KDE of TN, FP, FN, TP')
_ = ax.legend()
11.3. K-Nearest Neighbor
K-nearest neighbor (KNN) from PyOD can be used to detect outliers. However, we have to give KNN some hints about the proportion of contamination (proportion of outliers). Since we generated the data, we know the contamination is about 0.30.
[9]:
from pyod.models.knn import KNN
knn = KNN(contamination=0.30, n_jobs=-1)
knn.fit(X)
y_pred = knn.predict(X)
Let’s look at the distribution of predicted inliers and outliers by KNN.
[10]:
s = pd.Series(y_pred).value_counts().sort_index()
s.index = ['inlier', 'outlier']
fig, ax = plt.subplots(1, 2, figsize=(10, 4))
s.plot(kind='bar', ax=ax[0])
s.plot(kind='pie', ax=ax[1])
_ = ax[1].set_ylabel('')
_ = ax[0].set_title('Distribution of predicted outlier and inlier')
_ = ax[1].set_title('Proportion of predicted outlier and inlier')
plt.tight_layout()
What about the (non-validated) performance? Based on the average precision score, KNN performed slightly worse than Isolation Forest.
[11]:
y_p = y_pred
s = pd.Series([
accuracy_score(y, y_p),
f1_score(y, y_p),
average_precision_score(y, y_p),
roc_auc_score(y, y_p)
], index=['acc', 'f1', 'aps', 'roc'])
_ = s.plot(kind='bar', title='Performance of KNN')
Here’s the confusion matrix for the KNN predictions. Compared to Isolation Forest, KNN has a tougher time predicting the positive examples and Isolation Forest has a tougher time predicting the negative examples.
[12]:
pd.DataFrame(confusion_matrix(y, y_p), columns=['P_N', 'P_P'], index=['T_N', 'T_P'])
[12]:
P_N P_P
T_N 917 83
T_P 133 267
Let’s plot the KIDE of true vs predicted inliers and true vs predicted outliers for KNN.
[13]:
fig, ax = plt.subplots(1, 2, figsize=(15, 4))
_ = X[y == 0].x0.plot(kind='kde', bw_method=0.5, ax=ax[0], label='true_inlier')
_ = X[y_p == 0].x0.plot(kind='kde', bw_method=0.5, ax=ax[0], label='pred_inlier')
_ = X[y == 1].x0.plot(kind='kde', bw_method=0.5, ax=ax[1], label='true_outlier')
_ = X[y_p == 1].x0.plot(kind='kde', bw_method=0.5, ax=ax[1], label='pred_outlier')
_ = ax[0].legend()
_ = ax[1].legend()
_ = ax[0].set_title('KDE of true vs pred inliers')
_ = ax[1].set_title('KDE of true vs pred outliers')
plt.tight_layout()
These are the KDE plots for TN, FP, FN and TP generated by KNN.
[14]:
fig, ax = plt.subplots(figsize=(15, 4))
_ = X[(y== 0) & (y_p == 0)].x0.plot(kind='kde', label='TN', ax=ax)
_ = X[(y== 0) & (y_p == 1)].x0.plot(kind='kde', label='FP', ax=ax)
_ = X[(y== 1) & (y_p == 0)].x0.plot(kind='kde', label='FN', ax=ax)
_ = X[(y== 1) & (y_p == 1)].x0.plot(kind='kde', label='TP', ax=ax)
_ = ax.set_title('KDE of TN, FP, FN, TP')
_ = ax.legend()
What happens if we vary the contamination value in the range [0.1, 0.5]?
[15]:
r_df = []
for c in np.arange(0.1, 0.5+0.01, 0.01):
knn = KNN(contamination=c, n_jobs=-1)
knn.fit(X)
y_p = knn.predict(X)
r = {
'contamination': c,
'acc': accuracy_score(y, y_p),
'f1': f1_score(y, y_p),
'aps': average_precision_score(y, y_p),
'roc': roc_auc_score(y, y_p)
}
r_df.append(r)
r_df = pd.DataFrame(r_df)
r_df.index = r_df.contamination
We can see that as we move from 0.1 to the true contamination proportion, the performance measures increase, and, after which, the performances start to degrade. It seems getting the contamination parameter correct is crucial to optimal performance.
[16]:
_, ax = plt.subplots(figsize=(10, 4))
_ = r_df['acc'].plot(kind='line', ax=ax)
_ = r_df['f1'].plot(kind='line', ax=ax)
_ = r_df['aps'].plot(kind='line', ax=ax)
_ = r_df['roc'].plot(kind='line', ax=ax)
_ = ax.legend()
plt.tight_layout()
11.3.1. Combination
But, we will never know the amount of contamination or outliers in our data. No worries, since we can create an ensemble of KNN models and aggregate across them for a final prediction. We get two benefits; we no longer need to fine-tune the contamination parameter and we get an ensemble model that should do better than a single model.
In this approach, we create a large ensemble of KNN to detect outliers. The contamination and n_neighbors are both varied constantly in the ranges $$[0.1, 0.5]$$ by 0.01 and $$[10, 210]$$ by 10, respectively.
[17]:
import itertools
from pyod.utils.utility import standardizer
def do_learn(n, c):
knn = KNN(contamination=c, n_neighbors=n, n_jobs=-1)
knn.fit(X)
return knn.predict(X)
def get_key(n, c):
return f'n_{n:03}_c={c:.2f}'
n_neighbors = list(np.arange(10, 210, 10))
contaminations = list(np.arange(0.1, 0.5 + 0.01, 0.01))
r_df = pd.DataFrame({get_key(n, c): do_learn(n, c)
for n, c in itertools.product(n_neighbors, contaminations)})
r_mat = standardizer(r_df)
The ensemble KNN may then be aggregated by the following.
• aom: Take the average of the maximums (each sample has multiple predictions, these predictions are put into sub-groups, and the average of the maximum of these sub-groups is computed).
• moa: Maximization of average (each sample has multiple predictions, these predictions are put into sub-groups, and the maximum of these sub-groups is computed);
• average: Take the average of the outlier scores.
• maximization: Take the maximum of the outlier scores.
Now, we can compute the performance of these KNN ensembles. The KNN ensembles have average precision scores over 0.80, while KNN alone and Isolation forest have 0.60. The AUC-ROC performance is also improved; KNN alone and Isolation Forest have 0.80, while KNN ensemble has over 0.90.
[18]:
from pyod.models.combination import aom, moa, average, maximization
pd.DataFrame([{'score': f.__name__, 'roc': roc_auc_score(y, f(r_mat)), 'aps': average_precision_score(y, f(r_mat))}
for f in [average, moa, average, maximization]])
[18]:
score roc aps
0 average 0.930193 0.856301
1 moa 0.930248 0.856549
2 average 0.930193 0.856301
3 maximization 0.926509 0.824471 | 3,278 | 10,538 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-14 | longest | en | 0.517451 |
http://mathhelpforum.com/calculus/147022-problem-integration.html | 1,508,621,631,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00180.warc.gz | 224,910,631 | 10,651 | 1. ## Problem with Integration
Hi
I am having trouble trying to integrate this equation:
$\int \sqrt(1+2x+2x^3)$
how would i approach an expression like this?
Do i integrate the inside and then the outside?
P.S
2. Are you sure it's $\int{\sqrt{1+2x+3x^3}}\;{dx}$?
3. Well, the original question asked to find the length of arc of $f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})
$
$y=2-\frac{1}{2}x$ between x=0, y=3
I used the length of arc rule:
$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$
This is where i was stuck:
$\int_0^3 \sqrt{1+2x+2x^3}dx$
4. Originally Posted by Paymemoney
Well, the original question asked to find the length of arc of $f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})
$
$y=2-\frac{1}{2}x$ between x=0, y=3
I used the length of arc rule:
$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$
This is where i was stuck:
$\int_0^3 \sqrt{1+2x+2x^3}dx$
Dear Paymemoney,
I have some problems regarding your question. You want to find the arc length of $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$ is'nt?
What is the second equation $y=2-\frac{1}{2}x$ ??? How did you get this? Do you want to find the arc length of this curve too??
5. Originally Posted by Sudharaka
Dear Paymemoney,
I have some problems regarding your question. You want to find the arc length of $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$ is'nt?
What is the second equation $y=2-\frac{1}{2}x$ ??? How did you get this? Do you want to find the arc length of this curve too??
opps srry my mistake the equation is only $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$
6. Originally Posted by Paymemoney
opps srry my mistake the equation is only $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$
Dear Paymemoney,
You have simplified incorrectly,
$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$
$=\int_0^3\sqrt{1+4x^2(1+x^2)}dx$
$=\int_0^3\sqrt{1+4x^2+4x^4}dx$
$=\int_0^3\sqrt{(1+2x^2)^2}dx$
Can you continue from here??
7. Originally Posted by Sudharaka
Dear Paymemoney,
You have simplified incorrectly,
$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$
$=\int_0^3\sqrt{1+4x^2(1+x^2)}dx$
$=\int_0^3\sqrt{1+4x^2+4x^4}dx$
$=\int_0^3\sqrt{(1+2x^2)^2}dx$
Can you continue from here??
yes,
thank you, i worked it out | 860 | 2,173 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-43 | longest | en | 0.801387 |
https://math.stackexchange.com/questions/2606047/lebesgue-measure-and-limit-of-the-integral | 1,582,962,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148671.99/warc/CC-MAIN-20200229053151-20200229083151-00489.warc.gz | 450,137,693 | 31,630 | # Lebesgue measure and limit of the integral.
I have problem with this integral and limit : $\lim\limits_{n \to \infty } \int_{\mathbb{R}^+} \left( 1+ \frac{x}{n} \right) \sin ^n \left( x \right) d\mu_1$. Where $\mu_1$ is Lebesgue measure. I know, that I have to show that $(1+\frac{x}{n})\sin ^n(x)$ is bounded, yes?
• The function at hand is not bounded on $\mathbb{R}^{+}$. – TheOscillator Jan 15 '18 at 10:12
## 1 Answer
When $n$ is odd, $\sin^n x$ is periodic of period $2\pi$, is $\ge 0$ on $[2k\pi,(2k+1)\pi]$ and is $\le 0$ on $[(2k+1)\pi, 2(k+1)\pi]$ for any integer $k$.
Letting $f(x) = \left(1+\frac xn\right)\sin^n x$, by definition the integral of $f$ is$$\int_{\Bbb R_+} f\,d\mu := \int_{\Bbb R_+} f_+\,d\mu - \int_{\Bbb R_+} f_-\,d\mu$$ where $f_+(x) := \max\{f(x), 0\}$ and $f_-(x) := \max\{-f(x), 0\}$
But \begin{align}\int_{\Bbb R_+} f_+\,d\mu &= \sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\left(1+\frac xn\right)\sin^n x\,dx\\ &\ge \sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\sin^n x\,dx\\ &= \sum_{k=0}^\infty\int_0^{\pi}\sin^n x\,dx\\&=\infty\end{align} Similarly, \begin{align}\int_{\Bbb R_+} f_-\,d\mu &= \sum_{k=0}^\infty\int_{(2k+1)\pi}^{2(k+1)\pi}-\left(1+\frac xn\right)\sin^n x\,dx\\ &\ge \sum_{k=0}^\infty\int_{(2k+1)\pi}^{2(k+1)\pi}-\sin^n x\,dx\\ &= \sum_{k=0}^\infty\int_{\pi}^{2\pi}-\sin^n x\,dx\\&=\infty\end{align}
So $$\int_{\Bbb R_+} f\,d\mu = \infty - \infty$$
Which is undefined. Since you are trying to take the limit of an undefined sequence, the limit is also undefined. | 693 | 1,516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-10 | latest | en | 0.618554 |
https://web2.0calc.com/questions/geometry_16186 | 1,600,655,463,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00374.warc.gz | 727,059,942 | 6,464 | +0
# geometry
0
71
2
+111
In right triangle $$ABC,$$ $$\angle C = 90^\circ.$$ Median $$\overline{AM}$$ has a length of $$19$$, and median $$\overline{BN}$$ has a length of $$13$$. What is the length of the hypotenuse of the triangle?
Note: A median of a triangle is a line segment connecting a vertex of a triangle to the midpoint of the opposite side of the triangle.
Jun 20, 2020
#1
0
By the Pythagorean Theorem, the hypotenuse is 6*sqrt(11).
Jun 21, 2020
#2
0
I think it's wrong
Guest Jun 21, 2020 | 182 | 510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-40 | longest | en | 0.652938 |
http://cn.metamath.org/mpeuni/fucass.html | 1,659,954,529,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00293.warc.gz | 9,716,564 | 9,027 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > fucass Structured version Visualization version GIF version
Theorem fucass 16834
Description: Associativity of natural transformation composition. Remark 6.14(b) in [Adamek] p. 87. (Contributed by Mario Carneiro, 6-Jan-2017.)
Hypotheses
Ref Expression
fucass.q 𝑄 = (𝐶 FuncCat 𝐷)
fucass.n 𝑁 = (𝐶 Nat 𝐷)
fucass.x = (comp‘𝑄)
fucass.r (𝜑𝑅 ∈ (𝐹𝑁𝐺))
fucass.s (𝜑𝑆 ∈ (𝐺𝑁𝐻))
fucass.t (𝜑𝑇 ∈ (𝐻𝑁𝐾))
Assertion
Ref Expression
fucass (𝜑 → ((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)(⟨𝐹, 𝐺 𝐾)𝑅) = (𝑇(⟨𝐹, 𝐻 𝐾)(𝑆(⟨𝐹, 𝐺 𝐻)𝑅)))
Proof of Theorem fucass
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 eqid 2770 . . . . 5 (Base‘𝐷) = (Base‘𝐷)
2 eqid 2770 . . . . 5 (Hom ‘𝐷) = (Hom ‘𝐷)
3 eqid 2770 . . . . 5 (comp‘𝐷) = (comp‘𝐷)
4 fucass.r . . . . . . . . . 10 (𝜑𝑅 ∈ (𝐹𝑁𝐺))
5 fucass.n . . . . . . . . . . 11 𝑁 = (𝐶 Nat 𝐷)
65natrcl 16816 . . . . . . . . . 10 (𝑅 ∈ (𝐹𝑁𝐺) → (𝐹 ∈ (𝐶 Func 𝐷) ∧ 𝐺 ∈ (𝐶 Func 𝐷)))
74, 6syl 17 . . . . . . . . 9 (𝜑 → (𝐹 ∈ (𝐶 Func 𝐷) ∧ 𝐺 ∈ (𝐶 Func 𝐷)))
87simpld 476 . . . . . . . 8 (𝜑𝐹 ∈ (𝐶 Func 𝐷))
9 funcrcl 16729 . . . . . . . 8 (𝐹 ∈ (𝐶 Func 𝐷) → (𝐶 ∈ Cat ∧ 𝐷 ∈ Cat))
108, 9syl 17 . . . . . . 7 (𝜑 → (𝐶 ∈ Cat ∧ 𝐷 ∈ Cat))
1110simprd 477 . . . . . 6 (𝜑𝐷 ∈ Cat)
1211adantr 466 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝐷 ∈ Cat)
13 eqid 2770 . . . . . . 7 (Base‘𝐶) = (Base‘𝐶)
14 relfunc 16728 . . . . . . . 8 Rel (𝐶 Func 𝐷)
15 1st2ndbr 7365 . . . . . . . 8 ((Rel (𝐶 Func 𝐷) ∧ 𝐹 ∈ (𝐶 Func 𝐷)) → (1st𝐹)(𝐶 Func 𝐷)(2nd𝐹))
1614, 8, 15sylancr 567 . . . . . . 7 (𝜑 → (1st𝐹)(𝐶 Func 𝐷)(2nd𝐹))
1713, 1, 16funcf1 16732 . . . . . 6 (𝜑 → (1st𝐹):(Base‘𝐶)⟶(Base‘𝐷))
1817ffvelrnda 6502 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((1st𝐹)‘𝑥) ∈ (Base‘𝐷))
197simprd 477 . . . . . . . 8 (𝜑𝐺 ∈ (𝐶 Func 𝐷))
20 1st2ndbr 7365 . . . . . . . 8 ((Rel (𝐶 Func 𝐷) ∧ 𝐺 ∈ (𝐶 Func 𝐷)) → (1st𝐺)(𝐶 Func 𝐷)(2nd𝐺))
2114, 19, 20sylancr 567 . . . . . . 7 (𝜑 → (1st𝐺)(𝐶 Func 𝐷)(2nd𝐺))
2213, 1, 21funcf1 16732 . . . . . 6 (𝜑 → (1st𝐺):(Base‘𝐶)⟶(Base‘𝐷))
2322ffvelrnda 6502 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((1st𝐺)‘𝑥) ∈ (Base‘𝐷))
24 fucass.t . . . . . . . . . 10 (𝜑𝑇 ∈ (𝐻𝑁𝐾))
255natrcl 16816 . . . . . . . . . 10 (𝑇 ∈ (𝐻𝑁𝐾) → (𝐻 ∈ (𝐶 Func 𝐷) ∧ 𝐾 ∈ (𝐶 Func 𝐷)))
2624, 25syl 17 . . . . . . . . 9 (𝜑 → (𝐻 ∈ (𝐶 Func 𝐷) ∧ 𝐾 ∈ (𝐶 Func 𝐷)))
2726simpld 476 . . . . . . . 8 (𝜑𝐻 ∈ (𝐶 Func 𝐷))
28 1st2ndbr 7365 . . . . . . . 8 ((Rel (𝐶 Func 𝐷) ∧ 𝐻 ∈ (𝐶 Func 𝐷)) → (1st𝐻)(𝐶 Func 𝐷)(2nd𝐻))
2914, 27, 28sylancr 567 . . . . . . 7 (𝜑 → (1st𝐻)(𝐶 Func 𝐷)(2nd𝐻))
3013, 1, 29funcf1 16732 . . . . . 6 (𝜑 → (1st𝐻):(Base‘𝐶)⟶(Base‘𝐷))
3130ffvelrnda 6502 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((1st𝐻)‘𝑥) ∈ (Base‘𝐷))
325, 4nat1st2nd 16817 . . . . . . 7 (𝜑𝑅 ∈ (⟨(1st𝐹), (2nd𝐹)⟩𝑁⟨(1st𝐺), (2nd𝐺)⟩))
3332adantr 466 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑅 ∈ (⟨(1st𝐹), (2nd𝐹)⟩𝑁⟨(1st𝐺), (2nd𝐺)⟩))
34 simpr 471 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑥 ∈ (Base‘𝐶))
355, 33, 13, 2, 34natcl 16819 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → (𝑅𝑥) ∈ (((1st𝐹)‘𝑥)(Hom ‘𝐷)((1st𝐺)‘𝑥)))
36 fucass.s . . . . . . . 8 (𝜑𝑆 ∈ (𝐺𝑁𝐻))
375, 36nat1st2nd 16817 . . . . . . 7 (𝜑𝑆 ∈ (⟨(1st𝐺), (2nd𝐺)⟩𝑁⟨(1st𝐻), (2nd𝐻)⟩))
3837adantr 466 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑆 ∈ (⟨(1st𝐺), (2nd𝐺)⟩𝑁⟨(1st𝐻), (2nd𝐻)⟩))
395, 38, 13, 2, 34natcl 16819 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → (𝑆𝑥) ∈ (((1st𝐺)‘𝑥)(Hom ‘𝐷)((1st𝐻)‘𝑥)))
4026simprd 477 . . . . . . . 8 (𝜑𝐾 ∈ (𝐶 Func 𝐷))
41 1st2ndbr 7365 . . . . . . . 8 ((Rel (𝐶 Func 𝐷) ∧ 𝐾 ∈ (𝐶 Func 𝐷)) → (1st𝐾)(𝐶 Func 𝐷)(2nd𝐾))
4214, 40, 41sylancr 567 . . . . . . 7 (𝜑 → (1st𝐾)(𝐶 Func 𝐷)(2nd𝐾))
4313, 1, 42funcf1 16732 . . . . . 6 (𝜑 → (1st𝐾):(Base‘𝐶)⟶(Base‘𝐷))
4443ffvelrnda 6502 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((1st𝐾)‘𝑥) ∈ (Base‘𝐷))
455, 24nat1st2nd 16817 . . . . . . 7 (𝜑𝑇 ∈ (⟨(1st𝐻), (2nd𝐻)⟩𝑁⟨(1st𝐾), (2nd𝐾)⟩))
4645adantr 466 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑇 ∈ (⟨(1st𝐻), (2nd𝐻)⟩𝑁⟨(1st𝐾), (2nd𝐾)⟩))
475, 46, 13, 2, 34natcl 16819 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → (𝑇𝑥) ∈ (((1st𝐻)‘𝑥)(Hom ‘𝐷)((1st𝐾)‘𝑥)))
481, 2, 3, 12, 18, 23, 31, 35, 39, 44, 47catass 16553 . . . 4 ((𝜑𝑥 ∈ (Base‘𝐶)) → (((𝑇𝑥)(⟨((1st𝐺)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑆𝑥))(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑅𝑥)) = ((𝑇𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))((𝑆𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐻)‘𝑥))(𝑅𝑥))))
49 fucass.q . . . . . 6 𝑄 = (𝐶 FuncCat 𝐷)
50 fucass.x . . . . . 6 = (comp‘𝑄)
5136adantr 466 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑆 ∈ (𝐺𝑁𝐻))
5224adantr 466 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑇 ∈ (𝐻𝑁𝐾))
5349, 5, 13, 3, 50, 51, 52, 34fuccoval 16829 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)‘𝑥) = ((𝑇𝑥)(⟨((1st𝐺)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑆𝑥)))
5453oveq1d 6807 . . . 4 ((𝜑𝑥 ∈ (Base‘𝐶)) → (((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)‘𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑅𝑥)) = (((𝑇𝑥)(⟨((1st𝐺)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑆𝑥))(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑅𝑥)))
554adantr 466 . . . . . 6 ((𝜑𝑥 ∈ (Base‘𝐶)) → 𝑅 ∈ (𝐹𝑁𝐺))
5649, 5, 13, 3, 50, 55, 51, 34fuccoval 16829 . . . . 5 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((𝑆(⟨𝐹, 𝐺 𝐻)𝑅)‘𝑥) = ((𝑆𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐻)‘𝑥))(𝑅𝑥)))
5756oveq2d 6808 . . . 4 ((𝜑𝑥 ∈ (Base‘𝐶)) → ((𝑇𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))((𝑆(⟨𝐹, 𝐺 𝐻)𝑅)‘𝑥)) = ((𝑇𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))((𝑆𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐻)‘𝑥))(𝑅𝑥))))
5848, 54, 573eqtr4d 2814 . . 3 ((𝜑𝑥 ∈ (Base‘𝐶)) → (((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)‘𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑅𝑥)) = ((𝑇𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))((𝑆(⟨𝐹, 𝐺 𝐻)𝑅)‘𝑥)))
5958mpteq2dva 4876 . 2 (𝜑 → (𝑥 ∈ (Base‘𝐶) ↦ (((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)‘𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑅𝑥))) = (𝑥 ∈ (Base‘𝐶) ↦ ((𝑇𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))((𝑆(⟨𝐹, 𝐺 𝐻)𝑅)‘𝑥))))
6049, 5, 50, 36, 24fuccocl 16830 . . 3 (𝜑 → (𝑇(⟨𝐺, 𝐻 𝐾)𝑆) ∈ (𝐺𝑁𝐾))
6149, 5, 13, 3, 50, 4, 60fucco 16828 . 2 (𝜑 → ((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)(⟨𝐹, 𝐺 𝐾)𝑅) = (𝑥 ∈ (Base‘𝐶) ↦ (((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)‘𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐺)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))(𝑅𝑥))))
6249, 5, 50, 4, 36fuccocl 16830 . . 3 (𝜑 → (𝑆(⟨𝐹, 𝐺 𝐻)𝑅) ∈ (𝐹𝑁𝐻))
6349, 5, 13, 3, 50, 62, 24fucco 16828 . 2 (𝜑 → (𝑇(⟨𝐹, 𝐻 𝐾)(𝑆(⟨𝐹, 𝐺 𝐻)𝑅)) = (𝑥 ∈ (Base‘𝐶) ↦ ((𝑇𝑥)(⟨((1st𝐹)‘𝑥), ((1st𝐻)‘𝑥)⟩(comp‘𝐷)((1st𝐾)‘𝑥))((𝑆(⟨𝐹, 𝐺 𝐻)𝑅)‘𝑥))))
6459, 61, 633eqtr4d 2814 1 (𝜑 → ((𝑇(⟨𝐺, 𝐻 𝐾)𝑆)(⟨𝐹, 𝐺 𝐾)𝑅) = (𝑇(⟨𝐹, 𝐻 𝐾)(𝑆(⟨𝐹, 𝐺 𝐻)𝑅)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 382 = wceq 1630 ∈ wcel 2144 ⟨cop 4320 class class class wbr 4784 ↦ cmpt 4861 Rel wrel 5254 ‘cfv 6031 (class class class)co 6792 1st c1st 7312 2nd c2nd 7313 Basecbs 16063 Hom chom 16159 compcco 16160 Catccat 16531 Func cfunc 16720 Nat cnat 16807 FuncCat cfuc 16808 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1869 ax-4 1884 ax-5 1990 ax-6 2056 ax-7 2092 ax-8 2146 ax-9 2153 ax-10 2173 ax-11 2189 ax-12 2202 ax-13 2407 ax-ext 2750 ax-rep 4902 ax-sep 4912 ax-nul 4920 ax-pow 4971 ax-pr 5034 ax-un 7095 ax-cnex 10193 ax-resscn 10194 ax-1cn 10195 ax-icn 10196 ax-addcl 10197 ax-addrcl 10198 ax-mulcl 10199 ax-mulrcl 10200 ax-mulcom 10201 ax-addass 10202 ax-mulass 10203 ax-distr 10204 ax-i2m1 10205 ax-1ne0 10206 ax-1rid 10207 ax-rnegex 10208 ax-rrecex 10209 ax-cnre 10210 ax-pre-lttri 10211 ax-pre-lttrn 10212 ax-pre-ltadd 10213 ax-pre-mulgt0 10214 This theorem depends on definitions: df-bi 197 df-an 383 df-or 827 df-3or 1071 df-3an 1072 df-tru 1633 df-ex 1852 df-nf 1857 df-sb 2049 df-eu 2621 df-mo 2622 df-clab 2757 df-cleq 2763 df-clel 2766 df-nfc 2901 df-ne 2943 df-nel 3046 df-ral 3065 df-rex 3066 df-reu 3067 df-rab 3069 df-v 3351 df-sbc 3586 df-csb 3681 df-dif 3724 df-un 3726 df-in 3728 df-ss 3735 df-pss 3737 df-nul 4062 df-if 4224 df-pw 4297 df-sn 4315 df-pr 4317 df-tp 4319 df-op 4321 df-uni 4573 df-int 4610 df-iun 4654 df-br 4785 df-opab 4845 df-mpt 4862 df-tr 4885 df-id 5157 df-eprel 5162 df-po 5170 df-so 5171 df-fr 5208 df-we 5210 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-pred 5823 df-ord 5869 df-on 5870 df-lim 5871 df-suc 5872 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-riota 6753 df-ov 6795 df-oprab 6796 df-mpt2 6797 df-om 7212 df-1st 7314 df-2nd 7315 df-wrecs 7558 df-recs 7620 df-rdg 7658 df-1o 7712 df-oadd 7716 df-er 7895 df-map 8010 df-ixp 8062 df-en 8109 df-dom 8110 df-sdom 8111 df-fin 8112 df-pnf 10277 df-mnf 10278 df-xr 10279 df-ltxr 10280 df-le 10281 df-sub 10469 df-neg 10470 df-nn 11222 df-2 11280 df-3 11281 df-4 11282 df-5 11283 df-6 11284 df-7 11285 df-8 11286 df-9 11287 df-n0 11494 df-z 11579 df-dec 11695 df-uz 11888 df-fz 12533 df-struct 16065 df-ndx 16066 df-slot 16067 df-base 16069 df-hom 16173 df-cco 16174 df-cat 16535 df-func 16724 df-nat 16809 df-fuc 16810 This theorem is referenced by: fuccatid 16835
Copyright terms: Public domain W3C validator | 6,316 | 8,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-33 | latest | en | 0.225311 |
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# 12V car battery indicator Answered
Hi, I have been trying to design a circuit to monitor my 12V leisure car battery, I'm thinking this should be fairly simple, but its been years since I did any real electronics.
So the basic idea: Use the circular bargraph to indicate the level of the battery when a momentary push button is pressed
the circular LED bar graph: https://www.sparkfun.com/products/11492
So the range of the battery would be something like 10.5V to 13.5V, there are 32 LEDs in pairs, so essentially 16.
1. So if we take the first LED pair, this should illuminate when less than 10.5V, so that should be easy, as long as there is some life in the battery it should illuminate
2. The next LED should only illuminate at the next step, 16 LEDs, 3V range 16/3= steps of 0.2V So at the following steps:
10.5, 10.7, 10.9, 11.1, 11.3, 11.5, 11.7, 11.9, 12.1, 12.3, 12.5, 12.7, 12.9, 13.1, 13.3, 13.5.
So for 10.7V I guess I should use resistor before each LED.
3. The LEDs should light up after a momentary push switch is triggered, I would like this to be graceful, so the ring lights up smoothly one LED at a time as the voltage increases, I guess capacitors between the resistors and LEDs?
This is the simplest way I can think of doing it, I have seen circuits which use a LM339 chip, but that can only monitor 4 voltages, perhaps 4 of these in series could do the job? http://www.electroschematics.com/7068/lm339-lm239-lm2901-datasheet/ I think this is the way to go rather than using inline resistor/ capacitor pairs?
So I am a clear novice when it comes to electronics (but I'm enthusiastic), so if anyone can help me with this I would be more than happy to put together an instructables how to, something similar has been done before: https://www.instructables.com/id/Car-battery-tester/
In that example only one bar is lit, I want the other bars to stay lit.
I can use Eagle to make the PCB diagram and get it printed, but I need to figure out what components I need and how to do it!
Thanks for any help.
Gary
Tags:
## Discussions
Get the original 3914 datasheet: there is a worked example with cascaded devices.
Ok, so with extra searching I have come across this http://www.electro-tech-online.com/electronic-projects-design-ideas-reviews/108340-lm3914-voltmeter.html
So I can use a LM3914 for 10 LEDs, but I have 16, so two of these chips?
And put them in Bar Display mode.
I'm working through this tutorial: http://www.youtube.com/watch?v=iIKGvHjDQHs
This is much easier to do with a processor like the Arduino. You can make the scale suit precisely what you want - its impossible to make it unequal with a 3914
Thanks, I only want equal steps of 0.2V so unequal scales are not required so the 3914 should be best suited once the reference voltage is adjusted
https://www.instructables.com/files/deriv/FUZ/U2JX/HIPISGYE/FUZU2JXHIPISGYE.THUMB.jpg
Attached is a partial circuit, I got stuck.
So I've gone with the LM3914 IC set to Bar mode (pin 9)
On the wiring diagram I saw the LEDs were connected cathode to anode from the IC, with this ring LED setup they share 4 common cathodes, so how do I get around this?
I am returning to electronics after 15 years and a very bad teacher, so I'm learning as I go, any obvious problems?
The capacitors are going to be used to create a fade out effect, I guess this is where they should go?
You will notice I have only concentrated on the right side LEDS, I will use a second IC for them once I have this side working.
Thanks for any input
Gary
Oh yes, here is a resistor/capacitor/transistor solution indicating the fading effect: http://www.pcbheaven.com/circuitpages/LED_Fade_In_Fade_Out_Dimmer/ | 996 | 3,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-43 | longest | en | 0.93621 |
https://www.physicsforums.com/threads/an-olympic-long-jumper-question.1015131/ | 1,718,804,583,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00622.warc.gz | 817,370,828 | 18,206 | An olympic long jumper question
• Purp1eM0nsta
In summary: This would give you: It's a bit hard to read the handwritten algebra (@Purp1eM0nsta , images are for diagrams and textbook extracts; please type algebra into the posts), but it looks to me that the penultimate line finishes with ##\frac 12 a_xt^2##, and a squiggle in the final line could be a zero substituted for the ##a_x##. This would give you:##\frac {1}{9} \left( {a_xt^2} \right)##
Purp1eM0nsta
Homework Statement
Long jumping
Relevant Equations
https://ibb.co/5GHjb2c
Trying to answer this question I made and turn it into a slideshow presentation on how to solve it
I need to make a presentation on "the physics of long jumping" and one part I am struggling with is showing how to answer this question, I am struggling because i forget the important knowledge easily, and looking back at notes I've made confuses me even more, if i could have someone help me by taking a look at how I originally solved it, and explaining how i did actually do it? I am so sorry i just have a lot of learning struggles and i was paired with one of the most off-hands do it yourself teachers ever, so this has NOT been a good year for me :(
An olympic long jumper, initially going at a speed of 9 meters per second jumps with a take-off angle of 25 degrees.
Find:
-The amount of time the jumper stays in the air
-The horizontal distance the long jumper went
My work that I don't understand anymore/cant figure out how to convert into a slideshow presentation: https://ibb.co/5GHjb2c
if you need more context I'd be more than willing to give you chat logs i had with another tutor online, please help!
[Mentor Note -- Image pasted from external link into the thread]
Purp1eM0nsta said:
Homework Statement:: Long jumping
Relevant Equations:: https://ibb.co/5GHjb2c
An olympic long jumper, initially going at a speed of 9 meters per second
Why do you show a different number for ##V_{ix}## in the image of your work?
You say that the acceleration in the x-direction is zero and that is correct. However, the expression you have for x as a function of time does not reflect that observation and is incorrect.
kuruman said:
You say that the acceleration in the x-direction is zero and that is correct. However, the expression you have for x as a function of time does not reflect that observation and is incorrect.
It's a bit hard to read the handwritten algebra (@Purp1eM0nsta , images are for diagrams and textbook extracts; please type algebra into the posts), but it looks to me that the penultimate line finishes with ##\frac 12 a_xt^2##, and a squiggle in the final line could be a zero substituted for the ##a_x##.
What is the world record for the longest long jump in the Olympics?
The current world record for the longest long jump in the Olympics is held by Mike Powell, with a distance of 8.95 meters (29 feet 4 1/4 inches). This record was set in 1991 and still stands today.
What are the key techniques used by Olympic long jumpers?
Some key techniques used by Olympic long jumpers include the approach run, takeoff, flight, and landing. The approach run involves building up speed and momentum before the takeoff. The takeoff involves jumping off one foot and extending the body forward. During the flight, the athlete must maintain proper body position and control. The landing requires proper technique to avoid fouling and to maximize the distance of the jump.
How is the winner determined in Olympic long jump?
The winner in Olympic long jump is determined by the distance of their best jump. Each athlete has a total of six attempts, and the longest jump out of those six is their final score. If multiple athletes have the same distance, the winner is determined by the second-longest jump, and so on.
What are the common mistakes made by Olympic long jumpers?
Some common mistakes made by Olympic long jumpers include poor approach run technique, improper takeoff and flight technique, and faulty landing technique. Other mistakes include not properly pacing their jumps, not adjusting for wind or other environmental factors, and not maintaining proper body control throughout the jump.
How do Olympic long jumpers train for the event?
Olympic long jumpers typically train by practicing the key techniques mentioned earlier, including the approach run, takeoff, flight, and landing. They also focus on building strength, speed, and explosiveness through weight training and plyometric exercises. Endurance training is also important for maintaining proper form throughout the jump. Mental training and visualization techniques are also commonly used to prepare for the pressure of competition.
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# Title: Map between nodes and node numbers in a binary tree in C#
For more information about trees and other algorithmic topics, see my book Essential Algorithms: A Practical Approach to Computer Algorithms.
Suppose you have a binary tree with the nodes numbered as shown on the right. Note that it may have some missing nodes. For example, the binary tree shown here has nodes 10 and 13 but no other nodes on that level.
This example shows how you can map between node objects and numbers in the binary tree. In other words, if you have a reference to a particular node, you should be able to climb over the tree to find its number. Conversely if you know a node's number, you should be able to quickly climb through the tree to find the corresponding node.
If you move the mouse over a node, the program displays the node's label and its node number (which are the same in the sample binary tree) in the status message at the bottom of the form.
If you open the Data menu and select Find Node, a dialog appears. If you enter a node number and click OK, the program highlights the corresponding node.
Before I start explaining how to do this, you need to know how the tree is organized. Basically the TreeNode class represents a node in the binary tree. Each node has three key properties that refer to other TreeNode objects. Those properties are LeftChild, RightChild, and Parent. You can traverse the tree by using those properties.
# Finding a Node's Number
Given a node, how do you find its number? This is actually pretty straightforward if you notice one key fact about the node numbering: If a node has number N, then its children have numbers 2 × N + 1 and 2 × N + 2.
The following NodeNumber method uses that fact to find a node's number.
// Return this node's number. public int NodeNumber() { // If we're the root node, return 0. if (Parent == null) return 0; // We are not the root node. // See if we are the left or right child. if (this == Parent.LeftChild) return Parent.NodeNumber() * 2 + 1; else return Parent.NodeNumber() * 2 + 2; }
If this node has no parent, then it is the binary tree's root node so the method returns 1.
If this node has a parent, the code figures out whether this node is the parent's left or right child. It then calls the parent's NodeNumber method to get the parent's node number and then uses that to figure out this node's number.
# Finding a Number's Node
Given a node number, how do you find the corresponding node? This is more confusing than the other case. The key here is to notice that the bits in the number tell you which branches to take while traversing the binary tree.
First look at the number and, if it is 0, return the root node.
Next add one to the number and then consider the result's binary representation. Start at the root node and start at the second-to-leftmost bit. At each step, if the next bit you are considering is 0, move down the left branch. If the next bit is 1, move down the right branch.
For example, suppose you want to find node 10. Adding 1 to the number gives 11, which is 1011 in binary. Start at the root node and with the second bit from the left, which is a 0. That 0 tells us to move down the left branch to node number 1.
Now consider the next bit, which is the third from the left in 1011. That bit is a 1 so we move down the right branch to node 4.
The next bit, is the fourth from the left in 1011, is another 1 so we move down the right branch again to node 10.
At this point we have run out of bits so we have found the target node.
As you traverse the tree, if you ever follow a branch that doesn't lead to a node, then the target node isn't in the tree.
For another example, suppose you want to find node 13. Adding 1 to the number gives 14, which is 1110 in binary. Start at the root node. The second bit from the left is a 1 so we move down the right branch to node 2.
The third bit from the left is another 1 so we move down the right branch again to node 6.
The fourth bit is from the left is a 0 so we move down the left branch to node 13.
We're out of bits so we have found the desired node.
For a last example, suppose you want to find node 4. Adding 1 to the number gives 5, which is 101 in binary. Start at the root node. The second bit from the left is a 0 so we move down the left branch to node 1.
The third bit from the left is a 1 so we move down the right branch to node 4.
Again we're out of bits so we have found the desired node.
The following code shows how the program implements this algorithm.
// Find the indicated node in this subtree. public TreeNode<T> FindNodeByNumber(int number) { // If this is the root node, return it. if (number <= 0) return this; // Add 1 to the number so the numbers at each level // in the tree has the same number of bits. number++; // Find the least significant bit. int msb = FindMsb(number); // Climb down through the tree. TreeNode<T> node = this; for (int i = msb - 1; i >= 0; i--) { // Use this bit to decide which way to go. if ((number & (1 << i)) == 0) node = node.LeftChild; else node = node.RightChild; // See if we fell off the tree. if (node == null) return null; } // Return the current node. return node; } // Return the number's least significant bit. private int FindMsb(int number) { for (int i = 31; i >= 0; i--) if ((number & (1 << i)) != 0) return i; return -1; }
The FindNodeByNumber method searches the tree as described above. The FindMsb helper method searches a number from left-to-right to find its most significant digit and returns that digit's position in the number.
This program does a lot of other things besides traverse the tree. In particular, it lets you add and remove nodes and it displays the resulting tree. The program is based on the example Handle generic TreeNode mouse events in C# except it has been modified to work with binary trees instead of trees with arbitrary degree. See that example for information about how those other features work. | 1,443 | 6,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-18 | latest | en | 0.881433 |
https://www.popflock.com/learn?s=Symmetrical_components | 1,623,915,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629632.54/warc/CC-MAIN-20210617072023-20210617102023-00397.warc.gz | 825,177,627 | 22,946 | Symmetrical Components
Get Symmetrical Components essential facts below. View Videos or join the Symmetrical Components discussion. Add Symmetrical Components to your PopFlock.com topic list for future reference or share this resource on social media.
Symmetrical Components
In electrical engineering, the method of symmetrical components simplifies analysis of unbalanced three-phase power systems under both normal and abnormal conditions. The basic idea is that an asymmetrical set of N phasors can be expressed as a linear combination of N symmetrical sets of phasors by means of a complex linear transformation.[1] Fortescue's theorem (symmetrical components) is based on superposition principle,[2] so it is applicable to linear power systems only, or to linear approximations of non-linear power systems.
In the most common case of three-phase systems, the resulting "symmetrical" components are referred to as direct (or positive), inverse (or negative) and zero (or homopolar). The analysis of power system is much simpler in the domain of symmetrical components, because the resulting equations are mutually linearly independent if the circuit itself is balanced.[]
## Description
Set of three unbalanced phasors, and the necessary symmetrical components that sum up to the resulting plot at the bottom.
In 1918 Charles Legeyt Fortescue presented a paper[3] which demonstrated that any set of N unbalanced phasors (that is, any such polyphase signal) could be expressed as the sum of N symmetrical sets of balanced phasors, for values of N that are prime. Only a single frequency component is represented by the phasors.
In 1943 Edith Clarke published a textbook giving a method of use of symmetrical components for three-phase systems that greatly simplified calculations over the original Fortescue paper.[4] In a three-phase system, one set of phasors has the same phase sequence as the system under study (positive sequence; say ABC), the second set has the reverse phase sequence (negative sequence; ACB), and in the third set the phasors A, B and C are in phase with each other (zero sequence, the common-mode signal). Essentially, this method converts three unbalanced phases into three independent sources, which makes asymmetric fault analysis more tractable.
By expanding a one-line diagram to show the positive sequence, negative sequence, and zero sequence impedances of generators, transformers and other devices including overhead lines and cables, analysis of such unbalanced conditions as a single line to ground short-circuit fault is greatly simplified. The technique can also be extended to higher order phase systems.
Physically, in a three phase system, a positive sequence set of currents produces a normal rotating field, a negative sequence set produces a field with the opposite rotation, and the zero sequence set produces a field that oscillates but does not rotate between phase windings. Since these effects can be detected physically with sequence filters, the mathematical tool became the basis for the design of protective relays, which used negative-sequence voltages and currents as a reliable indicator of fault conditions. Such relays may be used to trip circuit breakers or take other steps to protect electrical systems.
The analytical technique was adopted and advanced by engineers at General Electric and Westinghouse, and after World War II it became an accepted method for asymmetric fault analysis.
As shown in the figure to the above right, the three sets of symmetrical components (positive, negative, and zero sequence) add up to create the system of three unbalanced phases as pictured in the bottom of the diagram. The imbalance between phases arises because of the difference in magnitude and phase shift between the sets of vectors. Notice that the colors (red, blue, and yellow) of the separate sequence vectors correspond to three different phases (A, B, and C, for example). To arrive at the final plot, the sum of vectors of each phase is calculated. This resulting vector is the effective phasor representation of that particular phase. This process, repeated, produces the phasor for each of the three phases.
## The three-phase case
Symmetrical components are most commonly used for analysis of three-phase electrical power systems. The voltage or current of a three-phase system at some point can be indicated by three phasors, called the three components of the voltage or the current.
This article discusses voltage; however, the same considerations also apply to current. In a perfectly balanced three-phase power system, the voltage phasor components have equal magnitudes but are 120 degrees apart. In an unbalanced system, the magnitudes and phases of the voltage phasor components are different.
Decomposing the voltage phasor components into a set of symmetrical components helps analyze the system as well as visualize any imbalances. If the three voltage components are expressed as phasors (which are complex numbers), a complex vector can be formed in which the three phase components are the components of the vector. A vector for three phase voltage components can be written as
${\displaystyle \mathbf {v} _{abc}={\begin{bmatrix}V_{a}\\V_{b}\\V_{c}\end{bmatrix}}}$
and decomposing the vector into three symmetrical components gives
${\displaystyle {\begin{bmatrix}V_{a}\\V_{b}\\V_{c}\end{bmatrix}}={\begin{bmatrix}V_{a,0}\\V_{b,0}\\V_{c,0}\end{bmatrix}}+{\begin{bmatrix}V_{a,1}\\V_{b,1}\\V_{c,1}\end{bmatrix}}+{\begin{bmatrix}V_{a,2}\\V_{b,2}\\V_{c,2}\end{bmatrix}}}$
where the subscripts 0, 1, and 2 refer respectively to the zero, positive, and negative sequence components. The sequence components differ only by their phase angles, which are symmetrical and so are ${\displaystyle \scriptstyle {\frac {2}{3}}\pi }$ radians or 120°.
### A matrix
Define a phasor rotation operator ${\displaystyle \alpha }$, which rotates a phasor vector counterclockwise by 120 degrees when multiplied by it:
${\displaystyle \alpha \equiv e^{{\frac {2}{3}}\pi i}}$.
Note that ${\displaystyle \alpha ^{3}=1}$ so that ${\displaystyle \alpha ^{-1}=\alpha ^{2}}$.
The zero sequence components have equal magnitude and are in phase with each other, therefore:
${\displaystyle V_{0}\equiv V_{a,0}=V_{b,0}=V_{c,0}}$,
and the other phase sequences have the same magnitude, but their phases differ by 120°:
{\displaystyle {\begin{aligned}V_{1}&\equiv V_{a,1}=\alpha V_{b,1}=\alpha ^{2}V_{c,1}\end{aligned}}},
{\displaystyle {\begin{aligned}V_{2}&\equiv V_{a,2}=\alpha ^{2}V_{b,2}=\alpha V_{c,2}\end{aligned}}},
meaning that
{\displaystyle {\begin{aligned}V_{b,1}=\alpha ^{2}V_{1}\end{aligned}}},
{\displaystyle {\begin{aligned}V_{c,1}=\alpha V_{1}\end{aligned}}},
{\displaystyle {\begin{aligned}V_{b,2}=\alpha V_{2}\end{aligned}}},
{\displaystyle {\begin{aligned}V_{c,2}=\alpha ^{2}V_{2}\end{aligned}}}.
Thus,
{\displaystyle {\begin{aligned}\mathbf {v} _{abc}&={\begin{bmatrix}V_{0}\\V_{0}\\V_{0}\end{bmatrix}}+{\begin{bmatrix}V_{1}\\\alpha ^{2}V_{1}\\\alpha V_{1}\end{bmatrix}}+{\begin{bmatrix}V_{2}\\\alpha V_{2}\\\alpha ^{2}V_{2}\end{bmatrix}}\\&={\begin{bmatrix}1&1&1\\1&\alpha ^{2}&\alpha \\1&\alpha &\alpha ^{2}\end{bmatrix}}{\begin{bmatrix}V_{0}\\V_{1}\\V_{2}\end{bmatrix}}\\&={\textbf {A}}\mathbf {v} _{012}\end{aligned}}}
where
${\displaystyle \mathbf {v} _{012}={\begin{bmatrix}V_{0}\\V_{1}\\V_{2}\end{bmatrix}},{\textbf {A}}={\begin{bmatrix}1&1&1\\1&\alpha ^{2}&\alpha \\1&\alpha &\alpha ^{2}\end{bmatrix}}}$
In reverse phase rotation systems, the following matrix can be similarly derived
${\displaystyle {\textbf {Aacb}}={\begin{bmatrix}1&1&1\\1&\alpha &\alpha ^{2}\\1&\alpha ^{2}&\alpha \end{bmatrix}}}$
### Decomposition
The sequence components are derived from the analysis equation
${\displaystyle \mathbf {v} _{012}={\textbf {A}}^{-1}\mathbf {v} _{abc}}$
where
${\displaystyle {\textbf {A}}^{-1}={\frac {1}{3}}{\begin{bmatrix}1&1&1\\1&\alpha &\alpha ^{2}\\1&\alpha ^{2}&\alpha \end{bmatrix}}}$
The above two equations tell how to derive symmetrical components corresponding to an asymmetrical set of three phasors:
• Sequence 0 is one-third the sum of the original three phasors.
• Sequence 1 is one-third the sum of the original three phasors rotated counterclockwise 0°, 120°, and 240°.
• Sequence 2 is one-third the sum of the original three phasors rotated counterclockwise 0°, 240°, and 120°.
Visually, if the original components are symmetrical, sequences 0 and 2 will each form a triangle, summing to zero, and sequence 1 components will sum to a straight line.
### Intuition
Napoleon's theorem: If the triangles centered on L, M, and N are equilateral, then so is the green triangle.
The phasors ${\displaystyle \scriptstyle V_{(ab)}=V_{(a)}-V_{(b)};\;V_{(bc)}=V_{(b)}-V_{(c)};\;V_{(ca)}=V_{(c)}-V_{(a)}}$ form a closed triangle (e.g., outer voltages or line to line voltages). To find the synchronous and inverse components of the phases, take any side of the outer triangle and draw the two possible equilateral triangles sharing the selected side as base. These two equilateral triangles represent a synchronous and an inverse system.
If the phasors V were a perfectly synchronous system, the vertex of the outer triangle not on the base line would be at the same position as the corresponding vertex of the equilateral triangle representing the synchronous system. Any amount of inverse component would mean a deviation from this position. The deviation is exactly 3 times the inverse phase component.
The synchronous component is in the same manner 3 times the deviation from the "inverse equilateral triangle". The directions of these components are correct for the relevant phase. It seems counter intuitive that this works for all three phases regardless of the side chosen but that is the beauty of this illustration. The graphic is from Napoleon's Theorem, which matches a graphical calculation technique that sometimes appears in older references books.[5]
## Poly-phase case
It can be seen that the transformation matrix above is a discrete Fourier transform, and as such, symmetrical components can be calculated for any poly-phase system.
## Contribution of harmonics to symmetrical components in 3-phase power systems
Harmonics often occur in power systems as a consequence of non-linear loads. Each order of harmonics contributes to different sequence components. The fundamental and harmonics of order ${\displaystyle \scriptstyle 1+3n}$ will contribute to the positive sequence component. Harmonics of order ${\displaystyle \scriptstyle 2+3n}$ will contribute to the negative sequence. Harmonics of order ${\displaystyle \scriptstyle 3+3n}$ contribute to the zero sequence.
Note that the rules above are only applicable if the phase values (or distortion) in each phase are exactly the same. Please further note that even harmonics are not common in power systems.
## Consequence of the zero sequence component in power systems
The zero sequence represents the component of the unbalanced phasors that is equal in magnitude and phase. Because they are in phase, zero sequence currents flowing through an n-phase network will sum to n times the magnitude of the individual zero sequence currents components. Under normal operating conditions this sum is small enough to be negligible. However, during large zero sequence events such as lightning strikes, this nonzero sum of currents can lead to a larger current flowing through the neutral conductor than the individual phase conductors. Because neutral conductors are typically not larger than individual phase conductors, and are often smaller than these conductors, a large zero sequence component can lead to overheating of neutral conductors and to fires.
One way to prevent large zero sequence currents is to use a delta connection, which appears as an open circuit to zero sequence currents. For this reason, most transmission, and much sub-transmission is implemented using delta. Much distribution is also implemented using delta, although "old work" distribution systems have occasionally been "wyed-up" (converted from delta to wye) so as to increase the line's capacity at a low converted cost, but at the expense of a higher central station protective relay cost.
## References
Notes
1. ^ Hadjsaïd, Nouredine; Sabonnadière, Jean-Claude (2013). Power Systems and Restructuring. John Wiley & Sons. p. 244. ISBN 9781118599921.
2. ^ Mathis, Wolfgang; Pauli, Rainer. "Network Theorems". Wiley Online Library. doi:10.1002/047134608X.W2507. [...] the results of Fortescue [...] are proven by the superposition theorem, and for this reason, a direct generalization to nonlinear networks is impossible.
3. ^ Charles L. Fortescue, "Method of Symmetrical Co-Ordinates Applied to the Solution of Polyphase Networks". Presented at the 34th annual convention of the AIEE (American Institute of Electrical Engineers) in Atlantic City, N.J. on 28 June 1918. Published in: AIEE Transactions, vol. 37, part II, pages 1027-1140 (1918). For a brief history of the early years of symmetrical component theory, see: J. Lewis Blackburn, Symmetrical Components for Power Engineering (Boca Raton, Florida: CRC Press, 1993), pages 3-4.
4. ^ Gabriele Kass-Simon, Patricia Farnes, Deborah Nash (ed), Women of Science: Righting the Record , Indiana University Press, 1993, ISBN 0253208130. pages 164-168
5. ^ Wagner, C. F.; Evans, R. D. (1933). Symmetrical Components. New York and London: McGraw Hill. p. 265.
Bibliography
• J. Lewis Blackburn Symmetrical Components for Power Systems Engineering, Marcel Dekker, New York (1993). ISBN 0-8247-8767-6
• William D. Stevenson, Jr. Elements of Power System Analysis Third Edition, McGraw-Hill, New York (1975). ISBN 0-07-061285-4.
• History article from IEEE on early development of symmetrical components, retrieved May 12, 2005.
• Westinghouse Corporation, Applied Protective Relaying, 1976, Westinghouse Corporation, no ISBN, Library of Congress card no. 76-8060 - a standard reference on electromechanical protective relays | 3,464 | 14,069 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-25 | latest | en | 0.919171 |
https://mathblog.com/book-review-e-the-story-of-a-number/ | 1,719,010,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00519.warc.gz | 336,355,569 | 21,117 | # Book Review: e: the Story of a Number
e: The Story of a Number
by Eli Maor
Princeton Science Library Series
Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540
Page Count: 248 pages
My Rating: 4/5
Introduction
e: The Story of a Number is a book about e (2.718281828459045…), sometimes known as Euler’s Number or Euler’s Constant after the great mathematician Leonhard Euler. e: The Story of a Number is an “accessible” math book, rather than a “popular” math book, that tries to teach an advanced topic (really first and second year calculus) to a general audience without requiring accompanying class lectures or even worked problems. This is a difficult undertaking, something we don’t really know how to do well at the calculus and beyond levels. The author, mathematician Eli Maor, tries to make the subject more engaging by including stories about famous mathematicians such as John Napier, usually identified as the discoverer of logarithms, and trying to avoid the dry, pedantic style of math textbooks which turns off so many students. He is partially successful but students who lack a solid grasp of the limit concept and its rigorous definition ( the epsilon $$\epsilon$$ and delta $$\delta$$ definition of limit attributed to Cauchy and Weierstrass ) may find key parts of the book hard to understand.
The Father of Logarithms
The book opens with a chapter on John Napier, generally credited with the invention of logarithms, truly a major breakthrough in mathematics and computation. I found this one of the stronger parts of the book. I had not realized that Napier’s “logarithms” were actually not logarithms in the modern sense. The base ten (10) logarithms that dominated practical calculation until the advent of electronic calculators in the 1970’s (Napier lived around 1600) are actually an improvement on Napier’s logarithms developed by a contemporary, Henry Briggs, and published in 1625.
The logarithm of a number is the power that a base, typically ten (10), e, or two (2), must be raised to to yield the original number. For example, the logarithm base ten (10) of one-hundred (100) is two (2) since ten squared is one-hundred. Logarithms using a base of ten (10) are known as “common logarithms” and logarithms that use e, the subject of the book, as the base are known as “natural logarithms.” Logarithms that use a base of two (2) are sometimes used in computer science because most computers use base two, binary arithmetic internally, even if results are reported using decimal, base ten (10), numbers.
An important property of logarithms is that the logarithm of the product of two numbers is the sum of the logarithms of each number in the sum.
$$\log ab = log(a) + log(b)$$
This and other convenient properties of the logarithm makes it possible to quickly multiply two numbers by adding their logarithms and then converting the sum back to the product of the original numbers using a precomputed table of logarithms or a mechanical slide rule. Tables of logarithms and mechanical slide rules became the mainstay of practical computation in science and engineering from the time of Napier until the 1970’s, much faster than the hand calculation with pen and paper that preceded Napier’s invention.
Many readers who have grown up with calculators and computers may have trouble appreciating the practical benefits of logarithms for over three centuries. I attended a dinner at the Udvar-Hazy Center, part of the Smithsonian National Air and Space Museum, at which one of the guests marveled that the nearly state of the art airplanes and rockets from the 1960’s and 1970’s hanging from the ceiling over head had been designed primarily with slide rules (and a few early computers comparable to 1980’s personal computers)! It is interesting to note how much was accomplished for three centuries with tables of logarithms and slide rules, and how limited progress in many fields such as aviation, rocketry, power and propulsion has been since modern digital computers replaced them.
Slide Rule Era Technology
Beckmann’s High Bar
The book is explicitly an attempt to do for e what Petr Beckmann’s now classic A History of $$\pi$$ does for the number $$\pi$$ (the ratio of the circumference of a circle to its diameter).
Beckmann’s book is a wonderful accessible discussion of $$\pi$$, that mostly avoids calculus until the last few chapters and makes careful limited use of calculus in those chapters. $$\pi$$ is defined in a visual geometric way and all of the ancient Greek work on $$\pi$$ and related topics is pure geoemtry that can be expressed visually and concretely with little abstraction.
I think it is probably possible to present $$e$$ and interrelated basic calculus concepts such as the limit in a visual, geometric way that is easy for students to understand, but Maor fails in a number of places to do this, tending to use the language and abstract symbols of math textbooks which often confuses and intimidates students.
The Elusive Limit
While $$\pi$$ can be defined in a purely geometric way without recourse to limits, $$e$$ is intimately associated with the limit concept:
$$e = \lim_{n \to +\infty} ( 1 + \frac{1}{n} )^n$$
This is hardly accessible to most students and it is actually hard to put the limit concept on a solid basis. The early mathematicians that Maor writes about such as Isaac Newton, Gottfried Leibniz, the Bernoulli brothers, Leonhard Euler and others all used hand-waving intuitive concepts of a limit that were not rigorous and sometimes yielded wrong results.
$$e$$ as the Limit of (1 + 1/N)^N
The book devotes a chapter to the limit concept: Chapter 4 – To the Limit, If It Exists. This has a good introductory verbal description of the limit for $$\frac{1}{n}$$, but that is about it.
When we say that a sequence of numbers $$a_1$$, $$a_2$$, $$a_3$$ . . . , $$a_n$$, tends to a limit $$L$$ as $$n$$ tends to infinity, we mean that as $$n$$ grows larger and larger, the terms of the sequence get closer and closer to the number $$L$$. Put in different words, we can make the difference (in absolute value) between $$a_n$$ and $$L$$ as small as we please by going out far enough in our sequence— that is, by choosing $$n$$ to be sufficiently large. Take, for example, the sequence 1, 1/ 2, 1/ 3, 1/ 4 whose general term is $$a_n$$ = 1/ $$n$$. As $$n$$ increases, the terms get closer and closer to 0. This means that the difference between 1/ $$n$$ and the limit 0 (that is , just 1/ $$n$$) can be made as small as we please if we choose $$n$$ large enough. Say that we want 1/ $$n$$ to be less than 1/ 1,000; all we need to do is make n greater than 1,000. If we want 1/ $$n$$ to be less than 1/ 1,000,000, we simply choose any n greater than 1,000,000. And so on. We express this situation by saying that 1/ $$n$$ tends to 0 as $$n$$ increases without bound, and we write 1/ $$n$$ → 0 as $$n$$ → ∞. We also use the abbreviated notation
$$\lim_{n \to +\infty} \frac{1}{n} = 0$$
Maor, Eli (2009-01-19). e: The Story of a Number (Princeton Science Library) (Kindle Locations 610-620). Princeton University Press – A. Kindle Edition.
This is actually a good explanation and some pre-calculus students may be able to use it to understand the rest of the book, but it is also brief and not entirely rigorous. It took a lot more for me to get the limit concept adequately when I first learned calculus. Specifically, my Advanced Placement BC Calculus course spent a full six weeks going over many simple examples of the epsilon $$\epsilon$$ and delta $$\delta$$ definition of the limit. I found both “accessible” math books that tried to teach calculus and formal high school and college calculus textbooks by themselves inadequate. Like e: the Story of a Number many accessible math books try to use an intuitive, non-rigorous definition of the limit which may at first seem clear to the student but quickly becomes problematic as the student tries to put the limit into practical and precise use. It was the extensive in-class and homework exercises working through the rigorous definition of the limit that proved essential to mastering the limit concept.
In calculus, both the derivative and integral (differentiation and integration) are defined as limits; it is impossible to understand or master calculus in its modern form without a solid understanding and mastery of the limit.
Conclusion
I really enjoyed e: the Story of a Number and learned a number of new things, both technically and about the history of mathematics, especially the chapter about John Napier and the first “logarithms.” I have however the advantage of having already taken Advanced Placement BC Calculus and four years of advanced calculus at Caltech and having a pretty good advanced understanding of the limit concept. Pre-calculus students will almost certainly find the book more difficult and may need to complement it with other sources to understand the use of the limit concept in the book, which is essential to understanding the material in the book.
Credits
The picture of the Apollo 11 launch (July 16, 1969) is from Wikimedia Commons and is in the public domain. Apollo 11 was the first successful landing of men on the Moon. | 2,125 | 9,256 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.938288 |
https://www.mathworks.com/examples/matlab/community/20270-test-e-reconstruction-from-gradients | 1,518,945,201,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811795.10/warc/CC-MAIN-20180218081112-20180218101112-00529.warc.gz | 949,723,432 | 13,915 | MATLAB Examples
# Test E: Reconstruction from Gradients
This script is supplemental material to the paper:
"An Algebraic Framework for the Real-Time Solution of Inverse Problems on Embedded Systems"
by: Matthew Harker, Christoph Gugg, Paul O'Leary and Gerhard Rath
This script require our Discrete orthogonal polynomial toolbox, it is available at:
http://www.mathworks.com/matlabcentral/fileexchange/41250
and the function to convert figures to eps:
http://www.mathworks.com/matlabcentral/fileexchange/42388
## Prepare to execute code.
close all clear all; % % Set some defaults % fontSize = 10; set(0,'DefaultFigureColor',[1,1,1]); set(0,'DefaultaxesFontName','Times'); set(0,'DefaultaxesFontSize',fontSize); set(0,'DefaulttextFontName','Times'); set(0,'DefaulttextFontSize',fontSize); set(0,'DefaultfigurePaperType','A4'); set(0,'DefaultTextInterpreter', 'latex'); % FigureSize=[1 1 10 6]; set(0,'DefaultFigureUnits','centimeters'); set(0,'DefaultFigurePosition',FigureSize); set(0,'DefaultFigurePaperUnits','centimeters'); set(0,'DefaultFigurePaperPosition',FigureSize); MyAxesPosition=[0.13 0.1 0.86 0.86]; set(0,'DefaultaxesPosition',MyAxesPosition); %
load BVPFitData; %
## Pre-calculations for the Reconstruction
generate the particular and homogeneous solutions.
ls = 11; % name = ['BVP',int2str(ls)]; % % generate a set of constrained basis functions. % noBfs = length(x) - 1; noBfs = 11; [yc, Bc, S] = dopGenConstrained( x, noBfs, T, ls, ls); % D = dopDiffLocal( x, ls, ls); % % define the linear differential aperator % L = D; % % Use the basis functions to generate a regularizing operator % M = Bc * pinv( L * Bc ); % % Compute the particular solution % yp = yc - M * L * yc; % % Compute the covariancde propagation % C = (M * M') ; % % The square root of the diagonal of the covariance matrix is the % unscaled standard deviation. % s = sqrt( diag( C ) ); %
Using a local approx for M
## Perofrm a Monte Carlo Simulation for the Reconstruction
noSims = 10000; % % Define a noise gain in terms of the maximum derivative, in this % example 1% of the maximum derivative. This corresponds to 1% noise % in the measurement signal. % noiseGain = 0.01; normDy = max( dy ); stdIn = noiseGain * normDy; % % Prepare stoorage for the solutions % sols = zeros( length(x), noSims ); G = zeros( length(x), noSims ); % % Run the Monte Sarlo Simulations % for k=1:noSims % % generate the perturbed forcing function % gn = stdIn * randn( length(x), 1); g = dy + gn; G(:,k) = g; % % Compute the solution % yrk = M * g + yp; sols(:,k) = yrk; end;
## Present the Results of the Monte Carlo Simulation
This fugure shows the analytical derivatives and the perturbations at each node.
FigureSize=[1 1 10 6]; set(0,'DefaultFigureUnits','centimeters'); set(0,'DefaultFigurePosition',FigureSize); set(0,'DefaultFigurePaperUnits','centimeters'); set(0,'DefaultFigurePaperPosition',FigureSize); MyAxesPosition=[0.15 0.15 0.81 0.8]; set(0,'DefaultaxesPosition',MyAxesPosition); % figGs = figure; errorbar( x, dy, stdIn * ones(size(x)), '-kx'); xlabel('$$x$$'); ylabel('Perturbed $$D y$$'); grid on; axis([0,1,-1,1]);
Compute the mean reconstruction and plot all the reconstruction points.
FigureSize=[1 1 10 6]; set(0,'DefaultFigureUnits','centimeters'); set(0,'DefaultFigurePosition',FigureSize); set(0,'DefaultFigurePaperUnits','centimeters'); set(0,'DefaultFigurePaperPosition',FigureSize); MyAxesPosition=[0.15 0.15 0.77 0.8]; set(0,'DefaultaxesPosition',MyAxesPosition); % % Compute the mean solution % sm = mean(sols')'; % % Scale the errors to make them visible. % scale = 10; % figRs = figure; plot( x, y, 'k-'); hold on; errorbar( x, sm, scale * stdIn*s, 'kx'); grid on; xlabel('$$x$$'); ylabel('Reconstructions'); axis([0,1,-0.15,0.25]); hold on; [n,m] = size(T); legend( 'Analitical', 'Reconstruction', 'Location', 'Southwest'); for k=1:n if T(k,1) == 0 % is a value constraint plot( T(k,2), T(k,3), 'ko', 'MarkerFaceColor', 'k'); end; end; %
Plot the Mean error of the reconstructions. this verifies if there is a systematic reconstruction error.
FigureSize=[1 1 10 6]; set(0,'DefaultFigureUnits','centimeters'); set(0,'DefaultFigurePosition',FigureSize); set(0,'DefaultFigurePaperUnits','centimeters'); set(0,'DefaultFigurePaperPosition',FigureSize); MyAxesPosition=[0.15 0.15 0.84 0.8]; set(0,'DefaultaxesPosition',MyAxesPosition); % scale = 1E5; % figME = figure; plot( x, scale*(y - sm), '-k.'); xlabel('$$x$$'); ylabel('$$\bar{\epsilon} \, \, \, (\times 10^5)$$'); grid on;
Plot the standard deviation of the reconstructions
Scale the standard deviation to make it visible
stdS = std( sols' )'; scale = 1E3; % figVer = figure; plot( x, scale * stdS, '-k.'); grid on; % xlabel('$$x$$'); ylabel('$$\mathrm{std}(\epsilon) \, \, \, (\times 10^3)$$'); hold on; plot( x, scale*stdIn*s, '-r.'); [n,m] = size(T); for k=1:n if T(k,1) == 0 % is a value constraint plot( T(k,2), 0, 'ko', 'MarkerFaceColor', 'k'); end; end; %
## Present the covariance matrix as an image.
figCov = figure; imagesc( C ); axis image; colorbar; xlabel('$$x$$'); ylabel('$$x$$'); %
figMStd = figure; subplot(2,1,1); plot( x, scale*(y - sm), '-k.'); ylabel('$$\bar{\epsilon} \, \, \, (\times 10^5)$$'); grid on; % hold on; [n,m] = size(T); for k=1:n if T(k,1) == 0 % is a value constraint plot( T(k,2), 0, 'ko', 'MarkerFaceColor', 'k'); end; end; % subplot(2,1,2); plot( x, scale * stdS, '-k.'); grid on; xlabel('$$x$$'); ylabel('$$\mathrm{std}(\epsilon) \, \, \, (\times 10^3)$$'); hold on; plot( x, scale*stdIn*s, 'ko'); [n,m] = size(T); for k=1:n if T(k,1) == 0 % is a value constraint plot( T(k,2), 0, 'ko', 'MarkerFaceColor', 'k'); end; end; axis([0,1,0,2]); legend( 'P','MC','location','northEast'); %
## Save the figures as .eps files for possible publication.
h = @figure2eps; hInfo = functions( h ); if isempty( hInfo.file ) disp( 'You will need to download the function figure2eps if you'); disp( 'wish to generate the eps files for the figures' ); disp( 'The function is available at:'); disp( 'http://www.mathworks.at/matlabcentral/fileexchange/42388'); else % figure2eps( figMStd, [name, 'BiasAndStd']); figure2eps( figGs, 'BVPGs'); figure2eps( figVer, [name, 'confid']); figure2eps( figRs, [name, 'Resonstrions']); figure2eps( figME, [name, 'meanE']); figure2eps( figCov, [name, 'CovL']); end; | 1,974 | 6,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-09 | latest | en | 0.399736 |
https://www.nrich.maths.org/5612 | 1,675,090,218,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00155.warc.gz | 934,275,651 | 5,335 | ### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### Sending Cards
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Dice and Spinner Numbers
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
##### Age 7 to 14Challenge Level
These printable resources may be useful: Methods in Multiplying Madness
If you had to work out $23 \times 21$ how would you do it?
What if you needed to work out $246 \times 34$?
Here are eight videos showing four different methods for working out the two multiplications. Can you make sense of them?
Grid Multiplication
$23 \times 21$
$246 \times 34$
Column Multiplication
$23 \times 21$
$246 \times 34$
Multiplying with Lines
$23 \times 21$
$246 \times 34$
Gelosia Multiplication
$23 \times 21$
$246 \times 34$
Once you have watched the videos, make up some multiplication calculations of your own and have a go at answering them using all four methods. Check that you get the same answer each time!
Here are some questions to consider:
Why does each method work?
What do the methods have in common? | 334 | 1,440 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-06 | latest | en | 0.910589 |
https://www.luogu.org/problemnew/show/P3420 | 1,555,628,141,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00511.warc.gz | 773,039,257 | 5,820 | # P3420 [POI2005]SKA-Piggy Banks
• 598通过
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• 评测方式 云端评测
• 标签 并查集 概率论,统计 连通块 POI 2005 高性能
• 难度 普及+/提高
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## 题目描述
Byteazar the Dragon has $N$ piggy banks. Each piggy bank can either be opened with its corresponding key or smashed. Byteazar has put the keys in some of the piggy banks - he remembers which key has been placed in which piggy bank. Byteazar intends to buy a car and needs to gain access to all of the piggy banks. However, he wants to destroy as few of them as possible. Help Byteazar to determine how many piggy banks have to be smashed.
reads from the standard input the number of piggy banks and the deployment of their corresponding keys,finds the minimal number of piggy banks to be smashed in order to gain access to all of them,writes the outcome to the standard output.
Byteazar the Dragon拥有N个小猪存钱罐。每一个存钱罐能够用相应的钥匙打开或者被砸开。Byteazar已经将钥匙放入到一些存钱罐中。现在已知每个钥匙所在的存钱罐，Byteazar想要买一辆小汽车，而且需要打开所有的存钱罐。然而，他想要破坏尽量少的存钱罐，帮助Byteazar去决策最少要破坏多少存钱罐。
## 任务：
写一段程序包括：
读入存钱罐的数量以及相应的钥匙的位置，求出能打开所有存钱罐的情况下，需要破坏的存钱罐的最少数量并将其输出。
## 输入输出格式
输入格式：
The first line of the standard input contains a single integer $N$ ($1\le N\le 1\ 000\ 000$) - this is the number of piggy banks owned by the dragon. The piggy banks (as well as their corresponding keys) are numbered from $1$ to $N$. Next, there are $N$ lines: the $(i+1)$'st line contains a single integer - the number of the piggy bank in which the $i$'th key has been placed.
第一行：包括一个整数N（1<=N<=1000000)，这是Byteazar the Dragon拥有的存钱罐的数量。
存钱罐（包括它们对应的钥匙）从1到N编号。
接下来有N行：第i+1行包括一个整数x，表示第i个存钱罐对应的钥匙放置在了第x个存钱罐中。
输出格式：
The first and only line of the standard output should contain a single integer - the minimal number of piggy banks to be smashed in order to gain access to all of the piggy banks.
仅一行：包括一个整数，表示能打开所有存钱罐的情况下，需要破坏的存钱罐的最少数量。
## 输入输出样例
输入样例#1： 复制
4
2
1
2
4
输出样例#1： 复制
2
提示
标程仅供做题后或实在无思路时参考。
请自觉、自律地使用该功能并请对自己的学习负责。
如果发现恶意抄袭标程，将按照I类违反进行处理。 | 824 | 1,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-18 | latest | en | 0.503078 |
http://www.expertsmind.com/questions/explain-avl-tree-30164516.aspx | 1,670,382,745,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711126.30/warc/CC-MAIN-20221207021130-20221207051130-00792.warc.gz | 65,056,280 | 12,800 | ## Explain avl tree, Data Structure & Algorithms
Assignment Help:
AVL tree
An AVL tree is a binary search tree in which the height of the left and right subtree of the root vary by at most 1 and in which the left and right subtrees are again AVL trees. With every node of an AVL tree is associated with a balanced factor that is left high, equal or right high, according, respectively, as the left subtrees has height greater than, equivalent to, or less than that of the right subtree.
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Write an algorithm to calculate a postfix expression. Execute your algorithm using the given postfix expression as your input : a b + c d +*f ↑ . T o evaluate a postfix expr | 597 | 2,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-49 | latest | en | 0.906995 |
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The volume = base times height, so if it gives you the volume and base, it would look like this
volume=10(im making an example up, bear with me)
base=5
height=x(x is variable)
10=5x
divide by five on both sides, and you should get an answer
:)hope this helped!
Thank you very much!:) | 152 | 617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-04 | latest | en | 0.932492 |
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# If the average (arithmetic mean) of n consecutive odd intege
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If the average (arithmetic mean) of n consecutive odd intege [#permalink]
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26 Nov 2009, 11:12
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If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?
(1) The range of the n integers is 14
(2) The greatest of the n integers is 17
[Reveal] Spoiler: OA
Last edited by Bunuel on 16 Dec 2012, 07:16, edited 1 time in total.
Edited the question and added the OA.
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Re: Quant Guide DS 66 [#permalink]
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26 Nov 2009, 13:13
JimmyWorld wrote:
Hi,
I am unsure about the explanation on DS #66 of Quant Guide.
Can someone explain the answer to 2) for me?
If the average(arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers.
1) The range of n integers is 14.
2) The greatest of the n integers is 17.
I understood 1) but didnt know how to get 2).
n is a set of consecutive odd integers and the last one being 17
17 + 15 = 22/2 = 11
17 + 15 + 13 = 45/3 = 15
working backwards you'll find 3 is the smallest number, 8 numbers total 80 and the average 10
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Re: Quant Guide DS 66 [#permalink]
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26 Nov 2009, 14:00
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JimmyWorld wrote:
Thanks for the response. I was just wondering what would be a faster way to do this problem. If I were to work backwards to find when the average is 10, it would take me longer then 2 minutes to do that. Is there a quick way to figure out that the least number would be 3?
in a group of consecutive integers the average is also the median
10 would be the median
17 the largest
that would have to make 3 the smallest
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Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink]
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15 Dec 2012, 23:05
Bunuel,
how to solve this using algebraic approach?
Regards,
Sach
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Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink]
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15 Dec 2012, 23:44
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Sachin9 wrote:
Bunuel,
how to solve this using algebraic approach?
Regards,
Sach
Let me give it a try.
Say the first term of the progression, i.e. the least of the integers is a. Hence n-th term of the progression, i.e. the largest of the integers will be [a + 2(n - 1)].
Therefore, Range = Max - Min = [a + 2(n - 1)] - a = 2(n - 1)
and, Arithmetic Mean = (Max + Min)/2 = [a + 2(n - 1) + a]/2 = [a + (n - 1)]
Now, [a + (n - 1)] = 10 => (a + n) = 11
Thus, we can determine the value of a, once we know the value of n.
Statement 1: Range = 14
Hence, 2(n - 1) = 14
=> n = 8 => We can determine the value of a.
Sufficient
Statement 2: [a + 2(n - 1)] = 17
We have two equations in two unknowns. Hence, we can determine the value of a.
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Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink]
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16 Dec 2012, 07:18
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Sachin9 wrote:
Bunuel,
how to solve this using algebraic approach?
Regards,
Sach
If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?
Odd consecutive integers is an evenly spaced set. For any evenly spaced set the mean equals to the average of the first and the last terms, so in our case $$mean=10=\frac{x_1+x_{n}}{2}$$ --> $$x_1+x_{n}=20$$. Question: $$x_1=?$$
(1) The range of the n integers is 14 --> the range of a set is the difference between the largest and smallest elements of a set, so $$x_{n}-x_1=14$$ --> solving for $$x_1$$ --> $$x_1=3$$. Sufficient.
(2) The greatest of the n integers is 17 --> $$x_n=17$$ --> $$x_1+17=20$$ --> $$x_1=3$$. Sufficient.
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-average-arithmetic-mean-of-n-consecutive-odd-106036.html
_________________
Kudos [?]: 135737 [12], given: 12706
Re: If the average(arithmetic mean) of n consecutive odd integer [#permalink] 16 Dec 2012, 07:18
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# If the average (arithmetic mean) of n consecutive odd intege
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,887 | 6,344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-51 | latest | en | 0.862201 |
https://opentextbc.ca/physicalgeologyh5p/chapter/metamorphic-facies-and-index-minerals/ | 1,719,056,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00549.warc.gz | 380,729,773 | 33,815 | # Metamorphic Facies
In any given metamorphic setting there can be a variety of protolith types exposed to metamorphism. While these rocks will be exposed to the same range of pressure and temperatures conditions within that setting, the metamorphic rock that results will depend on the protolith. A convenient way to indicate the range of possible metamorphic rocks in a particular setting is to group those possibilities into metamorphic facies. In other words, a given metamorphic facies groups together metamorphic rocks that form under the same pressure and temperature conditions, but which have different protoliths.
Figure 10.34 shows the different metamorphic facies as patches of different colours. The axes on the diagram are temperature and depth; the depth within the Earth will determine how much pressure a rock is under, so the vertical depth axis is also a pressure axis. Each patch of colour represents a range of temperature and pressure conditions where particular types of metamorphic rocks will form. Metamorphic facies are named for rocks that form under specific conditions (e.g., eclogite facies, amphibolite facies etc.), but those names don’t mean that the facies is limited to that one rock type.
Concept Check: Metamorphic Facies
Fill in the missing words to complete the summary.
Metamorphic group together metamorphic rocks that form under (hint: the same or different?) pressure and temperature conditions, but which have (hint: the same or different?) parent rocks.
The groups are named for a single metamorphic rock that forms under those specific conditions. They can include (hint: Many or no?) other metamorphic rocks.
To check your answers, navigate to the below link to view the interactive version of this activity.
Another feature to notice in the diagram are the many dashed lines. The yellow, green, and blue dashed lines represent the geothermal gradients in different environments. Recall that the geothermal gradient describes how rapidly the temperature increases with depth in Earth. In most areas (green dashed line), the rate of increase in temperature with depth is 30 °C/km. In other words, if you go 1,000 m down into a mine, the temperature will be roughly 30 °C warmer than the average temperature at the surface. In volcanic areas (yellow dashed line), the geothermal gradient is more like 40 to 50 °C/km, so the temperature increases much faster as you go down. Along subduction zones (blue dashed line), the cold ocean lithosphere keeps temperatures low, so the gradient is typically less than 10 °C/km.
The yellow, green, and blue dashed lines in Figure 10.34 tell you what metamorphic facies you will encounter for rocks from a given depth in that particular environment. A depth of 15 km in a volcanic region falls in the amphibolite facies. Under more typical conditions, this is the greenschist facies, and in a subduction zone it is the blueschist facies. You can make the connection more directly between the metamorphic facies and the types of metamorphism discussed in the previous section by matching up the letters a through e in Figure 10.34 with the labels in Figure 10.35.
One other line to notice in Figure 10.34 is the red dashed line on the right-hand side of the figure. This line represents temperatures and pressures where granite will begin to melt if water is present. Migmatite is to the right of the line because it forms when some of the minerals in a metamorphic rock begin to melt, and then cool and crystallize again.
Practice with Metamorphic Facies and Geothermal Gradients
Note: It’s okay to peek at the metamorphic facies diagram (Figure 10.34) if you need to.
Match the words into the correct boxes.
The geothermal gradient is in subduction zones, because temperatures are at depth than in other locations. Most subduction zone conditions fall within the facies, named for a uniquely coloured foliated rock.
The geothermal gradient is in volcanic regions because temperatures get at shallower depths than in other locations. At the highest pressures and temperatures, the volcanic region geothermal gradient falls within the facies.
Contact metamorphism falls within relatively pressure conditions in the facies. This is because as you go deeper, temperatures get too for there to be a big contrast between magma and other rocks.
Fill-in-the-blank options:
• hornfels
• shallowest
• low
• steepest
• hotter
• blueschist
• high
• cooler
• granulite
To check your answers, navigate to the below link to view the interactive version of this activity.
# Index Minerals
Some common minerals in metamorphic rocks are shown in Figure 10.36, arranged in order of the temperature ranges where they tend to be stable. The upper and lower limits of the ranges are intentionally vague because these limits depend on a number of different factors, such as the pressure, the amount of water present, and the overall composition of the rock.
Even though the limits of the stability ranges are vague, the stability range of each mineral is still small enough that the minerals can be used as markers for those metamorphic conditions. Minerals that make good markers of specific ranges of metamorphic conditions are called index minerals.
## The Meguma Terrane of Nova Scotia: An Example of How Index Minerals Are Used
The southern and southwestern parts of Nova Scotia were regionally metamorphosed during the Devonian Acadian Orogeny (around 400 Ma), when a relatively small continental block—the Meguma Terrane (Figure 10.37 top )—collided with the existing eastern margin of North America. The clastic sedimentary rocks within this terrane were variably metamorphosed.
Index minerals have been used to map areas of higher or lower metamorphic intensity, called metamorphic zones. A metamorphic zone is a region bounded by the first appearance of one index mineral and the first appearance of the next. In the Meguma Terrane, the biotite zone (darker green) begins in the east and north with the first appearance of biotite. The biotite zone ends toward the south and west where garnet first appears. Because index minerals can have overlapping stability conditions, a lower-intensity index mineral can still be present in a higher-intensity metamorphic zone.
Knowledge of metamorphic zones makes it possible to draw conclusions about the geological conditions in which metamorphic rocks formed. The highest-intensity metamorphism (highest metamorphic grade)—the sillimanite zone—is in the southwest. Progressively lower grades of metamorphism exist toward the east and north. The rocks of the sillimanite zone were likely heated to over 700 °C, and therefore must have been buried to depths between 20 km and 25 km. The surrounding lower-grade rocks were not buried as deeply, and the rocks within the peripheral chlorite zone were likely not buried to more than about 5 km depth.
A probable explanation for this pattern is that the area with the highest-grade rocks was buried beneath the central part of a mountain range formed by the collision of the Meguma Terrane with North America. The collision caused rocks to be folded, and to be faulted and stacked on top of each other. These mountain-building processes thickened Earth’s crust, and increased its mass locally as the mountains grew. The increased mass of the growing mountains caused the lithosphere to float lower in the mantle (Figure 10.38, left). As the mountains were eventually eroded over tens of millions of years, the crust floated higher and higher in the mantle, and erosion exposed metamorphic rocks that were deep within the mountains (Figure 10.38, right).
Building a narrative for the metamorphism in Nova Scotia’s Meguma Terrane is just one example of how index minerals can be used.
Try It Yourself: Meguma Terrane Index Minerals
#### References
Keppie, D., & Muecke, G. (1979). Metamorphic map of Nova Scotia. (Nova Scotia Department of Mines and Energy, Map 1979-006).
White, C. E., & Barr, S. M. (2012) Meguma terrane revisited: Stratigraphy, metamorphism, paleontology and provenance. Geoscience Canada, 39(1). https://journals.lib.unb.ca/index.php/GC/article/view/19450/21005
## License
Physical Geology - H5P Edition Copyright © 2021 by Karla Panchuk is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. | 1,811 | 8,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-26 | latest | en | 0.889175 |
https://efloridamortgagegroup.com/qa/question-how-many-dimensions-do-humans-see.html | 1,618,242,002,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00515.warc.gz | 299,638,784 | 7,557 | # Question: How Many Dimensions Do Humans See?
## What are the 12 dimensions of life?
Achieve Your Vision Of Success In All 12 Dimensions Of Your LifeHealth & Fitness.Intellectual.Emotional.Character.Spiritual.Love Relationship.Parenting.Social.More items….
## Why is time the 4th Dimension?
Light clocks A and B moving horizontally through space. According to length contraction, clock A should tick faster than clock B. … But in the 106 years since Einstein, the prevailing view in physics has been that time serves as the fourth dimension of space, an arena represented mathematically as 4D Minkowski spacetime.
## Do we live in 3d or 4d?
We see in 3 dimensions. We experience in 4. There are 3 spatial dimensions. when relevant time can be considered a dimension, but it is not a spatial dimension.
## Is there a 4th Dimension?
Most of us think of time as the fourth dimension, but modern physics theorizes that there is a fourth spatial dimension as well—not width, height, or length but something else that we can’t experience through our physical senses.
## How many dimensions do we live in?
In everyday life, we inhabit a space of three dimensions – a vast ‘cupboard’ with height, width and depth, well known for centuries. Less obviously, we can consider time as an additional, fourth dimension, as Einstein famously revealed.
## How many dimensions are proven?
In fact, the theoretical framework of Superstring Theory posits that the universe exists in ten different dimensions. These different aspects are what govern the universe, the fundamental forces of nature, and all the elementary particles contained within.
## How many dimensions are there?
The world as we know it has three dimensions of space—length, width and depth—and one dimension of time. But there’s the mind-bending possibility that many more dimensions exist out there. According to string theory, one of the leading physics model of the last half century, the universe operates with 10 dimensions.
## Do humans see in 4d?
Some believe that it is impossible for us to visualize 4D, since we are confined to 3D and therefore cannot directly experience it. … Even though we are 3D beings who live in a 3D world, our eyes actually only see in 2D. Our retina has only a 2D surface area with which it can detect light coming into our eye.
## What are the 26 dimensions?
In bosonic string theory, spacetime is 26-dimensional, while in superstring theory it is 10-dimensional, and in M-theory it is 11-dimensional. In order to describe real physical phenomena using string theory, one must therefore imagine scenarios in which these extra dimensions would not be observed in experiments. | 577 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.921683 |
https://socratic.org/questions/in-graphing-y-4-x-3-2-2-the-vertex-of-the-parabola-is-at-what-point | 1,638,443,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00334.warc.gz | 568,543,642 | 5,827 | # In graphing y=4(x-3)^2 -2, the vertex of the parabola is at what point?
May 6, 2016
Vertex$\to \left(x , y\right) \to \left(3 , - 2\right)$
#### Explanation:
Reading directly from the vertex form equation
${x}_{\text{vertex}} = \left(- 1\right) \times \left(- 3\right) = + 3$
${y}_{\text{vertex}} = - 2$
So Vertex$\to \left(x , y\right) \to \left(3 , - 2\right)$ | 142 | 370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-49 | latest | en | 0.587207 |
https://www.e-olymp.com/en/contests/11581/problems/107945 | 1,558,276,185,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00022.warc.gz | 777,845,809 | 2,771 | Competitions
# Vasya and matrix
Mother gave Vasya a rectangular matrix n by m. Each cell of the matrix contains one integer. Vasya played a lot of math games with it: he found its determinant and powered it into some degrees.
But such games made him a little tired, so he invented a new entertainment: he choose an integer k and tries to find a submatrix of maximal area with total sum of numbers in it not greater than k. The submatrix is a rectangular area of matrix.
#### Input
The first line contains three integers n, m and k (1n, m300, 1k109).
Each of the next n lines contains m nonnegative integers, each no more than 1000.
#### Output
Print the area of maximal submatrix, which sum of numbers is not greater than k.
Time limit 1 second
Memory limit 128 MiB
Input example #1
1 3 4
8 6 4
Output example #1
1
Input example #2
3 3 12
7 5 7
8 4 8
4 3 2
Output example #2
3 | 247 | 885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-22 | latest | en | 0.889629 |
http://fsconline.info/tag/1st-year-physics-notes/page/3/ | 1,513,615,796,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948618633.95/warc/CC-MAIN-20171218161254-20171218183254-00243.warc.gz | 114,626,571 | 22,626 | Latest
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### Author Topic: Which electromagnetic wavelength has more energy, and which has more power? (Read 2058 times)
#### taregg
• Sr. Member
• Posts: 168
##### Which electromagnetic wavelength has more energy, and which has more power?
« on: 30/10/2013 12:56:51 »
which electromagnetic has more energy and the other has more power.........for long waves and short waves ..
« Last Edit: 01/11/2013 22:15:43 by chris »
#### JP
• Neilep Level Member
• Posts: 3366
• Thanked: 2 times
##### Re: which electromagnetic has more energy and the other has more power.........
« Reply #1 on: 30/10/2013 16:00:42 »
For each photon, the energy is proportional to 1/wavelength, so short waves have higher energy photons. The power delivered by a wave depends on how many photons it transmits per second, which can vary. A low wavelength wave with many photons/second can deliver more energy than a high wavelength wave with only a small number of photons/second.
#### taregg
• Sr. Member
• Posts: 168
##### Re: which electromagnetic has more energy and the other has more power.........
« Reply #2 on: 30/10/2013 22:02:08 »
like if we have ultraviolt laser and microwave laser.....which one has more power....if ultraviolt has short wave and microwave has long wave..
#### JP
• Neilep Level Member
• Posts: 3366
• Thanked: 2 times
##### Re: which electromagnetic has more energy and the other has more power.........
« Reply #3 on: 30/10/2013 23:22:02 »
Lasers, like light bulbs in your house, can run at different powers. A 10 mW microwave laser has the same power as a 10 mW ultraviolet laser.
#### Supercryptid
• Hero Member
• Posts: 606
##### Re: which electromagnetic has more energy and the other has more power.........
« Reply #4 on: 01/11/2013 20:29:32 »
#### evan_au
• Neilep Level Member
• Posts: 4126
• Thanked: 247 times
##### Re: which electromagnetic has more energy and the other has more power.........
« Reply #5 on: 01/11/2013 20:52:58 »
A photon with long wavelength carries less energy than a photon with short wavelength.
Two lasers with the same optical power will deliver the same amount of energy every second, but the one with longer wavelength delivers more photons per second, so the total energy is the same.
#### The Naked Scientists Forum
##### Re: which electromagnetic has more energy and the other has more power.........
« Reply #5 on: 01/11/2013 20:52:58 » | 647 | 2,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-50 | latest | en | 0.852966 |
https://physics.stackexchange.com/questions/326504/this-vector-potential-gives-a-magnetic-monopole-field-whats-wrong-with-it | 1,585,598,893,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00116.warc.gz | 639,985,095 | 34,694 | This vector potential gives a magnetic monopole field, what's wrong with it?
$\mathbf{A} = \frac{g(1-\cos\theta)}{r\sin\theta} \mathbf{\hat\phi} \Rightarrow \mathbf{B} = g \mathbf{\hat r}/r^2$
But yet the existence of $\mathbf{A}$ itself hinges on the fact that there are no magnetic monopoles. Is the problem because the given $\mathbf{A}$ has singularities?
• Your $\mathbf{B}$ equals $\nabla\times \mathbf{A}$ except at the origin. – velut luna Apr 14 '17 at 4:09
• See another answer. – Ng Chung Tak Apr 14 '17 at 11:00
• If anyone's wondering, pages 146-149 of Sakurai's QM book has a nice discussion. – physcopath Apr 20 '17 at 3:08
Yes, you have problems with this potential due to the singularity. Notice that you do want the singularity in $r=0$, as you are talking about a point charge (and the electric potential is singular in the position of a particle).
There is also another enormous problem: you know that $\vec\nabla\cdot(\vec\nabla\times\vec A)=0$, as you are taking the gradient of a curl. Due to this fact, you cannot have magnetic charge: remember that, for a point charge at the origin, $\vec \nabla\cdot \vec E=q\delta^3(\vec x)$. That $\delta^3$ factor is what allows us to say that, in any set containing the origin, we have a total charge $q$. This does not work with the magnetic field, when we integrate in a naive way.
Those two problems are solvable, through the introduction of the concept of fiber bundles. I'll try to not use them, but know for future reference that modern gauge theory is formulated around the concept of fiber bundles, that allow you to describe things like magnetic monopoles in a correct way.
I will refer to Manton and Sutcliffe's Topological Solitons in answering. In chapter 8, they discuss magnetic monopoles.
Let's examine your potential. I assume that your azimutal coordinate $\theta$ goes from $0$ to $\pi$, as it should be the case for your potential to work. You can choose between two potentials: $$\vec A_+=\frac{g}{4\pi r}\frac{1-\cos\theta}{\sin\theta}\hat \phi,\quad\vec A_-=\frac{g}{4\pi r}\frac{-1-\cos\theta}{\sin\theta}\hat \phi.$$ You can verify that both potentials have $\vec B$ as curl, whenver $r\neq 0$, $\theta\neq 0$ and $\theta\neq\pi$. The additional $4\pi$ factor is just a redefinition of $g$, that is convenient as it makes the flux of the magnetic field equal to the magnetic charge, with no proportionality constant. Which potential will we use? The answer is "both". Notice that the first potential is nonsingular in $\theta=0$ (north pole, for definiteness), while the second one is nonsingular in $\theta=\pi$ (perform the limit: it exists and is 0, so you can extend the definition in a continous way).
Let us say that you want to find the magnetic field at a given distance from the monopole, $R$. In modern language, you are looking for the magnetic field on a 2-sphere of radius $R$, that I will call $S^2_R$, under the boundary condition that the flux of this magnetic field over the whole $S^2_R$'s boundary should equal to the magnetic charge: $$\int_{S^2_R}\vec B\cdot d\vec S=g,$$ where $\vec S$ is the vector pointing outwards from the sphere.
The key fact is that the sphere $S^2_R$ cannot be described by a simple set of coordinates $(\theta,\phi)$, without excluding one of the poles. In differential geometry language, you have that $S^2_R$ is not a trivial manifold, and you need at least two systems of coordinates to cover the whole sphere. Let $\theta_N$ and $\theta_S$ be angles such as $0<\theta_S<\theta_N<\pi$: you can use a system of coordinates $(\theta_+,\phi_+)$ where $0\leq\theta_+<\theta_N$ and another system of coordinates $(\theta_-,\phi_-)$, where $\theta_S<\theta_-\leq\pi$. Now, due to the fact that $\theta_S<\theta_N$, you have that those two coordinate systems cover the whole sphere, in the sense that any point is described by at least one of such sets of coordinates. When it is described by both sets (as it is the case for all points in the strip $\theta_S<\theta<\theta_N$ you must have a transition function, that associates a coordinate in a set to a coordinate in another (in this case, you just have to take the same $\theta$, but more complicated choices are possible).
A gauge theory over the sphere is (TRULY LOOSELY SPEAKING) an assignement of a gauge field $\vec A$ on any patch of the sphere. Now, we can say that $\vec A_+$ describes the potential in the $(\theta_+,\phi_+)$ system, so it extends to the north pole (where it is nonsingular). We assign the potential $\vec A_-$ to the south pole (where it is nonsingular). Now, what can we say on the overlap string? You can verify that, on the strip $$\vec A_-=\vec A_+-\vec\nabla\left(\frac{g}{2\pi}\phi\right).$$ I don't need to specify $\phi_1$ or $\phi_2$, as the transition map is the identity. This is exactly a gauge transformation! Due to the fact that the fields are related by a gauge transformation, you can say that they describe the same physical field.
How does this construction solve the problem of magnetic charge? Or, is the flux condition working here? A rigorous (and quick) explanation would require notions of integration in differential geometry, so I'll go with intuitive answer. If you take $\theta_N$ and $\theta_S$ such as the equator $\theta=\frac\pi2$ is in the overlapping region, you can divide the integral as $$\int_{S^2_R}\vec B\cdot d\vec S=\int_{NP}\vec\nabla\times\vec A_+\cdot d\vec S+\int_{SP}\vec\nabla\times\vec A_-\cdot d\vec S.$$ Here, $NP$ means $\theta<\frac{\pi}{2}$ and $SP$ means $\theta>\frac{\pi}{2}$. Notice that the equator does not belong to $NP$ or $SP$, but it is a line and integrals of well defined functions over that line are $0$. The equator is a boundary for both north pole and south pole, so we can use Stokes theorem: it is immediate then to be convinced that the previous integral is equal to two line integrals on the equator, taken once clockwise and the other time counterclockwise: $$\int_{S^2_R}\vec B\cdot d\vec S=\int_{equator}\vec A_{1}\cdot d\vec l-\int_{equator} \vec A_2\cdot d\vec l=2\pi R\left(\frac{g}{4\pi R}+\frac{g}{4\pi R}\right)=g.$$ Mind, this is not the proper way to proceed. Take it just for intuition.
To conclude, magnetic monopoles are theorically possible, and a potential for a magnetic monopole can be written. But you have to use the notions of coordinate charts to define the potential without ambiguity, and obtain an analogous of Gauss' law for magnetism. Your potential is part of the solution. If you really are interested in gauge theories, you will have to learn a lot of differential geometry and the basic of fiber bundles to be able to do the funnier things with gauge fields.
• can you check A plus and A minus for typo? on my mobile one is just the negative of the other so their curl if course doesnt give the same value. – lalala Apr 15 '17 at 15:15
• Tnx for spotting, definitively a typo. – Salvatore Baldino Apr 15 '17 at 15:36 | 1,891 | 6,949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-16 | latest | en | 0.88184 |
https://forum.allaboutcircuits.com/threads/capacitive-load-on-potentiometers.184098/ | 1,709,040,628,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.26/warc/CC-MAIN-20240227121318-20240227151318-00209.warc.gz | 258,535,827 | 22,415 | #### DaniKowa
Joined Sep 23, 2020
218
Happy New Year everyone!
today I was playing with an LCR tester, I often use it to check inductors and capacitors. This morning I had a "blue" 3590s multiturn potentiometer (wire wound) in my hand. I was testing the value in ohms, then curiosity made me set to measure C too, that is the capacity. I noticed that by rotating the potentiometer the capacitive value is inversely proportional to the value of R. That is, the more the resistance increases, the capacitance decreases. The question is: do all potentiometers have this variable capacity? Thank you.
#### AlbertHall
Joined Jun 4, 2014
12,328
Or is this just a product of the way the meter measures capacitance? Try measuring a resistor of the same value as the potentiometer on the C range.
#### MisterBill2
Joined Jan 23, 2018
17,792
What is the tester's frequency? All wirewound pots have some inductance and that might possibly be confusing the tester.
#### MrChips
Joined Oct 2, 2009
30,468
The C meter is measuring the inverse of reactance.
Xc = 1 / ωC
C = 1 / ωXc
Hence the meter displays low values of C for high values of R.
C ∝ 1 / R
#### DaniKowa
Joined Sep 23, 2020
218
Sorry for the late reply, so I understand correctly if I say that:
1) Is it normal to read a capacitive value on a resistive potentiometer?
2) Then by rotating the potentiometer, this capacitive value decreases as the resistive value increases and vice versa.
Thank you
#### drjohsmith
Joined Dec 13, 2021
850
Sorry for the late reply, so I understand correctly if I say that:
1) Is it normal to read a capacitive value on a resistive potentiometer?
2) Then by rotating the potentiometer, this capacitive value decreases as the resistive value increases and vice versa.
Thank you
There are many types of variable resistor,
you say you have a wire wound one, "most" are a metal film and a wiper.
Any two conductors, separated by an insulator, form a capacitance,
There is no inherent relationship between a variable resistors resistance and its capacitance,
I'd imagine if you change how your holding or not the parts, the capacitance will change.
You say this is a wire wound resistor type,
this is effectively coil, a coil by definition has resistance, capacitance and inductance.
a wire wound resistor is designed to be used "near DC "
at "near DC" the inductance and capacitance have negligible effect,
Your meter is using a wave form to measure the phase capacitance
QED, its not DC,
An interesting effect,
but not un expected side effect of this type of component and not repeatable to any degree
#### MisterBill2
Joined Jan 23, 2018
17,792
Happy New Year everyone!
today I was playing with an LCR tester, I often use it to check inductors and capacitors. This morning I had a "blue" 3590s multiturn potentiometer (wire wound) in my hand. I was testing the value in ohms, then curiosity made me set to measure C too, that is the capacity. I noticed that by rotating the potentiometer the capacitive value is inversely proportional to the value of R. That is, the more the resistance increases, the capacitance decreases. The question is: do all potentiometers have this variable capacity? Thank you.
One other consideration is that a meter set to read capacitance may not deliver a useful reading when connected to a resistance. An ohm meter will not often provide the inductance measurement when it is connected to an inductor. It will display the resistive part of the reactance.
In short, readings obtained with the "wrong" settings may not be right.
#### drjohsmith
Joined Dec 13, 2021
850
One other consideration is that a meter set to read capacitance may not deliver a useful reading when connected to a resistance. An ohm meter will not often provide the inductance measurement when it is connected to an inductor. It will display the resistive part of the reactance.
In short, readings obtained with the "wrong" settings may not be right.
Just out of interest,
wonder what a "LCR" bridge would give you
for info
https://www.hioki.com/global/learning/usage/lcr-meters_1.html
#### DaniKowa
Joined Sep 23, 2020
218
There are many types of variable resistor,
you say you have a wire wound one, "most" are a metal film and a wiper.
Any two conductors, separated by an insulator, form a capacitance,
There is no inherent relationship between a variable resistors resistance and its capacitance,
I'd imagine if you change how your holding or not the parts, the capacitance will change.
You say this is a wire wound resistor type,
this is effectively coil, a coil by definition has resistance, capacitance and inductance.
a wire wound resistor is designed to be used "near DC "
at "near DC" the inductance and capacitance have negligible effect,
Your meter is using a wave form to measure the phase capacitance
QED, its not DC,
An interesting effect,
but not un expected side effect of this type of component and not repeatable to any degree
Ok, the potentiometer I used is a 3590 multiturn model, so definitely wire wound. But for my purpose I want to use this vessel not in DC but in AC or RF. So in case of AC / RF purposes, wrapped wire may not be recommended. Correct? If so, what is the solution, can I always use multiturn?
Thank you guys!.
#### MrChips
Joined Oct 2, 2009
30,468
Ok, the potentiometer I used is a 3590 multiturn model, so definitely wire wound. But for my purpose I want to use this vessel not in DC but in AC or RF. So in case of AC / RF purposes, wrapped wire may not be recommended. Correct? If so, what is the solution, can I always use multiturn?
Thank you guys!.
It depends on the AC frequency. If it is 50 or 60Hz it should have little effect.
#### DaniKowa
Joined Sep 23, 2020
218
It depends on the AC frequency. If it is 50 or 60Hz it should have little effect.
No i mean some Mhz . At least up to 10Mhz.
#### MisterBill2
Joined Jan 23, 2018
17,792
Using wirewound pots at higher frequencies often produces unanticipated results because of the inductance. That has been the reason for carbon film and cermet variable resistors. A wirewound deevice is not going to do what you want, probably. Thus an alternate scheme is required. | 1,534 | 6,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-10 | latest | en | 0.910845 |
https://jhui.github.io/2018/02/09/PyTorch-Basic-operations/ | 1,719,036,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00037.warc.gz | 292,243,675 | 11,169 | This tutorial helps NumPy or TensorFlow users to pick up PyTorch quickly.
### Basic
By selecting different configuration options, the tool in the PyTorch site shows you the required and the latest wheel for your host platform. For example, on a Mac platform, the pip3 command generated by the tool is:
pip3 install http://download.pytorch.org/whl/torch-0.3.0.post4-cp36-cp36m-macosx_10_7_x86_64.whl
pip3 install torchvision
Run the following code and you should see an un-initialized 2x3 Tensor is printed out. Tensor is a data structure representing multi-dimensional array. It is similar to a NumPy ndarray. It’s size is equivalent to the shape of the NumPy ndarray.
import torch
x = torch.Tensor(2, 3) # Create an un-initialized Tensor of size 2x3
print(x) # Print out the Tensor
# 0.0000e+00 -2.0000e+00 0.0000e+00
# -2.0000e+00 1.8856e+31 4.7414e+16
# [torch.FloatTensor of size 2x3]
This printout represents the Tensor type and its size (dimension: 2x3).
[torch.FloatTensor of size 2x3]
### Sample programs:
import torch
# Initialize
x = torch.Tensor(2, 3) # An un-initialized Tensor object. x holds garbage data.
y = torch.rand(2, 3) # Initialize with random values
# Operations
z1 = x + y
z2 = torch.add(x, y) # Same as above
print(z2) # [torch.FloatTensor of size 2x3]
### Operations
The syntax on a tensor operation:
torch.is_tensor(obj)
#### In-place operation
All operations end with “_” is in place operations:
x.add_(y) # Same as x = x + y
#### out
We can assign the operation result to a variable. Alternatively, all operation methods have an out parameter to store the result.
r1 = torch.Tensor(2, 3)
It is the same as:
r2 = torch.add(x, y)
#### Indexing
We can use the NumPy indexing in Tensors:
x[:, 1] # Can use numpy type indexing
x[:, 0] = 0 # For assignment
#### Conversion between NumPy ndarray and Tensor
During the conversion, both ndarray and Tensor share the same memory storage. Change value from either side will affect the other.
# Conversion
a = np.array([1, 2, 3])
v = torch.from_numpy(a) # Convert a numpy array to a Tensor
b = v.numpy() # Tensor to numpy
b[1] = -1 # Numpy and Tensor share the same memory
assert(a[1] == b[1]) # Change Numpy will also change the Tensor
#### Tensor meta-data
Size of the Tensor and number of elements in Tensor:
### Basic Tensor operation
x.size() # torch.Size([2, 3])
torch.numel(x) # 6: number of elements in x
#### Reshape Tensor
Reshape a Tensor to different size:
### Tensor resizing
x = torch.randn(2, 3) # Size 2x3
y = x.view(6) # Resize x to size 6
z = x.view(-1, 2) # Size 3x2
### Create a Tensor
#### Creating and initializing a Tensor
### Create a Tensor
v = torch.Tensor(2, 3) # An un-initialized torch.FloatTensor of size 2x3
v = torch.Tensor([[1,2],[4,5]]) # A Tensor initialized with a specific array
v = torch.LongTensor([1,2,3]) # A Tensor of type Long
#### Create a random Tensor
To increase the reproducibility of result, we often set the random seed to a specific value first.
torch.manual_seed(1)
v = torch.rand(2, 3) # Initialize with random number (uniform distribution)
v = torch.randn(2, 3) # With normal distribution (SD=1, mean=0)
v = torch.randperm(4) # Size 4. Random permutation of integers from 0 to 3
#### Tensor type
x = torch.randn(5, 3).type(torch.FloatTensor)
#### Identity matrices, Fill Tensor with 0, 1 or values
eye = torch.eye(3) # Create an identity 3x3 tensor
v = torch.ones(10) # A tensor of size 10 containing all ones
v = torch.ones(2, 1, 2, 1) # Size 2x1x2x1
v = torch.ones_like(eye) # A tensor with same shape as eye. Fill it with 1.
v = torch.zeros(10) # A tensor of size 10 containing all zeros
# 1 1 1
# 2 2 2
# 3 3 3
v = torch.ones(3, 3)
v[1].fill_(2)
v[2].fill_(3)
#### Initialize Tensor with a range of value
v = torch.arange(5) # similar to range(5) but creating a Tensor
v = torch.arange(0, 5, step=1) # Size 5. Similar to range(0, 5, 1)
# 0 1 2
# 3 4 5
# 6 7 8
v = torch.arange(9)
v = v.view(3, 3)
#### Initialize a linear or log scale Tensor
v = torch.linspace(1, 10, steps=10) # Create a Tensor with 10 linear points for (1, 10) inclusively
v = torch.logspace(start=-10, end=10, steps=5) # Size 5: 1.0e-10 1.0e-05 1.0e+00, 1.0e+05, 1.0e+10
#### Initialize a ByteTensor
c = torch.ByteTensor([0, 1, 1, 0])
#### Summary
Creation Ops
~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: eye
.. autofunction:: from_numpy
.. autofunction:: linspace
.. autofunction:: logspace
.. autofunction:: ones
.. autofunction:: ones_like
.. autofunction:: arange
.. autofunction:: range
.. autofunction:: zeros
.. autofunction:: zeros_like
### Indexing, Slicing, Joining, Mutating Ops
We will prepare a Matrix that will be used in this section:
# 0 1 2
# 3 4 5
# 6 7 8
v = torch.arange(9)
v = v.view(3, 3)
#### Concatenate, stack
# Concatenation
torch.cat((x, x, x), 0) # Concatenate in the 0 dimension
# Stack
r = torch.stack((v, v))
#### Gather : reorganize data element
# Gather element
# torch.gather(input, dim, index, out=None)
# out[i][j][k] = input[index[i][j][k]][j][k] # if dim == 0
# out[i][j][k] = input[i][index[i][j][k]][k] # if dim == 1
# out[i][j][k] = input[i][j][index[i][j][k]] # if dim == 2
# 0 1
# 4 3
# 8 7
r = torch.gather(v, 1, torch.LongTensor([[0,1],[1,0],[2,1]]))
#### Split a Tensor
# Split an array into 3 chunks
# (
# 0 1 2
# [torch.FloatTensor of size 1x3]
# ,
# 3 4 5
# [torch.FloatTensor of size 1x3]
# ,
# 6 7 8
# [torch.FloatTensor of size 1x3]
# )
r = torch.chunk(v, 3)
# Split an array into chunks of at most size 2
# (
# 0 1 2
# 3 4 5
# [torch.FloatTensor of size 2x3]
# ,
# 6 7 8
# [torch.FloatTensor of size 1x3]
# )
r = torch.split(v, 2)
# Index select
# 0 2
# 3 5
# 6 8
indices = torch.LongTensor([0, 2])
r = torch.index_select(v, 1, indices) # Select element 0 and 2 for each dimension 1.
# 0 0 0
# 1 1 1
# 1 1 1
# Size 6: 3 4 5 6 7 8
#### Squeeze and unsqueeze
t = torch.ones(2,1,2,1) # Size 2x1x2x1
r = torch.squeeze(t) # Size 2x2
r = torch.squeeze(t, 1) # Squeeze dimension 1: Size 2x2x1
# Un-squeeze a dimension
x = torch.Tensor([1, 2, 3])
r = torch.unsqueeze(x, 0) # Size: 1x3
r = torch.unsqueeze(x, 1) # Size: 3x1
#### Non-zero elements
# Non-zero
# [torch.LongTensor of size 8x2]
# [i, j] index for non-zero elements
# 0 1
# 0 2
# 1 0
# 1 1
# 1 2
# 2 0
# 2 1
# 2 2
r = torch.nonzero(v)
#### take
# Flatten a TensorFlow and return elements with given indexes
# Size 3: 0, 4, 2
r = torch.take(v, torch.LongTensor([0, 4, 2]))
#### transpose
# Transpose dim 0 and 1
r = torch.transpose(v, 0, 1)
#### Summary
Indexing, Slicing, Joining, Mutating Ops
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: cat
.. autofunction:: chunk
.. autofunction:: gather
.. autofunction:: index_select
.. autofunction:: nonzero
.. autofunction:: split
.. autofunction:: squeeze
.. autofunction:: stack
.. autofunction:: t - Transpose a 2-D tensor
.. autofunction:: take
.. autofunction:: transpose
.. autofunction:: unbind - Removes a tensor dimension
.. autofunction:: unsqueeze
.. autofunction:: where - Select x or y Tensor elements based on condition Tensor c
### Distribution
#### Uniform, bernoulli, multinomial, normal distribution
# 2x2: A uniform distributed random matrix with range [0, 1]
r = torch.Tensor(2, 2).uniform_(0, 1)
# bernoulli
r = torch.bernoulli(r) # Size: 2x2. Bernoulli with probability p stored in elements of r
# Multinomial
w = torch.Tensor([0, 4, 8, 2]) # Create a tensor of weights
r = torch.multinomial(w, 4, replacement=True) # Size 4: 3, 2, 1, 2
# Normal distribution
# From 10 means and SD
r = torch.normal(means=torch.arange(1, 11), std=torch.arange(1, 0.1, -0.1)) # Size 10
#### Summary
Random sampling
----------------------------------
.. autofunction:: manual_seed - Set a manual seed
.. autofunction:: initial_seed - Randomize a seed by the system
.. autofunction:: get_rng_state
.. autofunction:: set_rng_state
.. autodata:: default_generator
.. autofunction:: bernoulli
.. autofunction:: multinomial
.. autofunction:: normal
.. autofunction:: rand
.. autofunction:: randn
.. autofunction:: randperm
In-place random sampling
~~~~~~~~~~~~~~~~~~~~~~~~
There are a few more in-place random sampling functions defined on Tensors as well. Click through to refer to their documentation:
- :func:torch.Tensor.bernoulli_ - in-place version of :func:torch.bernoulli
- :func:torch.Tensor.cauchy_ - numbers drawn from the Cauchy distribution
- :func:torch.Tensor.exponential_ - numbers drawn from the exponential distribution
- :func:torch.Tensor.geometric_ - elements drawn from the geometric distribution
- :func:torch.Tensor.log_normal_ - samples from the log-normal distribution
- :func:torch.Tensor.normal_ - in-place version of :func:torch.normal
- :func:torch.Tensor.random_ - numbers sampled from the discrete uniform distribution
- :func:torch.Tensor.uniform_ - numbers sampled from the continuous uniform distribution
### Point-wise operations
### Math operations
f= torch.FloatTensor([-1, -2, 3])
r = torch.abs(f) # 1 2 3
# Add x, y and scalar 10 to all elements
# Clamp the value of a Tensor
r = torch.clamp(v, min=-0.5, max=0.5)
# Element-wise divide
r = torch.div(v, v+0.03)
# Element-wise multiple
r = torch.mul(v, v)
#### Summary
Pointwise Ops
~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: abs
.. autofunction:: acos - arc cosine
.. autofunction:: addcdiv - element wise: t1 + s * t2/t3
.. autofunction:: addcmul - element wise: t1 + s * t2 * t3
.. autofunction:: asin - arc sin
.. autofunction:: atan
.. autofunction:: atan2
.. autofunction:: ceil - ceiling
.. autofunction:: clamp - clamp elements into a range
.. autofunction:: cos
.. autofunction:: cosh
.. autofunction:: div - divide
.. autofunction:: erf - Gaussian error functiom
.. autofunction:: erfinv - Inverse
.. autofunction:: exp
.. autofunction:: expm1 - exponential of each element minus 1
.. autofunction:: floor
.. autofunction:: fmod - element wise remainder of division
.. autofunction:: frac - fraction part 3.4 -> 0.4
.. autofunction:: lerp - linear interpolation
.. autofunction:: log - natural log
.. autofunction:: log1p - y = log(1 + x)
.. autofunction:: mul - multiple
.. autofunction:: neg
.. autofunction:: pow
.. autofunction:: reciprocal - 1/x
.. autofunction:: remainder - remainder of division
.. autofunction:: round
.. autofunction:: rsqrt - the reciprocal of the square-root
.. autofunction:: sigmoid - sigmode(x)
.. autofunction:: sign
.. autofunction:: sin
.. autofunction:: sinh
.. autofunction:: sqrt
.. autofunction:: tan
.. autofunction:: tanh
.. autofunction:: trunc - truncated integer
### Reduction operations
### Reduction operations
# Accumulate sum
# 0 1 2
# 3 5 7
# 9 12 15
r = torch.cumsum(v, dim=0)
# L-P norm
r = torch.dist(v, v+3, p=2) # L-2 norm: ((3^2)*9)^(1/2) = 9.0
# Mean
# 1 4 7
r = torch.mean(v, 1) # Size 3: Mean in dim 1
r = torch.mean(v, 1, True) # Size 3x1 since keep dimension = True
# Sum
# 3 12 21
r = torch.sum(v, 1) # Sum over dim 1
# 36
r = torch.sum(v)
#### Summary
Reduction Ops
~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: cumprod - accumulate product of elements x1*x2*x3...
.. autofunction:: cumsum
.. autofunction:: dist - L-p norm
.. autofunction:: mean
.. autofunction:: median
.. autofunction:: mode
.. autofunction:: norm - L-p norm
.. autofunction:: prod - accumulate product
.. autofunction:: std - compute standard deviation
.. autofunction:: sum
.. autofunction:: var - variance of all elements
### Comparison operation
### Comparison
# Size 3x3: Element-wise comparison
r = torch.eq(v, v)
# Max element with corresponding index
r = torch.max(v, 1)
#### Sort
# Sort
# Second tuple store the index
# (
# 0 1 2
# 3 4 5
# 6 7 8
# [torch.FloatTensor of size 3x3]
# ,
# 0 1 2
# 0 1 2
# 0 1 2
# [torch.LongTensor of size 3x3]
r = torch.sort(v, 1)
#### k-th and top k
# k-th element (start from 1) ascending order with corresponding index
# (1 4 7
# [torch.FloatTensor of size 3]
# , 1 1 1
# [torch.LongTensor of size 3]
# )
r = torch.kthvalue(v, 2)
# Top k
# (
# 2 5 8
# [torch.FloatTensor of size 3x1]
# ,
# 2 2 2
# [torch.LongTensor of size 3x1]
# )
r = torch.topk(v, 1)
Comparison Ops
~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: eq - Compare elements
.. autofunction:: equal - True of 2 tensors are the same
.. autofunction:: ge - Element-wise greater or equal comparison
.. autofunction:: gt
.. autofunction:: kthvalue - k-th element
.. autofunction:: le
.. autofunction:: lt
.. autofunction:: max
.. autofunction:: min
.. autofunction:: ne
.. autofunction:: sort
.. autofunction:: topk - top k
### Matrix, vector multiplication
#### Dot product of Tensors
# Dot product of 2 tensors
r = torch.dot(torch.Tensor([4, 2]), torch.Tensor([3, 1])) # 14
#### Matrix, vector products
### Matrix, vector products
# Matrix X vector
# Size 2x4
mat = torch.randn(2, 4)
vec = torch.randn(4)
r = torch.mv(mat, vec)
# Matrix + Matrix X vector
# Size 2
M = torch.randn(2)
mat = torch.randn(2, 3)
vec = torch.randn(3)
#### Matrix, Matrix products
# Matrix x Matrix
# Size 2x4
mat1 = torch.randn(2, 3)
mat2 = torch.randn(3, 4)
r = torch.mm(mat1, mat2)
# Matrix + Matrix X Matrix
# Size 3x4
M = torch.randn(3, 4)
mat1 = torch.randn(3, 2)
mat2 = torch.randn(2, 4)
#### Outer product of vectors
# Outer product of 2 vectors
# Size 3x2
v1 = torch.arange(1, 4) # Size 3
v2 = torch.arange(1, 3) # Size 2
r = torch.ger(v1, v2)
# Add M with outer product of 2 vectors
# Size 3x2
vec1 = torch.arange(1, 4) # Size 3
vec2 = torch.arange(1, 3) # Size 2
M = torch.zeros(3, 2)
#### Batch matrix multiplication
# Batch Matrix x Matrix
# Size 10x3x5
batch1 = torch.randn(10, 3, 4)
batch2 = torch.randn(10, 4, 5)
r = torch.bmm(batch1, batch2)
# Batch Matrix + Matrix x Matrix
# Performs a batch matrix-matrix product
# 3x4 + (5x3x4 X 5x4x2 ) -> 5x3x2
M = torch.randn(3, 2)
batch1 = torch.randn(5, 3, 4)
batch2 = torch.randn(5, 4, 2)
### Other
#### Cross product
m1 = torch.ones(3, 5)
m2 = torch.ones(3, 5)
v1 = torch.ones(3)
# Cross product
# Size 3x5
r = torch.cross(m1, m2)
#### Diagonal matrix
# Diagonal matrix
# Size 3x3
r = torch.diag(v1)
#### Histogram
# Histogram
# [0, 2, 1, 0]
torch.histc(torch.FloatTensor([1, 2, 1]), bins=4, min=0, max=3)
#### Renormalization
# Renormalize
# Renormalize for L-1 at dim 0 with max of 1
# 0.0000 0.3333 0.6667
# 0.2500 0.3333 0.4167
# 0.2857 0.3333 0.3810
r = torch.renorm(v, 1, 0, 1)
#### Summary
Other Operations
~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: cross - cross product
.. autofunction:: diag - convert vector to diagonal matrix
.. autofunction:: histc - histogram
.. autofunction:: renorm - renormalize a tensor
.. autofunction:: trace - tr(M)
.. autofunction:: tril - lower triangle of 2-D matrix
.. autofunction:: triu - uppser triangle
A summary of available operations:
Tensors
----------------------------------
.. autofunction:: is_tensor
.. autofunction:: is_storage
.. autofunction:: set_default_tensor_type
.. autofunction:: numel
.. autofunction:: set_printoptions
Serialization
----------------------------------
.. autofunction:: save - Saves an object to a disk file
.. autofunction:: load - Loads an object saved with torch.save() from a file
Parallelism
----------------------------------
.. autofunction:: get_num_threads - Gets the number of OpenMP threads used for parallelizing CPU operations
Spectral Ops
~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: stft - Short-time Fourier transform
.. autofunction:: hann_window - Hann window function
.. autofunction:: hamming_window - Hamming window function
.. autofunction:: bartlett_window - Bartlett window function
BLAS and LAPACK Operations
~~~~~~~~~~~~~~~~~~~~~~~~~~~
.. autofunction:: addbmm - Batch add and mulitply matrices nxp + b×n×m X b×m×p -> bxnxp
.. autofunction:: addmm - Add and mulitply matrices nxp + n×m X m×p -> nxp
.. autofunction:: addmv - Add and matrix, vector multipy n + nxm X m -> n
.. autofunction:: addr - Outer product of vectors
.. autofunction:: bmm - Batch mulitply matrices b×n×m X b×m×p -> b×n×p
.. autofunction:: btrifact - LU factorization
.. autofunction:: btrifact_with_info
.. autofunction:: btrisolve
.. autofunction:: btriunpack
.. autofunction:: dot - Dot product of 2 tensors
.. autofunction:: eig - Eigenvalues and eigenvectors ofsquare matrix
.. autofunction:: gels - Solution for least square or p-norm(AX - B)
.. autofunction:: geqrf
.. autofunction:: ger - Outer product of 2 vectors
.. autofunction:: gesv - Solve linear equations
.. autofunction:: inverse - Inverse of square matrix
.. autofunction:: det - Determinant of a 2D square Variable
.. autofunction:: matmul - Matrix product of tensors
.. autofunction:: mm - Matrix multiplication
.. autofunction:: mv - Matrix vector product
.. autofunction:: orgqr - Orthogal matrix Q
.. autofunction:: ormqr - Multiplies matrix by the orthogonal Q matrix
.. autofunction:: potrf - Cholesky decomposition
.. autofunction:: potri - Inverse of a positive semidefinite matrix with Cholesky
.. autofunction:: potrs - Solve linear equation with positive semidefinite
.. autofunction:: pstrf - Cholesky decomposition of a positive semidefinite matrix
.. autofunction:: qr - QR decomposition
.. autofunction:: svd - SVD decomposition
.. autofunction:: symeig - Eigenvalues and eigenvectors
.. autofunction:: trtrs - Solves a system of equations with a triangular coefficient | 5,611 | 18,604 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.767619 |
https://www.jiskha.com/display.cgi?id=1445803113 | 1,532,199,779,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592654.99/warc/CC-MAIN-20180721184238-20180721204238-00627.warc.gz | 942,002,232 | 3,646 | # MATH
posted by CHEY
A person who is 6 feet tall is standing 117 feet from the base of a tree, and the tree casts a 130 foot shadow. The person's shadow is 13 feet in length. What is the height of the tree?
1. Yasmine
its a proportion
6/13=x/130
2. j
60
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More Similar Questions | 629 | 2,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.90722 |
https://math.stackexchange.com/questions/2533476/explain-lim-n-to-infty-sup-x-in-bbb-r-left-sum-n-n-infty-a-nx-r?noredirect=1 | 1,566,685,331,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321786.95/warc/CC-MAIN-20190824214845-20190825000845-00115.warc.gz | 541,811,222 | 31,060 | # Explain $\lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right| =0$
$\sum_{ n=0}^{\infty} a_n(x)$ converges uniformly if and only if $$\lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right|=0$$
This proposition is not in the book I'm using - I'm only aware of the cauchy criterion and Weiserstrass m-test for uniform convergence. Can someone tell me whether this is a proof or definition, and how it is derived. In particular I'm not sure what $\sup_{x\in\Bbb R}$ means. Also an example of the proposition used in a short proof would be helpful.
UPDATE:
Seems the proposition does follow from this basic definition:
Only that I'm not sure why the $\sup_{x\in\Bbb R}$ comes up. Can anyone explain?
• That is trivial. just give us what definition your book has. By the way answer below is correct – Guy Fsone Nov 23 '17 at 7:18
• $\sup_{x \in \R}$ means the supremum over the reals. The supremum is the same as the least upper bound of a set of real numbers. In this case, for each N, you’re looking at the set of all absolute values of the tail ends of the series that start at N, each element depending on x, and you want to know what the least upper bound is of this set, i.e., the supremum of the set. – gorzardfu Nov 23 '17 at 7:23
• @gorzardfu Hi, I updated my post. I still have trouble with the $\sup_{x\in\Bbb R}$. I understand what you said in terms of what it means. But I can't see why its there - the definition I'm given has no mention of supremums. – helios321 Nov 23 '17 at 8:03
If you define $u_n(x) = \sum_{k=1}^n a_k(x)$ and the limit as $u(x) = \sum_{k=1}^\infty a_k(x) = \lim _{n\to\infty} u_n(x)$, then $u_n$ converges uniformly to $u$ in $\Bbb R$ if it converges in $\|\cdot\|_\infty$ norm on $\Bbb R$, i.e. $$0 = \lim_{n\to\infty} \|u_n - u\|_\infty = \lim_{n\to\infty} \sup_{x\in \Bbb R} \left|\sum_{k=1}^n a_k(x) - \sum_{k=1}^\infty a_{k}(x)\right| = \lim_{n\to\infty} \sup_{x\in \Bbb R} \left|\sum_{k=n+1}^\infty a_k(x)\right|,$$ and notice that the subscript $k=n+1$ can be displaced by $1$. | 710 | 2,070 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-35 | latest | en | 0.802703 |
https://amva4newphysics.wordpress.com/2016/09/20/hermione-had-become-a-bit-pink/ | 1,685,586,746,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00135.warc.gz | 119,736,529 | 35,655 | by Pietro Vischia
Hi there!
So, in a recent blog post I told a bit about me. However, this morning I was lazily re-reading it, when I suddenly remembered that a text is nothing more than a collection of elements, linked one to each other by some rules (grammar) and some standard associations: by standard associations, I mean that the topic you are writing about dictates the words that you are most likely to use. For example, if you are writing about baking a cake, you are more likely to use terms as oven, wheat, eggs, than if you are writing about a guidance system for missiles. Hopefully 😀
This lead me to remember two concepts: frequency tables and Markov Chains.
Frequency tables are very simple: you take an input text, you pick each word, and you count how many times it is followed by another word. For example, you take hot, and you might discover that 60% of the times it is followed by cake, and 40% of the times by topic. You do that for each word of the text.
And now we come to the awesome part, that is the Markov chain stuff: a Markov chain is, as Wikipedia puts it, a stochastic process that satisfies the Markov property. In layman’s terms, this means a process in which the evolution of the system depen
ds only on the current state of the system, and not on its past history. This means that you only need an initial state and a decision rule for any given state in order to perform a full evolution of the system, but that the future evolution will not depend on that initial state: it will depend only on the state immediately preceding any further state.
Let’s apply it to our text, to make it clear. Let’s say that from our input text we select as initial state the word hot. If we take our frequency table to be our decision rule, we see that we can generate the next word of our new text by picking up either cake with 60% probability or topic with 40% probability. Let’s say we picked cake. Now, to generate the next word, we look at the frequency table for the word cake, which might be constituted by made 90% of the times and from 10% of the times. We would obtain then either hot cake made or hot cake from.
As you can see, the probability of picking up made or from depends only on the frequency table (our decision rule) and not on the past state of the system: it is not a function of the frequency table for the word hot. If we modified the frequency table for hot, when eventually arriving at the word cake we would still have an evolution dictated by a 90% probability of finding made and 10% of finding from.
Good, we have all the ingredients for a little game, but we need one thing: some software to do that for us. Since it is Sunday and I have stuff to do (involving cleaning the house and having a nice walk), I did not write myself the code for such a method (I might do it in the next days, though). I just found some code from Andrew Plotkin, which is very nice, also because it enables to group words together.
Since the language rules are quite strict (both in terms of grammar and in terms of associations), instead of building a frequency table for single words we can build a frequency table for pairs of words. This will enable us to find groups of words that will be more likely to be following our usual grammar rules, resulting in a text that will somehow make more sense. Another very nice feature of Andrew’s code is that it includes punctuation and paragraph division in the count, thus making the output text much more realistic.
OK, what else do we need? Well, an input text, of course! Let’s take my presentation post and say that we want to generate 1000 words using frequency tables built by pairs of words. Here is the result, that I will comment below:
Hi!
Very shamefully, this is my first year, I looked at the University of Oviedo, in Spain. Here, my analysis work will be oriented to searches of new physics, mostly in the stat stuff” for the first pages, it already promises to be built in a future multivariate study.
During my Ph.D. in Lisbon, I started to read a Spanish series of books called “Las aventuras de capitán Alatriste” (The Adventures of Captain Alatriste) that reportedly is the Spanish region Oviedo is the heavier brother of the Tevatron accelerator, near Chicago), so there I started working at simple tasks as learning how to plot variables.
An additional internship experience, the following year, finally convinced me that the challenge of their energy. The results were somehow inconclusive, though, leaving just the hope of being able to exploit one particular variable in a future multivariate study.
During my summer student I had the patience of arriving here) and encourage you to drop by in the CMS detector.
I moved then to the Higgs decay topology.
Actually, digging up old emails: by removing QCD events via the Neural Network, the resolution on the value of some ongoing war, had been performed by Ohio University on improving the di-jet mass got down from 18.0% to 10.7%, according to some criteria, and finally you redo the whole process a few non-work-related bits about me!
I always loved reading (thanks to my notes!!!
For the master thesis, I switched to the measurement of the work is to distinguish jets from hadronically decaying taus. I exploited this final state also in the stat stuff” for the jets coming from quarks of type “b” have some peculiar characteristics, that can be exploited to classify events into “signal-like” events and “background-like” events. For this classification problem, I used a machine learning technique called Boosted Decision Trees, which is somehow an extension of the BDT permitted to enhance significantly (in some cases even by 50%) the ability of choosing, in the CMS detector.
I moved then to the CMS experiment at the University of Oviedo group that is joining the network: more on the society.
I went on to humanities high school and beyond. Nowadays, I would have chosen on the basis of the most recent years I started to play live roleplaying games: the difference is that you are not seated around a table describing what your characters do; you are roaming around a table describing what your characters do; you are roaming around a room, or a garden, or a town, effectively acting as if you were your character. This implies that you can forget stuff typical of a heavy charged Higgs decays mostly into a top quark mass as a function of the blog (Sabine, the Press Office of the network!), so I will summarize the rest quite briefly: also, I must only say that it is expected to find two jets is not known (there are neutrinos, and when we collide two proton beams we don’t know exactly the longitudinal momentum of the muon, and its signature characteristic is that it simply reflects my prior ranking of all the possible discriminant variables: this way, I would then skip mentioning the fact that this practice helped me with my general dialectic skills. If you drop by in the CMS experiment at Tevatron, where I am reading none of them: I just began my postdoc at the Conservatory. I was preparing for a direct search of a solid scientific result. It is actually a pity that a non-negligible part of what an experimental physicist should know is not really “work”.
During my Ph.D. last July, and am currently in the comments for any question. For now, talk to you in the event, the correct pair of jets for the production of a heavy charged Higgs boson is lower than 173 GeV), the charged Higgs boson decaying into a tau and a bottom quark!
Across the years of my Ph.D., so right now I am still working nowadays. There, I worked on something we call “VBF H->ZZ->mumubb” Higgs boson decaying into two muons and the other side (Then you do this many times, for many variables, according to some criteria, and finally you redo the whole process a few hundred or thousand times). So, what has the result been? Well, it turned out that the Higgs boson search: OK, what the hell does this mean? Let’s start with “H”: that is the coolest physicist you might think about, if you were your character. This implies that you can forget stuff typical of the LHC).
How on Earth did I end up doing these things? During primary school I already knew that I just began my postdoc at the LHC, where I was a typical search channel for a heavy charged Higgs boson with mass higher than the top quark pair events in the context of a solid scientific result. It is actually a pity that a non-negligible part of what an experimental physicist working in the near future. Uh, and I soon became fascinated by the idea of becoming a lawyer (not a judge. Lawyer is cooler: Perry Mason was a typical search channel for a heavy charged Higgs boson we discovered could have some delicate statistics issue: having direct access to tricky and delicate conversations on statistics boosted and is boosting my understanding and experience in this blog some posts from Giles and from my former supervisor João Varela), where I have probably started to think that most likely the most amiable professor. Back at the time gathering useful documentation on the network blog: I hope to make up for this by spamming writing a reasonable amount of
Well, what can we say? First of all, there are some scrambled bits that are quite funny (meaning that from a grammatical point of view they make sense, but they say unexpected things), but we can also see that there are entire phrases that appear exactly as they appeared in the original blog post! Why is that? Well, this is simple: the frequency tables are often composed by a 100% probability of going to a given additional word, and this is due to the fact that the original blog post is composed by roughly 3000 words; this is not enough to have multiple choices in most of the steps.
What if we tried to use a longer input text? Well, I tried using as an input Harry Potter and the Sorcerer’s Stone, for a total of 78451 words, and generated a 1000-words text out of it, using again a frequency table built for 2-words groups.
Here is the output, that I will comment on below:
Harry Potter and the Weasley twins insisted that nothing had been trying to get past that three-headed dog at Halloween! That’s where he was laden down with rock cakes were shapeless lumps with raisins that almost broke their teeth, but Harry and felt it. It was chalk white with glaring red eyes and slits for nostrils, like a toothless walnut. The low buzz of chatter stopped when they walked in. Everyone seemed to have forgotten that Malfoy had
gotten a detention, too.
“It’s tonight,” said Harry, “so there’s someone called Nicolas Flamel is, that’s all,” said Hermione.
“I saw you and me, that’s saying something. You know that night you won Norbert? What did the best morning he’d had a word to him. The goblin wrinkled his nose. Harry watched the goblin on their right weighing a pile of Chocolate Frogs from Hermione.
“Well, it’s best if I added powdered root of asphodel to an infusion of wormwood?” Powdered root of what they were going to be somewhere, we’ll see you later –”
Professor McGonagall Harry made a mistake. I don’t think I know I don’t know
what was good at drawing, had done the guarding, really.” Hermione went on. “We wondered who had her arm around him.
“They’ll all forget this in a letter? These people will never forget you!”
“Neville, “Ron exploded, “get away from Professor Flitwick, you know.” He put Hedwig inside first and then shot at Wood, who dived on top of all were Professor Snape’s classes down in her pocket, she scrambled back along the walls, keeping their eyes on his card and gave Hermione a rare smile.
“Las’ time I saw you, you just remember what I’m famous for. I don’t see why
first years never — you seen anythin’?”
Ronan didn’t answer Aunt Petunia kept looking around. Everything looked so grim and worried, or why they had been given a wild jerk and Harry stood quite still, both thinking the
same thing — did the stranger you were trying to reason with him.
“Very good,” said Wood.
“Er — okay,” said Harry.
“ALL WHAT?” Hagrid thundered. “Now wait jus’ one second!”
He walked forward and pointed at the lights of passing cars and wondering….
They didn’t meet anyone else until they spotted a notice pinned up in the school, not jus’ then, anyway.
Harry was glad school was over, Hagrid himself was in the Gryffindor Seeker, which
could happen to anyone, I’m sure, so a penalty because George Weasley really did fall off his thick black coat and threw them into a nightmare — this year, the third-floor corridor
— and the round-faced boy Harry had to be working with Hagrid it wouldn’t budge, not even dressed in Muggle clothes, swapping rumors.”
She threw a dirty look at him and he got his revenge, without even realizing he was a plump woman who was quite bald and looked up at Harry, who couldn’t understand why they looked like a cork shot out of a centaur. Ruddy stargazers. Not interested in a minute, I hope…
And now there were a bit nearer home, said Hagrid. “So yeh haven’t noticed anythin’ strange?”
Yet Harry Potter day in the furor over the points they’d lost. He half expected Voldemort to find out that dangerous nonsense?”
That third night he tried to get any points back if we win.”
“Just as long as it may, fighting is against Hogwarts rules, Hagrid,” said Snape silkily. “Five points from Gryffindor. He sprinted back upstairs.
“Did you see all your family standing around you. Ronald Weasley, who has always been overshadowed by his name. Please cheer up, Hagrid, we saved the Stone, well, I’ll have thirty … thirty…”
“Thirty-nine, sweetums,” said Aunt Petunia had sheared it off He had no choice. The cut had turned to the first time in his step — Quirrell seemed to end with him in we’d put a stop to that rubbish,” said Uncle Vernon, “swore we’d stamp it out with difficulty, because it said:
CHAPTER SIXTEEN
THROUGH THE TRAPDOOR
In the back of his half-moon glasses. “It would be back to normal next year, or as normal as it cringed away from Curses and Countercurses (Bewitch Your Friends and Befuddle Your Enemies with the bat to stop me! Voldemort killed my mother because she said impatiently.
“You’d think they’d be running around looking for it again. Anyway, this — this wizard, about twenty years ago now, started lookin’ fer followers. Got ’em, too — Beaters.”
“I found out about the special circumstances, Potter. And what model is it?”
“A minute –“
Hermione had become a bit pink and pointing to a baked potato when Professor Quirrell
came sprinting into the clearing came — was that their sudden appearance had taken one step toward them out of his newspaper as usual and Dudley were leaning right up to something. And Gryffindor really can’t afford to lose him.” On Christmas Eve, Harry went to bed with a hag — never been brought up to it.”
“Nothing?” said Ron sleepily.
“Knuts?”
“The big one,” said Hermione. “I know he’s not about ter steal it.”
Harry watched Snape for a moment later they stood blinking in the house, rolled up and we can’t win at Quidditch?”
“Blimey, Harry, I don’ s’pose it could hurt ter tell yeh — mind you, he’s usually tremblin’.”
“Is it — any questions?”
Harry had never seen anything like it. Was that the Dursleys almost ten years, ten miserable years, as long as he poured sugar on his shoulder and looked until a distant shout.
“Is that where -?” whispered Professor McGonagall.
She turned to the portrait of the Sorcerer’s Stone “You can’t –“
“Bet you could,” Ron muttered.
There we are. Behold the power of Markov Chains!
Grammatically, most of the text is OK, and the sentences kind of make sense, although in a bizarre way (well, the bizarre part was our purpose from the beginning, uh? 😉 ).
The most astounding characteristic, though, is that the output text is by construction statistically equivalent to the input one (at least, if we take the set of frequencies as an estimator of statistical equivalence)! Markov chains can actually be used to analyze texts in order to decide if they are similar, which could enable to make some inference about authorship.
For example, if we find a new Shakespeare theater piece, we can compare it statistically with the full Shakespeare corpus and determine if it is compatible or not! Of course this is not a 100% secure method (authors change style, for example, or write in a different way for different purposes), but still it is something I find quite impressive!
If you want to explore the world of computer-generated text, though, you could go on and check stuff based on context-free grammar, which is basically a set of rules that, given a formal grammar, can generate any possible string regardless of the context.
It might seem silly, but computer-generated texts have gone way beyond simple examples like the ones I gave you. A software called Mathgen had a computer-generated gibberish text accepted for publication in an advanced mathematics journal! Similarly, the authors of SCIgen managed to get invited to a conference by sending a computer-generated paper, as they describe in a Reddit AMA. In one of the answers, by the way, they discuss also Markov chain generators:
we explicitly avoided Markov chains or anything else that was technically challenging, in the service of trying to make the papers as funny as possible. With Markov chains, you might get something syntactically correct, but it is likely to be boring.
Well, it is time for me to start cooking: I hope you found Markov chains as awesome as I do think they are! If you want to play with them, you can download and compile Andrew’s code (just add a #include<string> on top, and compile with cc or gcc), or find some more generators online!
By the way, if you like Harry Potter, you might find some amusement in reading Harry Potter and the Methods of Rationality, which is a fan-fiction revisiting the original books by applying the scientific method to Harry’s world!
Have a nice Sunday (or whichever day this will be published 😛 )!
P.S. No Muggle has been hurt during the preparation of this post. | 4,103 | 18,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-23 | latest | en | 0.947379 |
http://sw3d.mff.cuni.cz/software/sw3dcd18/data/len/len-mod.dat | 1,675,125,931,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00510.warc.gz | 39,581,494 | 4,167 | 'Sample input data len-mod.dat: data for the model. 1999, March 21.' 0 1 1 / (Cartesian coordinates, velocities, loss factors) 0 60 -20 20 -25 10 (boundaries of the model) 4 surfaces 5 simple blocks: -1 2 -3 / -1 3 / 1 2 -4 / 4 / -2 / (free space) 3 complex blocks: 1 3 / 2 / 4 / 'SURFACE' 1 1 2 -3 0 / (i.e. W(X1,X2)-X3=0, tension=0) 3 3 (numbers of grid points) 0 20 40 (X1 grid coordinates) -20 0 20 (X2 grid coordinates) 0 0 0 0 -12 0 0 0 0 (X3 coordinates at grid points) 'SURFACE' 2 1 -3 0 0 / (i.e. W(X1)-X3=0, tension=0) 4 (numbers of grid points) 0 20 40 60 (X1 grid coordinates) 3 3 0 3.5 (X3 coordinates at grid points) 'SURFACE' 3 1 2 -3 0 / (i.e. W(X1,X2)-X3=0, tension=0) 3 3 (numbers of grid points) 0 20 30 (X1 grid coordinates) -20 0 20 (X2 grid coordinates) -12 -12 -12 -12 -5 -12 -12 -12 -12 (X3 coordinates at grid points) 'SURFACE' 4 1 -3 0 0 / (i.e. W(X1)-X3=0, tension=0) 5 (numbers of grid points) 0 10 25 40 60 (X1 grid coordinates) -16 -17 -20 -17 -15 (X3 coordinates at grid points) 'END OF SURFACES' / 'COMPLEX BLOCK' 1 'VP ' 1 (P wave velocity) 3 0 0 0 / (i.e. VP=W(X3), tension=0) 2 (numbers of grid points) 0 -20 (X3 grid coordinates) 4 6 (velocities at grid points) 'DEN' 1 (density) 4 0 0 0 / (i.e. density=W(VP), tension=0) 2 (numbers of grid points) 0 10 (VP grid coordinates) 1.7 3.7 (densities at grid points, density=1.7+0.2*VP) 'COMPLEX BLOCK' 2 'VP ' 1 (P wave velocity) 0 0 0 0 / (i.e. VP=constant, tension=0) 6 (the value of velocity) 'DEN' 1 (density) 0 0 0 0 / (i.e. density=constant, tension=0) 2.9 (the value of density) 'COMPLEX BLOCK' 3 'VP ' 1 (P wave velocity) 0 0 0 0 / (i.e. VP=constant, tension=0) 7 (the value of velocity) 'DEN' 1 (density) 0 0 0 0 / (i.e. density=constant, tension=0) 3.1 (the value of density) 'END OF COMPLEX BLOCKS, END OF THE INPUT DATA FOR THE MODEL' / ------------------------------------------------------------------------ This 3-D model consists of two layers and of the lenticular inclusion with edges, situated in the upper layer, see the schematic Figure 5 in V.Cerveny, L.Klimes, I.Psencik: Complete Seismic-Ray Tracing in 3-D Structures (In: D.Doornbos, ed.: Seismological Algorithms, Academic Press 1988). For the general description of the input data format see the subroutine packages 'model', 'srfc' and 'parm'. P wave velocity in the upper layer (Complex Block 1) increases linearly with depth and has no lateral variations. The lenticular inclusion (Complex Block 2) and the bottom layer (Complex Block 3) are homogeneous. S wave velocities and loss factors are not specified by these input data, the default values (VS=0.57735*VP, zero loss factors) are assumed. The comments in brackets (starting from the column 23) in these input data have no influence on reading the data. ------------------------------------------------------------------------ Debugging information: The input data are stored in the common blocks /MODELT/, /MODELC/ and /VALC/. For the description of the common blocks /MODELT/ and /MODELC/ refer to the subroutine package 'model', for the common block /VALC/ refer to the subroutine package 'val'. The arrays KSB and KCB of the common block /MODELC/ and the arrays IPAR and RPAR of the common block /VALC/ are filled by the parameters of the model in the following way: Arrays KSB and KCB of the common block /MODELC/: I-----------I-----------I I-----------I-----------I I I K KSB(K) I I I K KCB(K) I I-----------I-----------I I-----------I-----------I I NSB I 0 5 I I NCB I 0 3 I I-----------I-----------I I-----------I-----------I I S.Block 1 I 1 8 I I C.Block 1 I 1 5 I I S.Block 2 I 2 10 I I C.Block 2 I 2 6 I I S.Block 3 I 3 13 I I C.Block 3 I 3 7 I I S.Block 4 I 4 14 I I-----------I-----------I I S.Block 4 I 5 15 I I C.Block 1 I 4 1 I I-----------I-----------I I I 5 3 I I I 6 -1 I I-----------I-----------I I S.Block 1 I 7 2 I I C.Block 2 I 6 2 I I I 8 -3 I I-----------I-----------I I-----------I-----------I I C.Block 3 I 7 4 I I S.Block 2 I 9 -1 I I-----------I-----------I I I 10 3 I I-----------I-----------I I I 11 1 I I S.Block 3 I 12 2 I I I 13 -4 I I-----------I-----------I I S.Block 4 I 14 4 I I-----------I-----------I I S.Block 5 I 15 -2 I I-----------I-----------I Arrays IPAR and RPAR of the common block /VALC/: I--------------------------------------I------------I I Description I K IPAR(K) I I--------------------------------------I------------I I (A) Number of the classes I 0 2 I I-------------------------I------------I------------I I (B) Classes I Class 1 I 1 6 I I I Class 2 I 2 9 I I------------I------------I------------I------------I I I I Surface 1 I 3 10 I I I Class 1 I Surface 2 I 4 11 I I I I Surface 3 I 5 12 I I (C) Groups I I Surface 4 I 6 13 I I I------------I------------I------------I I I I C.Block 1 I 7 60 I I I Class 2 I C.Block 2 I 8 107 I I I I C.Block 3 I 9 154 I I------------I------------I------------I------------I I I Surface 1 I I 10 207 I I I------------I------------I------------I I I Surface 2 I I 11 242 I I I------------I------------I------------I I I Surface 3 I I 12 295 I I I------------I------------I------------I I I Surface 4 I I 13 337 I I I------------I------------I------------I I I I P velocity I 14 358 I I I I Density I 15 379 I I I C.Block 1 I No data I 16 379 I I (D) I I No data I 17 379 I I Functions I I No data I 18 379 I I I I . I . . I I I I : I : : I I I I No data I 60 379 I I I------------I------------I------------I I I I P velocity I 61 386 I I I I Density I 62 393 I I I C.Block 2 I No data I 63 393 I I I I No data I 64 393 I I I I No data I 65 393 I I I I . I . . I I I I : I : : I I I I No data I107 393 I I I------------I------------I------------I I I I P velocity I108 400 I I I I Density I109 407 I I I C.Block 3 I No data I110 407 I I I I No data I111 407 I I I I No data I112 407 I I I I . I . . I I I I : I : : I I I I No data I154 407 I I------------I------------I------------I------------I (E) Parameters of individual functions: Surface 1: Surface 2: Surface 3: Surface 4: I-------------I I-------------I I-------------I I-------------I I K IPAR(K) I I K IPAR(K) I I K IPAR(K) I I K IPAR(K) I I RPAR(K) I I RPAR(K) I I RPAR(K) I I RPAR(K) I I-------------I I-------------I I-------------I I-------------I I 155 1 I I 208 1 I I 243 1 I I 296 1 I I 156 1.000 I I 209 1.000 I I 244 1.000 I I 297 1.000 I I 157 1 I I 210 1 I I 245 1 I I 298 1 I I 158 2 I I 211 -3 I I 246 2 I I 299 -3 I I 159 -3 I I 212 0 I I 247 -3 I I 300 0 I I 160 0.000 I I 213 0.000 I I 248 0.000 I I 301 0.000 I I 161 3 I I 214 4 I I 249 3 I I 302 5 I I 162 3 I I 215 0.000 I I 250 3 I I 303 0.000 I I 163 0.000 I I 216 20.000 I I 251 0.000 I I 304 10.000 I I 164 20.000 I I 217 40.000 I I 252 20.000 I I 305 25.000 I I 165 40.000 I I 218 60.000 I I 253 30.000 I I 306 40.000 I I 166 -20.000 I I 219 3.000 I I 254 -20.000 I I 307 60.000 I I 167 0.000 I I 220 2.822 I I 255 0.000 I I 308 -16.000 I I 168 20.000 I I 221 -1.289 I I 256 20.000 I I 309 -10.691 I I 169 0.000 I I 222 3.500 I I 257 -12.000 I I 310 -14.537 I I 170 0.000 I I 223 .004 I I 258 -7.200 I I 311 -10.231 I I 171 0.000 I I 224 .000 I I 259 -12.000 I I 312 -15.000 I I 172 0.000 I I 225 .002 I I 260 -8.000 I I 313 .005 I I 173 -12.000 I I 226 .250 I I 261 2.200 I I 314 .000 I I 174 0.000 I I 227 .167 I I 262 -8.000 I I 315 .007 I I 175 0.000 I I 228 .000 I I 263 -12.000 I I 316 .345 I I 176 0.000 I I 229 -.007 I I 264 -7.200 I I 317 .257 I I 177 0.000 I I 230 .004 I I 265 -12.000 I I 318 .000 I I 178 .002 I I 231 .000 I I 266 .004 I I 319 -.020 I I 179 .000 I I 232 .250 I I 267 .000 I I 320 .008 I I 180 .002 I I 233 .004 I I 268 .004 I I 321 .000 I I 181 .167 I I 234 -.007 I I 269 .067 I I 322 .292 I I 182 .167 I I 235 .000 I I 270 .067 I I 323 .009 I I 183 .000 I I 236 .250 I I 271 .000 I I 324 -.014 I I 184 -.007 I I 237 .000 I I 272 -.015 I I 325 .005 I I 185 .000 I I 238 .002 I I 273 .000 I I 326 .151 I I 186 .000 I I 239 .000 I I 274 .000 I I 327 .345 I I 187 .000 I I 240 .004 I I 275 .000 I I 328 .006 I I 188 .002 I I 241 .167 I I 276 .005 I I 329 -.010 I I 189 .000 I I 242 .250 I I 277 .000 I I 330 .000 I I 190 .002 I I-------------I I 278 .005 I I 331 .229 I I 191 .167 I I 279 .333 I I 332 .000 I I 192 .167 I I 280 .333 I I 333 .003 I I 193 .002 I I 281 .002 I I 334 .000 I I 194 .000 I I 282 .000 I I 335 .009 I I 195 .002 I I 283 .002 I I 336 .117 I I 196 .167 I I 284 .167 I I 337 .151 I I 197 .167 I I 285 .167 I I-------------I I 198 .000 I I 286 .000 I I 199 -.007 I I 287 -.007 I I 200 .000 I I 288 .000 I I 201 .000 I I 289 .000 I I 202 .000 I I 290 .000 I I 203 .002 I I 291 .002 I I 204 .000 I I 292 .000 I I 205 .002 I I 293 .002 I I 206 .167 I I 294 .167 I I 207 .167 I I 295 .167 I I-------------I I-------------I C.Block 1: VP: C.B.1: Density: C.Block 2: VP: C.Block 3: VP: I-------------I I-------------I I-------------I I-------------I I K IPAR(K) I I K IPAR(K) I I K IPAR(K) I I K IPAR(K) I I RPAR(K) I I RPAR(K) I I RPAR(K) I I RPAR(K) I I-------------I I-------------I I-------------I I-------------I I 338 1 I I 359 3 I I 380 1 I I 394 1 I I 339 1.000 I I 360 1.000 I I 381 1.000 I I 395 1.000 I I 340 3 I I 361 4 I I 382 0 I I 396 0 I I 341 0 I I 362 0 I I 383 0 I I 397 0 I I 342 0 I I 363 0 I I 384 0 I I 398 0 I I 343 0.000 I I 364 0.000 I I 385 0.000 I I 399 0.000 I I 344 2 I I 365 2 I I 386 6.000 I I 400 7.000 I I 345 -20.000 I I 366 0.000 I I-------------I I-------------I I 346 0.000 I I 367 10.000 I I 347 6.000 I I 368 1.700 I C.B.2: Density: C.B.3: Density: I 348 4.000 I I 369 3.700 I I-------------I I-------------I I 349 .002 I I 370 1.000 I I K IPAR(K) I I K IPAR(K) I I 350 .000 I I 371 .000 I I RPAR(K) I I RPAR(K) I I 351 .000 I I 372 .000 I I-------------I I-------------I I 352 .167 I I 373 .167 I I 387 3 I I 401 3 I I 353 .000 I I 374 .000 I I 388 1.000 I I 402 1.000 I I 354 .000 I I 375 .000 I I 389 0 I I 403 0 I I 355 .000 I I 376 .000 I I 390 0 I I 404 0 I I 356 .002 I I 377 1.000 I I 391 0 I I 405 0 I I 357 .000 I I 378 .000 I I 392 0.000 I I 406 0.000 I I 358 .167 I I 379 .167 I I 393 2.900 I I 407 3.100 I I-------------I I-------------I I-------------I I-------------I Attention: The storage of the above model coefficients in the common block /VALC/ may be changed when reading additional input data, especially when defining auxiliary surfaces in the model. Example: If calling subroutine SRFC1 once again to read the data 'AUXILIARY SURFACE' 5 -1 0 0 0 (I.E. W-X1=0, TENSION=0) 0. (W=0) 'END OF AUXILIARY SURFACES' / in order to define additional surface 5, the above contents of arrays IPAR and RPAR of the common block /VALC/ would become modified in the following way: (A),(B),(C): Parameters (IPAR(K),K=1,6) are increased by 1. (C),(D): Parameters (IPAR(K),K=7,13) are increased by 2 and shifted to (IPAR(K),K=8,14). (D): Parameters (IPAR(K),K=14,154) are increased by 9 and shifted to (IPAR(K),K=16,156). (E): Parameters (*PAR(K),K=155,337) are shifted to (*PAR(K),K=157,339). Parameters (*PAR(K),K=338,407) are shifted to (*PAR(K),K=347,416). New data added to the common block /VALC/: I--------------------------------------I------------I I Description I K IPAR(K) I I--------------------------------------I------------I I (C) Groups I Class 1 I Surface 5 I 7 15 I I------------I------------I------------I------------I I (D) Funct. I Surface 5 I I 15 346 I I------------I------------I------------I------------I (E) Parameters of individual functions: Surface 5: I-------------I I K IPAR(K) I I RPAR(K) I I-------------I I 340 1 I I 341 1.000 I I 342 -1 I I 343 0 I I 344 0 I I 345 0.000 I I 346 0.000 I I-------------I ======================================================================== | 4,438 | 11,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.550108 |
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Apr20 awarded Informed Apr20 comment Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$ Additional link for sin 1. This is also a useful document Exact Values of the Sine and Cosine Functions in In crements of 3 degrees. Apr20 revised Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$ Edited for sqrt Apr20 revised Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$ Edited for sqrt Apr20 answered Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$ Apr7 awarded Organizer Apr7 revised A 10 digit positive number and ordered triplets I have changed the Tag from Liner Algebra to elementary-number-theory Apr7 suggested approved edit on A 10 digit positive number and ordered triplets Mar18 revised IMO programs of different nations? I rewrote to added clarity. I have used points clairified in the comments. Also a spelling correction. Mar18 suggested approved edit on IMO programs of different nations? Mar17 answered Calculate difference in days to reference date Mar9 revised Why does the least common denominator work? Just added a clarification for my earlier answer. Mar9 revised How to represent sign function in range $[-1, 1]$ Put a ",' between -1 & 1 Mar9 suggested approved edit on How to represent sign function in range $[-1, 1]$ Mar9 answered Why does the least common denominator work? Jan30 revised How can a beginner researcher or Ph.D. student efficiently and effectively learn new concepts while staying motivated? Replaced the word 'type consuming' by 'time consuming' Jan30 suggested approved edit on How can a beginner researcher or Ph.D. student efficiently and effectively learn new concepts while staying motivated? Sep24 awarded Autobiographer Oct25 awarded Supporter Oct14 answered Find by integrating the area of the triangle vertices $(5,1), (1,3)\;\text{and}\;(-1,-2)$ | 479 | 1,943 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2015-18 | longest | en | 0.852941 |
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# Higher Order Derivatives: The Value of a Fourth Derivative
APCALC-ABNVIU
If $f\left( x \right) ={ x }^{ 5 }$, then $f^{ 4 }\left( 3 \right) =$
A
$120$
B
$180$
C
$360$
D
$540$ | 84 | 205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-13 | latest | en | 0.569495 |
https://www.univerkov.com/what-voltage-must-be-applied-to-a-lead-wire-2-m-long-so-that-the-current-in-the-wire-is-2-a/ | 1,642,731,219,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00296.warc.gz | 1,104,696,736 | 6,289 | # What voltage must be applied to a lead wire 2 m long so that the current in the wire is 2 A?
What voltage must be applied to a lead wire 2 m long so that the current in the wire is 2 A? Wire cross-sectional area 0.3 mm2
Initial data: l (lead wire length) = 2 m; I (required current strength) = 2 A; S (cross-sectional area of lead wire) = 0.3 mm2.
Reference data: ρ (lead resistivity, n.o.) = 0.21 Ohm * mm2 / m.
1) Let’s calculate the resistance of the lead wire: R = ρ * l / S = 0.21 * 2 / 0.3 = 1.4 Ohm.
2) Determine the voltage that needs to be applied to the wire: U = I * R = 2 * 1.4 = 2.8 V.
Answer: A voltage of 2.8 V must be applied to the lead wire.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 279 | 978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-05 | latest | en | 0.903387 |
https://www.reference.com/math/convert-octal-decimal-6fa7088136172c97 | 1,539,919,668,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00294.warc.gz | 1,047,849,741 | 17,091 | How Do You Convert Octal to Decimal?
How Do You Convert Octal to Decimal?
Octal is a system of real numbers that was commonly used in older computing systems. The main use of the octal system today is in Unix applications. It's digits range from 0 to 7. The decimal system is the main number system used today. It has nine discrete digits ranging from 0 to 9. The main purpose of octal and the more popular hexadecimal system is because they are easier to convert to and from binary than can be done by the decimal system.
1. Count the number of digits
Starting from the right, count the number of digits beginning with the number 0. The number 427 would be numbered 210.
2. Multiply each digit by 8 to a power
The next step is to multiply each individual digit by 8 to the power of that digit's place in the number. For example, if you want to convert the number 427 base 8 to decimal, you will arrange the conversion like so: (4 * 8^2) + (2 * 8^1) + (7 * 8^0).
3. Add the products of each digit
The next step is to multiply each individual digit and then add the products together: 256 + 16 + 7 = (279)base 10. Thus, the answer would be the decimal number 279.
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https://www.chessvariants.org/40.dir/oblique.html | 1,709,410,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00614.warc.gz | 700,373,275 | 11,129 | Check out Symmetric Chess, our featured variant for March, 2024.
# Oblique
Object: This 40 square Chess variant is distinguished by its board. It's the regular Chess board set at 45 degrees, hence the name. The board was designed this way to make it 40 squares, but retain a symmetrical rectangular grid. The pieces move as they do in standard chess, but seem to move obliquely due to the orientation of the board grid. There are some exceptions: there is no castling for the King; the pawn moves along the diagonals, which are now straight, and may move forward, backward, and to the sides; the Pawn captures along the orthogonals, which are now oblique; the pawn does not have the special initial move, does not promote, and cannot capture "En-passant".
This Chess variant was designed for the Chess Variant Pages 40 square variant contest. My goal was to create something special in 40 squares. The oblique grid appeals to me because it retains something of the symmetry of a regular chess board, thus retaining the greatest useful area for the game. Its also appealing because this simple change alone is enough to give the game a very different feel without changing the rules very much. All the pieces moves as they normally do, with the exception of the Pawn. However, the feel of the pieces is very different because of the oblique grid. This is particularly true of the two pieces restricted to the orthogonals and diagonals, the Rook and the Bishop. The Pawn's move is made straight by changing its movement to the diagonals. This appealed to me because it adds another element of obliqueness and retains the forward movement of the Pawn, which is important for playability. I also enhanced its powers to give it a bigger role in the game. To compensate, I made the Pawn non-promotable.
In the opening, you will need to develop your pieces. This is not so easy because there is a wall of Pawns in front of them. So the initial effort will be to create avenues for the pieces by attacking with the Pawns, or moving them laterally to support other Pawns. The middle game develops quickly. The key here is to use careful tactics to gain a material advantage for the end game. The end game is similar to standard Chess, however, the Pawns are active combatants instead of time bombs ready to be promoted.
## Setup
Oblique has an unusual setup because of the orientation of the board. The coordinates include the nodes between squares, although they play no role in the game. There is complete symmetry. Each side places the white diagonal nearest the player, and places the pieces as follows:
``` (White Player)
(white diagonal) Bishop Queen King Knight
(black diagonal) Pawn Rook Bishop Rook Pawn
(white diagonal) Pawn Pawn Pawn Pawn
(black diagonal) empty Pawn Pawn Pawn empty
(white diagonal) empty empty empty empty
(black diagonal) empty Pawn Pawn Pawn empty
(white diagonal) Pawn Pawn Pawn Pawn
(black diagonal) Pawn Rook Bishop Rook Pawn
(white diagonal) Bishop Queen King Knight
(Black Player)
```
You can play this game with Zillions of Games, using the Zrf-file, made by Tony Quintanilla.
This is a submission in the contest to design a chess variant on a board with 40 squares.
Written by Tony Quintanilla. Webpage made by Hans Bodlaender.
WWW page created: August 11, 1999.  | 773 | 3,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.964898 |
http://arstechnica.com/science/2014/01/new-cyber-attack-model-helps-predict-timing-of-the-next-stuxnet/?comments=1 | 1,484,787,041,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00052-ip-10-171-10-70.ec2.internal.warc.gz | 16,518,077 | 22,380 | New cyber-attack model helps predict timing of the next Stuxnet
It accounts for the properties of vulnerability and the political situation.
Of the many tricks used by the world’s greatest military strategists, one usually works well—taking the enemy by surprise. It is an approach that goes back to the horse that brought down Troy. But surprise can only be achieved if you get the timing right. Timing which, researchers at the University of Michigan argue, can be calculated using a mathematical model—at least in the case of cyber-wars.
James Clapper, the director of US National Security, said cybersecurity is “first among threats facing America today,” and that’s true for other world powers as well. In many ways, it is even more threatening than conventional weapons, since attacks can take place in the absence of open conflict. And attacks are waged not just to cause damage to the enemy, but often to steal secrets.
Timing is key for these attacks, as the name of a common vulnerability—the zero-day attack—makes apparent. A zero-day attack refers to exploiting a vulnerability in a computer system on the same day that the vulnerability is recognized (aka when there are zero days to prepare for or defend against the attack). That is why cyber-attacks are usually carried out before an opponent has the time to fix its vulnerabilities.
As Robert Axelrod and Rumen Iliev at the University of Michigan write in a paper just published in PNAS, “The question of timing is analogous to the question of when to use a double agent to mislead the enemy, where it may be worth waiting for an important event but waiting too long may mean the double agent has been discovered.”
Equations are as good as weapons
Axelrod and Iliev decided the best way to answer the question of timing would be through the use of a simple mathematical model. They built the model using four variables:
1. Cyber-weapons exploit a specific vulnerability.
2. Stealth of the weapon measures the chance that an enemy may find out the use of the weapon and take necessary steps to stop its reuse.
3. Persistence of the weapon measures the chance that a weapon can still be used in the future, if not used now. Or, put another way, the chance that the enemy finds out their own vulnerability and fixes it, which renders the weapon useless.
4. Threshold defines the time when the stakes are high enough to risk the use of a weapon. Beyond the threshold you will gain more than you will lose.
Using their model, it is possible to calculate the optimum time of a cyber-attack:
When the persistence of a weapon increases, the optimal threshold increases—that is, the longer a vulnerability exists, the longer one can wait before using it.
When the stealth of a weapon increases, the optimal threshold decreases—the longer a weapon can avoid detection, the better it is to use it quickly.
Based on the stakes of the outcome, a weapon must be used soon (if stakes are constant) or later (if the stakes are uneven). In other words, when the gain from an attack is fixed and ramifications are low, it is best to attack as quickly as possible. When the gain is high or low and ramifications are high, it is best to be patient before attacking.
How to plan the next Stuxnet
Axelrod and Iliev’s model deserves merit, according to Allan Woodward, a cybersecurity expert at the University of Surrey, because it fits past examples well. Their model perfectly predicts timing of both the Stuxnet attack and Iran’s counter to it.
Stuxnet was a worm aimed at interfering with Iran’s attempts to enrich uranium to build nuclear weapons. So, from an American perspective, the stakes were very high. The worm itself remained hidden for nearly 17 months, which means its stealth was high and persistence was low. According to the model, US and Israel should have attacked as soon as Stuxnet was ready. And indeed that is what seems to have happened.
Iran may have responded to this attack by targeting the workstations of Aramco, an oil company in Saudi Arabia that supplied oil to the US. Although the US called this the “most destructive cyber-assault the private sector has seen to date," it achieved little. However, for Iran, the result mattered less than the speed of the response. In a high stakes case, the model predicts immediate use of a cyber-weapon, which is what happened in this case, too.
Although the model has been developed for cyber-attacks, it can be equally effective in modeling cyber-defense. Also, the model need not be limited to cyber-weapons; small changes in the variables can be made so that the model can be used to consider other military actions or economic sanctions.
Just like the atomic bomb
Eerke Boiten, a computer scientist at the University of Kent, said: “These models are a good start, but they are far too simplistic. The Stuxnet worm, for example, attacked four vulnerabilities in Iran’s nuclear enrichment facility. Had even one been fixed, the attack would have failed. The model doesn’t take that into account.”
In their book Cyber War: The Next Threat to National Security and What to Do About It, Richard Clarke and Robert Knake write:
It took a decade and a half after nuclear weapons were first used before a complex strategy for employing them, and better yet, for not using them, was articulated and implemented.
That transition period is what current cyber-weapons are going through. In that light, the simplicity of Axelrod and Iliev’s model may be more a strength than a weakness for now.
PNAS, 2014. DOI: 10.1073/pnas.1322638111 (About DOIs).
Quote:
James Clapper [snip] is “first among threats facing America today,”
Not that he'll be charged or resign any time soon for lying in front of congress...
Quote:
James Clapper [snip] is “first among threats facing America today,”
Not that he'll be charged or resign any time soon for lying in front of congress...
I've just taken to referring to him as "The Clap" -- you know: annoying as hell, dangerous in the long term, really tough to get rid of.
3. These models were probably fitted to match the data, which is why it matches the data so wel...
These sorts of things are hard to analyze because there isnt very much data to go off of in the first place.
4. ws3 wrote:
Quote:
James Clapper [snip] is “first among threats facing America today,”
Not that he'll be charged or resign any time soon for lying in front of congress...
I've just taken to referring to him as "The Clap" -- you know: annoying as hell, dangerous in the long term, really tough to get rid of.
DDG agrees:
https://duckduckgo.com/?q=the+clap
5. "
It took a decade and a half after nuclear weapons were first used before a complex strategy for employing them, and better yet, for not using them, was articulated and implemented
"
So basically, at this rate our entire smart-enabled power grid will be owned by the time we get to fixing it.
6. The predictive power of models like this is roughly zero.
7. M-S-G wrote:
"
It took a decade and a half after nuclear weapons were first used before a complex strategy for employing them, and better yet, for not using them, was articulated and implemented
"
WARNING!!! WILD TANGENT!!!!
I'm still pondering the stupidity of that statement... (which is no allusion to its use in the article)..
It (nuclear bombing) was instantly stratigized and not used again...
Had morality not won and the technological advantage been put into (it's fullest, smallest; anywhere in between except the "not again" we used) use, the U.S.A. and its, unequaled technological power would have been used in exactly the unrestrained use it employs today in its unequaled intelligence gathering. (as only one example)
IOW: we (the U.S.A.) had more restraint then, than do now...
They knew then(or at least those in power): just because one can, does not mean one should..
...and from my perspective: that made some powerful people that, then, lacked the power to dominate technological advantage for domination of the world mad and they have stratigized, since then, to never let a technological advantage "go to waste" again...
Yeah, that's wordy and sloppy: I hope you (all) get my drift...
*edit* was a little too sloppy.
Last edited by 'LilRedDog on Mon Jan 13, 2014 5:46 pm
8. Not related to the article, but rather to the picture, but if it had been taken 4 years ago it could have been a picture of me! Working with those tie down blocks used to be part of my job as a Tech controller. Among the others like running all sort of other things like our cisco network and various pieces of multiplexing equipment. We also had a lot of old legacy stuff like E&M phone lines, and some interesting stuff I can't talk about.
Our career field has changed somewhat since I got out, turning more towards a cyber defense angle.
Its really awesome to read a random article and see something like that. edit: It definitely feels a little bit unrelated, to be honest.
9. 'LilRedDog wrote:
M-S-G wrote:
"
It took a decade and a half after nuclear weapons were first used before a complex strategy for employing them, and better yet, for not using them, was articulated and implemented
"
WARNING!!! WILD TANGENT!!!!
I'm still pondering the stupidity of that statement... (which has nothing to do with its use in the article)..
Well, given the most vulnerable cyber-attack target in the US is our power grid and the power companies base their cyber-defense budgets based on threat modelling...I'd say it is pretty fucking relevant.
10. M-S-G wrote:
'LilRedDog wrote:
M-S-G wrote:
"
It took a decade and a half after nuclear weapons were first used before a complex strategy for employing them, and better yet, for not using them, was articulated and implemented
"
WARNING!!! WILD TANGENT!!!!
I'm still pondering the stupidity of that statement... (which has nothing to do with its use in the article)..
Well, given the most vulnerable cyber-attack target in the US is our power grid and the power companies base their cyber-defense budgets based on threat modelling...I'd say it is pretty fucking relevant.
You think it took a decade and a half to realize: "we should not nuke indiscriminately"?
....and I edited the parenthetical since then to be clear: I'm not dissing the author of this article.
11. Quote:
You think it took a decade and a half to realize: "we should not nuke indiscriminately"?
....and I edited the parenthetical since then to be clear: I'm not dissing the author.
I don't know man. I would like to be optimistic about this but it really does look like we are going to repeat all the same mistakes again.
Also, I don't know of a threat model that works for future technology ???
12. M-S-G wrote:
Quote:
You think it took a decade and a half to realize: "we should not nuke indiscriminately"?
....and I edited the parenthetical since then to be clear: I'm not dissing the author.
I don't know man. I would like to be optimistic about this but it really does look like we are going to repeat all the same mistakes again.
Also, I don't know of a threat model that works for future technology ???
I'll take responsibility: I was not clear.
We did it right with nukes. we used them twice and never again; too much concentrated power...
I wish we would make that "mistake" again...
2nd WILD TANGET!!!
The arms race that has followed( regardless the tech) was the tipping point where technology outpaced our morality...
13. One key here is how to estimate each parameter accurately. If the ramifications or threshold are underestimated the results could be disastrous. The importance of accuracy is evidenced by the underestimated risk and ramification estimates the NSA made in their assessments of their spying programs and the resulting firestorm after they were leaked.
14. What about all the cyber weapons the NSA has their hands on that are not being fixed because they want to exploit them instead? It's their own doing that is making us far more susceptible to future attacks by systematically weakening standards for their own twisted uses.
There is no model right now there can't be because our current is corrupted which makes any calculations based on that system completely worthless. When the shit hits the fan it's going to be our own fault and I hate that.
15. 'LilRedDog wrote:
[snip[
We did it right with nukes. we used them twice and never again; too much concentrated power... [snip]
We (the United States) really used atomic weapons once. Yes, there were two bombs and two targets, three days apart. But the message was that the United States was equipped and prepared to rain atomic destruction on the Japanese Home Islands until Japan surrendered. A single bomb would not have sent that message. Based on what I know, it's a good thing two of them did because I don't think we had a third one.
(For anyone who wants to debate the ethics of the atomic attack on Japan, I stand by "good thing" because the alternative was a ground invasion of the Japanese Home Islands.)
16. Bob.Brown wrote:
'LilRedDog wrote:
[snip[
We did it right with nukes. we used them twice and never again; too much concentrated power... [snip]
We (the United States) really used atomic weapons once. Yes, there were two bombs and two targets, three days apart. But the message was that the United States was equipped and prepared to rain atomic destruction on the Japanese Home Islands until Japan surrendered. A single bomb would not have sent that message. Based on what I know, it's a good thing two of them did because I don't think we had a third one.
(For anyone who wants to debate the ethics of the atomic attack on Japan, I stand by "good thing" because the alternative was a ground invasion of the Japanese Home Islands.)
IIRC, we didn't have a 4th (the first bomb was the test bomb at Alamogordo), bomb at that exact moment, but Groves said they could have had one ready by August 17th, and that if the war had continued there were plans to use up to a half dozen bombs as tactical weapons in support of the invasion.
Anyway, back on topic, it seems to this layman that the model isn't really much more than the codification of the various factors that go into a decision to use/not use cyberattacks.
The flaw I see for most people/organizations trying to use that model is that they don't have access to the all of the intelligence that went into the 'attack' decision until after the attack, if they ever have access to that particular information, that is.
17. Bob.Brown wrote:
'LilRedDog wrote:
[snip[
We did it right with nukes. we used them twice and never again; too much concentrated power... [snip]
We (the United States) really used atomic weapons once. Yes, there were two bombs and two targets, three days apart. But the message was that the United States was equipped and prepared to rain atomic destruction on the Japanese Home Islands until Japan surrendered. A single bomb would not have sent that message. Based on what I know, it's a good thing two of them did because I don't think we had a third one.
(For anyone who wants to debate the ethics of the atomic attack on Japan, I stand by "good thing" because the alternative was a ground invasion of the Japanese Home Islands.)
18. I have to agree with most of the comments here.
It seems to me that they were playing Jeopardy and the question was "These four factors can be used to 'predict' past/future cyber attacks."
There is far more that goes into planning and executing anything in the military. Now the 3Letters may not have such laid out framework for decision making that is used, but the armed forces use MDMP (Military Decision Making Process) for everything large or small and at every level. Even using the quicker Rapid MDMP has way more variables than this model can account for.
After reading the actual article, the arguments laid out and the 'simple mathematical formula' look to be just another version of cost benefit analysis. Their formula does nothing to give an actual date time group for a cyber attack, but simply shows a method for the attacker to determine if they should use a 'resource' sooner rather than later.
An even simpler formula would describe what they take 6 pages to describe. Below is my fully detailed 'For Dummies' version of the paper.
Terms
JUICE: resources gained/exploited, political statement, damage to X thing
SQUEEZE: possible loss of exploit for future use
Conclusion
If the JUICE is worth the SQUEEZE then attack, otherwise wait.
19. The next attack was triggered 35 minutes ago...
20. of course they know when the next stux net is
its whenever they release it
21. Bob.Brown wrote:
'LilRedDog wrote:
[snip[
We did it right with nukes. we used them twice and never again; too much concentrated power... [snip]
We (the United States) really used atomic weapons once. Yes, there were two bombs and two targets, three days apart. But the message was that the United States was equipped and prepared to rain atomic destruction on the Japanese Home Islands until Japan surrendered. A single bomb would not have sent that message. Based on what I know, it's a good thing two of them did because I don't think we had a third one.
(For anyone who wants to debate the ethics of the atomic attack on Japan, I stand by "good thing" because the alternative was a ground invasion of the Japanese Home Islands.)
I've always thought it would have been equally productive to blow up something with fewer people, but still enough to give the same overall picture of destruction. Or given warning to all urban centers of the impending doom.
Of course, life was cheap back then.
22. Is this a joke ?
23. Heck we have no desire to attack, but the model says we gotta do it within a week.
Attack.
You must to comment. | 3,883 | 17,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-04 | latest | en | 0.952777 |
http://accountlearning.com/secondary-distribution-of-overhead-criteria-bases-methods/ | 1,481,422,428,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00165-ip-10-31-129-80.ec2.internal.warc.gz | 5,039,651 | 13,589 | # Secondary distribution of Overhead | Criteria | Bases | Methods
The production departments do not produce any goods and services without getting any service from the service departments. Hence, there is logic that the cost of production should include the costs of service departments. Based on this logic, the total expenses of service departments are to be apportioned only to the production departments on suitable basis.
The process of redistribution of cost of service departments among production departments on suitable basis is known as secondary distribution. In making the secondary distribution, there is a distinction of production departments and service departments.
## Criteria for Secondary Distribution of Overhead
The following is the criteria for determining the basis for apportionment of cost of service departments among production departments.
2. Analysis of Survey or Survey of existing conditions.
3. The ability to pay basis.
4. Efficiency or Incentive method,
5. General use indices.
## Bases for Secondary Distribution of Overhead
The following bases are commonly used by many organizations for second distribution of overhead.
## Methods of Secondary Distribution of overhead (Redistribution)
First, the management should determine the basis for redistribution of service departments. Then, the actual redistribution can be done by any one of the following methods.
### 1. Direct Redistribution Method
Under this method, the costs of one service department are apportioned only to the production departments ignoring the service rendered by one service department to another. If this method is followed, the number of secondary distribution will be equal to number of service departments.
### 2. Step Method
This method is also known as non-reciprocal method. This method ignores the service rendered by one service department to another. Moreover, there is no two-way distribution of costs between two service departments. It means that the costs of one service department cannot be distributed to other service departments even though service rendered to them.
For example: Canteen expenses are distributed to stores department because canteen renders service to stores department employees. But, no part of stores department cost is not distributed to canteen even if stores department render service to canteen.
Sometimes, large service department costs are distributed first. Then, next large service department costs are distributed. In this way, least service department costs are distributed finally. The cost of last service department is distributed only among production departments. Some authors have the view of distribution of cost of service department with largest amount first.
### 3. Reciprocal Service Method
Under this method, the costs of one service department are distributed to other service department on the basis of rendering service. Here, there is a two-way distribution of costs between two service departments. It means that the costs of one service department can be distributed to other service departments if service rendered to them.
The full operating cost of a service department cannot be known till inter department transactions among the service department are taken into account. The service department costs may be controlled to some extent.
### 4. Trial and Error Method
Under this method, the cost of one service department is apportioned to another service department. Now, the cost of another service department plus the share of costs received from the first service department is again apportioned to first service department. In this way, this process is going on till the balancing figure becomes negligible.
### 5. Repeated Distribution Method
This method is otherwise called as continued distribution and attrition method. The service department costs are distributed to other service departments and production departments on agreed percentage. In this way, this process is repeated till the costs of service department are too small. Then, the costs of service department are distributed only among production department.
### 6. Simultaneous Equation Method
Under this method, the total costs of service departments are ascertained with the help of simultaneous equation. Then, the costs of service departments are apportioned on agreed percentage only among production departments.
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What is 11 squared times 2? How about the value of pi? Okay, let’s go with a simple one - what is 7 x (10 + 20) / 3? Does merely looking at these numbers make you palpitate and have sweaty palms? Welcome to the club.
Math anxiety is a common phenomenon. It occurs amongst a majority of kids, teens, and even adults. The source of anxiety arises from the fact that an individual is afraid that they won't be able to solve the math problem, which, in turn, leads to stress - causing them to solve it incorrectly.
It is a vicious cycle which people are unable to get out of.
Indeed, the fear of not being able to solve a math problem often leads to an individual getting more nervous about it than they should. This fear and anxiety then cause additional stress, and the cycle continues.
I have suffered from math anxiety since I was a child, and if I think about it, the root cause could be because I was berated for underperforming in school. Everyone's reasons for developing math anxiety can be different - from strict parents and teachers to a lack of self-confidence, bullying, and more.
However, you don't have to live in fear of numbers. Here are a couple of tips I discovered along the way that helped me overcome my math anxiety, and may assist you in doing the same as well.
#1 Take it easy: The first thing I did to overcome my perpetual math anxiety was to go slow, breathe, and remind myself that it wasn't the end of the world if I couldn't solve the problem. Sure, I might score low on a couple of tests or even fail, but still, it wasn't the end of the world. That's the most important thing you can remember.
#2 Tackle easy problems: Trying to solve complex math equations can be daunting - for anyone. I recommend starting with easy problems that you can tackle. Don't be afraid of taking help. Once you can solve the easier ones, you will gain more confidence to solve the bigger, "scarier" ones.
#3 Understand Math: Most people end up trying to memorize formulas and equations instead of breaking down each step to understand it. This is the biggest mistake you can make. Get help from a tutor, a friend, or someone you trust, and ask them to break down the steps into bite-sized pieces so you can understand the "why" s instead of following a pre-set method.
Have you had this problem? What steps did you take to overcome math anxiety?
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Well said. It is true, people tend to fear maths, which should not be the case if you put in the necessary preparation. I noticed like body excercises, the math muscle must be exercised regularly to ensure it grows. Once well practiced, most of the sums will be done subconsciously.
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On topic though, I think having anxiety with numbers is more common than people think. I know when I was in school, I had bad math anxiety from a teacher who told me if I don't learn multiplication, I will amount to nothing in life.
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Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
Q1. After replacing an old member by a new member, it was found that the average age of four members of a family is the same as it was 3 year ago. What is the difference between the ages of the replaced and the new member ?
10 years
11 years
12 years
8 years
None of these
Solution:
Let the sum of their present ages = x years.
After replacing, sum of their ages = x - (4 × 3) = x - 12 years
Thus, required difference in age of two members = 12 years.
Q2. Aditya and Anshuman started a business and invested Rs. 52,500and Rs. 75,000 respectively. At the end of year they decided to distribute 50% of profit equally among them and remaining in capital ratio. Had they distributed entire profit in capital ratio, Aditya got Rs. 2,250 less. Find the total profit amount.
Rs. 46,000
Rs. 51,000
Rs. 49000
Rs. 51650
None of these
Solution:
Q3. Pipe A can fill the tank in 4 hours, while pipe B can fill it in 6 hours working separately. Pipe C can empty whole the tank in 4 hours. He opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was half-filled, but he mistakenely adjusted his alarm at a time when his tank would be 3/4th filled. What is the time difference between both the cases, to fill the tank fully:
48 min.
54 min.
30 min.
44 min.
Can’t be determined
Solution:
Q4. Anoop sells a book to Mayank at a profit of 20% and Mayank sells this book to Siddharth at a profit of 25%. Now Siddharth sells this book at a loss of 10% to Shishir. At what percentage loss should Shishir sells this book now so that his SP becomes equal to Anoop’s CP?
36.68%
25.92%
48.66
Cannot be determined
None of these
Solution:
Q5. An equal sum is invested for seven years in Scheme A offering simple interest at x% p.a. and in scheme B for two years offering compound interest at 10% p.a. (compounded annually). The interest earned from scheme A is thrice of the interest earned from schemes B. Had the rate interest been (x-4)% simple interest per annum in scheme A, the difference in the interest earned from both the schemes would have been Rs. 700. What was the sum invested in each of the schemes ?
Rs. 8000
Rs. 5000
Rs. 6000
Rs. 4500
None of the above
Solution:
Directions (6-10): What should come in place of question mark (?) in the following questions ?
Q6.
3.5
2.5
1.28
4.6
5.3
Solution:
Q7. 48% of 360 + 27% of 920 - 38% of 880 = ?
10% of 868
15% of 868
100% of 868
8.68
None of these
Solution:
? = 172.8 + 248.4 - 334.4 = 86.8 or 10% of 868
Q8.
35/9
63/5
18/7
9/35
None of these
Solution:
Q9. 2000 × 473 ÷ 1000 – 245 = ?
651
701
951
851
541
Solution:
? = 946 - 245 = 701
Q10. 3939 ÷ 3 + 6363 ÷ 3 – 9696 ÷ 6 + 123 × 2 = ?
2064
1572
1464
1864
1564
Solution:
1313 + 2121 - 1616 + 246 = ?
? = 2064
Directions (11-15): Study the following table carefully and answer the questions that follow.
Marks obtained by seven students in different subjects in the Career Power Open Challenge Paper.
Note: M.M – Maximum marks, B.A – Banking awreness
Q11. If marks obtained in Banking awareness is not considered, then who got 2nd highest marks ?
Sanjeev
Abhishek
Kamaal
Rawat
Rajesh
Solution:
Rajesh = 531 – 105 = 426
Rawat = 477 – 98 = 379
Sanjeev = 590 – 112 = 478
Kamaal = 546 – 106 = 440
Abhishek = 556 – 119 = 437
Dharmendra = 485 – 107 = 378
Shailesh = 51 + 72 + 135 + 132 + 37 = 427
Thus, Kamaal got 2nd highest marks.
Q12. Approximately what is the minimum difference between the percentage of total marks obtained by any two students ?
5
3
1
0.1
4
Solution:
Q13. It was found later that marks obtained by Kamaal in R.C. and Reasoning were misread as 13 marks more and 31 marks less respectively than actual. Find the new average marks of Kamaal.
88
96
91
94
None of these
Solution:
Q14. Marks obtained by Sanjeev in Maths holds what approximate percent of total marks obtained by Shailesh in all subjects together ?
30%
27%
29%
25%
24%
Solution:
Q15. Find the difference between total marks scored by all students together in Maths and that in Banking awareness.
127
135
140
128
None of these
Solution:
Total marks in Maths = 880
Total marks in Banking awareness = 743
∴ Required difference = 137
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https://www.jiskha.com/display.cgi?id=1351276508 | 1,516,103,332,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00193.warc.gz | 953,210,509 | 3,895 | # Physics
posted by .
A ballistic pendulum was a device used in the past century to measure the speed of bullets. The pendulum consists of a large block of wood suspended from long wires. Initially, the pendulum is at rest. The bullet strikes the block horizontally and remains stuck in it. The impact of the bullet puts the block in motion, causing it to swing upward to a height h. If the bullet has a mass of 8.90 g, and the block of mass 2.20 kg swings up to a height of h = 7.70x101 cm, what was the speed of the bullet (in m/s) before impact ?
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More Similar Questions | 746 | 3,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-05 | latest | en | 0.909873 |
https://algosim.org/doc-3.0/%E2%88%8F.html | 1,696,090,229,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00249.warc.gz | 103,921,395 | 2,362 | ∏ – Algosim documentation
Algosim documentation: ∏
# ∏ (n-ary product)
Computes the product of all elements in a container or in a sequence. The elements may be numbers or square matrices.
## Syntax
• `∏(X)`
• `X` is a vector, matrix, list, or set
• `∏(expr, var, a, b)`
• `expr` is an expression in one variable `var`
• `var` is the variable in `expr`
• `a` is the lower bound
• `b` is the upper bound
## Description
### Computing the product of all elements in a container
If `X` is a container, then `∏(X)` is the product of all elements in `X`.
The following container types are supported:
• vectors
• the result is the product of all components
• matrices
• the result is the product of all entries
• lists containing numbers
• the result is the product of all numbers
• lists containing square matrices
• the result is the matrix product of all matrices
• sets containing numbers
• the result is the product of all numbers
Typewise, the factors may be either real or complex, and the result is complex iff at least one of the factors is complex.
### Computing the product of all elements in a sequence
`∏(expr, var, a, b)` computes the product of the expression `expr` in one variable `var` as `var` takes all integer values from `a` to `b` (inclusively). This precisely implements the pi notation for n-ary products.
`expr` must return a number or a matrix.
## Notes
These note apply to both modes of operation: computing the product of all elements in a container and computing the product of all elements in a sequence.
The empty product is `1`. Notice that the product of an empty collection of matrices is also equal to the scalar `1`, because the system cannot tell that the collection indeed is an empty collection of matrices. And even if it could, it wouldn’t know the size of the matrices.
When computing the product of a collection of objects, the objects must be of the same kind (numbers or matrices). If they are matrices, they must all be of the same size.
Notice that `∏` is a function, not a prefix operator. Hence, to compute the product of all elements in a container `X` you must write `∏(X)``∏X` would be a syntax error.
You cannot compute the product of all matrices in a set because matrix multiplication is not commutative and a set is not an ordered container.
## Examples
### Computing the product of all elements in a container
`∏(❨2, 1, 3, −4, 5, 2, 1, 3, −2❩)`
`1440`
`∏(❨❨4, 2, 1❩, ❨3, 2, 5❩❩)`
`240`
`∏('(❨❨1, 3❩, ❨2, 1❩❩, ❨❨−2, 1❩, ❨0, 3❩❩, ❨❨1, 0❩, ❨−1, i❩❩))`
```⎛−12 10⋅i⎞
⎝ −9 5⋅i ⎠
```
`∏(unique(PrimeFactors(19871226)))`
`6623742`
### Computing the product of all elements in a sequence
`abs(∏(1 − 1/prime(n)^2, n, 1, 100000) − 6/π^2)`
`3.11339510472⋅10^−8` | 826 | 2,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-40 | latest | en | 0.836464 |
https://testbook.com/question-answer/which-number-should-be-subtracted-from-23-30-57--63e101b8ea7b485a547fd989 | 1,721,223,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00447.warc.gz | 487,635,441 | 46,725 | # Which number should be subtracted from 23, 30, 57 and 78 so that remaining numbers are in proportion?
This question was previously asked in
Rajasthan CET (Senior Secondary) Official Paper (Held On: 05 Feb, 2023 Shift 2)
View all Rajasthan CET Senior Secondary Papers >
1. 9
2. 6
3. 7
4. 8
Option 2 : 6
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10 Questions 20 Marks 12 Mins
## Detailed Solution
Calculation:
Consider x be subtracted from each term
23 – x, 30 – x, 57 – x and 78 – x are proportional
It can be written as
23 – x: 30 – x :: 57 – x: 78 – x
⇒ $$\dfrac{(23 – x)}{(30 – x)}= \dfrac{(57 – x)}{(78 – x)}$$
⇒ (23 – x) (78 – x) = (30 – x) (57 – x)
⇒ 1794 − 23x - 78x + x2 = 1710 − 30x − 57x + x2
⇒ x2 -101 + 1794 - x2 + 87x - 1710
So we get
⇒ -14x + 84 = 0
⇒ 14x = 84
x = 84/14 = 6
∴ 6 is the number to be subtracted from each of the numbers.
Latest Rajasthan CET Senior Secondary Updates
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Rajasthan CET Senior Secondary Result announced on 16th May 2023. The Rajasthan CET Senior Secondary exam was conducted on the 4th, 5th, and 11th of February 2023. By qualifying for the Rajasthan CET Senior Secondary candidates will be eligible to apply for the posts of LDC, Forester, Junior Assistant, and more under the Government of Rajasthan. Candidates who have passed class 12th are eligible to appear for this exam. Prepare for the upcoming exam using Rajasthan CET Senior Secondary Previous Year Papers. | 495 | 1,460 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.853149 |
https://math.stackexchange.com/questions/2378663/lipschitz-smoothness-and-nesterovs-accelerated-gradient-descend | 1,708,855,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474594.56/warc/CC-MAIN-20240225071740-20240225101740-00453.warc.gz | 379,882,211 | 34,741 | # Lipschitz Smoothness and Nesterov's Accelerated Gradient Descend
I am reading about Nesterov's Accelerated Gradient Descend, and trying to apply it to my application (Computerised Tomography), which essentially means I am solving a linear problem of the form
$$\hat{x}=\underset{x}{\text{argmin}}\left\{ \lVert Ax-b\rVert^2\right\}$$
Regardless of the complexity of the proof, ultimately the algorithm seems fairly easy to implement, as it is just a slight modification from the classic gradient descend with some specific parameter updates. Nesterov's acceleration ends in an update that looks like:
$$y_{s+1}=x_{x}-\frac{1}{\beta}\nabla f(x_s)\\ x_{s+1}=(1-\gamma_s)y_{s+1}+\gamma_s y_s$$ for iteration number $s$ and $\gamma_s=\frac{1-\lambda_s}{\lambda_{s+1}}$, being $\lambda_S=\frac{1+\sqrt{1+4{\lambda_{s-1}^2}}}{2}$.
However it relies in the $\beta$-smoothness of the function $f$, and $\beta$ is used in as a parameter.
$f$ is $\lVert Ax-b\rVert^2$
What value should I give to $\beta$? How can I know how $\beta$-smooth my function is?
• If you're minimizing $\|Ax-b\|^2$, then isn't $f(x)=\|Ax-b\|^2$ in your case? And $\nabla f(x) = 2 A^T(Ax-b)$. Aug 1, 2017 at 12:07
• As a no-answer to my question: pretty much everyone just uses $\beta=1$ or some similar value. It apparently is not too important what value is given to that variable when doing numerical stuff. Aug 4, 2017 at 12:40
• @littleO the question has been clarified, you're welcome to take a shot at it (my answer wasn't corresponding to the actual question). Aug 4, 2017 at 12:45
• So you want know whether you can choose $\beta=1$ as an upper bound for the $\beta$ that satisfy $\Vert \nabla f(x) - \nabla f(y) \Vert = \Vert 2 A^t A (x-y) \Vert \leq \beta \Vert x-y\Vert \forall x,y$ ? Aug 4, 2017 at 12:56
The gradient of $f$ is $\nabla f(x) = 2 A^T(Ax-b)$. To use Nesterov's method we need a Lipschitz constant for $\nabla f$ (not $f$). Notice that \begin{align} \| \nabla f(x)-\nabla f(y) \| &= \|2 A^T A(x-y) \| \\ &\leq 2 \| A^T A \| \|x-y\| \\ &= 2 \| A\|^2 \|x-y\|^2. \end{align} So a Lipschitz constant for $\nabla f$ is $$\beta= 2 \|A\|^2.$$
• So this $\beta = 2\Vert A \Vert^2$ is an upper bound could it be that there might actually be a lower $\beta$ than your suggested one? Aug 8, 2017 at 14:28 | 764 | 2,295 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-10 | latest | en | 0.796511 |
http://mathhelpforum.com/advanced-statistics/74813-how-find-probability-type-ii-error-print.html | 1,527,241,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00244.warc.gz | 188,456,325 | 2,766 | # how to find the probability of Type II error
• Feb 21st 2009, 06:46 AM
szpengchao
how to find the probability of Type II error
$\displaystyle f(x : \theta)=\frac{\theta}{(x+\theta)^{2}} , \ \ \ H_{0}:\theta=1, \ \ \ H_{1}:\theta=2$
Mr F edit: Additional information given to me via pm:
Quote:
x>0, theta>0, is an unknown parameter.
find the likelihood ratio test .
and show the probability of TYPE II error is
19/21
• Feb 23rd 2009, 03:03 AM
mr fantastic
Quote:
Originally Posted by szpengchao
$\displaystyle f(x : \theta)=\frac{\theta}{(x+\theta)^{2}} , \ \ \ H_{0}:\theta=1, \ \ \ H_{1}:\theta=2$
Mr F edit: Additional information given to me via pm:
Quote:
x>0, theta>0, is an unknown parameter.
find the likelihood ratio test .
and show the probability of TYPE II error is
19/21
More information is needed here I think to calculator $\displaystyle \beta$. A value of $\displaystyle \alpha$, perhaps ....?
• Feb 23rd 2009, 08:12 PM
matheagle
Once we have alpha we can get Beta. | 313 | 995 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-22 | latest | en | 0.748888 |
https://www.codeproject.com/Articles/15680/How-to-write-a-Memory-Scanner-using-C?msg=1900028 | 1,508,206,717,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820556.7/warc/CC-MAIN-20171017013608-20171017033608-00705.warc.gz | 1,177,539,975 | 29,312 | 13,189,506 members (68,535 online)
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Posted 23 Sep 2006
# How to write a Memory Scanner using C#
, 23 Sep 2006
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Search a process' memory to find specified 16, 32 or 64 bit data values.
## Introduction
Have you ever had a problem completing a game because your health meter shows just 5% of your health available?
Have you ever used a game trainer to solve this problem?
Have you ever wondered how this trainer freezes your health meter to 100%?
These game trainers, do a simple job!
They write a little part of the Game's memory with the 100 `value`!
But how to find the exact part of memory?
## The Question and the Answer
Q: How to find the location in which a program stores a `value` in its memory?
A: Some programs named 'Memory Scanners', are written to read and search a program's memory for exact location and help freezing it! (And I have written this article to show you, how to write a 'Memory Scanner')
## Step 1: Where to begin?
Let's have a look at a program's memory.
Think that I have written a program and we can have a look at its memory.
To have a picture in mind, we can say, it looks something like this:
As you can see, the memory is made up of a huge number of small sectors, that hold a `value` in it. Our picture, just shows a small part of the memory, from sector 0 to sector 99 and a detail of sector 0 to sector 15. But as you know, a computer just knows the meaning of 0 and 1, so what do these Hexadecimal `value`s mean?
Let's take a deep look at the memory again.
As you can see, every byte is made up of 8 bits with each of them being just that 1 or 0, and in Binary mode, they can return the `value` stored in the byte((00010111)2 = (23)10 = (17)16).
As we saw, a computer's memory stores information by holding the 0 and 1s in the memory bits, and 8 bits of memory make a section of memory named byte, so a byte can hold `value`s up to (11111111)2 = (255)10 = (FF)16, but how about the bigger `value`s?
We usually work with `value`s greater than 255!
Ok, the answer is that, we have bigger units of memory to hold the bigger `value`s.
Let's look at another picture of memory that shows the bigger units and then, I will explain everything:
So we have 3 bigger memory units: 2 `Bytes` that make a 16 bit memory unit that we call 'short' in C#, 4 `Bytes` that make a 32 bit memory unit that we call it 'int' in C#, and 8 `Bytes` that make a 64 bit memory unit, and we call it 'long' in C#.
## Step 2: What to look for?
Now that we have a simple picture of the memory in our mind, let's go back to the first picture:
I know that, you are going to say: "Hey, it's just a row of `bytes`! How to find the memory units?" And then I'll tell you that, you asked the biggest question in writing a memory scanner!
Ok, let's think that it's a part of the memory of my program, and I know, where in the memory, I have stored the `value`s and I will show you that:
As you see, a memory unit, can be stored in any part of the memory and start from any memory sector. In this program for example, I have stored a 32 bit `value` in the 0 sector, and because a 32 bit `value` takes 4 `bytes` of memory, from the sector 0 to sector 3 is assigned for a 32 bit variable in the program, and after that, from the sector 4, there is a 16 bit variable that takes 2 `bytes` of memory, next is a 64 bit variable and at the end, there is again a 16 bit variable.
Now let's think that the `value`s of the sectors are the same, but the memory units start from different sectors:
Why everything changed?
Because, a variable could be stored in any memory sector number, and most of the time, even the programmer doesn't know, where the variable is stored in the memory, and just the program knows it!
## Step 3: Where to find it?
Now, let's think that we are playing a game, and the health meter shows 83%, and we don't know the location of the variable in the memory and we want to find the variable and we start from sector 0, so the memory looks like this:
So what? Is there any 83 in the memory?
First, we know that we have the hexadecimal `value`s of the memory `bytes`. Second, we should guess the variable type to look for.
Ok, let's say that the programmers of the game have used a 32 bit (int) variable, that is the most usual data type being used for storing the `value` of the health meter. So the `value` is stored in a 4 `bytes` long part of the memory. But, how to find it?
The only way to search the memory completely, is to start from the beginning, take 4 `bytes`, test them to see if the `value` equals our digit (here 83), and find the location. Like this:
Ok, now you know the main concept of memory scanning, but there are some other things that you should know to be able to write the Memory scanner:
1. Q: How long is a program's memory? (Where to begin and where to stop?)
A: As you know, Microsoft's first OS was DOS that was a 8 bit OS, after that, the Windows 3.1 became a 16 bit OS, and after that, the Windows OS became a 32 bit OS. (I'm a real fan of Apple Co. that developed the Apple Macintosh OS, a 64 bit OS, exactly when Microsoft was working on DOS (a 8 bit OS) and today, Microsoft is going to write a 64 bit Windows (and like the first Windows versions, it still looks like the Apple OSs) but, I still recommend Apple MacOS X (Ver. 10)).
So, in the DOS OS that was a 8 bit OS, programmers named 8 bits of memory, a "Byte". After that, When the Windows 3.1 OS was a 16 bit OS, they named 16 bits of memory(2 `Bytes`) a "WORD", and 32 bits of memory(4 `Bytes`) a "DWORD"(Double Word) and 64 bits of memory(8 `Bytes`) a "QWORD"(Quad Word).
As I experienced, the length of every program's memory in Windows XP, is from "0x00000000" to the maximum `value` of a "Int"("DWORD"), and equals to "0x7FFFFFFF".
I'm not sure, but I guess, it's because of that, the Windows is a 32 bit OS and the main memory unit for it, is a 32 bit memory unit, and so, the length of a program's memory, is the maximum `value` of a 32 bit memory unit!
Ok. So, we should start our search from "0x0000000" to "0x7FFFFFFF".
2. Q: Is the first found memory address, the exact answer of our search?
A: No! As you can see, there are "0x7FFFFFFF" sectors, and when you search it for a `value` like 83, you could find so many of them. So you need to hold the memory addresses and wait for the `value`s to be changed. Then search the addresses you have, for the new `value`, and do this, until you find, just one memory address that matches your `value`.
3. Q: How to read a program's memory and search it?
A: There are some functions in Windows API that make it possible for us to read and write the memory, from another program.
Thanks goes to "Arik Poznanski" for P/Invokes and methods needed to read and write the Memory, I just used his classes to do this and didn't do the P/Invokes myself.
You can search Codeproject.com, for "Minesweeper, Behind the scenes", to find his comments about these classes.
4. Q: How to convert these `bytes` to a 16, 32 or 64 bit `value`?
A: For this, we used .NET goods! There is a class with static methods that does this for us:
```byte[] bytes = new byte[] { 0x53, 0x00, 0x00, 0x00 };
int value = System.BitConverter.ToInt32(bytes, 0);
//The value is 83```
## Step 4: Let's do the final job
Ok, now you have all the information you need to write the memory scanner.
You just need to do these in your code:
1. Select a process to scan its memory.
2. Scan the whole memory for the specified `value` and hold the addresses.
3. Wait for the `value` to be changed and search the memory address list that you got from the first scan and again wait for the `value` to be changed and scan again, and do this until you find just the address that matches the `value`.
4. At the end, you can freeze the address with a new `value`, by using a timer to write the memory in every timer's tick.
If you download my code, you will find them with comments for every command and every line of code. But there it a little thing I should explain:
`ReadProcessMemory(IntPtr MemoryAddress, uint bytesToRead, out int bytesRead)` that is the most important function, returns an empty `bytes` array if reading your request's size is too big! So I had to read the memory in parts as big as 20480 `bytes` (20KB), and because, when you are searching these memory parts, byte by byte, at the end of the `bytes` array, there will be some `bytes` left! (for example, 3 `bytes` will be left when you are searching for 32 bit `value`s)!
```if (/*scan requirement is less than 20480 bytes*/)
{
}
else
{
/*Loop through blocks(of length 20480 bytes),
until the whole memory is read;
After the first loop, move the current address to
[Data type bytes count - 1] steps back in the memory,
to fix the previously told problem;
After the loops, check to see if any other memory addresses
are left outside of the loops and if so, read them;*/
}```
You can see it:
## That's all
Ok folks, that's all! Hope you like and enjoy it!
And now I'm working on a Enhanced Memory Scanner that could scan all types of data, including the `Bytes`, Signed Data Types and even Strings, it just takes some time!
I'll be back soon.
## Share
Founder Sojaner AB Sweden
UX designer and full stack developer mainly focused on .NET technologies.
Currently loving .NET Core 2.0.
## You may also be interested in...
Pro Pro
Really Nice! Coksnuss9-Feb-07 5:03 Coksnuss 9-Feb-07 5:03
Re: Really Nice! Sojaner18-Feb-07 0:15 Sojaner 18-Feb-07 0:15
scan game memory fail flash.lin1-Feb-07 14:18 flash.lin 1-Feb-07 14:18
Re: scan game memory fail Sojaner18-Feb-07 0:13 Sojaner 18-Feb-07 0:13
1) This is a Memory Scanner, not a game trainer! 2) The game could detect it because it invokes the platform API, so any other program that invokes the platform API to read an write the memory would be detected by this game! They have done this to stop `GAME CHEATERS`, are you? Sojaner!
CloseHandle Error ic3dwabit21-Dec-06 15:25 ic3dwabit 21-Dec-06 15:25
Re: CloseHandle Error Sojaner29-Dec-06 13:05 Sojaner 29-Dec-06 13:05
Re: CloseHandle Error ic3dwabit8-Jan-07 22:09 ic3dwabit 8-Jan-07 22:09
Error 299 Hall Maru28-Oct-06 3:15 Hall Maru 28-Oct-06 3:15
Re: Error 299 Sojaner29-Dec-06 13:08 Sojaner 29-Dec-06 13:08
Re: Error 299 [modified] The_Mega_ZZTer16-Nov-07 15:55 The_Mega_ZZTer 16-Nov-07 15:55
Last Visit: 31-Dec-99 18:00 Last Update: 16-Oct-17 16:18 Refresh « Prev123456789 Next » | 2,865 | 10,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-43 | latest | en | 0.934422 |
https://www.inchcalculator.com/convert/quarter-barrel-keg-to-half-barrel-keg/ | 1,719,350,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00568.warc.gz | 726,522,309 | 15,678 | # Convert Quarter Barrel Kegs of Beer to Half Barrel Kegs
Enter the beer volume in quarter barrel kegs below to get the value converted to half barrel kegs.
## Result in Half Barrel Kegs:
1 quarter barrel keg = 0.5 half barrel kegs
Do you want to convert half barrel kegs to quarter barrel kegs?
## How to Convert Quarter Barrel Kegs to Half Barrel Kegs
To convert a measurement in quarter barrel kegs to a measurement in half barrel kegs, divide the beer volume by the following conversion ratio: 2 quarter barrel kegs/half barrel keg.
Since one half barrel keg is equal to 2 quarter barrel kegs, you can use this simple formula to convert:
half barrel kegs = quarter barrel kegs ÷ 2
The beer volume in half barrel kegs is equal to the beer volume in quarter barrel kegs divided by 2.
For example, here's how to convert 5 quarter barrel kegs to half barrel kegs using the formula above.
half barrel kegs = (5 quarter barrel kegs ÷ 2) = 2.5 half barrel kegs
### How Many Half Barrel Kegs Are in a Quarter Barrel Keg?
There are 0.5 half barrel kegs in a quarter barrel keg, which is why we use this value in the formula above.
1 quarter barrel kegs = 0.5 half barrel kegs
Quarter barrel kegs and half barrel kegs are both units used to measure beer volume. Keep reading to learn more about each unit of measure.
## What Is a Quarter Barrel Keg?
The quarter-barrel keg is a keg equal to one-quarter the volume of a standard US beer barrel. A quarter-barrel keg contains 992 fl. oz. of beer, or just over 82 12 fl. oz. beers.
The quarter barrel keg is a US customary unit of beer volume. A quarter barrel keg is sometimes also referred to as a pony keg.
## What Is a Half Barrel Keg?
The half-barrel keg is a keg equal to one-half the volume of a standard US beer barrel. A half-barrel keg contains 1984 fl. oz. of beer, or just over 165 12 fl. oz. beers.
The half barrel keg is a US customary unit of beer volume. A half barrel keg is sometimes also referred to as a full sized keg.
## Quarter Barrel Keg to Half Barrel Keg Conversion Table
Table showing various quarter barrel keg measurements converted to half barrel kegs.
Quarter Barrel Kegs Half Barrel Kegs
1 0.5
2 1
3 1.5
4 2
5 2.5
6 3
7 3.5
8 4
9 4.5
10 5
11 5.5
12 6
13 6.5
14 7
15 7.5
16 8
17 8.5
18 9
19 9.5
20 10
21 10.5
22 11
23 11.5
24 12
25 12.5
26 13
27 13.5
28 14
29 14.5
30 15
31 15.5
32 16
33 16.5
34 17
35 17.5
36 18
37 18.5
38 19
39 19.5
40 20 | 757 | 2,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.850961 |
https://physics.stackexchange.com/questions/207302/flux-of-electric-field-through-a-closed-surface-with-no-charge-inside | 1,566,625,119,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319724.97/warc/CC-MAIN-20190824041053-20190824063053-00475.warc.gz | 581,830,693 | 28,942 | # Flux of electric field through a closed surface with no charge inside? [duplicate]
I'm reading the Feynman lectures on electromagnetism and in Vol II, Chapter 1.4 Vol. II, Chapter 1-4 he talks about the flux of the electric field and says that flux of $E$ through and closed surface is equal to the net charge inside divided by $\epsilon_0$.
If there are no charges inside the surface, even though there are charges nearby outside the surface, the average normal component of $E$ is zero, so there is no net flux through the surface
I cannot see why the net flux is zero here. Say we have a closed unit sphere at the origin with no charge inside it and at the point $(2, 0, 0)$ we have some charge $q$.
1. Well doesn't this charge then define the electric field $E$ for the system and it will flow into the unit sphere on the right hand side, and out of the unit sphere on the left hand side?
2. Furthermore, as the strength of the electric field decreases with distance from $q$ won't we have more flux going into the right hand side which is closer to the charge $q$, and less flux leaving through the left hand side as it is further away - and hence we should have a non-zero flux?
Can someone please explain what I am misinterpreting here?
## marked as duplicate by Rob Jeffries, John Rennie, Kyle Kanos, Martin, ACuriousMind♦Sep 17 '15 at 12:44
You have more flux per unit area going into the right side, but the area on the right side is smaller. These two balance out so that the total flux is the same going in as going out.
The part of the sphere which has electric flux going in, traced in red, is less than half the area of the sphere.
Incidentally, flux per unit area is just the electric field.
• Thanks...but although the flux "spreads out" on the left hand side means that less is going out there....is there not EVEN less again going out due to the fall-off in electrical force with distance from the source? – Riggs Sep 16 '15 at 15:59
• @Riggs I think you're double-counting the effect. The "spreading out" of the flux is exactly the same physical effect as the $1/r^2$ decrease of the electric field. – David Z Sep 16 '15 at 16:15
1. Yes.
2. No.
The strength of the field near the charge is higher (because it is closer to the charge) but this electric field is entering through a smaller area $S_1$ whereas the electric field leaving the sphere is relatively weaker (as it is further away from the charge) but leaves through a larger area $S_2$ as visualised below. Hence, the flux going in is exactly equal to the flux going out, and the net flux is 0.
To understand this better, you might see the proof of Gauss's Law using solid angles. You'll see that the area through which the field leaves or enters is proportional to $r^2$ whereas the field itself is proportional to $\frac{1}{r^2}$ and hence, the flux leaving or entering doesn't depend on the position of the charge. (I mean, it does not depend on where it is placed inside or where it is placed outside)
Note: This picture isn't that of the sphere and charge described exactly, it's sort of flipped but that won't be a problem, I think. | 752 | 3,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-35 | latest | en | 0.95756 |
https://www.physicsforums.com/threads/rocket-motion-problem.284986/ | 1,527,358,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867841.63/warc/CC-MAIN-20180526170654-20180526190654-00169.warc.gz | 836,513,090 | 15,938 | # Homework Help: Rocket motion problem
1. Jan 15, 2009
### pentazoid
1. The problem statement, all variables and given/known data
A rocket of initial mas M, of which M-m is fuel, burns its fuel at a constant rate in time tau and ejects the exhausts gases with constant speed u. The rocket starts from rest and moves vertically under uniform gravity . Show that the maximum speed achieved by the rocket is u ln($$\gamma$$)-g$$\tau$$ and that its height at burnout is
u$$\tau$$(1-ln($$\gamma$$)/($$\gamma$$-1) where $$\gamma$$=M/m[assume that the thrust is such that the rocket takes off immediately.)
2. Relevant equations
3. The attempt at a solution
I had no trouble finding v, I had trouble integrating v to obtain the height. v=u ln (gamma)-g*tau . h=$$\int$$v dt= $$\int$$u*ln(m0/m(t))-.5*gt^2
u is treated as a constant I think since I am integrating v with respect to dt. $$\int$$ln($$\gamma$$)=$$\gamma$$*ln($$\gamma$$)-$$\gamma$$. Now I am stuck on this part of the solution.
2. Jan 15, 2009
### chrisk
You are integrating a ln function. The integral is
int[ln(ax)] = xln(ax) - x.
The exhaust velocity, u, is treated as a constant. This integral is obtained by integrating by parts. An introductory calculus text will have the derivation for the intergral of ln(ax). Apply the limits of integration to each part of the solution.
3. Jan 15, 2009
### chrisk
Sorry about the previous post. Your equation for v is correct. Since v = dx/dt, separate the variables such that dx is on the left side and dt is on the right side. Recall, you are finding the maximum height so integrate dx from x = 0 to x = h and integrate (u ln (gamma) - gt)dt from t = 0 to t = tau. Then the remainder of the rocket flight is only under the force of gravity and becomes a vertically fired projectile with an initial velocity. Use v2 = vo2 - 2g(x - xo) to find the additional height by setting v2 = 0. Add this height, x - xo to the height at burnout.
4. Jan 15, 2009
### pentazoid
so I should still integrate (u ln (gamma)-gt)dt I am okay with that part of the problem and I am okay with integrating ln (gamma) because I can integrate ln(gamma) using integration by parts. How would the equation v2 = vo2 - 2g(x - xo) assist me in helping me find the height? Why do I even need the equation for v^2 for this problem?
5. Jan 15, 2009
### chrisk
The additional equation v2 = ... is used because the rocket doesn't suddenly stop once the fuel is gone.
6. Jan 15, 2009
### pentazoid
I am integrating u*ln (gamma)-gt)dt to obtain the height. I don't see any other use for the additional equation. couldn't I just plug in my initial conditions in u*ln (gamma)-gt)dt to find the maximum height?
7. Jan 15, 2009
### chrisk
Sorry, you do not need the additional equation. I missed the part "max height at burnout". Yes, just evaluate the integral using the initial conditions for height and time and the final time tau.
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https://dissertationwritingexperts.com/net-income/ | 1,701,176,502,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00878.warc.gz | 253,784,560 | 27,668 | Select Page
# Net income
A company desires to sell a sufficient quantity of products to earn a profit of \$300.00. If the unit sales price is \$20, unit variable cost is \$12, and total fixed costs are \$500,000, how many units must be sold to earn net income of \$300,000? | 72 | 276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-50 | longest | en | 0.905435 |
https://www.coursehero.com/file/5839306/chap1f/ | 1,487,581,898,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170434.7/warc/CC-MAIN-20170219104610-00216-ip-10-171-10-108.ec2.internal.warc.gz | 808,513,182 | 294,552 | # chap1f - Probability and Statistics with Reliability,...
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Copyright © 2003 by K.S. Trivedi 1 Probability and Statistics with Reliability, Queuing and Computer Science Applications second edition by K.S. Trivedi Publisher-John Wiley & Sons Chapter 1: Introduction Dept. of Electrical & Computer Engineering Duke University Email: kst@ee.duke.edu URL: www.ee.duke.edu/~kst
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Copyright © 2003 by K.S. Trivedi 2 Need to Model Random Phenomena ± Random Phenomena in a Computer/networking Environment ± Arrival of jobs/messages/requests. ± Execution (transmission/processing) time of jobs/messages/requests. ± Memory requirement of jobs/messages/requests. ± Failure or repair of components or resources. ± How to Quantify Randomness? ± Use probabilistic models. ± How to Estimate these Quantifiers? ± Use statistical techniques.
Copyright © 2003 by K.S. Trivedi 3 Modeling Random Phenomena Measurement Data Statistical Analysis Model Input Parameters Probability Model (PM) Input Output Validation Model Outputs
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Copyright © 2003 by K.S. Trivedi 4 Components of a Probability Model 1. Sample Space (S): A set of all possible observable “states” of a random phenomena. 2. Set of Events ( ): A set of all possible events of interest. 3. Probability of Events (P): A consistent description of the likelihood of observing an event. Thus a PM is a triple: PM = (S, , P).
Copyright © 2003 by K.S. Trivedi 5 Sample Space ± Probability of an event is meant to represent the relative likelihood that an outcome of an experiment will result in occurrence of that event. ± It implies random experiments. ± A random experiment can have many possible outcomes; each outcome is known as a sample point ( a.k.a. elementary event) has some probability assigned. This assignment may be based on measured data or guestimates (“equally likely” is a convenient and often made assumption). ± Sample Space S : a set of all possible outcomes (elementary events) of a random experiment. ± Finite (e.g., if statement execution; two outcomes) ± Countable (e.g., number of times a while statement is executed; Sample space is either finite or countably infinite) ± Continuous (e.g., time to failure of a component)
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Copyright © 2003 by K.S. Trivedi 6 Events ± An event E is a collection of zero or more sample points from S. Event E is a subset of S . ± S is the universal event and the empty set is represented by F F,E E E,F E E S Explained in next slide
Copyright © 2003 by K.S. Trivedi 7 Terminology and Definitions ± S and E are sets hence can use of set operations. ± E or F : : Union of two events ± E and , EF : Intersection ± Not E: Ē : Complement F E F E n i n i E ... E E E 1 2 1 = = n i n n i ...E E E E ... E E E 1 2 1 2 1 = = =
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Copyright © 2003 by K.S. Trivedi 8 Algebra of events ± Sample space is a set and events are the subsets of this (universal) set.
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## This note was uploaded on 04/08/2010 for the course COMPUTER E 409232 taught by Professor Mohammadabdolahiazgomiph.d during the Spring '10 term at Islamic University.
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chap1f - Probability and Statistics with Reliability,...
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Ask a homework question - tutors are online | 904 | 3,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-09 | longest | en | 0.88815 |
http://mathhelpforum.com/trigonometry/208281-inverse-function-identities-print.html | 1,527,206,214,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00054.warc.gz | 181,504,071 | 2,969 | # Inverse Function Identities
• Nov 23rd 2012, 08:39 PM
SirDinkledork
Inverse Function Identities
Hello,
I've been working on a problem I gave myself to work on over the holiday to curb my boredom, and I think I have almost solved half of it. I have this system of equations:
$\displaystyle Xsin(a) + Ysin(b) = d$
$\displaystyle Xcos(a) - Y cos(b) = h$
$\displaystyle phi = a + b$
The a's and b's are actually alpha's and beta's, respectively, in my notebook, I just don't know how type them here. $\displaystyle X$ and $\displaystyle Y$ are constants, $\displaystyle h$ and $\displaystyle d$ are given, and $\displaystyle d^2 = x^2 + y^2$.
Anyway I want to solve the system for phi. I tried this in a couple of ways, but my only successful attempt was solving the top two for alpha and beta and adding the two results together, yielding the following:
Attachment 25878
As you can see I already substituted for d since my answer had only d squared terms.
Now what I would like to do is simplify that further if at all possible. Any help would be much appreciated.
• Nov 23rd 2012, 11:57 PM
BobP
Re: Inverse Function Identities
Square each of the top two equations and add them.
Simplify and you finish up with
$\displaystyle \cos(a+b)=\frac{X^{2}+Y^{2}-d^{2}-h^{2}}{2XY}.$ | 357 | 1,278 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.947985 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/193/C5%5E2sQ8.html | 1,695,469,654,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00201.warc.gz | 501,004,078 | 5,815 | Copied to
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## G = C52⋊Q8order 200 = 23·52
### The semidirect product of C52 and Q8 acting faithfully
Aliases: C52⋊Q8, C52⋊C4.C2, C5⋊D5.3C22, SmallGroup(200,44)
Series: Derived Chief Lower central Upper central
Derived series C1 — C52 — C5⋊D5 — C52⋊Q8
Chief series C1 — C52 — C5⋊D5 — C52⋊C4 — C52⋊Q8
Lower central C52 — C5⋊D5 — C52⋊Q8
Upper central C1
Generators and relations for C52⋊Q8
G = < a,b,c,d | a5=b5=c4=1, d2=c2, ab=ba, cac-1=b-1, dad-1=a3, cbc-1=a, dbd-1=b2, dcd-1=c-1 >
25C2
2C5
2C5
2C5
25C4
25C4
25C4
10D5
10D5
10D5
25Q8
10F5
10F5
10F5
Character table of C52⋊Q8
class 1 2 4A 4B 4C 5A 5B 5C size 1 25 50 50 50 8 8 8 ρ1 1 1 1 1 1 1 1 1 trivial ρ2 1 1 -1 -1 1 1 1 1 linear of order 2 ρ3 1 1 1 -1 -1 1 1 1 linear of order 2 ρ4 1 1 -1 1 -1 1 1 1 linear of order 2 ρ5 2 -2 0 0 0 2 2 2 symplectic lifted from Q8, Schur index 2 ρ6 8 0 0 0 0 -2 3 -2 orthogonal faithful ρ7 8 0 0 0 0 3 -2 -2 orthogonal faithful ρ8 8 0 0 0 0 -2 -2 3 orthogonal faithful
Permutation representations of C52⋊Q8
On 10 points - transitive group 10T20
Generators in S10
```(1 2 3 4 5)(6 7 8 9 10)
(1 3 5 2 4)(6 9 7 10 8)
(2 3 5 4)(6 10 7 8)
(1 9)(2 6 5 7)(3 8 4 10)```
`G:=sub<Sym(10)| (1,2,3,4,5)(6,7,8,9,10), (1,3,5,2,4)(6,9,7,10,8), (2,3,5,4)(6,10,7,8), (1,9)(2,6,5,7)(3,8,4,10)>;`
`G:=Group( (1,2,3,4,5)(6,7,8,9,10), (1,3,5,2,4)(6,9,7,10,8), (2,3,5,4)(6,10,7,8), (1,9)(2,6,5,7)(3,8,4,10) );`
`G=PermutationGroup([[(1,2,3,4,5),(6,7,8,9,10)], [(1,3,5,2,4),(6,9,7,10,8)], [(2,3,5,4),(6,10,7,8)], [(1,9),(2,6,5,7),(3,8,4,10)]])`
`G:=TransitiveGroup(10,20);`
On 20 points - transitive group 20T47
Generators in S20
```(1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)
(1 3 5 2 4)(6 8 10 7 9)(11 14 12 15 13)(16 19 17 20 18)
(1 9)(2 6 5 7)(3 8 4 10)(11 18 12 16)(13 19 15 20)(14 17)
(1 14)(2 11 5 12)(3 13 4 15)(6 16 7 18)(8 20 10 19)(9 17)```
`G:=sub<Sym(20)| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20), (1,3,5,2,4)(6,8,10,7,9)(11,14,12,15,13)(16,19,17,20,18), (1,9)(2,6,5,7)(3,8,4,10)(11,18,12,16)(13,19,15,20)(14,17), (1,14)(2,11,5,12)(3,13,4,15)(6,16,7,18)(8,20,10,19)(9,17)>;`
`G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20), (1,3,5,2,4)(6,8,10,7,9)(11,14,12,15,13)(16,19,17,20,18), (1,9)(2,6,5,7)(3,8,4,10)(11,18,12,16)(13,19,15,20)(14,17), (1,14)(2,11,5,12)(3,13,4,15)(6,16,7,18)(8,20,10,19)(9,17) );`
`G=PermutationGroup([[(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20)], [(1,3,5,2,4),(6,8,10,7,9),(11,14,12,15,13),(16,19,17,20,18)], [(1,9),(2,6,5,7),(3,8,4,10),(11,18,12,16),(13,19,15,20),(14,17)], [(1,14),(2,11,5,12),(3,13,4,15),(6,16,7,18),(8,20,10,19),(9,17)]])`
`G:=TransitiveGroup(20,47);`
On 25 points: primitive - transitive group 25T17
Generators in S25
```(1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)
(1 25 9 12 19)(2 21 10 13 20)(3 22 6 14 16)(4 23 7 15 17)(5 24 8 11 18)
(2 25 5 19)(3 9 4 12)(6 7 15 14)(8 17 13 22)(10 23 11 16)(18 20 21 24)
(2 3 5 4)(6 24 15 20)(7 21 14 18)(8 23 13 16)(9 25 12 19)(10 22 11 17)```
`G:=sub<Sym(25)| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25), (1,25,9,12,19)(2,21,10,13,20)(3,22,6,14,16)(4,23,7,15,17)(5,24,8,11,18), (2,25,5,19)(3,9,4,12)(6,7,15,14)(8,17,13,22)(10,23,11,16)(18,20,21,24), (2,3,5,4)(6,24,15,20)(7,21,14,18)(8,23,13,16)(9,25,12,19)(10,22,11,17)>;`
`G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25), (1,25,9,12,19)(2,21,10,13,20)(3,22,6,14,16)(4,23,7,15,17)(5,24,8,11,18), (2,25,5,19)(3,9,4,12)(6,7,15,14)(8,17,13,22)(10,23,11,16)(18,20,21,24), (2,3,5,4)(6,24,15,20)(7,21,14,18)(8,23,13,16)(9,25,12,19)(10,22,11,17) );`
`G=PermutationGroup([[(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25)], [(1,25,9,12,19),(2,21,10,13,20),(3,22,6,14,16),(4,23,7,15,17),(5,24,8,11,18)], [(2,25,5,19),(3,9,4,12),(6,7,15,14),(8,17,13,22),(10,23,11,16),(18,20,21,24)], [(2,3,5,4),(6,24,15,20),(7,21,14,18),(8,23,13,16),(9,25,12,19),(10,22,11,17)]])`
`G:=TransitiveGroup(25,17);`
C52⋊Q8 is a maximal subgroup of D5≀C2⋊C2
C52⋊Q8 is a maximal quotient of (C5×C10).Q8
Polynomial with Galois group C52⋊Q8 over ℚ
actionf(x)Disc(f)
10T20x10-2x9-101x8+272x7+691x6-3142x5+3299x4+72x3-1939x2+978x-125233·37·58·234·292·1092·1272·12832
Matrix representation of C52⋊Q8 in GL8(ℤ)
1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 -1 -1 -1
,
0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 -1 -1 -1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 -1 -1 -1 -1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0
,
1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 -1 -1 -1 -1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 -1 -1 -1 -1
`G:=sub<GL(8,Integers())| [1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,-1,0,0,0,0,1,0,0,-1,0,0,0,0,0,1,0,-1,0,0,0,0,0,0,1,-1],[0,0,0,-1,0,0,0,0,1,0,0,-1,0,0,0,0,0,1,0,-1,0,0,0,0,0,0,1,-1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[0,0,0,0,1,-1,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,-1,0,1,0,0,0,0,0,-1,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0],[1,0,-1,0,0,0,0,0,0,0,-1,1,0,0,0,0,0,1,-1,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,1,0,0,-1,0,0,0,0,0,0,1,-1,0,0,0,0,0,0,0,-1,0,0,0,0,0,1,0,-1] >;`
C52⋊Q8 in GAP, Magma, Sage, TeX
`C_5^2\rtimes Q_8`
`% in TeX`
`G:=Group("C5^2:Q8");`
`// GroupNames label`
`G:=SmallGroup(200,44);`
`// by ID`
`G=gap.SmallGroup(200,44);`
`# by ID`
`G:=PCGroup([5,-2,-2,-2,-5,5,20,61,26,323,408,173,3004,409,1014]);`
`// Polycyclic`
`G:=Group<a,b,c,d|a^5=b^5=c^4=1,d^2=c^2,a*b=b*a,c*a*c^-1=b^-1,d*a*d^-1=a^3,c*b*c^-1=a,d*b*d^-1=b^2,d*c*d^-1=c^-1>;`
`// generators/relations`
Export
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𝔽 | 3,935 | 5,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.477538 |
https://oneclass.com/class-notes/us/ucsd/phys/phys-1b/1366366-phys-1b-lecture-6.en.html | 1,623,608,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610196.46/warc/CC-MAIN-20210613161945-20210613191945-00552.warc.gz | 421,491,424 | 103,944 | # PHYS 1B Lecture Notes - Lecture 6: Field Line, Electric Field, Surface Charge
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8 Feb 2017
School
Department
Course
Professor
PHYS1B Lecture 6 Notes
1/27/17
- For a single charge q, E = ke (q/r^2)r
o If multiple charges, the total electric field is the vector sum of the fields produced by
each charge qi where I = ,,…
- Clicker question: A positive charge q is located a distance d from a 2nd identical charge.
Is the electric field zero anywhere?
o YES, located in between the 2 charges
o What if there was one positive ad one negative charge?
YES, located on the outside of the charges
- Electric field of a dipole
o Electric dipole: consists of 2 point charges q and q separated by distance 2a
o Neutral atoms and molecules behave as dipoles when placed in an external electric
field (due to polarization)
o Some molecules (HCl) also permanent dipoles (H+ and Cl-)
o For r >> a, you find that E = ke(2qa/r^3)
Contrasts with a single charge
- Electric field lines
o Field lines represent the electric field pictorially
o BY DEFINITION: the lines point in the direction of the electric field vector at every
point in space
o The number of lines passing through a unit surface area perpendicular to the lines is
proportional to the magnitude of the electric field in that region
o Rules for drawing electric field lines:
Lines begin on positive charge and end on negative charge
If excess of one type of charge, some lines will begin/end infinitely far away
The number of lines drawn leaving the positive charge or approaching the
negative charge is proportional to the magnitude of the charge
2 field lines never cross
o ex of 2 equal positive charges:
same number of lines leave each charge since they are equal in magnitude
at great distance, field is 2q
o ex of 2 charges +2q and q
2 lines leave positive charge for each line that terminates on the negative charge
at great distance, field is q
- Continuous charge distributions
o Divide charged objects into small elements, each of which contains charge delta(q)
o Total field is the sum of fields due to each element
o Take the limit to change the sum to an integral
o *don’t actually need to do integral calculations
- Uniformly charged sphere (insulator)
o Charge q evenly distributed over the volume of sphere of radius a (must be an
insulator, or else charges would move and not be evenly distributed)
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https://www.mathcelebrity.com/community/threads/the-auto-repair-shop-took-2-5-hours-to-repair-victoria%E2%80%99s-car-the-cost-of-parts-was-93-and-the-tot.2736/ | 1,719,313,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00066.warc.gz | 757,014,491 | 8,887 | The auto repair shop took 2.5 hours to repair Victoria’s car. The cost of parts was \$93, and the tot
Discussion in 'Calculator Requests' started by math_celebrity, May 16, 2020.
Tags:
The auto repair shop took 2.5 hours to repair Victoria’s car. The cost of parts was \$93, and the total bill was \$248. What is the shops charge per hour.
Calculate Labor Cost:
Labor Cost = Total bill - Parts
Labor Cost = \$248 - \$93
Labor Cost = \$155
Calculate labor hourly rate:
Labor Hourly Rate = Labor Cost / Number of Labor Hours
Labor Hourly Rate = 155/2.5
Labor Hourly Rate = \$62 | 157 | 580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.940897 |
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Find the average distance between 2 points on a square
This is another excellent mathematical puzzle from the MindYourDecisions youtube channel. I like to try these without looking at the answer – and then to see how far I get. This one is pretty difficult (and the actual solution exceptionally difficult!) The problem is to take a square and randomly choose 2 points somewhere inside. If you calculate the distance between the 2 points, then do this trial approaching an infinite number of times what will the average distance be? Here is what I did.
Simplify the situation: 1×1 square
This is one of the most important strategies in tackling difficult maths problems. You simplify in order to gain an understanding of the underlying problem and possibly either develop strategies or notice patterns. So, I started with a unit square and only considered the vertices. We can then list all the possible lengths:
We can then find the average length by simply doing:
2×2 square
We can then follow the same method for a 2×2 square. This gives:
Which gives an average of:
Back to a 1×1 square
Now, we can imagine that we have a 1 x 1 square with dots at every 0.5. This is simply a scaled version of the 2×2 square, so we can divide our answer by 2 to give:
3×3 square
Following the same method we have:
This gives an average of:
Back to a 1×1 square
and if we imagine a 1×1 square with dots at every 1/3. This is simply a scaled version of the 3×3 square, so we can divide our answer by 3 to give:
We can then investigate what happens as we consider more and more dots inside our 1×1 square. When we have considered an infinite number then we will have our average distance – so we are looking the limit to infinity. This suggests using a graph. First I calculated a few more terms in the sequence:
Then I plotted this on Desmos. The points looked like they fit either an exponential or a reciprocal function – both which have asymptotes, so I tried both. The reciprocal function fit with an R squared value of 1. This is a perfect fit so I will use that.
This was plotted using the regression line:
And we can find the equation of the horizontal asymptote by seeing what happens when x approaches infinity. This will give a/c. Using the values provided by Desmos’ regression I got 0.515004887. Because I have been using approximate answers throughout I’ll take this as 0.52 (2sf). Therefore I predict that the average distance between 2 points in a 1×1 square will be approximately 0.52. And more generally, the average distance in an n x n square will be 0.52(n). This is somewhat surprising as a result – it’s not obvious why it would be a little over half the distance from 0 to 1.
Brute forcing using Python
We can also write a quick code to approximate this answer using Python (This is a Monte Carlo method). I generate 4 random numbers to represent the 2 x-coordinates and 2-y coordinates of 2 random points. I then work out the distance between them and repeat this 10 million times, then calculate the average distance. This gives:
Now for the moment of truth – and we watch the video to find out how accurate this is. The correct answer is indeed 0.52 (2sf) – which is great – our method worked! The exact answer is given by:
Our graphical answer is not quite accurate enough to 3 sf – probably because we relied on rounded values to plot our regression line. Our Python method with 10 million trials was accurate to 4 sf. Just to keep my computer on its toes I also calculated this with 100 million trials. This gave 0.5214126210834646 (now accurate to 5 sf).
We can also find the percentage error when using our graphical method. This is only:
Overall this is a decent result! If you are feeling extremely brave you might want to look at the video to see how to do this using calculus.
Extension: The average distance between 2 points in a unit circle
I modified the Python code slightly to now calculate the average distance between 2 points in a unit circle. This code is:
which returns an answer of 0.9054134561871364. I then looked up what the exact answer is. For the unit circle it is 128/(45 pi). This is approximately 0.9054147874. We can see that our computer method was accurate to 5 sf here. Again, the actual mathematical proof is extremely difficult.
Reflection
This is a nice example of important skills and techniques useful in mathematics – simplification of a problem, noticing patterns, graphical methods, computational power and perseverance!
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Further investigation of the Mordell Equation
This post carries on from the previous post on the Mordell Equation – so make sure you read that one first – otherwise this may not make much sense. The man pictured above (cite: Wikipedia) is Louis Mordell who studied the equations we are looking at today (and which now bear his name).
In the previous post I looked at solutions to the difference between a cube and a square giving an answer of 2. This time I’ll try to generalise to the difference between a cube and a square giving an answer of k. I’ll start with the same method as from the previous post:
In the last 2 lines we outline the 2 possibilities, either b = 1 or b = -1. First let’s see what happens when b = 1:
This will only provide an integer solution for a if we have:
Which generates the following first few values for k when we run through m = 1, 2,3..:
k = 2, 11, 26, 47
We follow the same method for b = -1 and get the following:
Which generates the following first few values for k when we run through m = 1, 2,3…:
k = 4, 13, 28, 49
These are the values of k which we will be able to generate solutions to. Following the same method as in the previous post this generates the following solutions:
Let’s illustrate one of these results graphically. If we take the solutions for k = 13, which are (17,70) and (17,-70), these points should be on the curve x cubed – y squared = 13.
This is indeed the case. This graph also demonstrates how all solutions to these curves will have symmetrical solutions (e, f) and (e, -f).
We can run a quick computer program to show that this method does not find all the solutions for the given values of k, but it does ensure solutions will be found for the k values in these lists.
In the code solutions above, results are listed k, x, y, x cubed, y squared. We can see for example that in the case of k = 11 our method did not find the solution x = 3 and y = 4 (though we found x = 15 and y = 58). So, using this method we now have a way of finding some solutions for some values of k – we’ve not cracked the general case, but we have at least made a start!
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
The Mordell Equation [Fermat’s proof]
Let’s have a look at a special case of the Mordell Equation, which looks at the difference between an integer cube and an integer square. In this case we want to find all the integers x,y such that the difference between the cube and the square gives 2. These sorts of problems are called Diophantine problems and have been studied by mathematicians for around 2000 years. We want to find integer solution to:
First we can rearrange and factorise, using the property of imaginary numbers.
Next we define alpha and beta such that:
For completeness we can say that alpha and beta are part of an algebraic number field:
Next we use an extension of the Coprime Power Trick, which ensures that the following 2 equations have solutions (if our original equation also has a solution). Therefore we define:
We can then substitute our definition for alpha into the first equation directly above and expand:
Next we equate real and imaginary coefficients to give:
This last equation therefore requires that either one of the following equations must be true:
If we take the case when b = 1 we get:
Therefore:
If we take the case when b = -1 we get
Therefore our solution set is (a,b): (1,1), (1,-1), (-1,1), (-1,-1. We substitute these possible answers into our definition for y to give the following:
We can then substitute these 2 values for y into the definition for x to get:
These therefore are the only solutions to our original equation. We can check they both work:
We can see this result illustrated graphically by plotting the graph:
and then seeing that we have our integer solutions (3,5) and (3,-5) as coordinate on this curve.
This curve also clearly illustrates why we have a symmetrical set of solutions, as our graph is symmetrical about the x axis.
This particular proof was first derived by Fermat (of Fermat’s Last Theorem fame) in the 1600s and is an elegant example of a proof in number theory. You can read more about the Mordell Equation in this paper (the proof above is based on that given in the paper, but there is a small mistake in factorization so that y = 7 and y = -7 is erroneously obtained).
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
Hollow Cubes investigation
Hollow cubes like the picture above [reference] are an extension of the hollow squares investigation done previously. This time we can imagine a 3 dimensional stack of soldiers, and so try to work out which numbers of soldiers can be arranged into hollow cubes.
Therefore what we need to find is what numbers can be formed from a3-b3
Python code
We can write some Python3 code to find this out (this can be run here):
``` for k in range(1,200):```
``` for a in range(0, 100): for b in range(0,100): if a**3-b**3 == k : print(k,a,b) ```
This gives the following: (the first number is the number of soldiers and the 2 subsequent numbers are the 2 cubes).
1 1 0
7 2 1
8 2 0
19 3 2
26 3 1
27 3 0
37 4 3
56 4 2
61 5 4
63 4 1
64 4 0
91 6 5
98 5 3
117 5 2
124 5 1
125 5 0
127 7 6
152 6 4
169 8 7
189 6 3
We could perhaps investigate any patterns in these numbers, or explore how we can predict when a hollow cube has more than one solution. I’ll investigate which numbers can be written as both a hollow square and also a hollow cube.
Hollow squares and hollow cubes
``` list1=[] for a in range(2, 50): for b in range(2,50): if a**2-b**2 !=0: if a**2-b**2 > 0: list1.append(a**2-b**2) list2=[] for j in list1: for c in range(2,50): for d in range(2,50): if c**3-d**3 == j: list2.append(c**3-d**3) print(list2) ```
This returns the following numbers which can all be written as both hollow squares and hollow cubes.
[56, 91, 19, 117, 189, 56, 208, 189, 217, 37, 279, 152, 117, 448, 513, 504, 448, 504, 387, 665, 504, 208, 875, 819, 936, 817, 61, 999, 988, 448, 728, 513, 189, 1216, 936, 784, 335, 469, 1323, 819, 1512, 1352, 1197, 992, 296, 152, 1519, 1512, 1197, 657, 1664, 1323, 1647, 1736, 1701, 1664, 936, 504, 2107, 1387, 1216, 1027, 91, 2015, 279, 2232]
Hollow squares, cubes and hypercubes
Taking this further, can we find any number which can be written as a hollow square, hollow cube and hollow hypercube (4 dimensional cube)? This would require our soldiers to be able to be stretch out into a 4th dimensional space – but let’s see if it’s theoretically possible.
Here’s the extra code to type:
``` list1=[] for a in range(2, 200): for b in range(2,200): if a**2-b**2 !=0: if a**2-b**2 > 0: list1.append(a**2-b**2) list2=[] for j in list1: for c in range(2,200): for d in range(2,200): if c**3-d**3 == j: list2.append(c**3-d**3) print(list2) for k in list2: for e in range(2,200): for f in range(2,200): if k == e**4-f**4: print(k) ```
Very pleasingly this does indeed find some solutions:
9919: Which can be formed as either 1002-92 or 223-93 or 104-34.
14625: Which can be formed as either 1212-42 or 253-103 or 114-24.
Given that these took some time to find, I think it’ll require a lot of computer power (or a better designed code) to find any number which is a hollow square, hollow cube, hollow hypercube and hollow 5-dimensional cube, but I would expect that there is a number out there that satisfies all criteria. Maybe you can find it?
Waging war with maths: Hollow squares
The picture above [US National Archives, Wikipedia] shows an example of the hollow square infantry formation which was used in wars over several hundred years. The idea was to have an outer square of men, with an inner empty square. This then allowed the men in the formation to be tightly packed, facing the enemy in all 4 directions, whilst the hollow centre allowed the men flexibility to rotate (and also was a place to hold supplies). It was one of the infantry formations of choice against charging cavalry.
So, the question is, what groupings of men can be arranged into a hollow square? This is a current Nrich investigation, so I thought I’d do a mini-investigation on this.
We can rethink this question as asking which numbers can be written as the difference between 2 squares. For example in the following diagram (from the Nrich task Hollow Squares)
We can see that the hollow square formation contains a larger square of 20 by 20 and a smaller hollow square of 8 by 8. Therefore the number of men in this formation is:
202-82 = 336.
The first question we might ask therefore is how many numbers from 1-100 can be written as the difference between 2 squares? These will all be potential formations for our army.
I wrote a quick code on Python to find all these combinations. I included 0 as a square number (though this no longer creates a hollow square, rather just a square!). You can copy this and run it in a Python editor like Repl.it.
``` for k in range(1,50):```
``` ```
``` for a in range(0, 100): for b in range(0,100): if a**2-b**2 == k : print(k,a,b) ```
This returned the following results:
1 1 0
3 2 1
4 2 0
5 3 2
7 4 3
8 3 1
9 3 0
9 5 4
11 6 5
12 4 2
13 7 6
15 4 1
15 8 7
16 4 0
16 5 3
17 9 8
19 10 9
20 6 4
21 5 2
21 11 10
23 12 11
24 5 1
24 7 5
25 5 0
25 13 12
27 6 3
27 14 13
28 8 6
29 15 14
31 16 15
32 6 2
32 9 7
33 7 4
33 17 16
35 6 1
35 18 17
36 6 0
36 10 8
37 19 18
39 8 5
39 20 19
40 7 3
40 11 9
41 21 20
43 22 21
44 12 10
45 7 2
45 9 6
45 23 22
47 24 23
48 7 1
48 8 4
48 13 11
49 7 0
49 25 24
Therefore we can see that the numbers with no solutions found are:
2,6,10,14,18,22,26,30,34,38,42,46,50
which are all clearly in the sequence 4n-2.
Thinking about this, we can see that this can be written as 2(2n-1) which is the product of an even number and an odd number. This means that all numbers in this sequence will require an odd factor in each of their factor pairs:
eg. 50 can be written as 10 (even) x 5 (odd) or 2 (even) x 25 (odd) etc.
But with a2-b2 = (a+b)(a-b), due to symmetries we will always end up with (a+b) and (a-b) being both even or both odd, so we can’t create a number with a factor pair of one odd and one even number. Therefore numbers in the sequence 4n-2 can’t be formed as the difference of 2 squares. There are some nicer (more formal) proofs of this here.
A battalion with 960 soldiers
Next we are asked to find how many different ways of arranging 960 soldiers in a hollow square. So let’s modify the code first:
``` for a in range(0, 1000): for b in range(0,1000): if a**2-b**2 == 960 : print(a,b) ```
Which gives us the following solutions:
31 1
32 8
34 14
38 22
46 34
53 43
64 56
83 77
122 118
241 239
General patterns
We can notice that when the number of soldiers is 1,3,5,7,9,11 (2n-1) we can always find a solution with the pair n and n-1. For example, 21 can be written as 2n-1 with n = 11. Therefore we have 10 and 11 as our pair of squares. This works because 112-102 = (11+10)(11-10) returns the factor pair 21 and 1. In general it always returns the factor pair, 2n-1 and 1.
We can also notice that when the number of soldiers is 4,8,12,16,20 (4n) we can always find a solution with the pair n+1 and n-1. For example, 20 can be written as 4n with n = 5. Therefore we have 6 and 4 as our pair of squares. This works because 62-42 = (6+4)(6-4) returns the factor pair 10 and 2. In general it always returns the factor pair, 2n and 2.
And we have already shown that numbers 2,6,10,14,18,22 (4n-2) will have no solution. These 3 sequences account for all the natural numbers (as 2n-1 incorporates the 2 sequences 4n-3 and 4n-1).
So, we have found a method of always finding a hollow square formation (if one exists) as well as being able to use some computer code to find other possible solutions. There are lots of other avenues to explore here – could you find a method for finding all possible combinations for a given number of men? What happens when the hollow squares become rectangles?
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Normal Numbers – and random number generators
Numberphile have a nice new video where Matt Parker discusses all different types of numbers – including “normal numbers”. Normal numbers are defined as irrational numbers for which the probability of choosing any given 1 digit number is the same, the probability of choosing any given 2 digit number is the same etc. For example in the normal number 0.12345678910111213141516… , if I choose any digit in the entire number at random P(1) = P(2) = P(3) = … P(9) = 1/10. Equally if I choose any 2 digit number at random I have P(10) = P(11) = P(12) = P(99) = 1/100.
It is incredibly hard to find normal numbers, but there is a formula to find some of them.
In base 10, we are restricted to choosing a value of c such that 10 and c are relatively prime (i.e share no common factors apart from 1). So if we choose c = 3 this gives:
We can now put this into Wolfram Alpha and see what number this gives us:
So we can put the first few digits into an online calculator to find the distributions
0.000333333444444444444448148148148148148148148148148148148148148149382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049827160493827160493827160479423863312 7572016460905349794238683127572016460905349794238683127572016460 9053497942386831275720164609053497942386831275720164609053497942
4: 61
1: 41
8: 40
3: 38
0: 36
2: 33
7: 33
9: 33
6: 32
5: 10
We can see that we are already seeing a reasonably similar distribution of single digits, though with 4 and 5 outliers. As the number progressed we would expect these distributions to even up (otherwise it would not be a normal number).
One of the potential uses of normal numbers is in random number generators – if you can use a normal number and specify a digit (or number of digits) at random then this should give an equal chance of returning each number.
To finish off this, let’s prove that the infinite series:
does indeed converge to a number (if it diverged then it could not be used to represent a real number). To do that we can use the ratio test (only worry about this bit if you have already studied the Calculus Option for HL!):
We can see that in the last limit 3 to the power n+1 will grow faster than 3 to the power n, therefore as n increases the limit will approach 0. Therefore by the ratio test the series converges to a real number.
Is pi normal?
Interestingly we don’t know if numbers like e, pi and ln(2) are normal or not. We can analyse large numbers of digits of pi – and it looks like it will be normal, but as yet there is no proof. Here are the distribution of the first 100,000 digits of pi:
1: 10137
6: 10028
3: 10026
5: 10026
7: 10025
0: 9999
8: 9978
4: 9971
2: 9908
9: 9902
Which we can see are all very close to the expected value of 10,000 (+/- around 1%).
So, next I copied the first 1 million digits of pi into a character frequency counter which gives the following:
5: 100359
3: 100230
4: 100230
9: 100106
2: 100026
8: 99985
0: 99959
7: 99800
1: 99758
6: 99548
This is even closer to the expected values of 100,000 with most with +/- 0.25 %.
Proving that pi is normal would be an important result in number theory – perhaps you could be the one to do it!
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Volume optimization of a cuboid
This is an extension of the Nrich task which is currently live – where students have to find the maximum volume of a cuboid formed by cutting squares of size x from each corner of a 20 x 20 piece of paper. I’m going to use an n x 10 rectangle and see what the optimum x value is when n tends to infinity.
First we can find the volume of the cuboid:
Next we want to find when the volume is a maximum, so differentiate and set this equal to 0.
Next we use the quadratic formula to find the roots of the quadratic, and then see what happens as n tends to infinity (i.e we want to see what the optimum x values are for our cuboid when n approaches infinity). We only take the negative solution of the + – quadratic solutions because this will be the only one that fits the initial problem.
Next we try and simplify the square root by taking out a factor of 16, and then we complete the square for the term inside the square root (this will be useful next!)
Next we make a u substitution. Note that this means that as n approaches infinity, u approaches 0.
Substituting this into the expression gives us:
We then manipulate the surd further to get it in the following form:
Now, the reason for all that manipulation becomes apparent – we can use the binomial expansion for the square root of 1 + u2 to get the following:
Therefore we have shown that as the value of n approaches infinity, the value of x that gives the optimum volume approaches 2.5cm.
So, even though we start with a pretty simple optimization task, it quickly develops into some quite complicated mathematics. We could obviously have plotted the term in n to see what its behavior was as n approaches infinity, but it’s nicer to prove it. So, let’s check our result graphically.
As we can see from the graph, with n plotted on the x axis and x plotted on the y axis we approach x = 2.5 as n approaches infinity – as required.
An m by n rectangle.
So, we can then extend this by considering an n by m rectangle, where m is fixed and then n tends to infinity. As before the question is what is the value of x which gives the maximum volume as n tends to infinity?
We do the same method. First we write the equation for the volume and put it into the quadratic formula.
Next we complete the square, and make the u substitution:
Next we simplify the surd, and then use the expansion for the square root of 1 + u2
This then gives the following answer:
So, we can see that for an n by m rectangle, as m is fixed and n tends to infinity, the value of x which gives the optimum volume tends to m/4. For example when we had a 10 by n rectangle (i.e m = 10) we had x = 2.5. When we have a 20 by n rectangle we would have x = 5 etc.
And we’ve finished! See what other things you can explore with this problem.
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Narcissistic Numbers
Narcissistic Numbers are defined as follows:
An n digit number is narcissistic if the sum of its digits to the nth power equal the original number.
For example with 2 digits, say I choose the number 36:
32 + 62 = 45. Therefore 36 is not a narcissistic number, as my answer is not 36.
For example with 3 digits, say I choose the number 124:
13 + 23 + 43 = 73. Therefore 124 is not a narcissistic number as my answer is not 124.
The question is how to find all the narcissistic numbers less than 1000, without checking 1000 different numbers? Let’s start with 1 digit numbers.
1 digit numbers
01 = 0
11 = 1
21 = 2 etc.
Therefore all numbers from 0-9 are narcissistic.
2 digit numbers
For 2 digit numbers in the form ab we need the following:
a2 + b2 = 10a + b.
Therefore
a2 – 10a + b2 – b = 0.
Next if we choose a = 1,2,3,4,5 we get the following simultaneous equations:
b2 – b -16 = 0
b2 – b -21 = 0
b2 – b -24 = 0
b2 – b -25 = 0
None of these factorise for integer solutions, therefore there are no 2 digit solutions from 11 to 59. Trying a = 6,7,8,9 we find that we get the same as the first four equations. This is because a and 10-a give equivalent solutions. In other words, when a = 1 we get the equation b2 – b -9 = 0 and when a = 9 we also get the equation b2 – b -9 = 0. This is because:
for:
a2 – 10a
if we substitute a = (10 – a) we get
(10 – a)2 – 10(10 – a) = a2 – 10a.
Therefore we prove that there are no 2 digit narcissistic numbers.
3 digit numbers
First we list the cube numbers:
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343, 83 = 512, 93 = 729.
and then consider 3 digit numbers of the form 1bc first. We need:
13+ b3 + c3 = 100 + 10b + c.
If our first digit is 1, then b3 + c3 need to add up to give us a number in the one hundreds, therefore:
99 ≤ b3 + c3≤ 198.
We can then check the cube numbers and see that the only possible combinations for a and b are 0 5, 5 0, 1 5, 5 1, 2 5, 5 2, 3 5, 5 3, 4 4, 4 5, 5 4. We can check these (only have to use the calculator for half as the reversed numbers give equivalent answers) and find that for 153 we do indeed get a narcissistic number i.e:
13+ 53 + 33 = 153.
Next we consider 3 digit numbers of the form 2bc first. We need:
192 ≤ b3 + c3≤ 292
This gives the following possibilities for b and c: 6 0, 0 6, 6 1, 16, 2 6, 6 2, 6 3, 3 6, 6 4, 4 6.
None of these give narcissistic numbers.
Next we consider 3 digit numbers of the form 3bc first. We need:
273 ≤ b3 + c3≤373
This gives the following possibilities for b and c: 6 4, 4 6, 6 5, 5 6, 7 1, 1 7, 7 2, 2 7, 7 3, 3 7, 7 0, 0 7.
Checking these we find 2 more narcissistic numbers:
370 = 33+ 73 + 03
371= 33+ 73 + 13
Using the same method, we can find that the only possibilities for 4bc are: 5 6, 6 5, 7 1, 1 7, 7 2, 2 7, 7 3, 3 7, 7 4, 4 7, 7 0, 0 7. Checking these gives us 1 more narcissistic number:
407= 43+ 03 + 73
We can carry on with this method checking up to the form 9ab (it gets easier as there are less combinations possible, but will find no more narcissistic numbers. Therefore we have all the narcissistic numbers less than 1000:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407.
Is there a limit to how many narcissistic numbers there are?
Surprisingly there is a limit – there are exactly 88 narcissistic numbers in base 10. To see why we can consider the following:
In 3 digits the biggest number we can choose is 999. That would give 93+ 93 + 93 (or 3(9)3). This needs to give a number in the hundreds (102) otherwise it would be impossible to achieve a narcissistic number. Therefore with an n digit number the largest number we can make is n(9)n and if we can’t make a number in the 10n-1, then a narcissistic number is not possible. If we can prove that the inequality:
n(9)n < 10n-1
is true for some values of n, then there will be an upper bound to the narcissistic numbers we can make. We could simply plot this directly, but let’s see if we can convince ourselves it’s true for some n without using graphical software first. Let’s see if we can find an equality:
n(9)n = 10n-1
First we take log base 10 of both sides
log n(9)n = n-1
log(n) + nlog(9) = n-1
n(log9 -1) + logn +1 = 0
Next we make the substitution logn = u and therefore 10u = n. This gives:
10u(log9 -1) + u + 1 = 0.
Now we can clearly see that 10u will grow much larger than u + 1, so any root must be for u is small. Let’s see, when u = 1 we get a positive number (as log9 -1 is a negative number close to 0), but when u = 2 we get a negative number. Therefore we have a root between u = 1 and u = 2. Given that we made the substitution logn = u, that means we have found the inequality n(9)n < 10n-1 will hold for n somewhere between 101 and 10 2.
Using Wolfram we can see that the equality is reached when u = 1.784, i.e when n = 101.784 or approx 60.8. Therefore we can see that when we have more than 60 digit numbers, it is no longer possible to make narcissistic numbers.
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).
How to Avoid The Troll: A Puzzle
In these situations it’s best to look at the extreme case first so you get some idea of the problem. If you are feeling particularly pessimistic you could assume that the troll is always going to be there. Therefore you would head to the top of the barrier each time. This situation is represented below:
The Pessimistic Solution:
Another basic strategy would be the optimistic strategy. Basically head in a straight line hoping that the troll is not there. If it’s not, then the journey is only 2km. If it is then you have to make a lengthy detour. This situation is shown below:
The Optimistic Solution:
The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.
The question is now, is there a better strategy than either of these? An obvious possibility is heading for the point halfway along where the barrier might be. This would make a triangle of base 1 and height 1/2. This has a hypotenuse of root (5/4). In the best case scenario we would then have a total distance of 2 x root (5/4). In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2. We find the expected value by multiply both by 0.5 and adding. This gives 2.63 (2 dp). But can we do any better? Yes – by using some algebra and then optimising to find a minimum.
The Optimisation Solution:
To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum. Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us. Wolfram Alpha is incredibly powerful -and also very easy to use. Here is what I entered:
and here is the output:
So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.
So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint. Not a huge difference – but nevertheless we’ll save ourselves a few seconds!
This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across? Maybe the troll appears only 1/3 of the time? Could you even generalise the results for when the troll is y distance away or appears z percent of the time?
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Zeno’s Paradox – Achilles and the Tortoise
This is a very famous paradox from the Greek philosopher Zeno – who argued that a runner (Achilles) who constantly halved the distance between himself and a tortoise would never actually catch the tortoise. The video above explains the concept.
There are two slightly different versions to this paradox. The first version has the tortoise as stationary, and Achilles as constantly halving the distance, but never reaching the tortoise (technically this is called the dichotomy paradox). The second version is where Achilles always manages to run to the point where the tortoise was previously, but by the time he reaches that point the tortoise has moved a little bit further away.
The first version we can think of as follows:
Say the tortoise is 2 metres away from Achilles. Initially Achilles halves this distance by travelling 1 metre. He halves this distance again by travelling a further 1/2 metre. Halving again he is now 1/4 metres away. This process is infinite, and so Zeno argued that in a finite length of time you would never actually reach the tortoise. Mathematically we can express this idea as an infinite summation of the distances travelled each time:
1 + 1/2 + 1/4 + 1/8 …
Now, this is actually a geometric series – which has first term a = 1 and common ratio r = 1/2. Therefore we can use the infinite summation formula for a geometric series (which was derived about 2000 years after Zeno!):
sum = a/(1-r)
sum = 1/(1-0.5)
sum = 2
This shows that the summation does in fact converge – and so Achilles would actually reach the tortoise that remained 2 metres away. There is still however something of a sleight of hand being employed here however – given an infinite length of time we have shown that Achilles would reach the tortoise, but what about reaching the tortoise in a finite length of time? Well, as the distances get ever smaller, the time required to traverse them also gets ever closer to zero, so we can say that as the distance converges to 2 metres, the time taken will also converge to a finite number.
There is an alternative method to showing that this is a convergent series:
S = 1+ 1/2 + 1/4 + 1/8 + 1/16 + …
0.5S = 1/2+ 1/4 + 1/8 + 1/16 + …
S – 0.5S = 1
0.5S = 1
S = 2
Here we notice that in doing S – 0.5S all the terms will cancel out except the first one.
Achilles and the Tortoise
The second version also makes use of geometric series. If we say that the tortoise has been given a 10 m head start, and that whilst the tortoise runs at 1 m/s, Achilles runs at 10 m/s, we can try to calculate when Achilles would catch the tortoise. So in the first instance, Achilles runs to where the tortoise was (10 metres away). But because the tortoise runs at 1/10th the speed of Achilles, he is now a further 1m away. So, in the second instance, Achilles now runs to where the tortoise now is (a further 1 metre). But the tortoise has now moved 0.1 metres further away. And so on to infinity.
This is represented by a geometric series:
10 + 1 + 0.1 + 0.01 …
Which has first time a = 10 and common ratio r = 0.1. So using the same formula as before:
sum = a/(1-r)
sum = 10/(1-0.1)
sum = 11.11m
So, again we can show that because this geometric series converges to a finite value (11.11), then after a finite time Achilles will indeed catch the tortoise (11.11m away from where Achilles started from).
We often think of mathematics and philosophy as completely distinct subjects – one based on empirical measurement, the other on thought processes – but back in the day of the Greeks there was no such distinction. The resolution of Zeno’s paradox by use of calculus and limits to infinity some 2000 years after it was first posed is a nice reminder of the power of mathematics in solving problems across a wide range of disciplines.
The Chess Board Problem
The chess board problem is nothing to do with Zeno (it was first recorded about 1000 years ago) but is nevertheless another interesting example of the power of geometric series. It is explained in the video above. If I put 1 grain of rice on the first square of a chess board, 2 grains of rice on the second square, 4 grains on the third square, how much rice in total will be on the chess board by the time I finish the 64th square?
The mathematical series will be:
1+ 2 + 4 + 8 + 16 +……
So a = 1 and r = 2
Sum = a(1-r64)/(1-r)
Sum = (1-264)/(1-2)
Sum = 264 -1
Sum = 18, 446,744, 073, 709, 551, 615
This is such a large number that, if stretched from end to end the rice would reach all the way to the star Alpha Centura and back 2 times.
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
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All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).
### New website for International teachers
I’ve just launched a brand new maths site for international schools – over 2000 pdf pages of resources to support IB teachers. If you are an IB teacher this could save you 200+ hours of preparation time.
Explore here!
### Free HL Paper 3 Questions
P3 investigation questions and fully typed mark scheme. Packs for both Applications students and Analysis students. | 14,302 | 56,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-21 | longest | en | 0.91433 |
https://math.stackexchange.com/questions/4166977/calculating-sheaf-cohomology-of-canonical-bundle-over-a-curve | 1,726,257,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00767.warc.gz | 346,868,072 | 37,105 | # Calculating Sheaf Cohomology of Canonical Bundle over a Curve
I'm trying to solve the following problem.
Let $$C$$ be a smooth projective curve and let $$\Omega^1_C$$ be the canonical bundle. Show that there is a canonical isomorphism, $$H^1(C, \Omega^1_C)\cong\mathbb{C}.$$
If I'm not mistaken, it's fairly straightforward to show that $$\text{dim }H^1(C, \Omega^1_C)=1$$ via the Riemann-Roch theorem. I'm getting a bit stuck with showing the canonical isomorphism however.
My attempt
Since $$C$$ is projective, we can define a closed embedding $$C\hookrightarrow\mathbb{P}^n$$. Let $$\mathcal{I}_C$$ be the corresponding ideal sheaf and consider the conormal short exact sequence $$0\to \mathcal{I}_C/\mathcal{I}_C^2\to \Omega^1_{\mathbb{P}^n}\otimes \mathcal{O}_C\to \Omega^1_C\to 0.$$ This induces a long exact sequence in cohomology $$\cdots\to H^1(C,\mathcal{I}_C/\mathcal{I}_C^2)\to H^1(C, \Omega^1_{\mathbb{P}^n}\otimes \mathcal{O}_C)\to H^1(C,\Omega^1_C)\to H^2(C,\mathcal{I}_C/\mathcal{I}_C^2)\to \cdots$$ But $$H^2(C,\mathcal{I}_C/\mathcal{I}_C^2)$$ vanishes (along with all other higher terms) for dimension reasons. Furthermore, I think I can show that $$H^1(C, \Omega^1_{\mathbb{P}^n}\otimes \mathcal{O}_C)\cong \mathbb{C}$$ via an Euler sequence argument.
If I could also show that $$H^1(C,\mathcal{I}_C/\mathcal{I}_C^2)=0$$ then I would be done, but it's this part I'm stuck on. I thought perhaps of using Serre duality but I didn't have much success.
Any help is much appreciated!
This is a direct consequence of Serre duality: $$H^1(C, \Omega^1_C) \cong H^0(C, \mathcal{O}_C)^{\vee} \cong \mathbb{C}$$ where the last isomorphism comes from the projectivity of $$C$$. The duality pairing also gives you that the isomorphism is canonical.
• It comes from the canonical morphism $\mathbb{C}\rightarrow \mathrm{Hom}(\mathcal{O}_C,\mathcal{O}_C)$. Commented Jun 8, 2021 at 22:46 | 649 | 1,901 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.795098 |
https://frequentlyasked.net/is-175-lbs-fat/ | 1,653,124,325,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00564.warc.gz | 322,756,200 | 17,066 | # Is 175 lbs fat?
It depends on your body composition. Muscle is more dense than fat and therefore 175lbs at 20% body fat is going to be different than 175lbs at 10% bodyfat. Based on your name, (Sorry I will be assuming your gender) the average female that doesn’t work out that weighs 175lbs at 5′7″ is yes in fact overweight.
## What is the remainder when 17 is divided by 5?
Using a calculator, if you typed in 17 divided by 5, you’d get 3.4. You could also express 17/5 as a mixed fraction: 3 2/5. If you look at the mixed fraction 3 2/5, you’ll see that the numerator is the same as the remainder (2), the denominator is our original divisor (5), and the whole number is our final answer (3).
## Is 174 pounds heavy?
Charts and online calculators 4 in. tall is considered overweight (BMI is 25 to 29) if she weighs between about 145 and 169 pounds. tall is considered overweight (BMI is 25 to 29) if he weighs between about 174 and 202 pounds, and is obese (BMI is 30 or more) if he is closer to 209 pounds or more.
## Is 175 pounds fat for a 13 year old?
According to the CDC, most 13-year-old girls weigh between 76 and 148 pounds (lb). The 50th percentile for weight in this group is around 101 lb. This means that about 50% of girls this age weigh less than 101 lb. If a 13-year-old girl weighs above the 95th percentile, the doctor may diagnose obesity.
## How do you solve 13 divided by 5?
Using a calculator, if you typed in 13 divided by 5, you’d get 2.6. You could also express 13/5 as a mixed fraction: 2 3/5. If you look at the mixed fraction 2 3/5, you’ll see that the numerator is the same as the remainder (3), the denominator is our original divisor (5), and the whole number is our final answer (2).
## What does 5 divided by 2 look like?
The number 5 divided by 2 is 2 with a remainder of 1 (5 / 2 = 2 R. 1). This is also written as 5 /2 = 2.5.
## IS 175 a perfect cube?
If we look at the number 175, we know that the cube root is 5.593444710407, and since this is not a whole number, we also know that 175 is not a perfect cube.
## What is the cube of 175?
Cube root of 175 can be represented as 3√175. The value of cube root of one is 175. The nearest previous perfect cube is 125 and the nearest next perfect cube is 216 . Cube root of 175 can be represented as 3√175.
## How do you get 225?
So we know that 15 × 15 = 225. In such cases, the number 225 is called a perfect square. Hence, the factors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, and 225.
## Is 150 pounds good for a 15 year old?
The “normal” weight for a 15-year old varies due to a number of factors and the best way to obtain a healthy weight range for your exact age, height and sex is to use the CDC’s calculator. For a girl of the same age and height, this weight range is 102 to 149 pounds.
## Is 200 pounds overweight for a 15 year old?
200 pounds is a lot for 15-year-old but in order for you to be medically declared obese, we must consider your height and gender to calculate BMI. If your BMI is over 30% you are considered obese.
## What BMI should a 15 year old have?
Normal BMI is between 18.5 and 24.9. BMI cut-offs for people of South Asian descent are different. Multiply your weight in pounds by 703. Divide that answer by your height in inches.
## What is an ideal body weight?
Normal ideal body weight (IBW) for patients can be calculated by the formulas21: • Healthy males: 106 lb for initial 5 feet, plus 6 lb for every inch over 5 feet, plus 10% if over 50 years old. • Healthy females: 100 lb for initial 5 feet, plus 5 lb for every inch over 5 feet, plus 10% if older than 50 years.
See also How long do you put a hot pocket in the air fryer?
## What can 5 be divided by?
A number is divisible by 5 if the number’s last digit is either 0 or 5. Divisibility by 5 – examples: The numbers 105, 275, 315, 420, 945, 760 can be divided by 5 evenly.
## What does divided by look like?
The division sign resembles a dash or double dash with a dot above and a dot below (÷). It is equivalent to the words “divided by.” This symbol is found mainly in arithmetic texts at the elementary-school level.
## How do you solve 5 divided by 1 2?
Using a calculator, if you typed in 5 divided by 12, you’d get 0.4167. You could also express 5/12 as a mixed fraction: 0 5/12.
## IS 196 a perfect square?
Is the number 196 a perfect square? As the factors of 196 are square of 2 and 7, 2² × 7². Hence, 196 is a perfect square. | 1,254 | 4,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-21 | latest | en | 0.952821 |
https://cgcookie.com/questions/15125-grid-fill-vs-simple-subdivision-modifier | 1,643,361,600,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00328.warc.gz | 221,246,260 | 12,529 | 6 answers · asked · Lesson: Topology Tools · Course: Blender Mesh Modeling Bootcamp
## Grid Fill vs. Simple Subdivision Modifier
I discovered that adding and then applying a subdivision surface modifier set to simple can sometimes do a better a job of filling in faces than grid fill. Is there a way in edit mode to grid fill in a way that aligns with how a simple subdivision surface modifier would fill in a face?
• I don't understand what you are doing there, but the use cases for Grid Fill and Subdivision Surface are completely diferent.
Grid fill fills a hole (needs an even number of Vertices), for instance:
Subdivision Surface won't do anything in this case, because there is no surface. If you first try to Fill this hole with an N-gon with 'F', and then use Subsurf, you'd get this kind of abomination:
Like I said, I don't know what you did, because Grid Fill doesn't work on a 5-sided circle (and is not meant to subdivide an N-gon anyway, although it might work when you have an even number of Verts...).)
• I'll try and be more clear. If I apply a subdivision surface modifier, set to simple, it seems to subdivides the face in a way that leaves a more symmetrical grid, with a central vertex, than if I were to grid fill a matching edge loop (see my example above). So, I am wondering if there is a way to end up with that same sort of symmetrical filling of an edge loop in edit mode.
• So what does that 'matching edge loop' (before using Grid fill) look like?
• Oh, I think I see what you mean...you can change the Blending of the Grid fill:
This does give a different result, but in cases like this, Subdiv would be the prefered method anyway (I think).
• crew
Good question! I think the one other thing to point out is that the subdiv method isn't technically creating a grid because it has that pole in the center, so grid fill could never create that because it's only trying to find solutions where each inside vertex is attached to exactly four edges. It is interesting though that the subdiv modifier shape can't be created with the subdivide operator in edit mode.
• Thanks spikeyxxx and jlampel ! I see now that the central pole is what is allowing for the different filling of the face. It would be cool if they added some way to do that in edit mode. | 521 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-05 | latest | en | 0.927707 |
https://www.slideserve.com/alvis/16-317-microprocessor-systems-design-i-powerpoint-ppt-presentation | 1,660,593,796,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00767.warc.gz | 851,523,992 | 19,462 | 16.317 Microprocessor Systems Design I
16.317 Microprocessor Systems Design I
16.317 Microprocessor Systems Design I
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Presentation Transcript
1. 16.317Microprocessor Systems Design I Instructor: Dr. Michael Geiger Fall 2013 Lecture 14 Compare instructions; conditional execution
2. Lecture outline • Announcements/reminders • HW 4 to be posted; due 10/16 • Exam 1 regrade requests due 10/11 • Today’s lecture • Compare instructions • CMOV • Set on condition • Early feedback forms Microprocessors I: Lecture 14
3. Compare Instructions • Compare 2 values; store result in ZF/SF • General format: CMP D,S • Works by performing subtraction (D) – (S) • D, S unchanged • ZF/SF/OF indicate result (signed values) • ZF = 1 D == S • ZF = 0, (SF XOR OF) = 1 D < S • ZF = 0, (SF XOR OF) = 0 D > S Microprocessors I: Lecture 14
4. Example—Initialization of internal registers with immediate data and compare. Example: MOV AX,1234H ;Initialize AX MOV BX,ABCDH ;Initialize BX CMP AX,BX ;Compare AX-BX Data registers AX and BX initialized from immediate data IMM16 (AX) = 1234H + integer IMM16 (BX) = ABCDH - integer Compare computation performed as: (AX) = 00010010001101002 (BX) = 10101011110011012 (AX) – (BX) = 00010010001101002 - 10101011110011012 ZF = 0 = NZ SF = 0 = PL ;treats as signed numbers CF = 1 = CY AF = 1 = AC OF = 0 = NV PF = 0 = PO Compare Instructions- Example Microprocessors I: Lecture 14
5. Condition codes • Conditional execution: result depends on value of flag bit(s) • Intel instructions specify condition codes • Condition code implies certain flag values • Opcodes written with cc as part of name • cc can be replaced by any valid code • Examples: CMOVcc, SETcc, Jcc • Specific examples: CMOVE, SETL, SETZ, JNE, JG Microprocessors I: Lecture 14
6. Condition codes (cont.) • Testing overflow • O (OF = 1), NO (OF =0) • Testing carry flag • C (CF = 1) • NC (CF = 0) • Testing sign flag • S (SF = 1), NS (SF = 0) • Testing parity flag • P or PE (PF = 1) • NP or PO (PF = 0) Microprocessors I: Lecture 14
7. Condition codes (cont.) • Testing equality/zero result • E or Z (ZF = 1) • NE or NZ (ZF = 0) • Signed comparison • L or NGE (SF XOR OF = 1) • NL or GE (SF XOR OF = 0) • LE or NG ((SF XOR OF) OR ZF = 1) • NLE or G ((SF XOR OF) OR ZF = 0) • Unsigned comparison • “Below” less than,“above” greater than • B, NAE (CF = 1) • NB, AE (CF = 0) • BE or NA (CF OR ZF = 1) • NBE or A (CF OR ZF = 0) Microprocessors I: Lecture 14
8. Conditional move (CMOV) • Only in Pentium Pro & later • Perform move only if condition is true • Examples: • CMOVZ AX, [SI] move if ZF == 1 • CMOVG EBX, EAX move if greater than Microprocessors I: Lecture 14
9. Byte Set on Condition Instruction • Byte set on condition instruction • Used to set byte based on condition code • Can be used for boolean results—complex conditions • General format: • SETcc D • cc = one of the supported conditional relationships Microprocessors I: Lecture 14
10. Byte Set on Condition Instruction • Operation: Flags tested for conditions defined by “cc” and the destination in a register or memory updated as follows If cc test True: 000000012 = 01H D If cc test False: 000000002 = 00H D • Examples of conditional tests: SETE = set byte if equal ZF = 1 SETC = set byte if carry CF =1 SETBE = set byte if below or equal CF = 1 +(or) ZF = 1 • Example: SETA AL = set byte if above if CF = 0 (and) ZF = 0 (AL) = 01H Otherwise, (AL) =00H Microprocessors I: Lecture 14
11. Example • Show the results of the following instructions, assuming that • “A” = DS:100H = 0001H • “B” = DS:102H = 0003H • “C” = DS:104H = 1011H • “D” = DS:106H = 1011H • “E” = DS:108H = ABCDH • “F” = DS:10AH = DCBAH • What complex condition does this sequence test? • MOV AX, [100H] • CMP AX, [102H] • SETLE BL • MOV AX, [104H] • CMP AX, [106H] • SETE BH • AND BL, BH • MOV AX, [108H] • CMP AX, [10AH] • SETNE BH • OR BL, BH Microprocessors I: Lecture 14
12. Example solution • Condition being tested: • To simplify, treat each word as a variable named “A” through “F” • ((A <= B) && (C == D)) || (E != F) • Source: http://www.arl.wustl.edu/~lockwood/class/cs306/books/artofasm/Chapter_6/CH06-4.html Microprocessors I: Lecture 14
13. Final notes • Next time: • Jump and loop instructions • Reminders: • HW 4 to be posted; due 10/16 • Exam 1 regrade requests due 10/11 Microprocessors I: Lecture 14 | 1,453 | 4,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-33 | latest | en | 0.659921 |
http://www.tutor2u.net/blog/index.php/business-studies/comments/qa-how-is-the-current-ratio-calculated-and-interpreted/ | 1,394,293,508,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999655040/warc/CC-MAIN-20140305060735-00066-ip-10-183-142-35.ec2.internal.warc.gz | 590,419,546 | 16,722 | Get Summer 2014 Right First Time with tutor2u Exam Coaching & Revision Workshops
# Q&A - How is the current ratio calculated and interpreted?
Saturday, January 08, 2011
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The current ratio is the classic measure of liquidity. It indicates whether the business can pay debts due within one year out of the current assets. The current ratio reveals how much “cover” the business has for every £1 that is owed by the firm. For example, a ratio of 1.5:1 would mean that a business has £1.50 of current assets for every £1 of current liabilities.
An example of this calculation is shown below:
The calculation can be illustrated as follows:
At 31 December 2010 current assets were 1.85 times the value of current liabilities. That ratio was more than the 1.7 times at the end of 2009, suggesting a slight improvement in the current ratio.
A current ratio of around 1.7-2.0 is pretty encouraging for a business. It suggests that the business has enough cash to be able to pay its debts, but not too much finance tied up in current assets which could be reinvested or distributed to shareholders.
A low current ratio (say less than 1.0-1.5 might suggest that the business is not well placed to pay its debts. It might be required to raise extra finance or extend the time it takes to pay creditors.
There is no such thing as an ideal current ratio (a good point to make in an exam). Different businesses and industries work with different levels of cover. However, a ratio of less than one is often a cause for concern, particularly if it persists for any length of time.
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Try some of these superb starter activities for Business Studies:
Business Studies Blog Economics Blog Geography Blog Politics Blog Give It A Go! Blog Sociology Blog | 746 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-10 | longest | en | 0.951817 |
https://networkx.github.io/documentation/latest/auto_examples/graph/plot_expected_degree_sequence.html | 1,601,369,783,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401632671.79/warc/CC-MAIN-20200929060555-20200929090555-00102.warc.gz | 466,315,983 | 4,473 | Note
This documents the development version of NetworkX. Documentation for the current release can be found here.
# Expected Degree SequenceΒΆ
Random graph from given degree sequence.
Out:
Degree histogram
degree (#nodes) ****
0 ( 0)
1 ( 0)
2 ( 0)
3 ( 0)
4 ( 0)
5 ( 0)
6 ( 0)
7 ( 0)
8 ( 0)
9 ( 0)
10 ( 0)
11 ( 0)
12 ( 0)
13 ( 0)
14 ( 0)
15 ( 0)
16 ( 0)
17 ( 0)
18 ( 0)
19 ( 0)
20 ( 0)
21 ( 0)
22 ( 0)
23 ( 0)
24 ( 0)
25 ( 0)
26 ( 0)
27 ( 0)
28 ( 0)
29 ( 0)
30 ( 0)
31 ( 0)
32 ( 0)
33 ( 0)
34 ( 0)
35 ( 2) **
36 ( 3) ***
37 ( 7) *******
38 ( 6) ******
39 ( 4) ****
40 (11) ***********
41 (16) ****************
42 (10) **********
43 (21) *********************
44 (14) **************
45 (20) ********************
46 (34) **********************************
47 (32) ********************************
48 (29) *****************************
49 (27) ***************************
50 (28) ****************************
51 (27) ***************************
52 (25) *************************
53 (29) *****************************
54 (27) ***************************
55 (21) *********************
56 (20) ********************
57 (16) ****************
58 (11) ***********
59 (11) ***********
60 (11) ***********
61 ( 9) *********
62 ( 4) ****
63 (10) **********
64 ( 2) **
65 ( 3) ***
66 ( 3) ***
67 ( 1) *
68 ( 2) **
69 ( 1) *
70 ( 1) *
71 ( 0)
72 ( 0)
73 ( 1) *
74 ( 1) *
import networkx as nx
from networkx.generators.degree_seq import expected_degree_graph
# make a random graph of 500 nodes with expected degrees of 50
n = 500 # n nodes
p = 0.1
w = [p * n for i in range(n)] # w = p*n for all nodes
G = expected_degree_graph(w) # configuration model
print("Degree histogram")
print("degree (#nodes) ****")
dh = nx.degree_histogram(G)
for i, d in enumerate(dh):
print(f"{i:2} ({d:2}) {'*'*d}")
Total running time of the script: ( 0 minutes 0.035 seconds)
Gallery generated by Sphinx-Gallery | 626 | 1,889 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-40 | latest | en | 0.388304 |
https://forum.mtasa.com/topic/121377-exp-bar-calculation/?tab=comments | 1,585,585,177,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00338.warc.gz | 487,574,114 | 14,886 | exp bar calculation
Recommended Posts
hello i need calculate between totalexp and requiredexp but i need reset exp bar when level up
this bar disturb and backward like that 3 or 4 level help me please
```function math.round(number)
return number - number % 1
end
function getRequiredEXP(level)
return math.round(math.pow(level, 2) * 300)
end
local level = 1
local currentEXP = 0
local totalEXP = 0
local requiredLevel = 0
local requiredEXP = getRequiredEXP(requiredLevel + 1)
function DRAW_HUD()
local percentageEXP = (currentEXP / requiredEXP) * 369 ---- formule
dxDrawRectangle(1539, 914, 369, 5, tocolor(0, 0, 0, 255))
dxDrawRectangle(1539, 914, percentageEXP, 5, tocolor(255, 255, 255, 255))
dxDrawBorderedText(2, "LEVEL:", 1539, 942, 0, 0, tocolor(255,255,255,255), 1.50, "bankgothic")
dxDrawBorderedText(2.5, level, 1722, 920, 0, 0, tocolor(255,255,255,255), 2.5, "bankgothic")
dxDrawBorderedText(1.5, "EXP:", 1538, 860, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
dxDrawBorderedText(1.5, totalEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
if requiredEXP <= 999 then
dxDrawBorderedText(1.5, " "..requiredEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
elseif requiredEXP <= 9999 then
dxDrawBorderedText(1.5, " "..requiredEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
elseif requiredEXP <= 99999 then
dxDrawBorderedText(1.5, " "..requiredEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
elseif requiredEXP <= 999999 then
dxDrawBorderedText(1.5, " "..requiredEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
elseif requiredEXP <= 9999999 then
dxDrawBorderedText(1.5, " "..requiredEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
elseif requiredEXP <= 99999999 then
dxDrawBorderedText(1.5, " "..requiredEXP, 1538, 885, 0, 0, tocolor(255,255,255,255), 1, "bankgothic")
end
end
function dxDrawBorderedText (outline, text, left, top, right, bottom, color, scale, font, alignX, alignY, clip, wordBreak, postGUI, colorCoded, subPixelPositioning, fRotation, fRotationCenterX, fRotationCenterY)
for oX = (outline * -1), outline do
for oY = (outline * -1), outline do
dxDrawText (text, left + oX, top + oY, right + oX, bottom + oY, tocolor(122, 122, 122, 255), scale, font, alignX, alignY, clip, wordBreak, postGUI, colorCoded, subPixelPositioning, fRotation, fRotationCenterX, fRotationCenterY)
end
end
dxDrawText (text, left, top, right, bottom, color, scale, font, alignX, alignY, clip, wordBreak, postGUI, colorCoded, subPixelPositioning, fRotation, fRotationCenterX, fRotationCenterY)
end
currentEXP = currentEXP + amount
totalEXP = totalEXP + amount
if totalEXP >= requiredEXP then
currentEXP = currentEXP - requiredEXP
level = level + 1
requiredLevel = requiredLevel + 1
requiredEXP = getRequiredEXP(requiredLevel + 1)
end
end
end)```
Edited by Burak5312
3 hours ago, Burak5312 said:
i need reset exp bar when level up
Debug line 50 and check if the value goes below 0.
yes goes below 0 but i dont know how to fix this i spend my all day :d problem calculate formule? i tryed other methods but when level up exp bar staying half
i need zero bar
Edited by Burak5312
3 minutes ago, Burak5312 said:
yes goes below 0 but i dont know how to fix this i spend my all day :d problem calculate formule?
Hmmm, maybe:
`currentEXP = totalEXP - requiredEXP`
this time not be full exp bar when level up
Edited by Burak5312 | 1,194 | 3,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.421597 |
https://read.somethingorotherwhatever.com/entry/Dividingbyzerohowbadisitreally | 1,721,439,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00326.warc.gz | 421,647,055 | 2,842 | # Dividing by zero - how bad is it, really?
• Published in 2016
In the collection
In computable analysis testing a real number for being zero is a fundamental example of a non-computable task. This causes problems for division: We cannot ensure that the number we want to divide by is not zero. In many cases, any real number would be an acceptable outcome if the divisor is zero - but even this cannot be done in a computable way. In this note we investigate the strength of the computational problem "Robust division": Given a pair of real numbers, the first not greater than the other, output their quotient if well-defined and any real number else. The formal framework is provided by Weihrauch reducibility. One particular result is that having later calls to the problem depending on the outcomes of earlier ones is strictly more powerful than performing all calls concurrently. However, having a nesting depths of two already provides the full power. This solves an open problem raised at a recent Dagstuhl meeting on Weihrauch reducibility. As application for "Robust division", we show that it suffices to execute Gaussian elimination.
## Other information
key
type
article
2016-06-17
date_published
2016-07-11
### BibTeX entry
@article{Dividingbyzerohowbadisitreally,
type = {article},
title = {Dividing by zero - how bad is it, really?},
author = {Takayuki Kihara and Arno Pauly},
abstract = {In computable analysis testing a real number for being zero is a fundamental
example of a non-computable task. This causes problems for division: We cannot
ensure that the number we want to divide by is not zero. In many cases, any
real number would be an acceptable outcome if the divisor is zero - but even
this cannot be done in a computable way.
In this note we investigate the strength of the computational problem "Robust
division": Given a pair of real numbers, the first not greater than the other,
output their quotient if well-defined and any real number else. The formal
framework is provided by Weihrauch reducibility. One particular result is that
having later calls to the problem depending on the outcomes of earlier ones is
strictly more powerful than performing all calls concurrently. However, having
a nesting depths of two already provides the full power. This solves an open
problem raised at a recent Dagstuhl meeting on Weihrauch reducibility.
As application for "Robust division", we show that it suffices to execute
Gaussian elimination.},
comment = {},
} | 556 | 2,489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.90509 |
https://web2.0calc.com/questions/2-more-questions-please | 1,561,003,718,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.98/warc/CC-MAIN-20190620024754-20190620050754-00030.warc.gz | 653,724,952 | 6,510 | +0
# 2 more questions please
0
304
1
+299
Question 1)
Question 2)
May 5, 2018
edited by Johnnyboy May 5, 2018
#1
+101228
+1
1. 16(x - 3) = (y - 1)^2
The vertex of this parabola is at ( 3, 1)
This parabola opens to the right
In the form 4p(x - 3) = (y - k)^2......the focus is ( 3 + p, 1)
We can solve for p as 4p =16 ⇒ p = 4
So....the focus is ( 3 + 4, 1) = (7,1)
The dirctrix lies 4 units to the left of the vertex and is given by x = ( 3 - p ) = (3 - 4 ) = -1
Here's a graph : https://www.desmos.com/calculator/iwubncnnan
2. y = -1/6 x^2 + 7x - 80 we need to complete the square on x
y = (-1/6) [ x^2- 42x + 480 ]
Take 1/2 of 42 = 21....square it = 441....add and subtract it within the brackets
y = (-1/6) [x^2 - 42x + 441 + 480 - 441 ] factor the frist three terms....simplify the rest
y = (-1/6) [ (x - 21)^2 + 39 ]
y = (-1/6)(x - 21)^2 - 39/6
y = (-1/6)(x - 21)^2 - 13/2 add 13/2 to both sides
(y + 13/2) = (-1/6)(x - 21)^2 multiply through by -6
-6(y + 13/2) = (x - 21)^2
Because of the presence of a "-" in front of the "6"......this parabola turns downward
The vertex is ( 21, - 13/2)
The focus is given by ( 21 , -13/2 + p )
And we can find p as ..... 4p = -6 ⇒ p = -6/4 = -3/2
So....the focus is (21, -13/2 - 3/2) = ( 21 , -16/2) = (21, -8)
The directerix is given as y = (-13/2 - p) = ( -13/2 - (-3/2) ) = ( -13/2 + 3/2) = -10/2 = -5
Here's a graph : https://www.desmos.com/calculator/g9zfw92yos
May 5, 2018
edited by CPhill May 5, 2018 | 744 | 1,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-26 | latest | en | 0.346023 |
https://tolstoy.newcastle.edu.au/R/help/05/12/16745.html | 1,585,707,020,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00240.warc.gz | 748,080,137 | 4,577 | # Re: [R] Solving Systems of Non-linear equations
From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Thu 01 Dec 2005 - 12:14:01 EST
Just one addition to this. I noticed that its not really true that the output can be used in R verbatim since the C output uses pow instead of ^; however, if one replaces the "code c" statement with the statement "list export" then it is valid R. That is the input to mathomatic should be:
mean = a/(a+b)
variance = (a*b)/(((a+b)^2) * (a+b+1))
eliminate b
a
simplify
list export
eliminate a
b
simplify
list export
On 11/30/05, Gabor Grothendieck <ggrothendieck@gmail.com> wrote:
> Sorry I seemed to have messed up the copying and pasting.
> Here it is again.
>
> ---
>
> Go to http://mathomatic.orgserve.de/math/ and install mathomatic
> (its free) or just connect to the online server and do this.
>
> The C output, i.e the result of the two code c commands,
> can be used verbatim in R.
>
> Note that mathomatic does not support logs but for simple
> problems like this its very useful.
>
> Note that 1-> and 2-> are the mathomatic prompts and what
> comes after them are what I typed in. The entry goes into
> the corresponding equation space, i.e. equation 1 or equation 2.
>
> This is what you enter:
>
> mean = a/(a+b)
> variance = (a*b)/(((a+b)^2) * (a+b+1))
>
> eliminate b
> a
> simplify
> code c
>
> eliminate a
> b
> simplify
> code c
>
> and this is the entire session:
>
>
> 1-> mean = a/(a+b)
>
> a
> #1: mean = -------
> (a + b)
>
> 1-> variance = (a*b)/(((a+b)^2) * (a+b+1))
>
> a*b
> #2: variance = -------------------------
> (((a + b)^2)*(a + b + 1))
>
> 2-> eliminate b
> Solving equation #1 for (b)...
>
> 1
> (a^2)*(---- - 1)
> mean
> #2: variance = ---------------------------------------------------
> 1 1
> (((a + (a*(---- - 1)))^2)*(a + (a*(---- - 1)) + 1))
> mean mean
>
> 2-> a
>
> mean*(1 - mean)
> #2: a = mean*(--------------- - 1)
> variance
>
> 2-> simplify
>
> ((mean^2) - (mean^3))
> #2: a = --------------------- - mean
> variance
>
> 2-> code c
> a = ((((mean * mean) - pow(mean, 3.0)) / variance) - mean);
>
> 2-> eliminate a
> Solving equation #1 for (a)...
>
> b*mean ((mean^2) - (mean^3))
> #2: ---------- = --------------------- - mean
> (1 - mean) variance
>
> 2-> b
>
> mean*(1 - mean)
> #2: b = (--------------- - 1)*(1 - mean)
> variance
>
> 2-> simplify
>
> ((mean^2) - mean)
> #2: b = (1 + -----------------)*(mean - 1)
> variance
>
> 2-> code c
> b = ((1.0 + (((mean * mean) - mean) / variance)) * (mean - 1.0));
>
>
>
>
> On 11/30/05, Gabor Grothendieck <ggrothendieck@gmail.com> wrote:
> > Go to http://mathomatic.orgserve.de/math/ and install mathomatic
> > (its free) or just connect to the online server and do this.
> >
> > The C output, i.e the result of the two code c commands,
> > can be used verbatim in R.
> >
> > Note that mathomatic does not support logs but for simply
> > problems like this its very useful.
> >
> > Note that 1-> and 2-> are the mathomatic prompts and what
> > comes after them are what I typed in. The entry goes into
> > the corresponding equation space, i.e. equation 1 or equation 2.
> >
> > 1-> mean = a/(a+b)
> >
> > a
> > #1: mean = -------
> > (a + b)
> >
> > 1-> variance = (a*b)/(((a+b)^2) * (a+b+1))
> >
> > a*b
> > #2: variance = -------------------------
> > (((a + b)^2)*(a + b + 1))
> >
> > 2-> eliminate b
> > Solving equation #1 for (b)...
> >
> > 1
> > (a^2)*(---- - 1)
> > mean
> > #2: variance = ---------------------------------------------------
> > 1 1
> > (((a + (a*(---- - 1)))^2)*(a + (a*(---- - 1)) + 1))
> > mean mean
> >
> > 2-> a
> >
> > mean*(1 - mean)
> > #2: a = mean*(--------------- - 1)
> > variance
> >
> > 2-> simplify
> >
> > ((mean^2) - (mean^3))
> > #2: a = --------------------- - mean
> > variance
> >
> > 2-> eliminate a
> > Solving equation #1 for (a)...
> >
> > b*mean ((mean^2) - (mean^3))
> > #2: ---------- = --------------------- - mean
> > (1 - mean) variance
> >
> > 2-> b
> >
> > mean*(1 - mean)
> > #2: b = (--------------- - 1)*(1 - mean)
> > variance
> > 2-> simplify
> >
> > ((mean^2) - mean)
> > #2: b = (1 + -----------------)*(mean - 1)
> > variance
> >
> >
> > 2-> code c
> > b = ((1.0 + (((mean * mean) - mean) / variance)) * (mean - 1.0));
> >
> > 2-> #1
> >
> > b*mean
> > #1: a = ----------
> > (1 - mean)
> >
> > 1-> code c
> > a = (b * mean / (1.0 - mean));
> >
> >
> >
> > On 11/30/05, Scott Story <sstory@montana.edu> wrote:
> > > I am trying to write a function that will solve a simple system of
> > > nonlinear equations for the parameters that describe the beta
> > > distribution (a,b) given the mean and variance.
> > >
> > >
> > > mean = a/(a+b)
> > > variance = (a*b)/(((a+b)^2) * (a+b+1))
> > >
> > > Any help as to where to start would be welcome.
> > >
> > >
> > >
> > > --
> > > Scott Story
> > > MSU Ecology Department
> > > 319 Lewis Hall
> > > Bozeman, Mt 59717
> > > 406.994.2670
> > > sstory@montana.edu
> > >
> > > ______________________________________________
> > > R-help@stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help | 1,773 | 5,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-16 | latest | en | 0.751394 |
http://www.algebra.com/algebra/homework/Functions/Functions.faq.question.38141.html | 1,371,726,378,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711441609/warc/CC-MAIN-20130516133721-00041-ip-10-60-113-184.ec2.internal.warc.gz | 300,180,388 | 5,561 | # SOLUTION: I have no idea how to do this problem type! Simplify. (Factor first) 1. 3y^2-3y/y-1 2. a^2-a-12/a^2-4a * 2a^2/a^2+2a-3
Algebra -> Algebra -> Functions -> SOLUTION: I have no idea how to do this problem type! Simplify. (Factor first) 1. 3y^2-3y/y-1 2. a^2-a-12/a^2-4a * 2a^2/a^2+2a-3 Log On
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Algebra: Functions, Domain, NOT graphing Solvers Lessons Answers archive Quiz In Depth
Question 38141: I have no idea how to do this problem type!
Simplify. (Factor first)
1. 3y^2-3y/y-1
2. a^2-a-12/a^2-4a * 2a^2/a^2+2a-3
You can put this solution on YOUR website!
Please only put one question in each submission, thanks!
Solved by pluggable solver: EXPLAIN simplification of an expression
Here's what you tried: Graphical form: simplifies to Text form: (3y^2-3y)/(y-1) simplifies to 3*yCartoon (animation) form: For tutors: `simplify_cartoon( (3y^2-3y)/(y-1) )` If you have a website, here's a link to this solution.
### DETAILED EXPLANATION
Look at .
Canceled out common factors (y-(1))
Look at .
Remove unneeded parentheses around factor ,
It becomes .
Look at .
Remove unneeded parentheses around factor
It becomes .
Look at .
Remove extraneous '1' from product
It becomes .
Result:
### Universal Simplifier and Solver
Done!
Hope this helps and feel free to contact me with further questions,
lyra | 470 | 1,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2013-20 | latest | en | 0.81193 |
https://rigidgeometricalgebra.org/wiki/index.php?title=Join_and_meet | 1,670,020,262,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00196.warc.gz | 525,193,512 | 6,453 | Join and meet
The join is a binary operation that calculates the higher-dimensional geometry containing its two operands, similar to a union. The meet is another binary operation that calculates the lower-dimensional geometry shared by its two operands, similar to an intersection.
The points, lines, and planes appearing in the following tables are defined as follows:
$$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$
$$\mathbf q = q_x \mathbf e_1 + q_y \mathbf e_2 + q_z \mathbf e_3 + q_w \mathbf e_4$$
$$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$
$$\mathbf g = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412} + g_w \mathbf e_{321}$$
$$\mathbf h = h_x \mathbf e_{423} + h_y \mathbf e_{431} + h_z \mathbf e_{412} + h_w \mathbf e_{321}$$
The join operation is performed by taking the wedge product between two geometric objects. The meet operation is performed by taking the antiwedge product between two geometric objects.
Formula Commutator Description Illustration
$$\begin{split}\mathbf p \wedge \mathbf q =\, &(p_wq_x - p_xq_w)\,\mathbf e_{41} + (p_wq_y - p_yq_w)\,\mathbf e_{42} + (p_wq_z - p_zq_w)\,\mathbf e_{43} \\ +\, &(p_yq_z - p_zq_y)\,\mathbf e_{23} + (p_zq_x - p_xq_z)\,\mathbf e_{31} + (p_xq_y - p_yq_x)\,\mathbf e_{12}\end{split}$$ $$[\mathbf p, \mathbf q]^{\Large\unicode{x27D1}}_-$$ Line containing points $$\mathbf p$$ and $$\mathbf q$$.
Zero if $$\mathbf p$$ and $$\mathbf q$$ are coincident.
$$\begin{split}\boldsymbol l \wedge \mathbf p =\, &(l_{vy} p_z - l_{vz} p_y + l_{mx} p_w)\,\mathbf e_{423} \\ +\, &(l_{vz} p_x - l_{vx} p_z + l_{my} p_w)\,\mathbf e_{431} \\ +\, &(l_{vx} p_y - l_{vy} p_x + l_{mz} p_w)\,\mathbf e_{412} \\ -\, &(l_{mx} p_x + l_{my} p_y + l_{mz} p_z)\,\mathbf e_{321}\end{split}$$ $$[\boldsymbol l, \mathbf p]^{\Large\unicode{x27D1}}_+$$ Plane containing line $$\boldsymbol l$$ and point $$\mathbf p$$.
Zero if $$\mathbf p$$ lies on the line $$\boldsymbol l$$.
$$\begin{split}\mathbf g \vee \mathbf h =\, &(g_zh_y - g_yh_z)\,\mathbf e_{41} + (g_xh_z - g_zh_x)\,\mathbf e_{42} + (g_yh_x - g_xh_y)\,\mathbf e_{43} \\ +\, &(g_xh_w - g_wh_x)\,\mathbf e_{23} + (g_yh_w - g_wh_y)\,\mathbf e_{31} + (g_zh_w - g_wh_z)\,\mathbf e_{12}\end{split}$$ $$[\mathbf g, \mathbf h]^{\Large\unicode{x27C7}}_-$$ Line where planes $$\mathbf g$$ and $$\mathbf h$$ intersect.
Direction $$\mathbf v$$ is zero if $$\mathbf g$$ and $$\mathbf h$$ are parallel.
$$\begin{split}\boldsymbol l \vee \mathbf g =\, &(l_{my} g_z - l_{mz} g_y + l_{vx} g_w)\,\mathbf e_1 \\ +\, &(l_{mz} g_x - l_{mx} g_z + l_{vy} g_w)\,\mathbf e_2 \\ +\, &(l_{mx} g_y - l_{my} g_x + l_{vz} g_w)\,\mathbf e_3 \\ -\, &(l_{vx} g_x + l_{vy} g_y + l_{vz} g_z)\,\mathbf e_4\end{split}$$ $$[\boldsymbol l, \mathbf g]^{\Large\unicode{x27C7}}_+$$ Point where line $$\boldsymbol l$$ intersects plane $$\mathbf g$$.
Weight $$w$$ is zero if $$\boldsymbol l$$ is parallel to $$\mathbf g$$.
$$\begin{split}\underline{\mathbf g_\smash{\unicode{x25CB}}} \wedge \mathbf p =\, &-g_xp_w \mathbf e_{41} - g_yp_w \mathbf e_{42} - g_zp_w \mathbf e_{43} \\ +\, &(g_yp_z - g_zp_y)\,\mathbf e_{23} + (g_zp_x - g_xp_z)\,\mathbf e_{31} + (g_xp_y - g_yp_x)\,\mathbf e_{12}\end{split}$$ $$[\mathbf p, \mathbf g]^{\Large\unicode{x27C7}}_+$$ Line perpendicular to plane $$\mathbf g$$ passing through point $$\mathbf p$$.
$$\begin{split}\underline{\boldsymbol l_\smash{\unicode{x25CB}}} \wedge \mathbf p =\, &-l_{vx} p_w \mathbf e_{423} - l_{vy} p_w \mathbf e_{431} - l_{vz} p_w \mathbf e_{412} \\ +\, &(l_{vx} p_x + l_{vy} p_y + l_{vz} p_z)\,\mathbf e_{321}\end{split}$$ $$-[\mathbf p, \boldsymbol l]^{\Large\unicode{x27C7}}_+$$ Plane perpendicular to line $$\boldsymbol l$$ containing point $$\mathbf p$$.
$$\begin{split}\underline{\mathbf g_\smash{\unicode{x25CB}}} \wedge \boldsymbol l =\, &(g_zl_{vy} - g_yl_{vz})\,\mathbf e_{423} + (g_xl_{vz} - g_zl_{vx})\,\mathbf e_{431} + (g_yl_{vx} - g_xl_{vy})\,\mathbf e_{412} \\ -\, &(g_xl_{mx} + g_yl_{my} + g_zl_{mz})\,\mathbf e_{321}\end{split}$$ $$[\boldsymbol l, \mathbf g]^{\Large\unicode{x27C7}}_-$$ Plane perpendicular to plane $$\mathbf g$$ containing line $$\boldsymbol l$$.
Normal direction is $$(0,0,0)$$ if $$\boldsymbol l$$ is perpendicular to $$\mathbf g$$. | 1,742 | 4,361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-49 | latest | en | 0.616825 |
https://pypi.org/project/frds/ | 1,695,901,604,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00695.warc.gz | 528,895,339 | 23,067 | Financial Research Data Services
# FRDS - Financial Research Data Services
frds is an open-sourced Python package for computing a collection of major academic measures used in the finance literature in a simple and straightforward way.
## Installation
pip install frds
## Note
This library is still under development and breaking changes may be expected.
## Supported measures
More to be added. For a complete list of supported built-in measures, please check frds.io/measures/.
## Examples
The primary purpose of frds is to offer ready-to-use functions.
For example, Kritzman, Li, Page, and Rigobon (2010) propose an Absorption Ratio that measures the fraction of the total variance of a set of asset returns explained or absorbed by a fixed number of eigenvectors. It captures the extent to which markets are unified or tightly coupled.
>>> import numpy as np
>>> from frds.measures import absorption_ratio
>>> data = np.array( # Hypothetical 6 daily returns of 3 assets.
... [
... [0.015, 0.031, 0.007, 0.034, 0.014, 0.011],
... [0.012, 0.063, 0.027, 0.023, 0.073, 0.055],
... [0.072, 0.043, 0.097, 0.078, 0.036, 0.083],
... ]
... )
>>> absorption_ratio.estimate(data, fraction_eigenvectors=0.2)
0.7746543307660252
Another example, Distress Insurance Premium (DIP) proposed by Huang, Zhou, and Zhu (2009) as a systemic risk measure of a hypothetical insurance premium against a systemic financial distress, defined as total losses that exceed a given threshold, say 15%, of total bank liabilities.
>>> from frds.measures import distress_insurance_premium
>>> # hypothetical implied default probabilities of 6 banks
>>> default_probabilities = np.array([0.02, 0.10, 0.03, 0.20, 0.50, 0.15]
>>> correlations = np.array(
... [
... [ 1.000, -0.126, -0.637, 0.174, 0.469, 0.283],
... [-0.126, 1.000, 0.294, 0.674, 0.150, 0.053],
... [-0.637, 0.294, 1.000, 0.073, -0.658, -0.085],
... [ 0.174, 0.674, 0.073, 1.000, 0.248, 0.508],
... [ 0.469, 0.150, -0.658, 0.248, 1.000, -0.370],
... [ 0.283, 0.053, -0.085, 0.508, -0.370, 1.000],
... ]
... )
0.28661995758
## Project details
### Source Distribution
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Uploaded cp38 | 1,764 | 4,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | longest | en | 0.811743 |
https://es.mathworks.com/matlabcentral/cody/players/496166-tim/solved | 1,575,675,782,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540491491.18/warc/CC-MAIN-20191206222837-20191207010837-00391.warc.gz | 356,874,393 | 22,295 | Cody
# Tim
Rank
Score
1 – 50 of 1,299
#### Problem 45220. Find triangles from edge
Created by: Nicolas Douillet
#### Problem 45219. Find edges from a vertex
Created by: Nicolas Douillet
#### Problem 45216. Rotate a matrix without using rot90
Created by: Asif Newaz
#### Problem 45218. Find a common edge
Created by: Nicolas Douillet
Tags index, indices, list
#### Problem 45217. Find a common vertex
Created by: Nicolas Douillet
#### Problem 45211. AES SubBytes() Transformation
Created by: David Hill
#### Problem 45210. another bullseye
Created by: Asif Newaz
#### Problem 45207. Calculate days until Christmas
Created by: David Motson
Tags days, christmas
#### Problem 45205. Gray code
Created by: Asif Newaz
Tags gray
#### Problem 45203. Check if a number belongs in the fibonacci squence
Created by: David Motson
Tags fibonacci
#### Problem 45200. Elliptic Curve Cryptography (ECC) Point Addition
Created by: David Hill
#### Problem 45196. Determine whether a given point is inside or outside a polygon
Created by: William
#### Problem 45192. check whether a number is a pentatope number
Created by: Asif Newaz
Tags pentatope
#### Problem 45191. generate nth pentatope number
Created by: Asif Newaz
#### Problem 45194. Cantor counting
Created by: Asif Newaz
#### Problem 45193. Fun with permutations
Created by: William
#### Problem 45189. Double Factorial
Created by: Mehmet OZC
#### Problem 45177. create a matrix y of size (n+1) whose last column's elements are the summation of the elements of all the other columns and last row's elements are the summation of the elements of all the other rows; where 'n' is the size of the input matrix 'x'.
Created by: Asif Newaz
Tags matrix
#### Problem 45182. Takuzu row
Created by: Xiaotao REN
#### Problem 45188. Repeat elements of a vector
Created by: Asif Newaz
Tags repeat
#### Problem 45187. Combinations using Stirling numbers of the second kind
Created by: David Hill
#### Problem 44899. D'Hondt Method of Proportional-Representation
Created by: Paul Morant
Tags sorting
#### Problem 45181. Take a binary number as input,rearrange it in any manner such that its value is greater than the original one.output should be all the possible values in decimal.
Created by: Asif Newaz
#### Problem 45180. All flights lead to Idaho Falls
Created by: David Hill
#### Problem 45171. Squares in Squares - Concentric Squares
Created by: Reece Blunt
#### Problem 44973. Create a "+" flag
Created by: Kiran Ramsaroop
#### Problem 44972. Area of Triangle with Oblique Coordinates
Created by: David Hill
#### Problem 44948. Calculate a Damped Sinusoid
Created by: Pooja Lalan
#### Problem 42651. Vector creation
Created by: ruta bhat
Tags vector, easy
#### Problem 1024. Doubling elements in a vector
Created by: Abdelhak ARESMOUK
Tags loop, vector
#### Problem 3076. Create a vector
Created by: Carlton
Tags vector, uab
#### Problem 2631. Flip the vector from right to left
Created by: Pritesh Shah
#### Problem 1107. Find max
Created by: Marco
Tags find, max, vector
#### Problem 44961. RSA decryption
Created by: David Hill
Created by: A C
#### Problem 1554. Calculate the derivative of a polynomial
Created by: Przemyslaw
#### Problem 44945. Calculate BMI
Created by: Debanjana Mukherjee
#### Problem 44959. RSA encryption using public key
Created by: David Hill
#### Problem 44956. Determine RSA keys (public and private) given two prime number character strings (p and q)
Created by: David Hill
#### Problem 44939. What time can I drive after drinking?
Created by: jack walker
#### Problem 44940. Get Next Combination
Created by: Mehmet OZC
#### Problem 2068. Determine the square root
Created by: Samantha Rollins
#### Problem 44933. Vogel-Dobbener entropy
Created by: Christian Schröder
#### Problem 44932. Compound Interest Rate Calculation With Yearly Deposition Into A Bank Account
Created by: Xuan Yee
Tags function, easy
#### Problem 44910. Find the centre of an arc and its radius
Created by: Reece Blunt
Tags geometry
#### Problem 44901. Rearrange the given matrix to have all its zeros climb up to the top of each column - using for loops.
Created by: Yoav Shallev
Tags loops
#### Problem 44904. Pascal's Equilateral triangle inside a Matrix!!!
Created by: Chirag Dudhat
#### Problem 44891. Invert a 3D rigid-body transformation
Created by: Peter Corke
#### Problem 44893. Create a 3D rotation matrix from roll-pitch-yaw angles
Created by: Peter Corke
1 – 50 of 1,299 | 1,192 | 4,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-51 | longest | en | 0.675497 |
https://www.coursehero.com/file/6561089/ch14/ | 1,498,384,415,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320476.39/warc/CC-MAIN-20170625083108-20170625103108-00188.warc.gz | 854,155,106 | 151,750 | # ch14 - Chapter 14 Fluids In this chapter we will explore...
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Chapter 14 Fluids In this chapter we will explore the behavior of fluids. In particular we will study the following: Static fluids: Pressure exerted by a static fluid Methods of measuring pressure Pascal’s principle Archimedes’ principle, buoyancy Real versus ideal Fluids in motion: fluids Equation of continuity Bernoulli’s equation (14 - 1)
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As the name implies a fluid is defined as a substance that can flow. Fluids conform to the boundaries of any container in which they are placed. A fluid cannot exert a force tangential to its Fluids. surface. It can only exert a force perpendicular to its surface. Liquids and gases are classified together as fluids to contrast them from solids. In crystalline solids the constituent atoms are organized in a rigid three dimensional regular array known as the "lattice". Density: Consider the fluid shown in the figure. It has a mass and volume . The density (symbol ) is defined as the ratio of m V ρ 3 the mass over the volume. If the fluid is homogeneous the above equati SI unit on has : k the fo / : g rm m m V m V = = Δ m Δ V m V = (14 - 2)
Consider the device shown in the insert of the figure which is immersed in a fluid filled vessel. The device can measure the normal force exerted on its piston from the compression of the spr F Pressure 2 ing attached to the piston. We assume that the piston has an area . The pressure exerted by the f N The SI unit luid on the for pressure is is known as the pasca pisto is defin l m ed ( as: A p F p A = 2 5 2 symbol: Pa). Other units are the atmosphere (atm), the torr, and the lb/in . The atm is defined as the average pressure of the atmosphere at sea level 1 atm = 1.01 10 Pa = 760 Torr = 14.7 lb/in Expe × rimentally it is found that the pressure p at any point inside the fluid has the same value regardless of the orientation of the cylinder. The assumption is made that the fluid is at rest. F p A = (14 - 3)
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1 2 Consider the tank shown in the figure. It contains a fluid of density at rest. We will determine the pressure difference between point 2 and point 1 whose y-coordinates are a p p y ρ - Fluids at rest 1 nd , respectively. Consider a part of the fluid in the form of a cylinder indicated by the dashed lines in the figure. This is our "system" and its is at equilibrium. The equilibrium condition is: y 2 1 2 1 1 1 2 2 1 0 Here and are the forces exerted by the rest of the fluid on the bottom and top faces of the cylinder, respectively. Each face has an area . , , ynet F F F mg F F A F p A F p A m V A y = - - = = = = = - ( 29 ( 29 ( 29 ( 29 2 2 1 1 2 2 1 1 2 1 2 1 2 If we substitute into the equilibrium conditon we get: 0 If we take 0 and then and The equation above takes the form: o o y p A p A gA y y p p g y y y h y p p p p p p gh - - - = - = - = = - = = = + o p p gh = + ( 29 ( 29 2 1 1 2 p p g y " o p p - Note : gauge pressure
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## This note was uploaded on 11/21/2011 for the course PHYS 2425 taught by Professor . during the Spring '11 term at San Jacinto.
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ch14 - Chapter 14 Fluids In this chapter we will explore...
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Ask a homework question - tutors are online | 954 | 3,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-26 | longest | en | 0.923437 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/how-to-find-valency-of-an-element/structure-of-the-atom/1681559 | 1,660,208,212,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00377.warc.gz | 108,905,894 | 9,199 | # how to find valency of an element?
• Valency of element is the no. of electron participating in chemical combination.
• Actually it is the valence shell which is involved in chemical combination.
• Therefore if electron present in valence shell is less than 3 than valency is directly equal to number of electron in valence shell.
• If electrons present in valence shell are greater than three than 8- no. of valence electrons.
• To find electrons in valence shell one should know atomic number. For eg for Al electronic configuration is 2,8,3 as atomic number is 13. Thus valency is 3 as valence shell electron equals to three. N electronic configuration 2,8,5 valency = 8-5 = 3.
• 20
Hi ,
The number of valence electrons is just how many electrons an atom has in its outer shell. It's easy to figure out if you've got a periodic table. (See the link to the left of this answer for a good periodic table).
All the elements in each column have the same number of electrons in their outer shells. All the elements in the first column all have a single valence electron (H, Li, Na, K, etc.).
The second column elements all have 2 valence electrons (Be, Mg, Ca, Sr, etc.).
Skipping over the gap, go to the Group 3 elements, which all have 3 valence electrons (B, Al, Ga, etc.).
The elements in the next column (C, Si, Ge, etc.) all have 4 valence electrons.
The elements in the next column (N, P, As, etc.) all have, yes, you guessed it, 5 valence electrons.
O, S, Se, and the others in this column have 6 valence electrons.
The halogens in the next-to-last column (F, Cl, Br, etc.) have 7 valence electrons.
The noble gases in the right-most column (Ne, Ar, Kr, etc.) all have 8 electrons in their out except for He, which only has 2 electrons.
If an atom is an ion, you must include the charge also:
For a positive ion, for each charge subtract one electron, *for instance, Na+ has 1-1 = 0, BUT it has 8 valence electrons because it has the same electron configuration as Ne. Just as K+ has the same configuration as Ar. Therefore, the Alkali metals will have 8 valence electrons.
For a negative ion, add one electron for each charge, for instance, O2- has 6+2 = 8 valence electrons
Hope this helps u.!!
cheers....!
• 81
Hi,
The number of valence electrons is just how many electrons an atom has in its outer shell. It's easy to figure out if you've got a periodic table. (See the link to the left of this answer for a good periodic table).
All the elements in each column have the same number of electrons in their outer shells. All the elements in the first column all have a single valence electron (H, Li, Na, K, etc.).
The second column elements all have 2 valence electrons (Be, Mg, Ca, Sr, etc.).
Skipping over the gap, go to the Group 3 elements, which all have 3 valence electrons (B, Al, Ga, etc.).
The elements in the next column (C, Si, Ge, etc.) all have 4 valence electrons.
The elements in the next column (N, P, As, etc.) all have, yes, you guessed it, 5 valence electrons.
O, S, Se, and the others in this column have 6 valence electrons.
The halogens in the next-to-last column (F, Cl, Br, etc.) have 7 valence electrons.
The noble gases in the right-most column (Ne, Ar, Kr, etc.) all have 8 electrons in their out except for He, which only has 2 electrons.
If an atom is an ion, you must include the charge also:
For a positive ion, for each charge subtract one electron, *for instance, Na+ has 1-1 = 0, BUT it has 8 valence electrons because it has the same electron configuration as Ne. Just as K+ has the same configuration as Ar. Therefore, the Alkali metals will have 8 valence electrons.
For a negative ion, add one electron for each charge, for instance, O2- has 6+2 = 8 valence electrons
hope this helps u.....!
cheers....!
• 20
valency of a metal=no. of electrons it looses for getting noble gas configuration
whereas
valency of an non-metal=no. of electrons an shell can accomodate-no. of electrons present in that particular shell
• 16
What are you looking for? | 1,049 | 4,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-33 | latest | en | 0.885285 |
https://www.mathworks.com/matlabcentral/cody/problems/1401-implement-a-bubble-sort-technique-and-output-the-number-of-swaps-required/solutions/225803 | 1,495,600,781,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607786.59/warc/CC-MAIN-20170524035700-20170524055700-00487.warc.gz | 931,553,629 | 12,846 | Cody
# Problem 1401. Implement a bubble sort technique and output the number of swaps required
Solution 225803
Submitted on 31 Mar 2013 by Raghu
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### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [5 1 4 2 8]; y_correct = 4; assert(isequal(bubblesort(x),y_correct))
``` x = 1 4 2 5 8 x = 1 2 4 5 8 ```
2 Pass
%% x = [7 9 17 4 19 3 17 13 20 18]; y_correct = 14; assert(isequal(bubblesort(x),y_correct))
``` x = 7 9 4 17 3 17 13 19 18 20 x = 7 4 9 3 17 13 17 18 19 20 x = 4 7 3 9 13 17 17 18 19 20 x = 4 3 7 9 13 17 17 18 19 20 x = 3 4 7 9 13 17 17 18 19 20 ```
3 Pass
%% x = [73 72 27 56 16 12 21 12 19 1 9 52 44]; y_correct = 54; assert(isequal(bubblesort(x),y_correct)) % x = [4 15 2 5 1 5 8 6 12 13 22 7]; y_correct = 18; assert(isequal(bubblesort(x),y_correct)) % x = [48 15 16 86 89 29 56 2 59 57 2 6 27 53 91 29 52 32 4 61 67 18 99 12 37 76 71 85 91 9]; y_correct = 180; assert(isequal(bubblesort(x),y_correct))
``` x = 72 27 56 16 12 21 12 19 1 9 52 44 73 x = 27 56 16 12 21 12 19 1 9 52 44 72 73 x = 27 16 12 21 12 19 1 9 52 44 56 72 73 x = 16 12 21 12 19 1 9 27 44 52 56 72 73 x = 12 16 12 19 1 9 21 27 44 52 56 72 73 x = 12 12 16 1 9 19 21 27 44 52 56 72 73 x = 12 12 1 9 16 19 21 27 44 52 56 72 73 x = 12 1 9 12 16 19 21 27 44 52 56 72 73 x = 1 9 12 12 16 19 21 27 44 52 56 72 73 x = 4 2 5 1 5 8 6 12 13 15 7 22 x = 2 4 1 5 5 6 8 12 13 7 15 22 x = 2 1 4 5 5 6 8 12 7 13 15 22 x = 1 2 4 5 5 6 8 7 12 13 15 22 x = 1 2 4 5 5 6 7 8 12 13 15 22 x = Columns 1 through 16 15 16 48 86 29 56 2 59 57 2 6 27 53 89 29 52 Columns 17 through 30 32 4 61 67 18 91 12 37 76 71 85 91 9 99 x = Columns 1 through 16 15 16 48 29 56 2 59 57 2 6 27 53 86 29 52 32 Columns 17 through 30 4 61 67 18 89 12 37 76 71 85 91 9 91 99 x = Columns 1 through 16 15 16 29 48 2 56 57 2 6 27 53 59 29 52 32 4 Columns 17 through 30 61 67 18 86 12 37 76 71 85 89 9 91 91 99 x = Columns 1 through 16 15 16 29 2 48 56 2 6 27 53 57 29 52 32 4 59 Columns 17 through 30 61 18 67 12 37 76 71 85 86 9 89 91 91 99 x = Columns 1 through 16 15 16 2 29 48 2 6 27 53 56 29 52 32 4 57 59 Columns 17 through 30 18 61 12 37 67 71 76 85 9 86 89 91 91 99 x = Columns 1 through 16 15 2 16 29 2 6 27 48 53 29 52 32 4 56 57 18 Columns 17 through 30 59 12 37 61 67 71 76 9 85 86 89 91 91 99 x = Columns 1 through 16 2 15 16 2 6 27 29 48 29 52 32 4 53 56 18 57 Columns 17 through 30 12 37 59 61 67 71 9 76 85 86 89 91 91 99 x = Columns 1 through 16 2 15 2 6 16 27 29 29 48 32 4 52 53 18 56 12 Columns 17 through 30 37 57 59 61 67 9 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 15 16 27 29 29 32 4 48 52 18 53 12 37 Columns 17 through 30 56 57 59 61 9 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 15 16 27 29 29 4 32 48 18 52 12 37 53 Columns 17 through 30 56 57 59 9 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 15 16 27 29 4 29 32 18 48 12 37 52 53 Columns 17 through 30 56 57 9 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 15 16 27 4 29 29 18 32 12 37 48 52 53 Columns 17 through 30 56 9 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 15 16 4 27 29 18 29 12 32 37 48 52 53 Columns 17 through 30 9 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 15 4 16 27 18 29 12 29 32 37 48 52 9 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 6 4 15 16 18 27 12 29 29 32 37 48 9 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 15 16 18 12 27 29 29 32 37 9 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 15 16 12 18 27 29 29 32 9 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 15 12 16 18 27 29 29 9 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 12 15 16 18 27 29 9 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 12 15 16 18 27 9 29 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 12 15 16 18 9 27 29 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 12 15 16 9 18 27 29 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 12 15 9 16 18 27 29 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 12 9 15 16 18 27 29 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 x = Columns 1 through 16 2 2 4 6 9 12 15 16 18 27 29 29 32 37 48 52 Columns 17 through 30 53 56 57 59 61 67 71 76 85 86 89 91 91 99 ``` | 2,724 | 4,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-22 | longest | en | 0.369171 |
https://www.quizover.com/key/terms/tonicity-bis2a-09-1-passive-transport-by-openstax | 1,532,263,157,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593223.90/warc/CC-MAIN-20180722120017-20180722140017-00616.warc.gz | 988,641,827 | 18,861 | # 0.32 Bis2a 09.1 passive transport (Page 6/31)
Page 6 / 31
To illustrate this, imagine two full glasses of water. One has a single teaspoon of sugar in it, whereas the second one contains one-quarter cup of sugar. If the total volume of the solutions in both cups is the same, which cup contains more water? Because the large amount of sugar in the second cup takes up much more space than the teaspoon of sugar in the first cup, the first cup has more water in it.
Returning to the beaker example, recall that it has a mixture of solutes on either side of the membrane. A principle of diffusion is that the molecules move around and will spread evenly throughout the medium if they can. However, only the material capable of getting through the membrane will diffuse through it. In this example, the solute cannot diffuse through the membrane, but the water can. Water has a concentration gradient in this system. Thus, water will diffuse down its concentration gradient, crossing the membrane to the side where it is less concentrated. This diffusion of water through the membrane—osmosis—will continue until the concentration gradient of water goes to zero or until the hydrostatic pressure of the water balances the osmotic pressure. Osmosis proceeds constantly in living systems.
## Tonicity
Tonicity describes how an extracellular solution can change the volume of a cell by affecting osmosis. A solution's tonicity often directly correlates with the osmolarity of the solution. Osmolarity describes the total solute concentration of the solution. A solution with low osmolarity has a greater number of water molecules relative to the number of solute particles; a solution with high osmolarity has fewer water molecules with respect to solute particles. In a situation in which solutions of two different osmolarities are separated by a membrane permeable to water, though not to the solute, water will move from the side of the membrane with lower osmolarity (and more water) to the side with higher osmolarity (and less water). This effect makes sense if you remember that the solute cannot move across the membrane, and thus the only component in the system that can move—the water—moves along its own concentration gradient. An important distinction that concerns living systems is that osmolarity measures the number of particles (which may be molecules) in a solution. Therefore, a solution that is cloudy with cells may have a lower osmolarity than a solution that is clear, if the second solution contains more dissolved molecules than there are cells.
## Hypotonic solutions
Three terms—hypotonic, isotonic, and hypertonic—are used to relate the osmolarity of a cell to the osmolarity of the extracellular fluid that contains the cells. In a hypotonic situation, the extracellular fluid has lower osmolarity than the fluid inside the cell, and water enters the cell. (In living systems, the point of reference is always the cytoplasm, so the prefix hypo - means that the extracellular fluid has a lower concentration of solutes, or a lower osmolarity, than the cell cytoplasm.) It also means that the extracellular fluid has a higher concentration of water in the solution than does the cell. In this situation, water will follow its concentration gradient and enter the cell.
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
what does post-translational control refer to?
Bioremediation includes | 1,351 | 6,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-30 | longest | en | 0.926291 |
https://ru.scribd.com/document/109081525/Computer-Controlled-Instruments | 1,610,912,998,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00445.warc.gz | 552,584,865 | 103,826 | Вы находитесь на странице: 1из 4
# Q1. a) The sensitivity of DC uses half wave rectifier. The sensitivity of an AC voltmeter uses a full wave rectifier.
## For the AC voltmeter sensitivity is given by:
c)
Wheatstone's bridge is the basic dc bridge that used for accurate measurement of resistance. Kelvins Bridge is employed when the resistance to be measured is of the order of magnitude of bridge contact and lead resistance, therefore its a modified form of Wheatstone's bridge. Kelvins bridge is a modification of Wheatstone's bridge and is used to measure values of resistance below 1. In low resistance measurement, the resistance of the leads connecting the unknown resistance to the terminal of the bridge circuit may affect the measurement.
d)
The bridge is balanced when there is no current through the galvanometer. When the potential difference at points C and D is equal, i.e. the potential across the galvanometer is zero.
Q2. a) Resistive transducers The change in the value of the resistance with a change in the length of the conductor can be used to measure displacement. Strain gauges work on the principle that the resistance of a conductor or semiconductor changes when strained. This can be used for the measurement of displacement, force and pressure. The resistivity of materials changes with changes in temperature. This property can be used for the measurement of temperature. Inductive transducer There are two common type inductive transducers: simple inductance type and twocoil mutual inductance type. Both have been described below along with their circuits. The inductive transducers work on the principle of the magnetic induction of magnetic material. Just as the resistance of the electric conductor depends on number of factors, the induction of the magnetic material depends on a number of variables like the number of turns of the coil on the material, the size of the magnetic material, and the permeability of the flux path. In the inductive transducers the magnetic materials are used in the flux path and there are one or more air gaps. The change in the air gap also results in change in the inductance of the circuit and in most of the inductive transducers it is used for the working of the instrument. Simple Inductance Type Inductive Transducers In the simple inductance type of the inductive transducers simple single coil is used as the transducer. When the mechanical element whose displacement is to be measured is moved, it changes the presence of the flux path generated by the circuit, which changes the inductance of the circuit and the corresponding output. The output from the circuit is calibrated directly against the value of the input, thus it directly gives the valve of the parameter to be measured. Two-Coil Mutual Inductance Type Inductive Transducer In the two coil arrangement there are two different coils. In the first coil the excitation is generated by external source of the power and in the second coil the output is obtained. The output is proportional to the mechanical input.
Capacitive Transducer The capacitive transducer is used extensively for the measurement of displacement, pressure etc. Let us see the principle of working of capacitive transducer or sensor also called as variable capacitance transducer
The capacitance C between the two plates of capacitive transducers is given by:
Where C = the capacitance of the capacitor or the variable capacitance transducer o = the absolute permittivity
## r = the relative permittivity
The product of o & r is also called as the dielectric constant of the capacitive transducer. A = the area of the plates D = the distance between the plates b) Thermocouple-based temperature measurement is often used in pharmaceutical validation projects. While the use of thermocouples is somewhat entrenched, the process of using them is well known to be complex and costly. A thermistor is a type of resistor with resistance varying according to its temperature. Output voltages are usually higher and they are quite often used without an amplifier. No compensation is needed as they measure absolute temperatures but they are quite non-linear. On the other hand thermistors come with a number of significant advantages as a tool for temperature measurement, such as: A large output signal that results in better precision. Greater stability, providing accurate performance for longer periods of time. Higher accuracy than thermocouples in mid-range temperatures.
c) Photo-emissive are photo devices which release free electrons from a light sensitive material such as caesium when struck by a photon of sufficient energy. The amount of energy the photons have depends on the frequency of the light and the higher the frequency, the more energy the photons have converting light energy into electrical energy. Photo-conductive photo devices vary their electrical resistance when subjected to light. Photoconductivity results from light hitting a semiconductor material which controls the current flow through it. Thus, more light increase the current for a given applied voltage. The most common photoconductive material is Cadmium Sulphide used in LDR photocells. Photo-voltaic generates an emf in proportion to the radiant light energy received and is similar in effect to photoconductivity. Light energy falls on to two semiconductor materials sandwiched together creating a voltage of approximately 0.5V. The most common photovoltaic material is Selenium used in solar cells.
Q3.
START
## Is it too hot or cold?
The cold and hot taps have to be adjusted to get reasonable warm water WARM WATER
N
Wait for two minutes or less
N
Bath tub is filled or not?
Y
Close the tap of hot and cold water
END | 1,152 | 5,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-04 | latest | en | 0.894117 |
https://www.arxiv-vanity.com/papers/1709.03765/ | 1,679,412,813,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00068.warc.gz | 775,820,298 | 25,179 | # Characterizations of o-polynomials by the Walsh transform
Claude Carlet and Sihem Mesnager LAGA, Department of Mathematics, University of Paris 8
(and Paris 13 and CNRS), Saint–Denis cedex 02, France.
E-mail: 1
1email: ,
July 27, 2022
###### Abstract
The notion of o-polynomial comes from finite projective geometry. In 2011 and later, it has been shown that those objects play an important role in symmetric cryptography and coding theory to design bent Boolean functions, bent vectorial Boolean functions, semi-bent functions and to construct good linear codes. In this note, we characterize o-polynomials by the Walsh transform of the associated vectorial functions.
Keywords: o-polynomial, vectorial function, Walsh–Hadamard transform.
## 1 Introduction
In all this paper, is a positive integer. The projective space is a point-line incidence structure whose points are the one-dimensional vector subspaces of , whose lines are the two-dimensional vector subspaces of , and where a point is incident with a line if and only if the one-dimensional vector subspace corresponding to the point is a subspace of the two-dimensional vector subspace corresponding to the line. A set of points such that no three of them lie in a common line is called a -arc. The maximum cardinality of an arc in is . An oval of is an arc of cardinality . A hyperoval of is an arc of maximum cardinality . A hyperoval of containing the fundamental quadrangle (that is, the set of points , , and ) can be described as for some polynomial that is called an o-polynomial. Those polynomials can be characterized as follows:
###### Theorem 1.1
([3]) A polynomial over is an -polynomial if and only is a permutation and each of the polynomials , , is a permutation where
Gs(t)=⎧⎪⎨⎪⎩G(t+s)+G(s)tif% t≠00if t=0.
Recently, in [1], has been discovered a relation between Niho bent functions and o-polynomials when is even. Niho bent functions are those Boolean functions over whose restrictions to are linear. It has been shown that each Niho bent function corresponds to an -polynomial and that each -polynomial gives rise to Niho bent functions. Further, it has been shown in [4], that o-polynomials also give rise to infinite families of semi-bent functions in even dimension. Very recently, new connections between bent vectorial functions and the hyperovals of the projective plane (extending the link provided in [1] between bent Boolean functions and the hyperovals) as well as a connection between o-polynomials in even characteristic and -ary simplex codes have been established in [5]. Furthermore, Ding has provided a nice article [2] where o-polynomials are employed to construct linear codes.
In this note, we present a new characterization of -polynomials by means of their Walsh transform.
## 2 Preliminaries
Given a finite set , denotes the cardinality of . Recall that for any positive integers , and dividing , the trace function from to , denoted by , is the mapping defined for every as:
Trkr(x):=kr−1∑i=0x2ir=x+x2r+x22r+⋯+x2k−r.
In particular, the absolute trace occurs for . We shall denote the absolute trace function . Let be a positive integer and be a vectorial Boolean function. The Walsh transform of at equals by definition the Walsh transform of the so-called component function at , that is:
WF(u,v):=∑x∈F2n(−1)trm(v(F(x))+trn(ux).
## 3 A characterization of o-polynomials
Consider a polynomial over and its associated function from to . We shall denote the polynomial by and the function by . It is well-known (see e.g. [1]) that is an -polynomial if and only if, for every and in , the equation has 0 or 2 solutions (and we know that is then a permutation). Consider the polynomial over equal to . It takes value 0 when equals 0 or 2 and takes strictly positive value when is in . We have then that
|{x∈F2n;F(x)+bx+a=0}|3−4|{x∈F2n;F(x)+bx+a=0}|2
+4|{x∈F2n;F(x)+bx+a=0}|≥0,
and is an o-polynomial if and only if this inequality is an equality for every nonzero and every . Equivalently, we have
∑a,b∈F2n,b≠0(|{x∈F2n;F(x)+bx+a=0}|3−4|{x∈F2n;F(x)+bx+a=0}|2
+4|{x∈F2n;F(x)+bx+a=0}|)≥0,
and is an o-polynomial if and only if this inequality is an equality.
We shall now characterize this condition by means of the Walsh transform. We have:
and therefore, for :
∑a,b∈F2n,b≠0|{x∈F2n;F(x)+bx+a=0}|j=
2−jn∑a,b∈F2n,b≠0∑x1,…,xj∈F2nv1,…,vj∈F2n(−1)∑ji=1trn(vi(F(xi)+bxi+a)),
and using that is nonzero for only and takes then value , we deduce that
∑a,b∈F2n,b≠0|{x∈F2n;F(x)+bx+a=0}|j=
2−(j−1)n∑b∈F∗2n∑v1,…,vj∈F2n∑ji=1vi=0j∏i=1WF(bvi,vi)
and that
∑a,b∈F2n,b≠0(|{x∈F2n;F(x)+bx+a=0}|3−4|{x∈F2n;F(x)+bx+a=0}|2
+4|{x∈F2n;F(x)+bx+a=0}|)=
2−2n∑b∈F∗2n∑v1,v2∈F2nWF(bv1,v1)WF(bv2,v2)WF(b(v1+v2),v1+v2)
−2−n+2∑b∈F∗2n∑v∈F2nW2F(bv,v)+(2n−1)2n+2.
We have . We deduce:
###### Theorem 3.1
Let be any -function and be the associated polynomial over . Then:
∑b∈F∗2n∑v1,v2∈F2nWF(bv1,v1)WF(bv2,v2)WF(b(v1+v2),v1+v2)
+2n+2∑v∈F∗2nW2F(0,v)−24n+2+23n+2≥0, (1)
and this inequality is an equality if and only if is an o-polynomial.
The sum is equal to
2n∑v1,v2∈F2n∑(x1,x2,x3)∈F32nv1x1+v2x2+(v1+v2)x3=0(−1)v1F(x1)+v2F(x2)+(v1+v2)F(x3).
We deduce:
###### Corollary 3.2
Let be any -function and be the associated polynomial over . Then:
2n∑v1,v2∈F2n∑(x1,x2,x3)∈F32nv1x1+v2x2+(v1+v2)x3=0(−1)v1F(x1)+v2F(x2)+(v1+v2)F(x3)
−∑v1,v2∈F2nWF(0,v1)WF(0,v2)WF(0,v1+v2)
+2n+2∑v∈F∗2nW2F(0,v)−24n+2+23n+2≥0, (2)
and this inequality is an equality if and only if is an o-polynomial.
###### Remark 3.3
Observe that, when is an -polynomial, we have:
∑v∈F2nW2F(bv,v) = ∑v∈F2n∑x1,x2∈F2n(−1)trn(v(F(x1)+F(x2)+b(x1+x2) = ∑v∈F2n∑γ,z∈F2n(−1)trn(v(F(z+γ)+F(γ)+bz) = 2n∑γ∈F2n|{z∈F2n∣F(z+γ)+F(γ)+bz=0}|.
Now, for every according to Theorem 1.1. And thus, if is an -polynomial,
∑v∈F2nW2F(bv,v)=22n+1.
One can then deduce from Theorem 3.1 that A polynomial is an -polynomial if and only if
∑b∈F⋆2n∑v∈F2nW2F(bv,v)=(2n−1)22n+1.
and
∑b∈F∗2n∑v1,v2∈F2nWF(bv1,v1)WF(bv2,v2)WF(b(v1+v2),v1+v2)=(2n−1)23n+2. | 2,164 | 6,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | latest | en | 0.908251 |
https://www.doubtnut.com/qa-hindi/simplified-value-of-the-following-exppression-is-1-sqrt11-2sqrt30-2-sqrt7-2sqrt10-4-sqrt8-4sqrt3-20887335/ | 1,638,590,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00586.warc.gz | 759,618,301 | 68,759 | # Simplified value of the following exppression is <br> 1/(sqrt(11-2sqrt(30)))-2/(sqrt(7-2sqrt(10)))-4/(sqrt(8-4sqrt(3)))
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-2sqrt(2)1sqrt(2)sqrt(3)
1/(sqrt(11-2sqrt(30)))-3/(sqrt(7)-2sqrt(10)) <br> 4/(sqrt(8-4sqrt(3))) <br> =1/((sqrt(sqrt(6)-sqrt(5))^(2)))-3/(sqrt((sqrt(5)-sqrt(2))^(2))) <br> -4/(sqrt(2(4-2sqrt(3))) <br> =1/(sqrt(6)-sqrt(5))-3/(sqrt(5)-sqrt(2)) <br> -4/((sqrt(2))xxsqrt((sqrt(3)-1)^(2))) <br> implies 1/(sqrt(6)-sqrt(5))-3/(sqrt(5)-sqrt(2))-(2sqrt(2))/(sqrt(3)-1) <br> Rationalising all the factors <br> =sqrt(6)+sqrt(5)-(3(sqrt(5)+sqrt(2)))/(5-2) <br> -(2sqrt(2)(sqrt(3)+1))/(3-1) <br> =sqrt(6)+sqrt(5)-sqrt(5)-sqrt(2)-sqrt(6)-sqrt(2) <br> =-sqrt(2) | 407 | 941 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-49 | latest | en | 0.448099 |
http://mathoverflow.net/questions/66400/probability-of-having-a-perfect-game-of-set | 1,469,807,745,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831769.86/warc/CC-MAIN-20160723071031-00032-ip-10-185-27-174.ec2.internal.warc.gz | 158,308,574 | 17,273 | # Probability of having a “perfect” game of Set
The card game Set has very simple rules (see here for rules), and it has prompted mathematicians to ask several questions. I will describe one of these questions. When the game ends, there are usually some left over cards, none of which form a Set; but occassionally it happens that the remaining cards all match up, and there are no left over cards. Call such a situation a "perfect" game. My question is:
What is the probability of having a perfect game? What are the best known upper and lower bounds for this probability?
Here we assume that if there are multiple sets available, then one is selected at random (with uniform probability among all distinct available sets). According to this handout the exact value for the probability of a perfect game is "very open". Anecdotally, I've probably played Set hundreds (maybe thousands?) of times, and only a handful have been perfect games (one of which was earlier today!).
A related question one could ask is whether or not the actual gameplay changes the probability. What I mean is:
Say all 81 cards are laid down face up, and Sets are removed randomly until no Sets remain. Is the probability of having no left over cards the same as it is when the game is played normally?
I could remark that the actual game of Set is a special case of a larger class of games, where the cards have n attributes each with r possible values; the usual game is the case n=4 and r=3. One could ask the same questions for this larger class of games. Also, these questions are more naturally viewed as questions about configurations of lines in affine space, but I've chosen to stick with the card game terminology.
-
You have to make some choices about which nice properties to drop if you want to extend Set to a deck with more than $3$ possible values. – Douglas Zare May 30 '11 at 0:06
As for the question, why not use a Monte Carlo test? – Douglas Zare May 30 '11 at 0:10
Because I personally find provable bounds more interesting.. – Anonymous May 30 '11 at 0:55
Looking at $12$ cards, and then extending this to $15$ or $18$ or $21$ if there are no Sets, is quite messy. I would be surprised if you could not get much more accurate estimates from a simulation than you would from pure deduction. – Douglas Zare May 30 '11 at 2:04
One way to get rigorous bounds for the main problem is to note that you have to go through some state with $18$ cards left (with some $12$, $15$, or all $18$ on the table). So, if you analyze all subsets with $18$ cards left which sum to $\vec{0}$ and choose a visible subset then you can get bounds on the probabilities. However, determining the probability of a perfect game for each one seems expensive, even $81 \choose 18$ is huge, and I think the bounds you get are weak. – Douglas Zare May 30 '11 at 14:28
I have simulated playing 10 million games of Set, and for those simulations 1.2% of the games were perfect games (when selecting randomly among the available Sets the whole game). The full results from the simulations are here.
-
At the end of a game there will be 0,6,9,12,15 or 18 cards left. Here are the results of 10000 trials (assuming that Maple's random processes are random enough and I wrote it correctly)
• $[[0, 88], [6, 4126], [9, 4732], [12, 1036], [15, 18]]$ with a full deck.
• $[[0, 119], [6, 4454], [9, 4592], [12, 824], [15, 11]]$ with the standard rules.
I guess I would lean toward there not being a difference although it is not as convincing as it could be.
In a game with 9 cards (say red diamonds) there is no way to avoid taking all the sets. If you want to do an analysis then perhaps first try 27 cards (say the red cards). If you can't analyze that then there is little chance of handling 81 cards.
An analysis of the structure of the cards left behind might also be useful. For example if there are 6 cards left then there are 15 cards (counting multiplicity) not present which would create a set with two of the 6 cards. Is that usually 15 distinct cards or 5 cards three times each?
-
Thanks for the comments. Those numbers are certainly about what I would expect.. you'd probably have to run it quite a few more times to be confident in whether or not there's a difference. And precise statements about the bounds still seem out of reach.. – Anonymous May 30 '11 at 16:09 | 1,060 | 4,355 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-30 | latest | en | 0.964024 |
https://www.geeksforgeeks.org/gate-gate-cs-2014-set-3-question-15/ | 1,632,413,943,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057424.99/warc/CC-MAIN-20210923135058-20210923165058-00643.warc.gz | 803,939,043 | 19,704 | Related Articles
# GATE | GATE-CS-2014-(Set-3) | Question 15
• Last Updated : 28 Jun, 2021
If V1 and V2 are 4-dimensional subspaces of a 6-dimensional vector space V, then the smallest possible dimension of V1 ∩ V2 is ______.
(A) 1
(B) 2
(C) 3
(D) 4
Explanation: First, note that V1+V2 is still in V, so dim(V1+V2)≤ 6.
We know that
dim(V1+V2)=dimV1+dimV2−dim(V1∩V2).
So
6≥dim(V1+V2)=dimV1+dimV2−dim(V1∩V2)
dim(V1∩V2)≥4+4−6=2. | 191 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-39 | latest | en | 0.7022 |
https://stats.stackexchange.com/questions/191319/how-to-write-unnormalized-posterior-when-prior-is-a-mixture-of-continuous-and-di | 1,571,803,219,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00508.warc.gz | 705,438,385 | 30,259 | # How to write unnormalized posterior when prior is a mixture of continuous and discrete
Suppose I want to do bayesian inference on the regression problem $\beta$ for Y = X$\beta$ + $\epsilon$ for $\epsilon_i \sim N(0,\sigma^2)$. The complication is that the prior for each component $\beta_i$ is a mixture of a point mass at 0 and a normal (I think this is called a spike and slab prior).
For example, $\beta_j \sim p\delta_{0} + (1-p)G$ for $G \sim N(0,\tau^2)$.
I want to write down the product of the prior and the likelihood but I find it difficult to do this with the dirac delta function in the prior. I tried introducing a latent variable to indicate which component of the prior I'm drawing from, but I wonder if there is a more principled way to write down the product of the likelihood and the prior.
I seem to have this problem generally whenever there is a dirac delta function in a likelihood or prior. Any general tips would be appreciated.
Thanks. | 235 | 968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.917622 |
https://forum.allaboutcircuits.com/threads/mixing-two-signals.67964/ | 1,571,353,430,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00533.warc.gz | 500,936,417 | 26,051 | # Mixing Two Signals
Discussion in 'Homework Help' started by inneedofhelp, Mar 25, 2012.
1. ### inneedofhelp Thread Starter New Member
Mar 25, 2012
3
0
A circuit that combines two or more signals is called a mixer. In this lab, your goal is to build a mixer that combines the signals generated by two voltage sources, V1 and V2.
V1 is a 1 kHz square wave that varies between 0V and +1V.
V2 is a 5 kHz sine wave that varies between -1V and +1V.
Please design a circuit that mixes V1 and V2 to produce Voutput such that
Voutput ~=1/2V1 +1/6V2
The resulting output should be similar to that shown in Figure 1. The maximum value of the output is approximately 667mv and the minimum value is approximately -167mv
Figure 1. Desired output waveform
Last edited: Mar 25, 2012
2. ### inneedofhelp Thread Starter New Member
Mar 25, 2012
3
0
Hint: Figure 2 shows a simple resistive mixer for combining two signals.
Figure 2. Simple resistive mixer Enter your circuit below, using the appropriate configuration of resistors. Please do not modify the wiring or parameters of the voltage sources -- your goal is to take the signals they generate and combine them, not to change what is generated. Run a 5ms transient analysis to verify the correct operation of your circuit.
Mar 25, 2012
3
0
4. ### crutschow Expert
Mar 14, 2008
23,116
6,853
I assume your problem is to determine the correct resistor values. So what do you have so far? | 385 | 1,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-43 | latest | en | 0.890056 |
https://oeis.org/A348286 | 1,642,696,769,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00673.warc.gz | 461,040,291 | 3,725 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A348286 Numbers k such that there exists 0 < a, b, c, d, e, f < k such that a + b = c + d = e + f and a*b = c*k + d , c*d = e*k + f. 0
26, 41, 46, 49, 61, 66, 81, 82, 101, 109, 129, 141, 145, 163, 170, 177, 186, 201, 204, 209, 217, 218, 225, 244, 248, 253, 257, 266, 280, 289, 290, 302, 316, 325, 329, 341, 342, 343, 352, 353, 356, 361, 375, 378, 381, 386, 389, 401, 409, 417, 419, 431, 433, 436, 442, 449, 451, 465 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS EXAMPLE For k = 26, a = 10, b = 15, c = 5, d = 20, e = 3, f = 22, and 10*15 = 5*26 + 20, 5*20 = 3*26 + 22, 10 + 15 = 5 + 20 = 3 + 22. CROSSREFS Sequence in context: A105997 A075288 A320255 * A125218 A174990 A285513 Adjacent sequences: A348283 A348284 A348285 * A348287 A348288 A348289 KEYWORD nonn,easy AUTHOR Naohiro Nomoto, Oct 10 2021 STATUS approved
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Last modified January 20 10:55 EST 2022. Contains 350472 sequences. (Running on oeis4.) | 528 | 1,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.65639 |
https://pressbooks.nscc.ca/chemistryatoms/chapter/formal-charges-and-resonance/ | 1,718,735,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00699.warc.gz | 404,251,462 | 33,005 | # 27 Formal Charges and Resonance
[latexpage]
### Learning Objectives
By the end of this section, you will be able to:
• Compute formal charges for atoms in any Lewis structure
• Use formal charges to identify the most reasonable Lewis structure for a given molecule
• Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule
In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
### Calculating Formal Charge
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows:
$$\text{formal charge}=\text{# valence shell electrons (free atom)}-\text{# lone pair electrons}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{# bonding electrons}$$
We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen ion $${\text{ICl}}_{4}{}^{\text{−}}.$$
Solution
1. We divide the bonding electron pairs equally for all I–Cl bonds:
2. We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
3. Subtract this number from the number of valence electrons for the neutral atom:
I: 7 – 8 = –1
Cl: 7 – 7 = 0
The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
Check Your Learning Calculate the formal charge for each atom in the carbon monoxide molecule:
C −1, O +1
Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen molecule BrCl3.
Solution
1. Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond:
2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
3. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:
Br: 7 – 7 = 0
Cl: 7 – 7 = 0
All atoms in BrCl3 have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
Check Your Learning Determine the formal charge for each atom in NCl3.
N: 0; all three Cl atoms: 0
### Using Formal Charge to Predict Molecular Structure
The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:
1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO2. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS, NCS, or CSN. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:
Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Using Formal Charge to Determine Molecular Structure Nitrous oxide, N2O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?
Solution Determining formal charge yields the following:
The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:
The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.
Check Your Learning Which is the most likely molecular structure for the nitrite $$\left({\text{NO}}_{2}{}^{\text{−}}\right)$$ ion?
ONO
### Resonance
You may have noticed that the nitrite anion in (Figure) can have two possible structures with the atoms in the same positions. The electrons involved in the N–O double bond, however, are in different positions:
If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in $${\text{NO}}_{2}{}^{\text{−}}$$ have the same strength and length, and are identical in all other properties.
It is not possible to write a single Lewis structure for $${\text{NO}}_{2}{}^{\text{−}}$$ in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in $${\text{NO}}_{2}{}^{\text{−}}$$ is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms. Thus, the electronic structure of the $${\text{NO}}_{2}{}^{\text{−}}$$ ion is shown as:
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, $${\text{CO}}_{3}{}^{\text{2−}},$$ provides a second example of resonance:
One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
### Key Concepts and Summary
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).
### Key Equations
• $$\text{formal charge}\phantom{\rule{0.2em}{0ex}}=\text{# valence shell electrons (free atom)}\phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\text{# one pair electrons}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}\text{# bonding electrons}$$
### Chemistry End of Chapter Exercises
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) selenium dioxide, OSeO
(b) nitrate ion, $${\text{NO}}_{3}{}^{\text{−}}$$
(c) nitric acid, HNO3 (N is bonded to an OH group and two O atoms)
(d) benzene, C6H6:
(e) the formate ion:
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) sulfur dioxide, SO2
(b) carbonate ion, $${\text{CO}}_{3}{}^{\text{2−}}$$
(c) hydrogen carbonate ion, $${\text{HCO}}_{3}{}^{\text{−}}$$ (C is bonded to an OH group and two O atoms)
(d) pyridine:
(e) the allyl ion:
(a)
(b)
(c)
(d)
(e)
Write the resonance forms of ozone, O3, the component of the upper atmosphere that protects the Earth from ultraviolet radiation.
Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, $${\text{NO}}_{\text{2}}{}^{\text{–}}\text{.}$$
In terms of the bonds present, explain why acetic acid, CH3CO2H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown:
Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
(a) CO2
(b) CO
(a)
(b)
CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO2 has double bonds.
Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
Determine the formal charge of each element in the following:
(a) HCl
(b) CF4
(c) PCl3
(d) PF5
(a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0
Determine the formal charge of each element in the following:
(a) H3O+
(b) $${\text{SO}}_{4}{}^{\text{2−}}$$
(c) NH3
(d) $${\text{O}}_{2}{}^{\text{2−}}$$
(e) H2O2
Calculate the formal charge of chlorine in the molecules Cl2, BeCl2, and ClF5.
Cl in Cl2: 0; Cl in BeCl2: 0; Cl in ClF5: 0
Calculate the formal charge of each element in the following compounds and ions:
(a) F2CO
(b) NO
(c) $${\text{BF}}_{4}{}^{\text{−}}$$
(d) $${\text{SnCl}}_{3}{}^{\text{−}}$$
(e) H2CCH2
(f) ClF3
(g) SeF6
(h) $${\text{PO}}_{4}{}^{\text{3−}}$$
Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:
(a) O3
(b) SO2
(c) $${\text{NO}}_{2}{}^{\text{−}}$$
(d) $${\text{NO}}_{3}{}^{\text{−}}$$
(a)
(b)
(c)
(d)
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH?
HOCl
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
Draw the structure of hydroxylamine, H3NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
The structure that gives zero formal charges is consistent with the actual structure:
Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:
(a) IF
(b) IF3
(c) IF5
(d) IF7
Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound.
NF3;
Which of the following structures would we expect for nitrous acid? Determine the formal charges:
Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H2SO4, which has two oxygen atoms and two OH groups bonded to the sulfur.
### Glossary
formal charge
charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)
molecular structure
arrangement of atoms in a molecule or ion
resonance
situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed
resonance forms
two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons
resonance hybrid
average of the resonance forms shown by the individual Lewis structures | 3,681 | 15,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-26 | latest | en | 0.857541 |
https://puzzling.stackexchange.com/questions/tagged/slitherlink | 1,720,976,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00834.warc.gz | 427,360,217 | 46,681 | Slitherlink is a logic puzzle where the goal is to draw a loop so each number has that many edges of the loop adjacent to it. Use with [grid-deduction] and [logical-deduction].
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### 🐍 In the Loop 🐍
Follow the ouroboros down valleys and over mountains. Is it even going anywhere? EDIT (7 days after initial posting): The grid and Penpa+ link have been updated to an easier version, with more numbers ...
• 1,054
222 views
The puzzle here is to analyze the following Slitherlink configuration and show that it is impossible. ...
• 470
1k views
### Continue a hard Loopy/Slitherlink puzzle without brute-force
This is a classic Slitherlink puzzle from Simon Tatham's Portable Puzzle puzzle collection. I was stuck at the position shown for quite a while until I found a way to continue by logical deduction. I ...
• 1,863
714 views
### Turn Back the Clocks
This is part 45 of the puzzle series Around the World in Many Days. Each part is solvable on its own. Dear Puzzling, This is a Slitherlink puzzle. Draw a single continuous loop along grid lines that ...
• 78.7k
234 views
### Help me with Slitherink puzzle (honeycomb) solution
I got stuck on this level (Honeycomb/Large/Hard/#12) and I always end up having issues with the part that I circled with red because it either wouldn't add up or there would be a space that's not ...
• 23
260 views
### SlitherLITS revisited - Do colour outside the lines
This is a follow-up to SlitherLITS: Don't colour outside the lines After creating that puzzle, I've been wondering whether it was possible to create a SlitherLITS puzzle where all the regions outside ...
• 16.6k
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This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. This is a Slitherlink-Yajilin hybrid. Shade some cells in the grid and draw one non-crossing, non-branching line ...
• 78.7k
373 views
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. A watermill is going to be built next to a cozy town. You are the designer of the aforementioned mill/slitherlink, and ...
• 2,466
391 views
### Double trouble - When slitherloops collide
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. What happens when you're working on a couple of puzzles for the Monthly Topic Challenge and try to merge them? Well, in ...
• 16.6k
359 views
### Walking a twisted path on a donut - a Möbius Slitherlink
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. Have you ever wondered why there has to be an "inside" and an "outside" to every slitherlink loop? ...
• 16.6k
840 views
### Slither-Numberlink: A monstrosity nature never intended
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. Rules of hexagonal Slither-Numberlink: Some numbers in the grid are slitherlink clues, some are numberlink clues. Zeroes ...
• 78.7k
524 views
### An X x X Slitherlink related to X
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. The final answer is a six letter word. Prefer no partial answers. penpa
• 27.7k
730 views
### SlitherLITS: Don't colour outside the lines
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. Adding a bit of colour to enhance the loopy goodness of a traditional Slitherlink. Draw a single, non-intersecting loop ...
• 16.6k
1k views
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. I find something strangely satisfying about a well designed slitherlink puzzle, even one which is oddly coloured and ...
• 16.6k
1k views
### Slitherlink? I hardly know her!
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. Rules of hexagonal Slitherlink: Draw one continuous loop in the grid along dotted lines. The line cannot cross itself or ...
• 78.7k
1k views
This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks. Below is a grid containing multiple small slitherlinks. These can be independently solved to form clues for the large ...
• 6,348
402 views
### Help with slitherlink honeycomb [closed]
This is the best I can do: It makes me think that I'm missing some simple detail.
• 27
2k views
### Next step in this Slitherlink puzzle (Honeycomb)
I have applied all of the techniques that I know of, and can't figure out the next logical step. I even turned on the visual clues, but it didn't help. What would be the next move without resorting to ...
• 41
954 views
This puzzle is part of the Puzzling Stack Exchange Advent Calendar 2022. The accepted answer to this question will be awarded a bounty worth 50 reputation.< Previous DoorNext Door > 'Twas the ...
• 6,348
841 views
### Next step in this Slitherlink puzzle?
I am playing this Slitherlink app on Android, and get stuck on this puzzle at the following state: I made a lot of progress on the right side, but after this I couldn't find anything that I can do, ...
• 6,042
2k views
### Is the "rocket" pattern inversible?
The rocket pattern in slitherlink is when a 3 sees a 1 in a diagonal, with two crosses (the exhaust) on the opposite side. 3 x 1 x two lines can be added ...
214 views
I recently learned what a Slitherlink puzzle is; from here, so I tried my hand at making one! This is a normal 7x7 slitherlink, but with the added restriction that the loop must form the number 2 in ...
• 3,765
205 views
### Slitherlink #1 - Very Simple [closed]
This is a simple slitherlink, which is very easy. The rules are as follows (suggested by @bobble and @PuzzlingFerret): There can be multiple loops. The final solution should be a shape. Congrats to ...
941 views
Yesterday, I got my pen-pal to teach me the rules of Slitherlink. I had to use a translator to understand what she wrote though, and as a result I suspect I've misunderstood some of her points. I've ...
• 408
297 views
### How to get started on this Slitherlink (No Swiss)?
This is a puzzle from Round 4 of WPF's Puzzle Grand Prix this year. It's a Slitherlink with the additional rule that "Every Swiss cross-shaped arrangement of five cells must contain at least one ...
• 133
699 views
This is a variation on the classic Slitherlink Puzzle. However, instead of creating one loop, you must create 2 separate loops. The two loops can not share any edges, however they can cross or meet on ...
• 647
431 views
I previously created a combined slitherlink and sukoku puzzle, but I was unsatisfied because the grid was smaller then a regular sudoku. So I made another, larger one. Below is a 9 by 9 grid. This is ...
• 647
268 views
I like combinations of puzzles, so I decided to combine a Sudoku and a Slitherlink. Below is a 6 by 6 grid. This is filled in with numbers based on the usual rules of sudoku. All of the numbers from 1 ...
• 647
732 views
Rules of Slitherlink: Connect adjacent dots with vertical or horizontal lines to make a single loop. The numbers indicate how many lines surround it, while empty cells may be surrounded by any number ...
• 34.3k
669 views
An entry in Fortnightly Topic Challenge #48: Unusual tag mix After pondering what my next puzzle would be about I decided to change my base towards the realm of slitherlink puzzles. I found this ...
• 6,348
571 views
### Slitherlink - A0, B1, C2, D3
The standard Slitherlink rules apply. However, first replace $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$, $D_1$, and $D_2$; such that the Slitherlink is uniquely solvable. Moreover, these equations must ...
• 34.3k
1k views
### A slitherlink with only 3's
As we know, 3's seems to be very rare in slitherlinks. This is the slitherlink with only 3's.
• 2,315
286 views
### Lost in this Loopy puzzle
I got this Loopy (Cairo) puzzle, but got stuck here: I have no idea how to proceed without guessing with backtracking. Are there any local deductions possible here?
• 1,236
380 views
### Next step/strategies for an Area 51 (Slitherlink variant) puzzle
I've been doing quite a few Slitherlinks recently (and puzzles containing Slitherlink elements, such as Area 51). I'm familiar with all the patterns described in the Wiki article but I feel like there'...
• 5,775
693 views
### How to solve this Loopy puzzle efficiently with little backtracking?
I would like to solve this Loopy (Cairo) puzzle: But I would like to do so with as little backtracking as possible. By this I mean that I want to have at most one pending guess at a time; I do not ...
• 1,236
632 views
### Solve this algebraic equation
Wait what? Is there any equation?
• 2,315
1k views
### What's going on with this puzzle?
What is going on with this slitherlink puzzle with clues $2^3$?
• 2,315
893 views
### Slitherlink with so many 2's
• 2,315
436 views
Rules: The lower part of the puzzle is a sudoku. Basic sudoku rules applied. The upper and right of the puzzle are suko puzzles. The numbers in circle and to the selected square has the same number ...
• 2,315
402 views
Both Masyu and Slitherlink puzzles have as their target a single closed loop that does not touch or intersect itself. So it seems natural to create a hybrid that uses both Masyu and Slitherlink clues. ...
• 28.7k
785 views
Apply Slitherlink rules. In addition, all sheep must be inside the loop and all wolves must be outside the loop.
• 34.3k
3k views
### This new puzzle type needs a name {EXTREME EDITION}
A companion puzzle to the 'This new puzzle type needs a name' series, and inspired in part by @JeremyDover's answer to a recent question. Usually, puzzles in this series comprise a mash-up of two ...
• 146k
270 views
### Another logic study
I'm working on another idea for a combination logic puzzle, and as before I've been gearing up by creating some studies. This first one came out decently, I think, so I figured I'd share it. This ...
• 28.7k
3k views
### Q: What kind of logic puzzle would you like? A: Yes
A recent post and Stiv's answer provided inspiration for a new puzzle. I posted a study for this puzzle earlier; this is the bigger effort envisioned. I hope you enjoy! This puzzle consists of four ...
• 28.7k
788 views
Yesterday's post and Stiv's answer provided inspiration for a new puzzle. What I envision will be a big effort, so I spent part of yesterday creating a study, a sample part of the bigger puzzle to get ...
• 28.7k
1k views
### How to union a slitherlink and a nonogram?
I would like to union two puzzles in one. The first puzzle is a slitherlink, the second is a nonogram. The main idea is: the slitherlink's solution is the entrance for the rows of the nonogram. In the ...
• 1,699
536 views
Here's a variant of a slitherlink puzzle. Each color (yellow, orange, green, blue) represents one of the numbers 0, 1, 2 or 3. Enjoy! Bonus: Can we be sure there is only one solution?
• 8,870
126 views
### It's all in the 2's follow-up question
This is in regards to the solution in Slitherlink - It's all in the 2s I know it may seem dumb, but I don't understand the logic behind the second 2 chain in the solution. To help provide clarity ...
• 31
733 views
### I'm stuck on this game of Slitherlink
I'm stuck on this game of Slitherlink. What deduction (or deductions) can be made from here? Am I missing something obvious? I verified that everything I have so far is correct by looking at the ...
• 223 | 2,919 | 11,336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.927679 |
https://oundleprimary.com/slide/gases-kawameeh-middle-school-dn2qj2 | 1,590,968,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00521.warc.gz | 489,500,119 | 8,593 | # Gases - Kawameeh Middle School
Gases KMS 8th Grade Science Ms. Bormann The Nature of Gases The first gas to be studied was air & it was a long time before it was discovered that air was actually a mixture of particles rather than a single gas. Although air is a mixture of several different gases, it behaves much the same as any single gas. Regardless of their chemical identity, gases tend to exhibit similar physical behaviors
Characteristics of Gases The Kinetic Molecular Theory Kinetic = Motion Molecular = Molecules The theory states the tiny particles in all forms of matter are in constant motion. Boyles Law Robert Boyle was among the first to note the relationship between pressure and
of avolume gas. During his experiments temperature and theNOT amount of gas were allowed to change. Boyles Law states The pressure exerted by a gas held at constant temperature varies inversely with the volume of the gas.
For example: If the volume is halved, the pressure would be doubled Pressure Pressure amount of force exerted per unit of area. Pa = Pascal. SI unit for pressure. One Pascal of pressure is 1 Newton per square meter therefore a Pascal is a very small unit. kPa = kilopascal. A kilopascal is 1,000 Pa. Example: At sea level atmospheric pressure is 101.3 kPa. This means that at Earths surface, the atmosphere exerts a force of 101,300 Newtons on every square meter (this is about the weight of a large truck)
Atm = standard atmosphere. 101.3 kPa or 101,300 Pa Boyles Mathematical Law P1V1 = P2V2 A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? Boyles Law 1. Determine which variables you have
P1 = 2 atm V1 = 3.0 L P2 = 4 atm V2 = ? 2. Determine which law is being represented P P and
and V V= = Boyles Boyles Law Law 3. Rearrange the equation for the unknown variable. 4. Plug in the variables and solve. Charless Law
Jacques Charles was among the first to note the and volume relationship between temperatureof a gas. During his experiments pressure and NOT allowed toof change. the amount
gas are Charless Law states At constant pressure, the volume of a fixed number of particles of gas is directly proportional to the absolute (Kelvin) temperature For example: If the temperature is increased, the pressure will increase Temperature Charles' Law must be used with the Kelvin temperature scale.
This scale is an absolute temperature scale. At 0 K, there is no kinetic energy (Absolute Zero). According to Charles' Law, there would also be no volume at that temperature. This condition cannot be fulfilled because all known gases will liquefy or solidify before reaching 0 K. oC 273 Kelvin Charless Mathematical Law Eg: A gas has a volume of 3.0 L at
127C. What is its volume at 227 C? Charless Law 1. Determine which variables you have 2. TT11 = = 127C 127C + + 273 273 =
= 400K 400K V V11 = = 3.0 3.0 LL TT22 = = 227C 227C + + 273
273 = = 5ooK 5ooK which law is Determine V V22 = = ?? being represented. 3. Plug in variables
4. Cross multiply and solve
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https://rd.springer.com/article/10.1186/s13660-018-1680-4 | 1,537,403,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156314.26/warc/CC-MAIN-20180919235858-20180920015858-00235.warc.gz | 571,221,830 | 28,987 | # Variable selection in generalized random coefficient autoregressive models
• Zhiwen Zhao
• Yangping Liu
• Cuixin Peng
Open Access
Research
## Abstract
In this paper, we consider the variable selection problem of the generalized random coefficient autoregressive model (GRCA). Instead of parametric likelihood, we use non-parametric empirical likelihood in the information theoretic approach. We propose an empirical likelihood-based Akaike information criterion (AIC) and a Bayesian information criterion (BIC).
## Keywords
Empirical likelihood Akaike information criterion Bayesian information criterion Generalized random coefficient autoregressive model Variable selection
62M10 91B62
## 1 Introduction
Consider the following p-order generalized random coefficient autoregressive model:
$$Y_{t}=\Phi_{t}^{\tau }Y(t-1)+ \varepsilon_{t},$$
(1)
where τ denotes the transpose of a matrix or vector, $$\Phi_{t}=( \Phi_{t1},\ldots ,\Phi_{tp})^{\tau }$$ is a random coefficient vector, $$Y(t-1)=(Y_{t-1}, \ldots ,Y_{t-p})^{\tau }$$, and $$\{\binom{\Phi_{t}}{\varepsilon_{t}}, t=0,\pm 1, \pm 2, \ldots \}$$ is a sequence of i.i.d. random vectors with $$E(\Phi_{t})=\phi =(\phi_{1},\ldots ,\phi_{p})$$, $$E(\varepsilon _{t})=0$$, and $$\operatorname{Var}\binom{\Phi_{t}}{\varepsilon_{t}} = \bigl( {\scriptsize\begin{matrix}{} V_{\phi }& \sigma_{\Phi \varepsilon }\cr \sigma_{\Phi \varepsilon }^{\tau }& \sigma_{\varepsilon }^{2} \end{matrix}}\bigr)$$.
As a generalization of the usual autoregressive model, the random coefficient autoregressive (RCAR) model (cf. [1, 2]), the Markovian bilinear model and its generalization, and the random coefficient exponential autoregressive model (cf. [3, 4, 5]), model (1) was first introduced by Hwang and Basawa [6]. GRCA has become one of the important models in the nonlinear time series context. In recent years, GRCA has been studied by many authors. For instance, Hwang and Basawa [7] established the local asymptotic normality of a class of generalized random coefficient autoregressive processes. Carrasco and Chen [8] provided the tractable sufficient conditions that simultaneously imply strict stationarity, finiteness of higher-order moments, and β-mixing with geometric decay rates. Zhao and Wang [9] constructed confidence regions for the parameters of model (1) by using an empirical likelihood method. Furthermore, Zhao et al. [10] also considered the problem of testing the constancy of the coefficients in the stationary one-order generalized random coefficient autoregressive model. In this paper, we consider the variable selection problem of the GRCA based on the empirical likelihood method.
Many model selection procedures have been proposed in the statistical literature, including the adjusted $$R^{2}$$ (see Theil [11]), the AIC (see Akaike [12]), BIC (see Schwarz [13]), Mallow’S $$C_{p}$$ (see Mallows [14]). Other criteria in the literature include Hannan and Quinn’s criterion [15], Geweke and Meese’s criterion [16], Cavanaugh’s Kullback information criterion [17], and the deviance information criterion of Spiegelhalter et al. [18]. Also, Tsay [19], Hurvich and Tsai [20] and Pötscher [21] have studied model selection methods in time series models. Recently, the model selection problem has been extended to moment selection as in Andrews [22], Andrews and Lu [23] and Hong et al. [24]. These model selection methods are concerned with parsimony, as was stressed in Zellner et al. [25], as well as accuracy or power in choosing models.
In this paper, we develop an information theoretic approach to variable selection problem of GRCA. Specifically, instead of parametric likelihood, we use non-parametric empirical likelihood (see Owen [26, 27]) in the information theoretic approach. We propose an empirical likelihood-based Akaike information criterion (EAIC) and a Bayesian information criterion (EBIC).
The paper proceeds as follows. The next section is concerned with the methodology and the main results. Section 3 is devoted to the proofs of the main results.
Throughout the paper, we use the symbols “$$\stackrel{d}{\longrightarrow}$$” and “$$\stackrel{p}{\longrightarrow }$$” to denote convergence in distribution and convergence in probability, respectively. We abbreviate “almost surely” and “independent identical distributed” to “a.s.” and “i.i.d.”, respectively. $$o_{p}(1)$$ means a term which converges to zero in probability. $$O_{p}(1)$$ means a term which is bounded in probability. Furthermore, the Kronecker product of the matrices A and B is denoted by $$A \otimes B$$, and $$\Vert M \Vert$$ denotes the $$L_{2}$$ norm for vector or matrix M.
## 2 Methods and main results
In this section, we will first propose the empirical likelihood-based information criteria for choice of a GRCA, then we investigate the asymptotic properties of the new variable selection method.
### 2.1 Empirical likelihood-based information criteria
Hwang and Basawa [6] derived the conditional least-squares estimator ϕ̂ of ϕ, which is given by
\begin{aligned}& \hat{\phi } = \Biggl(\sum_{t=1}^{n}Y(t-1)Y^{\tau }(t-1) \Biggr)^{-1} \Biggl(\sum_{t=1} ^{n}Y_{t}Y(t-1) \Biggr). \end{aligned}
By using the estimating equation of the conditional least-squares estimator, we can obtain the following score function:
$$\sum_{t=1}^{n} \bigl(Y_{t}Y(t-1)-Y(t-1)Y^{\tau }(t-1) \phi \bigr)=\sum_{t=1}^{n}G_{t}( \phi ),$$
where $$G_{t}(\phi )=Y_{t}Y(t-1)-Y(t-1)Y^{\tau }(t-1)\phi$$. Following Owen [26], the empirical likelihood statistic for ϕ is defined as
$$\tilde{l}(\phi )=-2\max_{\sum_{t=1}^{n}p_{t}G_{t}(\phi )=0}\sum _{t=1} ^{n}\log (np_{t}),$$
where $$p_{1},\ldots , p_{n}$$ are all sets of nonnegative numbers summing to 1. By using the Lagrange multiplier method, let
$$G=\sum_{t=1}^{n}\log (np_{t})-n \lambda^{\tau }\sum_{t=1} ^{n}p_{t}G_{t}( \phi )+\gamma \Biggl(\sum_{t=1}^{n}p_{t}-1 \Biggr).$$
After simple algebraic calculation, we have
$$\frac{\partial G}{\partial p_{t}}=\frac{1}{p_{t}}-n\lambda^{\tau }G _{t}(\phi )+\gamma , \quad t=1,\ldots , n.$$
Note that $$\sum_{t=1}^{n}p_{t}=1$$ and $$\sum_{t=1}^{n}p _{t}G_{t}(\phi )=0$$. So we have $$\gamma =-n$$ and $$p_{t}=\frac{1}{n(1+ \lambda^{\tau }G_{t}(\phi ))}$$, which implies that
$$\tilde{l}(\phi )=2\sum_{t=1}^{n} \log \bigl(1+\lambda^{\tau }G_{t}( \phi ) \bigr),$$
(2)
where λ is the solution of the equation
$$\frac{1}{n}\sum_{t=1}^{n} \frac{G_{t}(\phi )}{1+\lambda^{\tau }G _{t}(\phi )}=0.$$
(3)
The definition of $$\tilde{l}(\phi )$$ relies on finding a positive $${p_{t}}'s$$ such that $$\sum_{t=1}^{n}p_{t}G_{t}(\phi )=0$$ for each ϕ. The solution exists if and only if the convex hull of the $$G_{t}(\phi )$$, $$t=1, 2, \ldots , n$$ contains zero as an inner point. When the model is correct, the solution exists with probability tending to 1 as the sample size $$n\rightarrow \infty$$ for ϕ in a neighborhood of $$\phi_{0}$$. However, for finite n and at some ϕ value, the equation often does not have a solution in $$p_{t}$$. To avoid this problem, we introduce the adjusted empirical likelihood.
Further let $$\bar{G}_{n}=n^{-1}\sum_{t=1}^{n}p_{t}G_{t}(\phi )$$ and define $${G}_{n+1}=-a_{n}\bar{G}_{n}$$ for some positive constant $$a_{n}$$. We adjust the profile empirical log-likelihood ratio function to
\begin{aligned} l(\phi )&=-2\max_{\sum_{t=1}^{n+1}p_{t}G_{t}(\phi )=0} \sum_{t=1}^{n+1} \log \bigl({(n+1)}p_{t} \bigr) \\ &=2\sum_{t=1}^{n+1}\log \bigl\{ 1+\tilde{ \lambda }^{\tau }G_{t}( \phi ) \bigr\} \end{aligned}
(4)
with $$\tilde{\lambda }=\tilde{\lambda }(\phi )$$ being the solution of
$$\frac{1}{n+1}\sum_{t=1}^{n+1} \frac{G_{t}(\phi )}{1+\lambda^{ \tau }G_{t}(\phi )}=0.$$
(5)
Since 0 always lies on the line connecting $$\bar{G}_{n}$$ and $${G}_{n+1}$$, the adjusted empirical log-likelihood ratio function is well defined after adding a pseudo-value $${G}_{n+1}$$ to the data set. The adjustment is particularly useful so that a numerical program does not crash simply because some undesirable ϕ is assessed.
A full GRCA assumes that $$y_{t}$$ relates to $$\Phi_{t}^{\tau }Y(t-1)$$ with $$E(\Phi_{t})=\phi$$ being unknown parameter of size p. Let s be a subset of $$\{1, 2, \ldots , p\}$$, and $$Y^{[s]}(t-1)$$ and $$\phi^{[s]}$$ be subvectors of $$Y(t-1)$$ and ϕ containing entries in positions specified by s. Consider the pth-order GRCA specified by $$E(G_{t}(\phi ))=0$$ and a submodel specified by $$E(G^{[s]}_{t}( \phi^{[s]}))=0$$, where $$G^{[s]}_{t}(\phi^{[s]})=Y_{t}Y^{[s]}(t-1)-Y ^{[s]}(t-1)(Y^{[s]}(t-1))^{\tau }\phi^{[s]}$$. For a given s, let $$G^{[s]}_{t}=Y_{t}Y^{[s]}(t-1)-Y^{[s]}(t-1)(Y^{[s]}(t-1))^{\tau } \phi^{[s]}$$, $$\bar{G}_{n}^{[s]}=n^{-1}\sum_{t=1}^{n}G^{[s]} _{t}$$ and $$G^{[s]}_{n+1}=-a_{n}\bar{G}^{[s]}_{n}$$ for some positive constant $$a_{n}$$. The adjusted empirical log-likelihood ratio becomes
\begin{aligned} l \bigl(\phi^{[s]} \bigr)&=-2\max_{\sum_{t=1}^{n+1}p_{t}G^{[s]}_{t}=0} \sum _{t=1}^{n+1}\log \bigl({(n+1)}p_{t} \bigr) \\ &=2\sum_{t=1}^{n+1}\log \bigl\{ 1+\tilde{ \lambda }^{\tau }G^{[s]} _{t} \bigr\} \end{aligned}
(6)
with $$\tilde{\lambda }=\tilde{\lambda }(\phi )$$ being the solution of
$$\frac{1}{n+1}\sum_{t=1}^{n+1} \frac{G^{[s]}_{t}}{1+\lambda^{ \tau }G^{[s]}_{t}}=0.$$
(7)
We define the adjusted profile empirical log-likelihood ratio as
$$l(s)=\inf \bigl\{ l \bigl(\phi^{[s]} \bigr): \phi^{[s]} \bigr\} .$$
(8)
The empirical likelihood versions of AIC and BIC are then defined as
\begin{aligned}& \operatorname{EAIC}=l(s)+2k, \end{aligned}
(9)
\begin{aligned}& \operatorname{EBIC}=l(s)+k\log (n), \end{aligned}
(10)
where k is the cardinality of s.
After $$l(s)$$ is evaluated for all s, we select the model with the minimum EAIC or EBIC value.
### 2.2 Asymptotic properties
It is well known that under some mild conditions the parametric BIC is consistent for variable selection while the parametric AIC is not. Similarly, we can prove that, when p is constant, EBIC is consistent but EAIC is not.
For purposes of illustration, in what follows, we rewrite the model in the following matrix form (see Hwang and Basawa [6]): let $$U_{t}=(\varepsilon_{t},0,0,\ldots ,0)^{\tau }$$ are $$p\times 1$$ vectors, $$\tilde{\Phi }_{tj}={\Phi }_{tj}-\phi_{j}$$, $$j=1,\ldots ,p$$,
\begin{aligned} B=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} \phi_{1} & \phi_{2} & \cdots & \phi_{p-1} & \phi_{p} \\ 1 & 0 & \cdots & 0 & 0 \\ 0& 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 1 & 0 \end{array}\displaystyle \right ) _{p\times p},\qquad C_{t}= \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} \tilde{\Phi }_{t1} & \tilde{\Phi }_{t2} & \cdots & \tilde{\Phi }_{tp} \\ 0 & 0 & \cdots & 0 \\ 0& 0 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 0 \end{array}\displaystyle \right ) _{p\times p} . \end{aligned}
Then model (1) can be written as
$$Y(t)=(B+C_{t})Y(t-1)+ U_{t}.$$
(11)
In order to obtain our theorems, we need the following regularity conditions:
$$\mathbf{(A_{1})}$$
All the eigenvalues of the matrix $$E(C_{t}\otimes C _{t})+(B\otimes B)$$ are less than unity in modulus.
$$\mathbf{(A_{2})}$$
$$EY_{t}^{6}<\infty$$.
### Remark 1
As for the condition $$\mathbf{(A_{1})}$$ and the sufficient condition for $$E\vert y_{t} \vert ^{2m}<\infty$$ ($$m=1, 2, \ldots$$), we refer to Hwang and Basawa [6].
### Theorem 2.1
Let$$A=E(G_{t}(\phi_{0})G^{\tau }_{t}(\phi_{0}))$$and$$B=E((\partial G_{t}(\phi )/\partial \phi )|_{\phi =\phi_{0}})$$. If$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold, then there exists a sequence of adjusted empirical likelihood estimatesϕ̃ofϕsuch that
$$\sqrt{n}(\tilde{\phi }-\phi )\rightarrow N \bigl(0, \bigl(B^{\tau }A^{-1}B \bigr)^{-1} \bigr)$$
(12)
and
$$\sqrt{n}(\tilde{\lambda }-\lambda )\rightarrow N(0,U),$$
(13)
where$$U=A^{-1}-A^{-1}B(B^{\tau }A^{-1}B)^{-1}B^{\tau }A^{-1}$$.
Note that when a submodel s is a true model, it implies $$\phi_{0} ^{[\bar{s}]}=0$$. That is, components of $$\phi_{0}$$ not in s are zero. Therefore, $$Y_{t}$$ only relates to the variables in positions specified by s. The following theorem shows that when $$\phi_{0}^{[\bar{s}]}=0$$ is true, then adjusted empirical log-likelihood ratio statistic has a chi-squared limiting distribution with k fewer degrees of freedom.
### Theorem 2.2
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold and$$\phi_{0} ^{[\bar{s}]}=0$$for a submodelsof size k. Then when$$a_{n}=o _{p}(n^{\frac{1}{2}})$$, we have$$l(s)\rightarrow \chi^{2}_{p-k}$$in distribution as$$n\rightarrow \infty$$.
When the null hypothesis of $$\phi_{0}^{[\bar{s}]}=0$$ is not true, the likelihood ratio go to ∞ as $$n\rightarrow \infty$$. We state the following theorem in terms of the adjusted empirical likelihood which also applies to the usual empirical likelihood.
### Theorem 2.3
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold and$$a_{n}=o_{p}(n ^{\frac{1}{2}})$$. Then for any$$\phi \neq \phi_{0}$$such that$$E(G_{t}(\phi ))\neq 0$$, $$l(s)\rightarrow \infty$$in probability as$$n\rightarrow \infty$$.
The following theorem indicates that, when p is constant, EBIC is consistent but EAIC is not.
### Theorem 2.4
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold and if there exists a subset$$s_{0}$$of$$1, 2, \ldots , p$$such that, for any other subsets, $$E(G^{[s]}_{t}(\phi^{[s]}))=0$$for someϕif and only ifscontains$$s_{0}$$. Then, EBIC is consistent and EAIC is not consistent.
## 3 Proofs of the main results
In order to prove Theorem 2.1, we first present several lemmas.
### Lemma 3.1
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold. ThenAis positive definite andBhas rank p.
### Proof
After simple algebra calculation, we have, for any nonzero vector $$c=(c_{1}, \ldots , c_{p})\in R^{P}$$,
\begin{aligned}& c^{\tau }Ac =E \bigl(c^{\tau }G_{t}(\phi )G^{\tau }_{t}(\phi )c \bigr)=E \bigl( \bigl(c^{\tau }Y(t-1) \bigr)^{2}\operatorname{Var} \bigl(Y_{t}| Y(t-1) \bigr) \bigr). \end{aligned}
Note that the conditional distribution of $$Y_{t}$$, given $$Y(t-1)$$, is not a degenerate distribution, which implies that $$\operatorname{Var}(Y_{t}| Y(t-1))>0$$ a.s. It follows that $$(c^{\tau }Y(t-1))^{2}\operatorname{Var}(Y_{t}| Y(t-1)) \geq 0$$ a.s. Therefore, $$c^{\tau }Ac=0$$ if and only if $$c^{\tau }Y(t-1)=0$$ a.s. Without loss of generality, suppose that the first component $$c_{1}$$ of c is 1, so $$Y_{t-1}=-c_{2}Y_{t-2}-\cdots -c_{p}Y_{t-p}$$, which is contradictory with the fact that the conditional distribution of $$Y_{t-1}$$, given $$(Y_{t-2}, \ldots , Y _{t-p})$$, is not degenerate. Hence $$c^{\tau }Ac>0$$. That is, A is positive definite.
Similarly, we can also prove that B has rank p. The proof of Lemma 3.1 is thus complete. □
### Lemma 3.2
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold. Then when$$a_{n}=o(n^{\frac{1}{2}})$$, we have
\begin{aligned}& \sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}(\phi)G^{\tau }_{t}(\phi ) \Biggr\Vert =O(1)\quad (a.s.), \end{aligned}
(14)
uniformly about$$\phi \in \{\phi | \Vert \phi -\phi_{0} \Vert \leq n^{-\frac{1}{3}}\}$$.
### Proof
Note that
\begin{aligned} &\sup_{\phi } \Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi)G^{\tau }_{t}(\phi ) \Biggr\Vert \\ &\quad \leq \sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum _{t=1}^{n}G_{t}( \phi)G^{\tau }_{t}(\phi ) \Biggr\Vert +\sup _{\phi } \frac{1}{n+1}a_{n} ^{2} \Biggl\Vert \frac{1}{n} \sum_{t=1}^{n}G_{t}( \phi ) \Biggr\Vert ^{2} \\ &\quad \leq \sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum _{t=1}^{n}G_{t}( \phi)G^{\tau }_{t}(\phi )-\frac{1}{n+1}\sum _{t=1}^{n}G_{t}(\phi_{0})G^{\tau }_{t}( \phi_{0}) \Biggr\Vert \\ &\quad \quad {}+\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n}G_{t}( \phi_{0})G^{\tau }_{t}(\phi_{0}) \Biggr\Vert +\sup_{\phi }\frac{1}{n+1}a_{n}^{2} \Biggl\Vert \frac{1}{n}\sum_{t=1}^{n}G_{t}( \phi ) \Biggr\Vert ^{2} \\ &\quad \triangleq L_{n1}+L_{n2}+L_{n3}. \end{aligned}
(15)
First, note that
\begin{aligned} L_{n1} &=\sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n} \bigl(Y(t-1)Y^{\tau }(t-1) \bigl(Y^{\tau }(t-1) (\phi - \phi_{0}) \bigr) \bigr) \Biggr\Vert \\ &\leq \frac{1}{n+1}\sum_{t=1}^{n} \bigl\Vert Y(t-1) \bigr\Vert ^{3} \sup_{\phi } \Vert \phi -\phi_{0} \Vert . \end{aligned}
By the ergodic theorem, we have
\begin{aligned}& \frac{1}{n+1}\sum_{t=1}^{n} \bigl\Vert Y(t-1) \bigr\Vert ^{3}=O(1) \quad (a.s.). \end{aligned}
(16)
Further, note that
\begin{aligned}& \sup_{\phi }\Vert \phi -\phi_{0} \Vert =O(1). \end{aligned}
(17)
This, together with (16), proves that
\begin{aligned}& L_{n1} =O \bigl(n^{-\frac{1}{3}} \bigr) \quad (a.s.). \end{aligned}
(18)
Again by the ergodic theorem, we can prove that
\begin{aligned}& L_{n2} =O(1)\quad (a.s.). \end{aligned}
(19)
Finally, we prove that
\begin{aligned}& L_{n3} =O \bigl(n^{-\frac{1}{3}} \bigr) \quad (a.s.). \end{aligned}
(20)
Note that
\begin{aligned}& \sup_{\phi } \Biggl\Vert \frac{1}{n}\sum _{t=1}^{n}G_{t}(\phi ) \Biggr\Vert \leq \sup_{\phi }\Biggl\Vert \frac{1}{n}\sum_{t=1}^{n} \bigl(G_{t}(\phi )-G^{\tau }_{t}(\phi_{0}) \bigr) \Biggr\Vert + \Biggl\Vert \frac{1}{n}\sum _{t=1}^{n}G_{t}(\phi_{0}) \Biggr\Vert . \end{aligned}
Similar to the proof of (18), we can show that
\begin{aligned}& \sup_{\phi }\Biggl\Vert \frac{1}{n}\sum _{t=1}^{n} \bigl(G_{t}( \phi)-G^{\tau }_{t}(\phi_{0}) \bigr) \Biggr\Vert =O \bigl(n^{-\frac{1}{3}} \bigr)\quad (a.s.). \end{aligned}
(21)
In what follows, we consider $$\Vert \frac{1}{n}\sum_{t=1}^{n}G_{t}(\phi_{0}) \Vert$$.
Denote the ith component of $$G_{t}(\phi_{0})$$ by $$G_{ti}(\phi_{0})$$. Then $$\{G_{ti}(\phi_{0}), 1\leq i\leq p\}$$ is a stationary ergodic martingale difference sequence with $$E(G_{ti}(\phi_{0}))=0$$ and $$E((G_{ti}(\phi_{0}))^{2})<\infty$$. By the law of the iterated logarithm of martingale difference sequence, we have, for $$1\leq i \leq p$$,
$$\frac{1}{n}\sum_{t=1}^{n}G^{\tau }_{ti}( \phi_{0})= O \bigl(n^{- \frac{1}{2}} \bigl(\log_{2}^{n} \bigr)^{\frac{1}{2}} \bigr) \quad (a.s.).$$
It follows that
\begin{aligned}& \frac{1}{n}\sum_{t=1}^{n}G^{\tau }_{t}( \phi_{0})= O \bigl(n^{- \frac{1}{2}} \bigl(\log_{2}^{n} \bigr)^{\frac{1}{2}} \bigr) \quad (a.s.). \end{aligned}
(22)
Then, by (21) and (22), we have
\begin{aligned}& \sup_{\phi } \Biggl\Vert \frac{1}{n}\sum _{t=1}^{n}G_{t}(\phi ) \Biggr\Vert =O \bigl(n ^{-\frac{1}{3}} \bigr)\quad (a.s.). \end{aligned}
(23)
Therefore
\begin{aligned} L_{n3} &=O \bigl(n^{-1} \bigr)o \bigl(n^{\frac{1}{2}} \bigr)o \bigl(n^{\frac{1}{2}} \bigr)O \bigl(n^{- \frac{1}{3}} \bigr)O \bigl(n^{-\frac{1}{3}} \bigr)\quad (a.s.) \\ &=o \bigl(n^{-\frac{2}{3}} \bigr)\quad (a.s.). \end{aligned}
(24)
This, together with (18) and (19), proves Lemma 3.2. □
### Lemma 3.3
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold. Then when$$a_{n}=o(n^{\frac{1}{2}})$$, we have
$$\max_{1\leq t\leq {n+1}}\sup_{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert =o \bigl(n ^{\frac{1}{3}} \bigr) \quad (a.s.),$$
(25)
uniformly about$$\phi \in \{\phi | \Vert \phi -\phi_{0} \Vert \leq n^{-\frac{1}{3}}\}$$.
### Proof
Note that
\begin{aligned} \max_{1\leq t\leq {n+1}}\sup_{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert &\leq\max_{1\leq t\leq {n}}\sup_{\phi } \bigl\Vert G_{t}( \phi ) \bigr\Vert + \sup_{\phi }\Biggl\Vert a_{n} \frac{1}{n}\sum_{t=1}^{n}G_{t}( \phi ) \Biggr\Vert \\ &\triangleq K_{n1}+K_{n2}. \end{aligned}
From (23), together with $$a_{n}=o(n^{\frac{1}{2}})$$, it follows immediately that
\begin{aligned}& K_{n2} =o \bigl(n^{\frac{1}{3}} \bigr)\quad (a.s.). \end{aligned}
(26)
The next step in the proof is to show that
\begin{aligned}& K_{n1} =o \bigl(n^{\frac{1}{3}} \bigr) \quad (a.s.). \end{aligned}
(27)
By the Fubini theorem, we have, for any positive integer k,
\begin{aligned} \infty&>E \Bigl(\sup_{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert \Bigr)^{3} \\ &=\int_{0}^{\infty }P \Bigl( \Bigl(\sup _{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert \Bigr)^{3}>s \Bigr)\,ds \\ &=\sum_{n=1}^{\infty } \int_{(n-1)k^{3}}^{nk^{3}} P \Bigl( \Bigl(\sup _{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert \Bigr)^{3}>s \Bigr)\,ds \\ &\geq\sum_{n=1}^{\infty } \int_{(n-1)k^{3}}^{nk^{3}} P \Bigl( \Bigl(\sup _{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert \Bigr)^{3}>nk^{3} \Bigr)\,ds \\ &=\sum_{n=1}^{\infty }P \Bigl( \Bigl(\sup_{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert \Bigr)^{3}>nk ^{3} \Bigr)k^{3}\,ds. \end{aligned}
Thus, using the ergodic theorem,
$$\sum_{n=1}^{\infty }P \Bigl(\sup _{\phi } \bigl\Vert G_{n}(\phi ) \bigr\Vert >n ^{\frac{1}{3}}k \Bigr) < \infty .$$
(28)
By the Borel–Cantelli lemma, we know that
$$P \Bigl(\sup_{\phi } \bigl\Vert G_{n}( \phi ) \bigr\Vert >n^{\frac{1}{3}}k \mbox{ i.o.}\Bigr)=0,$$
(29)
so that
$$\sup_{\phi } \bigl\Vert G_{n}(\phi ) \bigr\Vert \leq n^{\frac{1}{3}}k \quad (a.s.).$$
(30)
Take $$k=\frac{1}{m}$$, then there exists $$Q_{m}$$ with $$P(Q_{m})=0$$, such that, for any $$\omega \in Q_{m}^{c}$$,
$$\frac{\sup_{\phi }\Vert G_{n}(\phi ) \Vert }{n^{\frac{1}{3}}}\leq \frac{1}{m}.$$
(31)
Further, let $$Q=\bigcup_{m=1}^{\infty }Q_{m}$$. Then
$$\lim_{n\rightarrow \infty }\frac{\sup_{\phi }\Vert G_{n}(\phi ) \Vert }{n^{\frac{1}{3}}}=0,$$
(32)
which, together with the fact that $$P(Q)=0$$, implies that
$$\max_{1\leq t\leq {n}}\sup_{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert =o \bigl(n ^{\frac{1}{3}} \bigr) \quad (a.s.).$$
(33)
The proof is complete. □
### Lemma 3.4
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold. Then when$$a_{n}=o(n^{\frac{1}{2}})$$, we have
$$\sup_{\phi } \bigl\Vert \lambda (\phi ) \bigr\Vert =O \bigl(n^{-\frac{1}{3}} \bigr)\quad (a.s.),$$
(34)
uniformly about$$\phi \in \{\phi | \Vert \phi -\phi_{0} \Vert \leq n^{-\frac{1}{3}}\}$$.
### Proof
Write $$\Vert \lambda (\phi ) \Vert =\rho (\phi )\theta (\phi )$$, where $$\rho (\phi )>0$$ and $$\Vert \theta (\phi ) \Vert =1$$. Further let
$$Q_{1,n+1}(\phi ,\lambda )=\frac{1}{n+1}\sum _{t=1}^{n+1}\frac{G _{t}(\phi )}{1+\lambda^{\tau }(\phi ) G_{t}(\phi )}.$$
(35)
Then
\begin{aligned} 0&= \bigl\Vert Q_{1,n+1}(\phi , \lambda ) \bigr\Vert \\ &\geq \Biggl\vert \frac{1}{n+1}\sum _{t=1}^{n+1}\frac{\theta^{\tau }(\phi )G_{t}(\phi )}{1+\lambda^{\tau }(\phi ) G_{t}(\phi )} \Biggr\vert \\ &\geq \Biggl\vert \frac{1}{n+1}\rho (\phi )\sum _{t=1}^{n+1}\frac{\theta^{\tau }(\phi )G_{t}(\phi )G^{\tau }_{t}(\phi )\theta (\phi )}{1+\rho (\phi )\theta^{\tau }(\phi ) G_{t}(\phi )} \Biggr\vert - \Biggl\vert \frac{1}{n+1}\sum_{t=1}^{n+1} \theta^{\tau }(\phi )G_{t}(\phi ) \Biggr\vert \\ &\geq \frac{\rho (\phi )\theta^{\tau }(\phi )(\frac{1}{n+1}\sum_{t=1} ^{n+1}G_{t}(\phi )G^{\tau }_{t}(\phi ))\theta (\phi )}{\max_{1\leq t\leq n}\{1+\rho (\phi )\theta^{\tau }(\phi ) G_{t}( \phi )\}}-\Biggl\vert \frac{1}{n+1}\sum _{t=1}^{n+1}\theta^{\tau }(\phi )G_{t}( \phi ) \Biggr\vert , \end{aligned}
which implies that
\begin{aligned} \frac{\rho (\phi )\theta^{\tau }(\phi )(\frac{1}{n+1}\sum_{t=1} ^{n+1}G_{t}(\phi )G^{\tau }_{t}(\phi ))\theta (\phi )}{\max_{1\leq t\leq n}\{1+\rho (\phi )\theta^{\tau }(\phi ) G_{t}( \phi )\}} &\leq\Biggl\vert \frac{1}{n+1}\sum_{t=1}^{n+1} \theta^{\tau }(\phi)G_{t}(\phi ) \Biggr\vert \\ &\leq \Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi ) \Biggr\Vert . \end{aligned}
(36)
Further, by the ergodic theorem, we have
$$\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi_{0})G^{\tau }_{t}(\phi_{0})-A \Biggr\Vert =o(1)\quad (a.s.),$$
(37)
where $$A=E(G_{t}(\phi_{0})G^{\tau }_{t}(\phi_{0}))$$.
Since
\begin{aligned} 0&\leq\sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi )G^{\tau }_{t}(\phi )-A \Biggr\Vert \\ &\leq \sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi)G^{\tau }_{t}(\phi )-\frac{1}{n+1}\sum _{t=1}^{n+1}G_{t}(\phi_{0})G^{\tau }_{t}( \phi_{0}) \Biggr\Vert \\ &\quad {}+\Biggl\Vert \frac{1}{n+1}\sum _{t=1}^{n+1}G_{t}(\phi_{0})G^{\tau }_{t}( \phi_{0})-A \Biggr\Vert , \end{aligned}
we have from (18) and (37)
$$\frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi )G^{\tau }_{t}(\phi )=A+o(1)\quad (a.s.),$$
(38)
which implies that
$$\theta^{\tau }(\phi ) \Biggl(\frac{1}{n+1}\sum _{t=1}^{n+1}G_{t}( \phi )G^{\tau }_{t}( \phi ) \Biggr)\theta (\phi )\geq \sigma_{\min }+o(1)\quad (a.s.),$$
(39)
where $$\sigma_{\min }$$ is the smallest eigenvalue and the largest eigenvalue of A. This, together with Lemma 3.1 and (36), proves that
\begin{aligned}& \sup_{\phi }\Biggl\Vert \frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}( \phi ) \Biggr\Vert \\& \quad \geq \sup_{\phi }\rho (\phi ) \Biggl( \sigma_{\min }+o(1)- \Bigl(\max_{1\leq t\leq {n+1}}\sup _{\phi } \bigl\Vert G_{t}(\phi ) \bigr\Vert \Bigr) \Biggl( \sup_{\phi } \Biggl\Vert \frac{1}{n+1}\sum _{t=1}^{n+1}G_{t}(\phi ) \Biggr\Vert \Biggr) \Biggr). \end{aligned}
Combined with (23) and Lemma 3.3, this establish (34) and completes the proof. □
### Lemma 3.5
Assume that$$\mathbf{(A_{1})}$$and$$\mathbf{(A_{2})}$$hold, and$$a_{n}=o _{(}n^{\frac{1}{2}})$$. Then, as$$n\rightarrow \infty$$, with probability 1, $$l(\phi )$$attains its minimum value at some pointϕ̃in the interior of the ball$$\Vert \phi -\phi_{0} \Vert \leq n^{-\frac{1}{3}}$$andϕ̃and$$\tilde{\lambda }=\lambda (\tilde{\phi })$$satisfy$$Q_{1,n+1}(\tilde{\phi },\tilde{\lambda })=0$$and$$Q_{2,n+1}( \tilde{\phi },\tilde{\lambda })=0$$, where$$Q_{1,n+1}(\phi ,\lambda )$$is defined in (35) and
$$Q_{2n}(\phi ,\lambda ) =\frac{1}{n+1}\sum _{t=1}^{n+1}\frac{1}{1+ \lambda^{\tau }G_{t}(\phi )} \biggl( \frac{\partial G_{t}(\phi )}{\partial \phi } \biggr)^{\tau }\lambda .$$
(40)
The proof is similar to the proof of Lemma 1 of Qin and Lawless [28], so we omit the details.
### Proof of Theorem 2.1
In what follows, we omit $$(\phi ,\lambda )$$ in the notation if a function is evaluated at $$(\phi_{0}, 0)$$. Expanding $$Q_{1,n+1}(\tilde{\phi }, \tilde{\lambda })$$, $$Q_{2,n+1}(\tilde{\phi },\tilde{\lambda })$$ at $$(\phi_{0}, 0)$$ leads to
$$0=Q_{1,n+1}(\tilde{\phi },\tilde{\lambda })=Q_{1,n+1}+ \biggl\{ \frac{\partial Q_{1,n+1}}{\partial \phi } \biggr\} ( \tilde{\phi }- \phi_{0})+ \biggl\{ \frac{\partial Q_{1,n+1}}{\partial \lambda } \biggr\} \tilde{\lambda }+o_{p}(\delta_{n})$$
(41)
and
$$0=Q_{2,n+1}(\tilde{\phi },\tilde{\lambda }) =Q_{2,n+1}+ \biggl\{ \frac{\partial Q_{2,n+1}}{\partial \phi } \biggr\} ( \tilde{\phi }- \phi_{0})+ \biggl\{ \frac{\partial Q_{2,n+1}}{\partial \lambda } \biggr\} \tilde{\lambda }+o_{p}(\delta_{n}),$$
(42)
where $$\delta_{n}=\Vert \tilde{\phi }-\phi_{0} \Vert ^{2}+\Vert \tilde{\lambda } \Vert ^{2}=O_{p}(n^{-\frac{2}{3}})$$.
Note that
\begin{aligned}& \frac{\partial Q_{1,n+1}}{\partial \phi } = \frac{1}{n+1}\sum _{t=1}^{n+1}\frac{\partial G_{t}}{\partial \phi }=B+o_{p}(1), \end{aligned}
(43)
\begin{aligned}& \frac{\partial Q_{1,n+1}}{\partial \lambda } = -\frac{1}{n+1}\sum_{t=1}^{n+1}G_{t}G^{\tau }_{t}=-A+o_{p}(1), \end{aligned}
(44)
\begin{aligned}& \frac{\partial Q_{2,n+1}}{\partial \phi } = 0, \end{aligned}
(45)
and
$$\frac{\partial Q_{1,n+1}}{\partial \lambda } =\frac{1}{n+1}\sum _{t=1}^{n+1}\frac{\partial G_{t}}{\partial \phi }=B^{\tau }+o _{p}(1).$$
(46)
These, combined with (41) and (42), give
$$\tilde{\lambda } =- \bigl\{ A^{-1}-A^{-1}B \bigl(B^{\tau }A^{-1}B \bigr)^{-1}B^{\tau }A ^{-1} \bigr\} Q_{1,n+1}+o_{p} \bigl(n^{-\frac{1}{2}} \bigr)$$
(47)
and
$$\tilde{\phi }-\phi_{0} = \bigl(B^{\tau }A^{-1}B \bigr)^{-1}B^{\tau }A^{-1}Q_{1,n+1}+o _{p} \bigl(n^{-\frac{1}{2}} \bigr).$$
(48)
Further, applying the central limit theorem to $$Q_{1,n+1}$$ and using Slustzky’s theorem, we can prove Theorem 2.1. □
### Proof of Theorem 2.2
Let λ̃ be the Lagrange multiplier corresponding to $$\tilde{\phi }^{[s]}$$, the maximum point of $$l(\phi^{[s]})$$. With this notation, we may write
$$l(s) =2\sum_{t=1}^{n+1}\log \bigl\{ 1+\tilde{\lambda }^{\tau }G^{[s]} _{t} \bigl( \tilde{\phi }^{[s]} \bigr) \bigr\} .$$
(49)
Note that
$$\tilde{\lambda }^{\tau }G^{[s]}_{t} \bigl(\tilde{\phi }^{[s]} \bigr)= \tilde{\lambda }^{\tau }G^{[s]}_{t}+ \tilde{\lambda }^{\tau } \biggl\{ \frac{ \partial G^{[s]}_{t}}{\partial \phi^{[s]} } \biggr\} ^{\tau } \bigl(\tilde{\phi } ^{[s]}-\phi_{0}^{[s]} \bigr)+o_{p}(1).$$
(50)
This, together with (49), yields
\begin{aligned} \l (s) &=2\tilde{\lambda }^{\tau }\sum_{t=1}^{n+1}G^{[s]}_{t}+2 \tilde{\lambda }^{\tau } \Biggl\{ \sum_{t=1}^{n+1}\frac{\partial G^{[s]}_{t}}{\partial \phi^{[s]} } \Biggr\} \bigl(\tilde{\phi }^{[s]}- \phi_{0}^{[s]} \bigr) - \tilde{\lambda }^{\tau }\sum_{t=1}^{n+1}G^{[s]}_{t} \bigl(G^{[s]} _{t} \bigr)^{\tau }+o_{p}(1) \tilde{\lambda } \\ &=n^{-1}Q_{1,n+1}^{\tau } \bigl\{ A^{-1}-A^{-1}B \bigl(B^{\tau }A^{-1}B \bigr)^{-1}B ^{\tau }A^{-1} \bigr\} Q_{1,n+1}+o_{p}(1). \end{aligned}
Further note that $$Q_{1,n+1}$$ is asymptotic normal with covariance matrix A and $$\{A^{-1}-A^{-1}B(B^{\tau }A^{-1}B)^{-1}B^{\tau }A^{-1} \}A\{A^{-1}-A^{-1}B(B^{\tau }A^{-1}B)^{-1}B^{\tau }A^{-1}\}=\{A^{-1}-A^{-1}B(B^{\tau }A^{-1}B)^{-1}B^{\tau }A^{-1}\}$$. Therefore, we have $$\l (s)\rightarrow \chi^{2}(p-k)$$ in distribution as $$n\rightarrow \infty$$. The proof is complete. □
### Proof of Theorem 2.3
Since $$E(G_{t}(\phi ))\neq 0$$, it follows that there exists $$\delta >0$$, such that
$$\frac{1}{n}\sum_{t=1}^{n}G_{t}^{\tau }(\phi )\frac{1}{n} \sum_{t=1}^{n}G_{t}(\phi )-\delta^{2}=o_{p}(1).$$
(51)
Furthermore, note that $$E(G_{t}^{\tau }(\phi ))^{2}<\infty$$. Thus, by a similar method to the proof of (27), we can prove that
$$\max_{1\leq t\leq n+1} \bigl\Vert G_{t}^{\tau }( \phi ) \bigr\Vert =o_{p} \bigl(n^{ \frac{1}{2}} \bigr).$$
(52)
Let $$\check{\lambda }=n^{-\frac{2}{3}}(\frac{1}{n}\sum_{t=1} ^{n}G_{t}(\phi ))\log n$$. Then
$$\max_{1\leq t\leq n+1} \bigl\vert \check{\lambda }^{\tau }G_{t}(\phi ) \bigr\vert =o _{p}(1).$$
(53)
Thus, with probability going to 1, $$1+\check{\lambda }^{\tau }G_{t}( \phi )>0$$ for $$i=1, \ldots , n+1$$. Using the duality of the maximization problem and (51)–(53), we have
\begin{aligned} l(\phi ) &=\sup_{\lambda } \Biggl( 2\sum_{t=1}^{n+1}\log \bigl\{ 1+ \lambda^{\tau }G_{t}( \phi ) \bigr\} \Biggr)\geq 2\sum_{t=1}^{n+1}\log \bigl\{ 1+\check{ \lambda }^{\tau }G_{t}( \phi ) \bigr\} \\ &=2\sum_{t=1}^{n}\log \bigl\{ 1+\check{ \lambda }^{\tau }G_{t}( \phi ) \bigr\} +o_{p}(1)=2n^{\frac{1}{3}}\delta^{2}\log (n)+o_{p}(1), \end{aligned}
which implies that $$l(s)\rightarrow \infty$$ in probability as $$n\rightarrow \infty$$. The proof is complete. □
### Proof of Theorem 2.4.
First, we consider EAIC. Consider the situation when $$s_{0}$$ is empty. Let $$s=\{1\}$$ which contains a single covariant. Based on expansion in the proof of Theorem 2.2, we can prove that $$l(s_{0})-l(s)\rightarrow \chi^{2}_{1}$$, which implies that $$\lim_{n\rightarrow \infty }P(l(s_{0})-l(s)>2)>0$$. Therefore, EAIC is not consistent.
Next, we consider EBIC. Suppose s is a model which does not contain $$s_{0}$$. Then, $$E(G^{[s]}_{t}(\phi^{[s]}))\neq 0$$ for any $$\phi^{[s]}$$. Therefore, we have $$l(s)\geq 2n^{\frac{1}{3}}\delta^{2}\log (n)+o _{P}(1)$$. This order implies that
$$P \bigl(\operatorname{EBIC}(s)< \operatorname{EBIC}(s_{0}) \bigr)\leq P \bigl(l(s)-l(s_{0})+p \log n \bigr)\rightarrow 0.$$
That is, EBIC will not select any model s that does not contain $$s_{0}$$.
Furthermore, if s contains $$s_{0}$$, and $$k>0$$ additional insignificant variables, by Theorem 2.2, we have
$$l(s_{0})-l(s)\rightarrow \chi^{2}_{k},$$
which implies that
$$P \bigl(\operatorname{EBIC}(s)< \operatorname{EBIC}(s_{0}) \bigr)= P \bigl(l(s)-l(s_{0})>k \log n \bigr)\rightarrow 0,$$
as $$n\rightarrow \infty$$. Thus, the model s will not be selected by EBIC as $$n\rightarrow \infty$$. Because p is finite, there are only finite number of scompeting against $$s_{0}$$, and each of them has $$o(1)$$ probability being selection. So EBIC is consistent. The proof is complete. □
## 4 Conclusions
It should be pointed out that variable selection has always been an important problem for our statistician. Many variable selection methods have been proposed in the statistical literature. But for the variable selection method of GRCA, so far it has not been provided by statistician. In this paper, instead of parametric likelihood, we further propose an Akaike information criterion (EAIC) and a Bayesian information criterion (EBIC) for the variable selection problem of GRCA based on the empirical likelihood method. Moreover, we also prove that under some mild conditions the parametric EBIC is consistent, while the parametric EAIC is not when p is constant.
## Notes
### Acknowledgements
This work is supported by National Natural Science Foundation of China (No. 11571138, 11671054, 11301137, 11271155, 11371168, J1310022, 11501241), the National Social Science fund of China (16BTJ020), Science and Technology Research Program of Education Department in Jilin Province for the 12th Five-Year Plan (440020031139). “Thirteenth Five-Year Plan” Science and Technology Research Project of the Education of Jilin Province (Grant No. 2016103) and Jilin Province Natural Science Foundation (20130101066JC, 20130522102JH, 20150520053JH).
### Authors’ contributions
All the authors read and approved the final manuscript.
### Competing interests
The authors declare that they have no competing interests.
## References
1. 1.
Nicholls, D.F., Quinn, B.G.: Random Coefficient Autoregressive Models: An Introduction. Springer, New York (1982)
2. 2.
Tong, H.: Nonlinear Time Series. Oxford University Press, Oxford (1990) Google Scholar
3. 3.
Tong, H.: A note on a Markov bilinear stochastic process in discrete time. J. Time Ser. Anal. 2, 279–284 (1981)
4. 4.
Feigin, P.D., Tweedie, R.L.: Random coefficient autoregressive processes: a Markov chain analysis of stationarity and finiteness of moments. J. Time Ser. Anal. 6, 1–14 (1985)
5. 5.
Hwang, S.Y., Basawa, I.V.: Asymptotic optimal inference for a class of nonlinear time series models. Stoch. Process. Appl. 46, 91–113 (1993)
6. 6.
Hwang, S.Y., Basawa, I.V.: Parameter estimation for generalized random coefficient autoregressive processes. J. Stat. Plan. Inference 68, 323–327 (1998)
7. 7.
Hwang, S.Y., Basawa, I.V.: The local asymptotic normality of a class of generalized random coefficient autoregressive processes. Stat. Probab. Lett. 34, 165–170 (1997)
8. 8.
Carrasco, M., Chen, X.: β-Mixing and moment properties of RCA models with application to $$\operatorname{GARCH}(p, q)$$. C. R. Acad. Sci., Sér. 1 Math. 331, 85–90 (2000).
9. 9.
Zhao, Z.W., Wang, D.H.: Statistical inference for generalized random coefficient autoregressive model. Math. Comput. Model. 56, 152–166 (2012)
10. 10.
Zhao, Z.W., Wang, D.H., Peng, C.X.: Coefficient constancy test in generalized random coefficient autoregressive model. Appl. Math. Comput. 219, 10283–10292 (2013)
11. 11.
Theil, H.: Economic Forecasts and Policy. North-Holland, Amsterdam (1961) Google Scholar
12. 12.
Akaike, H.: A new look at the statistical model identification. IEEE Trans. Autom. Control 19, 716–723 (1974)
13. 13.
Schwarz, G.: Estimating the dimension of a model. Ann. Stat. 6, 461–464 (1978)
14. 14.
Mallows, C.L.: Some comments on $$C_{p}$$. Technometrics 15, 661–675 (1973)
15. 15.
Hannan, E.J., Quinn, B.G.: The determination of the order of an autoregression. J. R. Stat. Soc., Ser. B, Stat. Methodol. 41, 190–195 (1979)
16. 16.
Geweke, J., Meese, R.: Estimating regression models of finite but unknown order. Int. Econ. Rev. 16, 55–70 (1981)
17. 17.
Cavanaugh, J.E.: A large-sample model selection criterion based on kull-backs symmetric divergence. Stat. Probab. Lett. 42, 333–343 (1999)
18. 18.
Spiegelhalter, D.J., Best, N.G., Carlin, B.P., Linde, A.V.D.: Bayesian measures of model complexity and fit. J. R. Stat. Soc., Ser. B, Stat. Methodol. 64, 583–639 (2002)
19. 19.
Tsay, R.S.: Order selection in nonstationary autoregressive models. Ann. Stat. 12, 1151–1596 (1984)
20. 20.
Hurvich, C.M., Tsai, C.L.: Regression and time series model selection in small samples. Biometrika 76, 297–307 (1989)
21. 21.
Pötscher, B.M.: Model selection under nonstationarity: autoregressive models and stochastic linear regression models. Ann. Stat. 17, 1257–1274 (1989)
22. 22.
Andrews, D.W.K.: Consistent moment selection procedures for generalized method of moments estimation. Econometrica 67, 543–564 (1999)
23. 23.
Andrews, D.W.K., Lu, B.: Consistent model and moment selection criteria for GMM estimation with applications to dynamic panel models. J. Econom. 101, 123–164 (2001)
24. 24.
Hong, H., Preston, B., Shum, M.: Generalized empirical likelihood-based model selection criteria for moment condition models. Econ. Theory 19, 923–943 (2003)
25. 25.
Zellner, A., Keuzenkamp, H.A., Mcaleer, M.: Simplicity, Inference and Modelling: Keeping It Sophisticatedly Simple. Cambridge University Press, Cambridge (2001)
26. 26.
Owen, A.B.: Empirical likelihood ratio confidence intervals for a single functional. Biometrika 75, 237–249 (1988)
27. 27.
Owen, A.B.: Empirical Likelihood. Chapman and Hall, New York (2001)
28. 28.
Qin, J., Lawless, J.: Empirical likelihood and general estimating equations. Ann. Stat. 22, 300–325 (1994) | 14,383 | 36,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-39 | longest | en | 0.682556 |
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