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Conical Pendulum proof of constant frequency

1. Jan 17, 2012

tpb77

1. The problem statement, all variables and given/known data
A pendulum consists of a particle of mass m at the end of a light rigid rod of
length l, the other end of the rod being freely attached to a stationary point
0. Let e(t) be a unit vector pointing along the rod, so that the position
vector relative to 0 of the particle at time t is le(t).
(a) Show from the principle of linear momentum that

e x e'' - (g/l) e x k = 0

where g is the acceleration due to gravity, and k is a unit vector pointing
vertically upwards. ( e'' represents the second derivative of the unit vector wrt time )

(b) If e(t) is confined to a vertical plane containing 0, the pendulum is
called simple. Suppose that

e(t) = sin θ(t) i - cos θ(t)k,
where i is a horizontal unit vector. Deduce that

θ'' + (g/l) sinθ = 0

(c) If e(t) rotates about k, the pendulum is called conical. Suppose that

e' = w(t) (k x e(t))

Deduce that t →w(t) and t →e(t).k are constant throughout the motion.

If w(0) = w > 0 and e(0).k = -cosθ with 0 < 0 < pi/2, show that

w = sqrt( g secθ/l)

Hint. The following identity holds for the triple vector product:

u x ( v x w) = (u.w)v + (u.v)w

3. The attempt at a solution

First two parts are alright, e x e'' + g/l e x k = 0
so e'' must be = - g/l k
so le'' = -g k
a = -gk
so F = -gm k which is just the force due to gravity, so the original statement is true

The second part i can do just by working through the differentiation of sinθ i - cosθj, then subbing in and cancelling all the terms.

the last bit though i think should be
e' = w( k x e)
multiply both sides by e
so e x e' = w( e x (k x e))
= 1/m ( e x p) = w( e x ( k x e))
the d/dt of the left is = 0, since the angular momentum is constant,

then w( (e.e)k -(e.k)e) from the hint must be constant in time
so d/dt [wk - w(e.k)e ]= 0.

I can see how this would imply that dw/dt = 0, but not how d(e.k)/dt must also be 0

Have i done enough and just dont see it, or should i need another proof that (e.k) is constant.


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## Monday, April 24, 2006 ... //

### Translation and related news

Just a technical detail: I've added two utilities to the web pages of individual articles:

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# Finding the relationship given by the generating elements of a non-abelian group (for a maximum case) where they correspond to finding cycles

Let $$G=(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$$, where $$p$$ and $$q$$ are odd distinct primes. Let $$G$$ be generated by the elements $$s=(g_1,e_q)$$ and $$t=(e_p,g_3)$$,

$$g_1,e_p \in \mathbb{Z}_p \times \mathbb{Z}_p$$,

$$g_3,e_q \in \mathbb{Z}_q$$,

$$e_p,e_q$$ are identity elements. $$|s|=p, |t|=q$$.

$$\phi:\mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$$

$$e_q \rightarrow \phi_{e_q}$$

$$0 \rightarrow \phi_{0}$$

$$1 \rightarrow \phi_{1}$$

.

.

.

$$q-1 \rightarrow \phi_{q-1}$$

So for any $$\phi_k, 0 \geq k \geq q-1$$, we can determine $$\phi_k(g_1)=g$$, where $$g \in \mathbb{Z}_p \times \mathbb{Z}_p$$ ($$g$$ is some element and it will change according to $$\phi_k$$).

Let $$\mathbb{Z}_p \times \mathbb{Z}_p$$ be generated by $$g_1$$ and $$g_2$$. Let $$|g_1|=|g_2|=p$$. Then any $$g \in \mathbb{Z}_p \times \mathbb{Z}_p$$ can be expressed in terms of $$g_1$$ and $$g_2$$.

For, $$\phi_k(g_1)=g$$ let, $$\phi_k(g_1)=g=g_1^{m_1} g_2^{m_2} \rightarrow (1)$$

For a product of elements, say $$sts$$,

$$(g_1,e_q)(e_p,g_3)(g_1,e_q)$$ we can simplify as,

$$(g_1 \phi_{e_q}(e_p), e_q g_3)(g_1,e_q)$$

$$(g_1 \phi_{e_q}(e_p) \phi_{g_3}(g_1), g_3 e_q)$$

$$(g_1 \phi_{e_q}(e_p) \phi_{g_3}(g_1), g_3) \rightarrow (2)$$

Now as mentioned by (1), $$\phi_{g_3}(g_1)$$ also can be written in terms of $$g_1, g_2$$ and $$\phi_{e_q}(e_p)=e_p$$. Therefore, the first coordinate of the above ordered pair in (2) will simplify as some power of $$g_1$$ and $$g_2$$. Let it be as,

$$(g_1^{m_3}g_2^{m_4}, g_3)$$

Now suppose the group elements satisfies relationships such as,

$$stst^{-1}s=e \rightarrow (3)$$ (No. of elements in L.H.S. =5)

$$ssts^{-1}t^{-1}st=e \rightarrow (4)$$ (No. of elements in L.H.S. =7)

.

.

.

$$sts^{-1}ts...ts=e \rightarrow (5)$$ (No. of elements in L.H.S. =$$p^2q$$)

(These relationships correspond to finding cycles in Cayley graphs if we draw the Cayley graph of the group $$G$$ with repect to $$s,t$$.)

Having $$p^2q$$ no. of elements in the L.H.S. is the maximum possible no. of elements that can be present such that the product is equal to the identity $$e$$, where $$e=(e_p,e_q) \in (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$$.

In (3)-(5) also if we simplify in the way we obtained (2), we will get something similar.

Suppose we get as,

$$(g_1^{m_5}g_2^{m_6}, g_3^{m_7})=(e_p,e_q)$$ for (5). Can we solve for $$m_5 , m_6 , m_7$$ values if we know $$p,q,g_1,g_2,g_3, \phi$$? (May be using the data that (5) gives the maximum case?)

What is the method of obtaining the solutions? Can we solve for $$m_5 , m_6 , m_7$$ by a system of linear congruence relations?

Thanks a lot in advance.

Note: we don't know (5) but only knows that (5) gives maximum case and $$p,q,g_1,g_2,g_3, \phi$$ values. That's why I'm asking for a way of solving algebraically.

Even and idea or guidance regarding this is great. Many thanks again.

• Assuming you know p,q, then (g1^a g2^b, g3^c) = (g1^d g2^e, g3^f) if and only if (1) a-d is divisible by p, (2) b-e is divisible by p, and (3) c-f is divisible by q. Therefore these numbers are usually reduced mod p and q, so that m5=0, m6=0, m7=0 after reduction. You may find the setup easier to work with if you replace Zp x Zp with GF(p^2) and Zq with a cyclic subgroup of the multiplicative group of GF(p^2) of order q. (In other words, choose an irreducible quadratic divisor of the polynomial x^q -1 over GF(p).) – Jack Schmidt Jun 17 '20 at 13:22
• Thank you very very much @JackSchmidt is it possible to connect it with the maximum nature of the case as well by considering GF(p^2) ? – Buddhini Angelika Jun 20 '20 at 9:05


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# Properties

 Label 13.4.e.b Level 13 Weight 4 Character orbit 13.e Analytic conductor 0.767 Analytic rank 0 Dimension 2 CM no Inner twists 2

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$13$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 13.e (of order $$6$$, degree $$2$$, minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$0.767024830075$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( 1 + \zeta_{6} ) q^{2} + ( -2 + 2 \zeta_{6} ) q^{3} -5 \zeta_{6} q^{4} + ( 1 - 2 \zeta_{6} ) q^{5} + ( -4 + 2 \zeta_{6} ) q^{6} + ( -16 + 8 \zeta_{6} ) q^{7} + ( 13 - 26 \zeta_{6} ) q^{8} + 23 \zeta_{6} q^{9} +O(q^{10})$$ $$q + ( 1 + \zeta_{6} ) q^{2} + ( -2 + 2 \zeta_{6} ) q^{3} -5 \zeta_{6} q^{4} + ( 1 - 2 \zeta_{6} ) q^{5} + ( -4 + 2 \zeta_{6} ) q^{6} + ( -16 + 8 \zeta_{6} ) q^{7} + ( 13 - 26 \zeta_{6} ) q^{8} + 23 \zeta_{6} q^{9} + ( 3 - 3 \zeta_{6} ) q^{10} + ( 8 + 8 \zeta_{6} ) q^{11} + 10 q^{12} + ( 39 + 13 \zeta_{6} ) q^{13} -24 q^{14} + ( 2 + 2 \zeta_{6} ) q^{15} + ( -1 + \zeta_{6} ) q^{16} -117 \zeta_{6} q^{17} + ( -23 + 46 \zeta_{6} ) q^{18} + ( -132 + 66 \zeta_{6} ) q^{19} + ( -10 + 5 \zeta_{6} ) q^{20} + ( 16 - 32 \zeta_{6} ) q^{21} + 24 \zeta_{6} q^{22} + ( 78 - 78 \zeta_{6} ) q^{23} + ( 26 + 26 \zeta_{6} ) q^{24} + 122 q^{25} + ( 26 + 65 \zeta_{6} ) q^{26} -100 q^{27} + ( 40 + 40 \zeta_{6} ) q^{28} + ( 141 - 141 \zeta_{6} ) q^{29} + 6 \zeta_{6} q^{30} + ( -90 + 180 \zeta_{6} ) q^{31} + ( -210 + 105 \zeta_{6} ) q^{32} + ( -32 + 16 \zeta_{6} ) q^{33} + ( 117 - 234 \zeta_{6} ) q^{34} + 24 \zeta_{6} q^{35} + ( 115 - 115 \zeta_{6} ) q^{36} + ( -83 - 83 \zeta_{6} ) q^{37} -198 q^{38} + ( -104 + 78 \zeta_{6} ) q^{39} -39 q^{40} + ( 157 + 157 \zeta_{6} ) q^{41} + ( 48 - 48 \zeta_{6} ) q^{42} -104 \zeta_{6} q^{43} + ( 40 - 80 \zeta_{6} ) q^{44} + ( 46 - 23 \zeta_{6} ) q^{45} + ( 156 - 78 \zeta_{6} ) q^{46} + ( -174 + 348 \zeta_{6} ) q^{47} -2 \zeta_{6} q^{48} + ( -151 + 151 \zeta_{6} ) q^{49} + ( 122 + 122 \zeta_{6} ) q^{50} + 234 q^{51} + ( 65 - 260 \zeta_{6} ) q^{52} + 93 q^{53} + ( -100 - 100 \zeta_{6} ) q^{54} + ( 24 - 24 \zeta_{6} ) q^{55} + 312 \zeta_{6} q^{56} + ( 132 - 264 \zeta_{6} ) q^{57} + ( 282 - 141 \zeta_{6} ) q^{58} + ( -328 + 164 \zeta_{6} ) q^{59} + ( 10 - 20 \zeta_{6} ) q^{60} -145 \zeta_{6} q^{61} + ( -270 + 270 \zeta_{6} ) q^{62} + ( -184 - 184 \zeta_{6} ) q^{63} -307 q^{64} + ( 65 - 91 \zeta_{6} ) q^{65} -48 q^{66} + ( -454 - 454 \zeta_{6} ) q^{67} + ( -585 + 585 \zeta_{6} ) q^{68} + 156 \zeta_{6} q^{69} + ( -24 + 48 \zeta_{6} ) q^{70} + ( 1220 - 610 \zeta_{6} ) q^{71} + ( 598 - 299 \zeta_{6} ) q^{72} + ( -265 + 530 \zeta_{6} ) q^{73} -249 \zeta_{6} q^{74} + ( -244 + 244 \zeta_{6} ) q^{75} + ( 330 + 330 \zeta_{6} ) q^{76} -192 q^{77} + ( -182 + 52 \zeta_{6} ) q^{78} + 1276 q^{79} + ( 1 + \zeta_{6} ) q^{80} + ( -421 + 421 \zeta_{6} ) q^{81} + 471 \zeta_{6} q^{82} + ( 456 - 912 \zeta_{6} ) q^{83} + ( -160 + 80 \zeta_{6} ) q^{84} + ( -234 + 117 \zeta_{6} ) q^{85} + ( 104 - 208 \zeta_{6} ) q^{86} + 282 \zeta_{6} q^{87} + ( 312 - 312 \zeta_{6} ) q^{88} + ( -564 - 564 \zeta_{6} ) q^{89} + 69 q^{90} + ( -728 + 208 \zeta_{6} ) q^{91} -390 q^{92} + ( -180 - 180 \zeta_{6} ) q^{93} + ( -522 + 522 \zeta_{6} ) q^{94} + 198 \zeta_{6} q^{95} + ( 210 - 420 \zeta_{6} ) q^{96} + ( 232 - 116 \zeta_{6} ) q^{97} + ( -302 + 151 \zeta_{6} ) q^{98} + ( -184 + 368 \zeta_{6} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + 3q^{2} - 2q^{3} - 5q^{4} - 6q^{6} - 24q^{7} + 23q^{9} + O(q^{10})$$ $$2q + 3q^{2} - 2q^{3} - 5q^{4} - 6q^{6} - 24q^{7} + 23q^{9} + 3q^{10} + 24q^{11} + 20q^{12} + 91q^{13} - 48q^{14} + 6q^{15} - q^{16} - 117q^{17} - 198q^{19} - 15q^{20} + 24q^{22} + 78q^{23} + 78q^{24} + 244q^{25} + 117q^{26} - 200q^{27} + 120q^{28} + 141q^{29} + 6q^{30} - 315q^{32} - 48q^{33} + 24q^{35} + 115q^{36} - 249q^{37} - 396q^{38} - 130q^{39} - 78q^{40} + 471q^{41} + 48q^{42} - 104q^{43} + 69q^{45} + 234q^{46} - 2q^{48} - 151q^{49} + 366q^{50} + 468q^{51} - 130q^{52} + 186q^{53} - 300q^{54} + 24q^{55} + 312q^{56} + 423q^{58} - 492q^{59} - 145q^{61} - 270q^{62} - 552q^{63} - 614q^{64} + 39q^{65} - 96q^{66} - 1362q^{67} - 585q^{68} + 156q^{69} + 1830q^{71} + 897q^{72} - 249q^{74} - 244q^{75} + 990q^{76} - 384q^{77} - 312q^{78} + 2552q^{79} + 3q^{80} - 421q^{81} + 471q^{82} - 240q^{84} - 351q^{85} + 282q^{87} + 312q^{88} - 1692q^{89} + 138q^{90} - 1248q^{91} - 780q^{92} - 540q^{93} - 522q^{94} + 198q^{95} + 348q^{97} - 453q^{98} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/13\mathbb{Z}\right)^\times$$.

 $$n$$ $$2$$ $$\chi(n)$$ $$\zeta_{6}$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
4.1
 0.5 + 0.866025i 0.5 − 0.866025i
1.50000 + 0.866025i −1.00000 + 1.73205i −2.50000 4.33013i 1.73205i −3.00000 + 1.73205i −12.0000 + 6.92820i 22.5167i 11.5000 + 19.9186i 1.50000 2.59808i
10.1 1.50000 0.866025i −1.00000 1.73205i −2.50000 + 4.33013i 1.73205i −3.00000 1.73205i −12.0000 6.92820i 22.5167i 11.5000 19.9186i 1.50000 + 2.59808i
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
13.e even 6 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 13.4.e.b 2
3.b odd 2 1 117.4.q.a 2
4.b odd 2 1 208.4.w.b 2
13.b even 2 1 169.4.e.a 2
13.c even 3 1 169.4.b.d 2
13.c even 3 1 169.4.e.a 2
13.d odd 4 2 169.4.c.h 4
13.e even 6 1 inner 13.4.e.b 2
13.e even 6 1 169.4.b.d 2
13.f odd 12 2 169.4.a.i 2
13.f odd 12 2 169.4.c.h 4
39.h odd 6 1 117.4.q.a 2
39.k even 12 2 1521.4.a.o 2
52.i odd 6 1 208.4.w.b 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
13.4.e.b 2 1.a even 1 1 trivial
13.4.e.b 2 13.e even 6 1 inner
117.4.q.a 2 3.b odd 2 1
117.4.q.a 2 39.h odd 6 1
169.4.a.i 2 13.f odd 12 2
169.4.b.d 2 13.c even 3 1
169.4.b.d 2 13.e even 6 1
169.4.c.h 4 13.d odd 4 2
169.4.c.h 4 13.f odd 12 2
169.4.e.a 2 13.b even 2 1
169.4.e.a 2 13.c even 3 1
208.4.w.b 2 4.b odd 2 1
208.4.w.b 2 52.i odd 6 1
1521.4.a.o 2 39.k even 12 2

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{2} - 3 T_{2} + 3$$ acting on $$S_{4}^{\mathrm{new}}(13, [\chi])$$.

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$1 - 3 T + 11 T^{2} - 24 T^{3} + 64 T^{4}$$
$3$ $$1 + 2 T - 23 T^{2} + 54 T^{3} + 729 T^{4}$$
$5$ $$1 - 247 T^{2} + 15625 T^{4}$$
$7$ $$1 + 24 T + 535 T^{2} + 8232 T^{3} + 117649 T^{4}$$
$11$ $$1 - 24 T + 1523 T^{2} - 31944 T^{3} + 1771561 T^{4}$$
$13$ $$1 - 91 T + 2197 T^{2}$$
$17$ $$1 + 117 T + 8776 T^{2} + 574821 T^{3} + 24137569 T^{4}$$
$19$ $$1 + 198 T + 19927 T^{2} + 1358082 T^{3} + 47045881 T^{4}$$
$23$ $$1 - 78 T - 6083 T^{2} - 949026 T^{3} + 148035889 T^{4}$$
$29$ $$1 - 141 T - 4508 T^{2} - 3438849 T^{3} + 594823321 T^{4}$$
$31$ $$( 1 - 308 T + 29791 T^{2} )( 1 + 308 T + 29791 T^{2} )$$
$37$ $$1 + 249 T + 71320 T^{2} + 12612597 T^{3} + 2565726409 T^{4}$$
$41$ $$1 - 471 T + 142868 T^{2} - 32461791 T^{3} + 4750104241 T^{4}$$
$43$ $$1 + 104 T - 68691 T^{2} + 8268728 T^{3} + 6321363049 T^{4}$$
$47$ $$1 - 116818 T^{2} + 10779215329 T^{4}$$
$53$ $$( 1 - 93 T + 148877 T^{2} )^{2}$$
$59$ $$1 + 492 T + 286067 T^{2} + 101046468 T^{3} + 42180533641 T^{4}$$
$61$ $$1 + 145 T - 205956 T^{2} + 32912245 T^{3} + 51520374361 T^{4}$$
$67$ $$1 + 1362 T + 919111 T^{2} + 409639206 T^{3} + 90458382169 T^{4}$$
$71$ $$1 - 1830 T + 1474211 T^{2} - 654977130 T^{3} + 128100283921 T^{4}$$
$73$ $$1 - 567359 T^{2} + 151334226289 T^{4}$$
$79$ $$( 1 - 1276 T + 493039 T^{2} )^{2}$$
$83$ $$1 - 519766 T^{2} + 326940373369 T^{4}$$
$89$ $$1 + 1692 T + 1659257 T^{2} + 1192807548 T^{3} + 496981290961 T^{4}$$
$97$ $$1 - 348 T + 953041 T^{2} - 317610204 T^{3} + 832972004929 T^{4}$$


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with —

Second Annual Data Science Bowl

Mon 14 Dec 2015
– Mon 14 Mar 2016 (10 months ago)

Dashboard

Data Files

File Name Available Formats
validate .csv (2.77 kb)
train .zip (12.71 gb)
train.csv .zip (3.05 kb)
sample_submission_validate.csv .zip (3.12 kb)
test .zip (11.84 gb)
sample_submission_test .csv (1.02 mb)
solution .csv (22.26 kb)

Each has the same contents but use different packaging methods.

In this dataset, you are given hundreds of cardiac MRI images in DICOM format. These are 2D cine images that contain approximately 30 images across the cardiac cycle. Each slice is acquired on a separate breath hold. This is important since the registration from slice to slice is expected to be imperfect.

The competition task is to create an automated method capable of determining the left ventricle volume at two points in time: after systole, when the heart is contracted and the ventricles are at their minimum volume, and after diastole, when the heart is at its largest volume.

The volumes at systole, $V_S$, and diastole, $V_D$, form the basis of an important clinical measurement known as the ejection fraction:

$$100 * \frac{V_D - V_S}{V_D}.$$

This quantity represents the fraction of outbound blood pumped from the heart with each heartbeat. An ejection fraction that is too low can signify a wide range of cardiac problems.

Variations in anatomy, function, image quality, and acquisition make automated quantification of left ventricle size a challenging problem. You will encounter this variation in the competition dataset, which aims to provide a diverse representation of cases. It contains patients from young to old, images from numerous hospitals, and hearts from normal to abnormal cardiac function. A computational method which is robust to these variations could both validate and automate the cardiologists' manual measurement of ejection fraction.

This is a two-stage competition. In the first stage, you are building models based on the training dataset, and testing your models by submitting predictions on the validation set. Two weeks before the final deadline, you will submit your model to Kaggle. At this point, the second stage of the competition starts. Kaggle will release the final test dataset, on which you will run your models. The final standings are based on this final test set.

File descriptions

Each case has an associated directory of DICOM files. The exact number of images will differ from case to case, either varying in the number of slices, the views which are captured, or the number of frames in the time sequences.

The main view for assessing ventricle size is the short axis stack, which contains images taken in a plane perpendicular to the long axis of the left ventricle. These have the prefix "sax_" in the competition dataset. Most cases also have alternative views, which you should feel free to incorporate into your methodology.

The structure is as follows:

• train.zip - the train set directory, contains cases where you will have the associated systolic and diastolic volumes
• validate.zip - the validation set directory, used for the leaderboard in stage one of the competition. You should predict the volumes for these cases during stage one.
• test.zip - the test set, used for the leaderboard in stage two of the competition (a.k.a. the final standings). You should predict the volumes for these cases during stage two. This file will not be released until the second stage.
• train.csv - contains the systolic and diastolic volumes for the cases in the training set.
• sample_submission_validate.csv - a sample submission file in the correct format for stage one
• sample_submission_test.csv - a sample submission file in the correct format for stage two. This file will not be released until the second stage.

DICOM

The DICOM standard is complex and there are a number of different tools to work with DICOM files. You may find the following resources helpful for managing the competition data:

• The lite version of OsiriX is useful for viewing images on OSX
• pydicom - a package for working with images in python
• oro.dicom - a package for working with images in R
• Mango is a useful DICOM viewer for Windows users

FAQ

We will add to this section as relevant common questions arise.

How do I know where the left ventricle is? How do I compute its volume?

Watch this video for a primer on the anatomy and process used by clinicians:

I see more than one series at the same slice location. How should we deal with those cases?

Generally, a slice location is repeated if there is an artifact on the images. You can use either slice but the odds are that the last slice at a given slice location is the best the technologist could acquire.

Some MRI images are not consistent (in size, shape, or structure). What should we do about these?

We have opted to include as many cases as possible in this dataset. As this is real data from many sources, it is bound to have some amount of unwanted variability. You should do your best to handle these files. Since this is a two stage competition and the test set may have unseen abnormalities, we recommend including some form of error catching as you write your code.

Citation

The data for the Data Science Bowl is available for research and academic pursuits. Please cite as ‘Data Science Bowl Cardiac Challenge Data’.


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# Polarity of feedback for a mosfet circuit

I came across this question in "fundamentals of microelectronics" The objective is to find out the polarity of the feedback. Could someone help me with this?

My approach to this question is as follows:

If we increase input current then the voltage at point x increases due to which the drain current of M1 increases causing a decrease in the Vout.This in turn decreases the drain current of M2 causing an decrease in the voltage at point x.Therefore the feedback is negative. Is this approach correct?

No, you are wrong.

You wrote: "...This in turn decreases the drain current of M2 causing a decrease in the voltage at point x.".

The right statement would be:

This in turn increases the M2 drain-source channel resistance causing an increase of the voltage at point x. (This voltage, $V_x = I_{in} \cdot R_{channel}$, while both $I_{in}$ and $R_{channel}$ are increasing in this case).

Don't forget the input source is a current source, not a voltage one. The only path for its current is the drain of M2 -> ground, no current can flow into the M1 gate. When you increase the input current, it can't decrease as it is an independent current source (it is under your full control). So your claim: ...This in turn decreases the drain current of M2... is a contradiction.

The feedback is positive.

No - your last statement "...decrease in the voltage at point x" is NOT correct. A decrease in drain current of M2 is not possible because of the constant current source. Instead, the resistance of the D-C path increases and causes a corresponding increase in the voltage at x.

Hence, we have positive feedback.


text:
# How many hours will there be in a day 5,000,000,000 years from now?

It is known that the moon is moving away from Earth 2cm a year, and in doing so makes the days longer.

I want to know how many hours will have one day, when our planet is near its end.

First off, we should take the wisdom of Jack Wisdom to heart: "Calculation of the history of the lunar orbit is fraught with difficulties." While calculating the history may well be fraught with difficulties, calculating the future is hugely problematic. For the sake of argument, I'll ignore that issue.

The OP suggests the end of the planet is about 5 billion years into the future. The end is much sooner than that. The Sun will be about 10% more luminous then than it is now about a billion years in the future. That should be enough to trigger a moist greenhouse, followed soon by a runaway greenhouse. That moist greenhouse / runaway greenhouse pretty much spells the end of the lunar recession. While some of the tidal dissipation does result from the Earth tides, that is but a tiny contributor to the lunar recession rate. Almost all of the dissipation that leads to the lunar recession results from the ocean tides. Once the oceans vanish, the tidal recession of the Moon will pretty much stop in its tracks.

Assuming the mean lunar recession rate over the next billion years is somewhere between one to four centimeters per year means the Moon will have receded an additional 10,000 km to 40,000 km beyond the current 385,000 km from the center of the Earth in one billion years. The total angular momentum of the Earth-Moon system is $$L_{\text{tot}} = \left(\frac {m_e m_m}{m_e+m_m} R^2 +I_m \right)\Omega + I_e \omega$$ where $m_e$ and $m_m$ are the masses of the Earth and the Moon, $I_m$ and $I_e$ are the moments of inertia of the Earth and Moon about their centers of mass, $R$ is the mean distance between the center of the Earth and the center of the Moon, $\Omega = \sqrt{G(m_e + m_m)/R^3}$ is the Earth-Moon orbit rate, and $\omega$ is the Earth's rotation rate.

Assuming this quantity is conserved, the length of a day in a billion years will be between 25.5 hours (1 cm/year recession rate) and 31.7 hours (4 cm/year recession rate). A recession rate of 2 cm/year will result in a day of 27.3 hours.

The above assumed a constant recession rate. While that is not a good assumption, that range of one to four centimeters per year is probably okay. The lunar recession rate has ranged from over 1.1 to under 3.9 centimeters per year over last several hundred million years. The key culprit in the variations is the shape and orientation of the continents. The current rate of 3.82 centimeters per year is anomalously high thanks to the Americas and Afro-Eurasia, which together create two huge north-south barriers to the tides. This alignment also creates for some marked resonances that increase tidal friction even more than that suggested by the barriers. At other times, the orientation and shape of the continents has impeded the flow of the tides to a much lesser extent.

Once the oceans vanish, it's only the atmosphere and solid Earth tides that can result in the Moon receding from the Earth. The atmosphere will be thicker, which will greatly magnify the atmospheric contribution to tidal recession. However, that contribution is currently extremely tiny. Greatly magnifying a tiny number results in a still tiny number. Moreover, plate tectonics will stop shortly after the oceans disappear. Mountains will disappear. Friction between the atmosphere and the Earth will be much reduced. The other factor, Earth tides, is also very small. That range of a 25.5 to 31.7 hour long day might grow a bit in the four billion years between the end of the oceans and the five billion year figure cited in the question, but not by much.

• Hang on, won't the Sun become a red giant and engulf Earth in 5 billion years? Wouldn't that put an end to Earth as well as the length of the day? – Jim Nov 19 '14 at 15:56
• @Jim - Different authors give different times as to when the Sun will become a red giant. Some say a bit less than five billion years from now, others over seven billion years. Whether the Sun engulfs the Earth also depends on which author you read. Both are a bit irrelevant here because the OP didn't ask about what happens after the Sun becomes a red giant. The question is about the length of a day when our planet is near the end of its life five billion years from now; i.e. before the Sun becomes a red giant. The Earth's rotation rate will be more or less constant once the oceans disappear. – David Hammen Nov 19 '14 at 16:12
• That works for me – Jim Nov 19 '14 at 16:13

By definition the day (mean solar day) has 24 hours. Ever.

The statement 'and in doing so makes the days longer' is not supported by the last 44 years or so OF official most precise satelite mesurements .

As can be seen in this official graph the Length of Day as compared to a more constant reference (86400 atomic seconds) is shortening indicated by a downward trend.

The naive mistake we observe in the media is this: One more leap second? Then the LOD is enlarging.

• As I know, yes it is supported by satellite measurements. The OP has thought on the rotation period of the Earth on the day. – peterh says reinstate Monica Mar 28 '17 at 19:16
• @peterh I included the official data of the actual shortening of the LOD (satellite data since 1973) – Helder Velez Mar 30 '17 at 12:00
• Ok-ok :-) Actually, I've found your post by search for the most controversial (i.e most downvoted) users of the site. :-) You lead the list. :-) – peterh says reinstate Monica Mar 30 '17 at 12:34
• @peterh Yes I know and I'm honoured because it is all about the same issue: Is it that the space expands or the atom is shrinking? None was able to provide a valid argument against my viewpoint (downvotes are not arguments) . – Helder Velez Mar 30 '17 at 12:48
• Centuries are needed, perhaps millennia. Such data exist. A few centuries by measurements of the length of a day, and quite a few millennia by where ancients reported seeing solar eclipses. – David Hammen May 21 '18 at 16:04

## protected by ACuriousMind♦Mar 21 '17 at 14:59

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


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# Category:Commutativity

Let $\circ$ be a binary operation.
Two elements $x, y$ are said to commute (with each other) if and only if:
$x \circ y = y \circ x$


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4_Source_Wavelet_Inversion.tex 9.2 KB
Newer Older
 Tilman Steinweg committed Oct 01, 2015 1 2 3 %------------------------------------------------------------------------------------------------% \chapter{\label{cha:STF-Inversion}Source Time Function Inversion} \textbf{Introduction:}\\  Florian Wittkamp committed Jan 25, 2016 4 To remove the contribution of the unknown source time function (STF) from the waveform residuals, it is necessary to design a filter which minimizes the misfit to the field recordings and raw synthetics. The library libstfinv from Thomas Forbriger was exported from TFSoftware and can be used with a C API in IFOS. The purpose of this library is to provide methods for the derivation of source-time-functions in approaches to full waveform inversion. Given a set of recorded data and a set of synthetic data (typically, but not necessarilly the impulse response of the subsurface) a source time function is obtained due to some optimization citerion. The synthetic waveforms are convolved with this wavelet and the convolved synthetics as well as the wavelet itself are returned to the user.  Tilman Steinweg committed Oct 01, 2015 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172  The source time wavelet in this context not necessarily is the actual force time history of the source used in the experiment or a similar quantity of physical meaning. The source time wavelet simply is the wavelet which minimizes the misfit between synthetic and recorded waveforms due to some misfit condition, if the synthetics are concolved with this wavelet. In particular this implies that the synthetics not necessarily must be the impulse response (Greens function) of the subsurface, they may simply be synthetic waveform computed for some generic source wavelet (like a Ricker wavelet). The derived source time function then have to be understood with respect to this generic wavelet.\\ \newline The library provides different engines to find an optimal source time wavelet. The basic steps of operation are: \begin{enumerate} \item An engine is initialized. At this step pointers to arrays are passed to the engine together with some header information. The engines memorizes these pointers and expects to find the recorded data as well as the synthetics at the inidcated locations in memory. \item The run()-function of the engine is called. The engine takes the recorded and synthetic data currently found at the memory arrays, calculates an optimzed wavelet and returns the wavelet together with the convolved synthetics by copying them to the memory locations inidicated by the initializer of the engine. This step is repeated after each computation of synthetic data. \item The engine is removed once the iteration of inversion is terminated. \end{enumerate} \textbf{How to construct parameter strings:}\\ A specific engine is selected by passing a parameter string to the library interface. This parameter string may further contain parameters to control the execution mode of the engine. The parameter string starts with an ID-sequence identifying the desired engine. In the parameter string the ID-sequence is terminated by a colon (:).\\ After selecting the desired engine, the interface function strips of the ID-sequence as well as the colon from the parameter string and initializes the engine, passing the references to user workspace as well as the rest of the parameter string. The rest of the parameter string may consist of several control parameters being separated by colons (:). Each control parameter may just be a flag (switch to turn an option on) or may come along with a parameter value. The value of the parameter is separated by an equal sign (=). Examples: \begin{itemize} \item To select frequency domain least squares and shift the returned source time function by 0.4s and switch on verbose mode, pass the following parameter string:\\ \textit{fdlsq:tshift=0.4:verbose} \item To select frequency domain least squares, apply offset dependent weights and use a power of two to speed up the FFT:\\ \textit{fdlsq:pow2:exp=1.4} \end{itemize} \textbf{Detailed description of the engine 'Fourier domain least squares (fdlsq)'}\\ If \begin{itemize} \item $d_{lk}$ is the Fourier coefficient of recorded data at Frequency $f_l$ and receiver $k$ at offset $r_k$, \item $s_{lk}$ is the Fourier coefficient of the corresponding synthetics and \item $q_l$ is that of the sought source time function, \end{itemize} then this engine will minimize the objective function\\ E=\sum\limits_{l,k}\left|w_{lk}\, \left(d_{lk}-s_{lk}q_l\right) \right|^2+\sum\limits_{l}\lambda^2\left|q_l\right|^2 =\chi^2+\psi^2 with respect to the real part $q_l^\prime$ and the imaginary part $q_l^{\prime\prime}$ of q_l=q_l^\prime+i\,q_l^{\prime\prime}. In the above expression \chi^2=\sum\limits_{l,k}\left|w_{lk}\, \left(d_{lk}-s_{lk}q_l\right) \right|^2 is the data misfit with weights $w_{lk}$ and \psi^2=\sum\limits_{l}\lambda^2\left|q_l\right|^2 is used for regularization and will introduce a water-level in the deconvolution. $\lambda$ will balance both contributions. The conditions \frac{\partial E}{\partial q_l^\prime}\stackrel{!}{=}0 \quad\wedge\quad \frac{\partial E}{\partial q_l^{\prime\prime}}\stackrel{!}{=}0 result in (\cite{Forbriger:01}, appendix A.3) q_l=\frac{ \eta^2\sum\limits_{k}f_k^2\,s_{kl}^\ast\,d_{kl} }{ \lambda^2+\eta^2\sum\limits_{k}f_k^2\,s_{kl}^\ast\,s_{kl} } \quad\forall\, l where w_{lk}=\eta\,f_k and $f_k$ is a receiver specific weighting factor. Now $\eta$ and $\lambda$ have to be used to balance the regularization. We aim to specify a waterlevel as a fraction of synthetic data energy.\\ \newline \textbf{Setting up the waterlevel:}\\ The misfit equals one if the scaled energy of the residual $d_{lk}-s_{lk}q_l$ equals the scaled energy of the synthetics $s_{lk}$ and \eta^2=\frac{1}{\sum\limits_k f_k^2\sum\limits_l \left|s_{lk}\right|^2} is the reciprocal of the scaled energy of the synthetics. If we then choose \frac{\lambda^2}{\eta^2}=\frac{\epsilon^2}{N\eta^2}= \frac{\epsilon^2}{N}\sum\limits_k f_k^2\sum\limits_{l=0}^{N-1} \left|s_{lk}\right|^2 where $N$ is the number of frequencies, then $\epsilon^2$ will specify a waterlevel as a fraction of the scaled energy of the synthetics.\\ \newline \textbf{Using Parceval's Theorem to calculate signal energy:}\\ Parceval's Theorem for a signal $a(t)$ and its Fourier transform $\tilde{a}(\omega)$ is \int\limits_{-\infty}^{+\infty}\bigl|a(t)\bigr|^2\,\textrm{d} t= \int\limits_{-\infty}^{+\infty}\bigl|\tilde{a}(\omega)\bigr|^2\, \frac{\textrm{d} \omega}{2\pi}. If $S_{jk}$ are the time series samples corresponding to the Fourier coefficients $\tilde{s}_{lk}$ and $\Delta t$ is the sampling interval then \sum\limits_{k=0}^{M-1}\left|S_{jk}\right|^2\,\Delta t= \sum\limits_{l=0}^{M-1}\left|\tilde{s}_{lk}\right|^2\,\frac{1}{M\,\Delta t}, where $M=2N$ is the number of samples in the time series. In the above calculation the energy sum only uses the positive frequencies and \sum\limits_k f_k^2\sum\limits_{l=0}^{N-1}\left|\tilde{s}_{lk}\right|^2 = N\,(\Delta t)^2\, \sum\limits_k f_k^2 \sum\limits_{j=0}^{2N-1}\left|S_{jk}\right|^2. Fourier coefficients $s_{lk}$ calculated by the stfinv::STFFourierDomainEngine are not scaled (see documentation of libfourierxx and libfftw3), such that \Delta t\,s_{lk}=\tilde{s}_{lk} (both, $s_{lk}$ and $\tilde{s}_{lk}$ are Fourier coefficients). Consequently \sum\limits_k f_k^2\sum\limits_{l=0}^{N-1}\left|s_{lk}\right|^2 = N\, \sum\limits_k f_k^2 \sum\limits_{j=0}^{2N-1}\left|S_{jk}\right|^2. \textbf{Final calculation recipe:}\\ The solution to our problem is q_l=\frac{ \sum\limits_{k}f_k^2\,s_{lk}^\ast\,d_{lk} }{ \epsilon^2\,\sum\limits_k f_k^2 \sum\limits_{j=0}^{2N-1}\left|S_{jk}\right|^2 +\sum\limits_{k}f_k^2\,s_{lk}^\ast\,s_{lk} } \quad\forall\, l, where \sum\limits_{j=0}^{2N-1}\left|S_{jk}\right|^2 is the sum of the squared sample values $S_{jk}$ of the synthetic time series for receiver $k$, $f_k$ are the scaling factors, and $\epsilon^2$ is the water level parameter. \newline Author: Thomas Forbriger. \newline For more information see \href{http://www.opentoast.de/Data_analysis_code_soutifu_and_libstfinv.php}{http://www.opentoast.de/Data\_analysis\_code\_soutifu\_and\_libstfinv.php} .


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# Hartshorne's weird definition of right derived functors and prop. III 2.6

There is something very weird with the way Hartshorne defines right derived functors.

Hartshorne p 204 Let $\mathfrak A$ be an abelian category with enough injectives, and let $F \colon \mathfrak A \to \mathfrak B$ be a covariant left exact functor. Then we construct the right derived functors $R^iF, i ≥ 0$, of $F$ as follows. For each object $A$ of $\mathfrak A$, choose once and for all an injective resolution $I^.$ of $A$. Then we define $R^iF(A) = h^i(F(I^.))$

With Prop 1.2A. Hartshorne claims that one can also compute the right exact functor $R^iF$ with a resolution $J^i$ that is acyclic for F.

But then, in prop. 2.6 (and similarly in prop. 8.4) Hartshorne states what was already true by definition: that the derived functors of $\Gamma(X,^.)$ coincide with the cohomology functors $H^i(X,^.)$. The proof goes around a circle to show the obvious fact that an injective resolution is equivalent to itself. Namely, that every injective is flasque (2.4), and flasques are acyclic (2.5), so by 1.2A. (above) we can use this very injective resolution(!).

What do we need to go around like that?

If you look closely, Hartshorne defines the cohomology functors as the right-derived functors of

$$\Gamma(X, -) : \mathfrak{Ab}(X) \to \mathfrak{Ab}$$

and then claims in Prop 2.6 that these agree with the right-derived functors of

$$\Gamma(X, -) : \mathfrak{Mod}(X) \to \mathfrak{Ab}$$

There's some abuse of notation here -- in the first definition $$X$$ is just a topological space, whereas in the second definition $$X$$ is a ringed space, and of course we're also using $$\Gamma$$ for two different functors.

So Proposition 2.6 says that the derived functors of $$\Gamma(X, -) : \mathfrak{Mod}(X) \to \mathfrak{Ab}$$, which a priori depend on the structure sheaf of $$X$$, are actually independent of it. What's being checked in the proof is that an injective resolution in the category $$\mathfrak{Mod}(X)$$ is still at least an acyclic resolution when we forget the $$\mathcal{O}_X$$-module structure and regard it as a chain complex in $$\mathfrak{Ab}(X)$$.

You might want to try finding (or asking for) an example of an injective $$\mathcal{O}_X$$-module which fails to be injective as a sheaf of abelian groups if this doesn't make sense.

• Very nice answer. I also remember I found these parts of the wonderful Hartshorne's book to be extremely tricky... – Piotr Pstrągowski Jul 19 '13 at 9:56
• To be honest, when I read the question originally I was as confused as the OP. Hartshorne could really stand a revised edition cleaning up all these things. – Daniel McLaury Jul 19 '13 at 17:14
• @DanielMcLaury from the bottom of my heart: Clean up all the things!! – guest Dec 30 '14 at 6:24


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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# Decomposition Reaction Questions

Atoms of one element do not transform into those of another during a chemical reaction. Atoms do not vanish from the mixture or arrive from somewhere else. In chemical reactions, in fact, bonds between atoms are broken and formed during reactions generating new substances.

 Definition: When a single reactant breaks down to give simpler products then the reaction is said to be a decomposition reaction.

## Decomposition Reaction Chemistry Questions with Solutions

Q-1: Decomposition reactions are generally

1. Exothermic
2. Oxidative
3. Reductive
4. Endothermic

Explanation:

Energy is required to break bonds, whereas energy is released when bonds are formed. A decomposition process requires breaking the bonds of a big molecule to generate two or more smaller molecules, making the reaction endothermic.

Q-2: Which of the following is produced upon thermal decomposition of zinc sulphate?

1. Zinc oxide
2. Sulphur trioxide
3. Both of the above
4. Sulphur dioxide

Answer: c) Both of the above

Explanation: Zinc sulphate undergoes thermal decomposition in the following way:

$$\begin{array}{l}ZnSO_{4}\overset{\Delta }{\rightarrow}ZnO + SO_{3}\end{array}$$

Q-3: What are the three different types of decomposition reactions? State one example for each.

Answer: The three types of decomposition reactions are:

1. Thermal Decomposition Reaction: A thermal decomposition reaction necessitates the delivery of heat energy to the reactants. When calcium carbonate is heated, it decomposes into calcium oxide and carbon dioxide, which is a common example of a thermal decomposition reaction.
2. Photolytic Decomposition Reaction: A photodecomposition reaction is a sort of decomposition reaction in which the reactant absorbs energy from photons and breaks down into its parts. Decomposition of ozone into dioxygen and an oxygen radical is an example of a photo decomposition reaction.
3. Electrolytic Decomposition Reaction:The activation energy for decomposition is delivered in the form of electrical energy in an electrolytic decomposition reaction. The electrolysis of water is one type of electrolytic breakdown reaction.

Q-4: Which reaction is the opposite of the decomposition reaction?

1. Displacement reaction
2. Disproportionation Reaction
3. Combination Reaction
4. Precipitation Reaction

Explanation: A combination reaction involves the formation of a single product from two or more reactants.

The general reaction for such reactions is:

A+B→ AB

Q-5: State a reaction that is involved in black and white photography.

The reaction that is most commonly employed in black and white photography is:

$$\begin{array}{l}2AgCl(s)\overset{sunlight}{\rightarrow}2Ag(s)+Cl_{2}(g)\end{array}$$

Q-6: Chemical decomposition causes the compound to separate

a) physically

b) chemically

c) electrically

d) thermally

Explanation: By breaking bonds, a single molecule is degraded into two or more products in these reactions. Because chemical bonds are breaking and new ones are forming, atoms in a molecule are chemically separated.

Q-7: Which of the following conditions is not required for a decomposition reaction?

1. Only one reactant is present.
2. There are two or more products available.
3. Electricity is always required
4. All of the above conditions are required.

Answer: c) Electricity is always required.

Explanation: Electricity is not always required. Many decomposition reactions take place thermally and as well as photolytically.

Q-8: Carefully observe the activity shown below and then answer the following questions.

1. Identify the type of the reaction
2. What do you observe? Note down the change,if any.
3. Explain the observation with the help of a chemical reaction.

1. Thermal decomposition reaction
2. The emission of brown fumes will be observed. These fumes are nitrogen dioxide (NO2 ).
3. The reaction that takes place is:

2Pb(NO3)2(s) + heat → 2PbO(s) + 4NO2(g) +O2(g)

Q-9: All decomposition reactions are redox reactions. State whether the statement is true or false.

Explanation:

A redox reaction is one in which the simultaneous reduction and oxidation process takes place.

Heating of zinc carbonate to produce zinc oxide and carbon dioxide is a decomposition reaction but not a redox reaction. There is no reduction(gain of electron) and oxidation(loss of electron) taking place.

ZnCO3→ ZnO + CO2

Taking another example of a decomposition reaction

2NaH→2Na +H2

This reaction is redox and as well as decomposition.

Hence, decomposition reactions can be redox as well but not always.

Q-10: Identify the substances reduced or oxidised in the following decomposition reactions.

1. 2KClO3 + heat → 2KCl + 3O2
2. 2KMnO4 + heat → K2MnO4 + MnO2 + O2

A specie that undergoes an increase in oxidation state is said to be oxidised.

A specie that undergoes a decrease in oxidation state is said to be reduced.

a. Oxidised specie: Oxygen atom

Reduced specie: Chlorine atom

b. Oxidised specie: Oxygen atom

Reduced specie: Manganese atom

Q-11: Convert the following statements into chemical equations and determine the type of chemical reaction.

1. Aluminium oxide on heating produces aluminium and oxygen gas.
2. Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

Answer: Chemical equations are symbolic representations of chemical reactions that express the reactants and products in terms of their chemical formulae.

1. 2Al2O3 + heat → 4Al + 3O2 (Thermal Decomposition reaction)
2. 2H2S +3O2 → 2H2O + 2SO2 ( Redox Reaction)

Q-12: What are applications of decomposition reactions?

Answer: Decomposition reaction plays an important role in the following ways:

1. Metals are extracted from their chloride or oxide ores through electrolytic breakdown.

2. A catalytic breakdown reaction occurs when starch and other carbs are broken down into simple sugars like glucose in the presence of enzymes during digestion.

3. Photographers use decomposition reactions. The photographic film features a silver bromide layer that splits into silver and bromine when exposed to light.

Q-13: In the following decomposition reaction,

2N2O5 → 4NO2 +O2

Oxygen gas is produced at the average rate of 9.1×10-4 molL-1s-1, find the average rate of appearance of nitrogen dioxide and disappearance of nitrogen pentoxide.

Given:

$$\begin{array}{l} \frac{\Delta O_{2}}{\Delta t}\end{array}$$
= 9.1×10-4 molL-1s-1
Rate=
$$\begin{array}{l}-\frac{1}{2}\frac{\Delta N_{2}O_{5}}{\Delta t} = +\frac{1}{4}\frac{\Delta NO_{2}}{\Delta t}=\frac{\Delta O_{2}}{\Delta t}\end{array}$$

Rate of the appearance of nitrogen dioxide =

$$\begin{array}{l}+\frac{1}{4}\frac{\Delta NO_{2}}{\Delta t}\end{array}$$
=9.1×10-4 molL-1s-1

$$\begin{array}{l}\frac{\Delta NO_{2}}{\Delta t}\end{array}$$
= 4×9.1×10-4

= 3.6×10-3molL-1s-1

Rate of disappearance of nitrogen pentoxide =

$$\begin{array}{l}-\frac{1}{2}\frac{\Delta N_{2}O_{5}}{\Delta t}\end{array}$$
= 9.1×10-4 molL-1s-1

$$\begin{array}{l}\frac{\Delta N_{2}O_{5}}{\Delta t}\end{array}$$
= 2×9.1×10-4

= 1.8×10-3molL-1s-1

Q-14: In which of the following decomposition reactions, the evolving gas does not have the same number of sigma and pi bonds.

1. BeCO3 + heat →
2. FeSO4 + heat→
3. (NH4)2Cr2O7 + heat →
4. ZnSO4 + heat →

Explanation:

1. BeCO3 + heat → BeO +CO2

CO2(g) has same number of sigma and pi bond equal to 2

1. FeSO4 + heat→ Fe2O3 +SO2+SO3

SO2 and SO3 have the same number sigma and pi bond equal to 2 and 3 respectively.

1. (NH4)2Cr2O7 + heat → N2 + Cr2O3 + 4H2O

N2 has one sigma and two pi bonds.

1. ZnSO4 + heat → ZnO + SO3

SO3 has the same number of sigma and pi bonds equal to 3.

Q-15: In a 500 mL evacuated vessel, two moles of ammonia are added at high temperature. 2NH3(g) ⇋ N2(g) + 3H2(g) is the decomposition reaction. NH3 becomes 1 mole at equilibrium. Determine the nitrogen and hydrogen gas equilibrium concentration.

 2NH3(g) ⇋ N2(g) + 3H2(g) Initial moles 2 – – At equilibrium 2-2x x 3x

At equilibrium, moles of ammonia =1

This implies, 2-2x =1

x= 0.5

Equilibrium moles of N2 = x= 0.5mol

Equilibrium concentration of N2 = Moles of N2/Volume( in L)

= 0.5 mol/0.5L

= 1M

Equilibrium moles of H2 = 3x= 3×0.5mol

=1.5 mol

Equilibrium concentration of H2 = Moles of H2/Volume( in L)

= 1.5 mol/0.5L

= 3M

## Practise Questions on Decomposition Reaction

Q-1: When a china dish containing 2g of silver bromide is placed in sunlight as shown in the below figure, the colour changes. What is its colour?

Q-2: What form(s) of energy causes decomposition reaction?

1. Thermal
2. Light
3. Electrical
4. All of these

Q-3: Which reaction shows the physical decomposition?

1. CuCO3 +heat → CuO(s) + CO2 (g)
2. 2H2O + electricity → H2(g) + O2(g)
3. CuSO4.5H2O(s) + heat→ CuSO4 + 5H2O
4. 2FeSO4 + heat → Fe2O3 + SO2 + SO3

Q-4: Based on the experiment shown, answer the following questions?

1. Comment on the experiment.
2. Identify the gases that evolved.
3. Give a chemical test for distinguishing the two gases.
4. Write the chemical reactions taking place on anode and cathode.
5. Is the volume of the gas collected the same in both the test tubes?

Q-5: What is a double decomposition reaction?

Click the PDF to check the answers for Practice Questions.


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# Is the Stefan-Boltzmann law an upper limit?

The Stefan-Boltzmann law says that a black body radiates energy at a rate $$\sigma T^4$$ per unit surface area, where $$\sigma$$ is the Stefan-Boltzmann constant.

I have always assumed this is an upper limit, i.e. no body can ever lose heat faster than this due to radiation. However, the Wikipedia article linked above claims that this limit can be exceeded for some metamaterials. The reference cited for this seems not to be super-famous, and since it's such a surprising result I'd like to check I've understood it.

To make my question as concrete as possible, let me set up a hypothetical experiment: suppose I have an infinitely conductive metal sphere of radius $$r_1$$, which contains a nuclear reactor that keeps its surface always at a fixed temperature $$T$$. Surround this physical sphere with an imaginary sphere of radius $$r_2>r_1$$. Inside this imaginary sphere we can place any kind of apparatus we want, but it's not allowed to stick out of the imaginary sphere, and it must operate in steady state, i.e. it can't have any power source besides the temperature difference between the metal sphere and its surroundings.

This whole thing is to be placed in empty space, far away from any stars, and the goal is to extract energy from the metal sphere and radiate it away at as high a rate as possible. (Any spectrum of radiation is acceptable - it doesn't have to be black body radiation.)

The obvious solution is to surround the metal sphere with an infinitely conductive sphere of radius $$r_2$$, with a completely black surface. This will extract heat from the metal sphere and radiate it away at a rate $$4\pi r_2^2 \sigma T^4$$. The question is, is it possible, in principle, to do better than this?

If it's not possible, then is Wikipedia's reference wrong, or is it just claiming something else (and if so what)? If it is possible, then how can this be achieved, and is there any fundamental limit to how fast a body can radiate?

The Wikipedia article is correct, and moreover, its statements are relatively well known and uncontroversial. To see why, let's recap the physical intuition for why $$\sigma T^4$$ is an upper limit.

1. Begin with some object in the middle of free space. The object has some quantum energy levels, and the space itself has associated electromagnetic field modes. The object and the electromagnetic field can exchange energy by absorption and emission.
2. Assume that both the object and the surrounding electromagnetic field are at thermal equilibrium at temperature $$T$$. For each mode of the electromagnetic field, given a fixed shape for the object, there is a maximal rate of absorption, which is achieved when the object is completely black.
3. In thermal equilibrium, detailed balance is achieved: the average power absorbed by the object from each field mode precisely matches the average power the object puts into it by emission. Thus step 2 implies an upper bound on the emission rate per mode.
4. Summing over all of the electromagnetic field modes and setting the emission rate for each to the maximum quantitatively gives an emission intensity of $$\sigma T^4$$.
5. When the electromagnetic field isn't at temperature $$T$$, this bound on emission still applies, giving a maximum rate at which the object can cool off.

Almost all of this reasoning is correct in general. The only subtlety is in step 4. The point is that a metamaterial can have a different distribution of electromagnetic mode energies (i.e. density of states) than free space. When the sum over modes is performed, this can lead to an enhancement of the emission rate.

These kinds of effects are well known in applied physics. For example, in cavity QED, emission rates can be suppressed or enhanced by orders of magnitude, a result known as the Purcell effect.

Note that none of this discussion is about emission from metamaterials, it's about emission from objects embedded inside metamaterials, which see this modified density of states. If you insist your outer sphere is surrounded by vacuum, then as far as I know, there's no way to do better than $$\sigma T^4$$.

• Thanks, this I think must be the right answer. The key point is your last paragraph - Wikipedia's reference isn't making the claim it sounded like it was making. I'll probably accept this later on, though I'm still keen to know if the Stefan-Boltzmann law can be proven to be an upper bound if the apparatus is in empty space. Dec 14 '19 at 17:34
• Probably Wikipedia needs to be updated - the way it's worded strongly implies that metamaterials can radiate faster than $\sigma T^4$ into empty space. Would you update it (since you have greater knowledge), or should I? Dec 14 '19 at 17:37
• @Nathaniel Not me! I’d never be confident enough to edit Wikipedia, this isn’t my subfield. Dec 14 '19 at 19:28


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Control charts can show distribution of data and/or trends in data. Variable control charts for measured data. and preliminary (or present) samples, each of size $$n$$. Shewhart variables control charts. The Control_Chart in 7 QC Tools is a type of run_chart used for studying the process_variation over time. $$. Then $$n/2 = 7/2 = 3.5$$ The proportion of technical support calls due to installation problems is another type of discrete data. A disadvantage of control charts for variables and attributes is that they only use data from the most recent measurement to draw conclusions about the process. â This data can be used to create many different charts for process capability study analysis. Quality control charts represent a great tool for engineers to monitor if a process is under statistical control. Only with a stable process can an organization use process capability to compare the voice of the customer (VOC) with the voice of the process (VOP). These are often refered to as Shewhart control charts because they were invented by Walter A. â¦ --------------------------------------------- These include: The type of data being charted (continuous or attribute) The required sensitivity (size of the change to be detected) of the chart manuf. +1 Sigma Quality Assurance is the act of inspecting quality - assuring that the parts, process or system is making good quality. Various advantages of control charts for variables are as follows: (1) Control charts warn in time, if required rectification is done, well in time the scrap and percentage rejection can be reduced. variable is the control parameter because it influences the behavior of. Within these two categories there are seven standard types of control charts. We find that there are several situations in which CUSUM control charts have an economic advantage over $$\bar X$$ charts. 2. Variable Control Charts have limitations must be able to measure the quality characteristics in numbers may be impractical and uneconomical e.g. We also have to deal with the fact that $$\sigma$$ A popular method of implementing stock control is through the use of inventory (stock) control charts and algorithms that automate the process. ascertaining whether the process is in control. {\left( \frac{n-1}{2}-1 \right)!} p Control Charts â This attribute-type chart is effective when elements are not equal. For the X-bar chart, the center line can be entered directly or estimated from the With a control chart, you can monitor a process variable over time. what they are actually willing to pay for. Voice of Customer (VoC) tells us what the customer expects the product to do/achieve, i.e. can benefit from a single source construction and measurement control system. of Control Line Quality improvement methods have been applied in the last few 10 years to fulfill the needs of consumers. Continuous data is essentially a measurement such as length, amount of time, temperature, or amount of money. This function will be discussed shortly. = Benefits of using Control Charts They are simple graphical tools that enable process performance monitoring They are designed to identify which type of variation exists within a process They highlight areas of performance that may require further investigation They are easy to construct and interpret © 2017 Copyright ISC Ltd.$$ \sigma_s = \sigma \sqrt{1 - c_4^2} \, . Benefits from control charting are derived from both attributes and variables charts. 1. These situations are: 1. when there are high costs of false alarms and high costs of repairing a process; 2. when there are restrictions on sample size and sampling interval; 3. when there are several components of variance, and; 4. when there are statistical constraints on ARL . If you knew what a p chart and c chart were you would have your answer. We can also call it as process behavior chart. of Control Line  We The format of the control charts is fully customizable.


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# Hexed Hex

Geometry Level 4

An equilateral hexagon, that is one with all sides equal to one another, has $\angle BCD=122^\circ,\ \angle DEF=138^\circ,\ \angle FAB=100^\circ.$ Find $$\angle CDE$$ in degrees.

Inspiration: Calvin Lin's Problems of the Week, March 20

×


text:
Estimate on partial sum $\sum_{n=1}^x \frac{\sin^2(n)}{n}$ using Abel Plana Summation Formula :

I'm trying to get estimate on the following partial summation using Abel-Plana Summation formula:

$$\sum_{n=1}^x \frac{\sin^2(n)}{n}$$

I can handle the first integral in the formula but I'm stuck at the following functional:

$$T(x)=\int_0^\infty\frac{ (\frac{\sin^2(x+iy)}{(x+iy)}-\frac{\sin^2(x-iy)}{(x-iy)})}{(e^{2πy}-1)}dy$$

Questions:

(1) Sharp upper and lower bounds on $$T(x)$$.

(2) Graph of $$T(x)$$ (need an idea about it's growth).

Note: I tried to expand $$\sin$$ as complex variable in terms of hyperbolic functions ; but then I can't handle integral after expansion .

• Close votes !? What more clarity can I add? – Bambi Feb 24 at 10:03

If I am not mistaken, the difference in the numerator is $$\frac{{e^{2y} (2y\cos ^2 x + x\sin (2x) - y) + e^{ - 2y} (2y\cos ^2 x - x\sin (2x) - y) - 2y}}{{2(x^2 + y^2 )}}$$ (two times the imaginary part of the first term). Now for large positive $$y$$, this is $$= \mathcal{O}(1)\frac{{e^{2y} (x + y)}}{{x^2 + y^2 }} = \mathcal{O}(1)\frac{{e^{2y} }}{{\sqrt {x^2 + y^2 } }} = \mathcal{O}(1)\frac{1}{x}\frac{{e^{2y} }}{y},$$ and for small $$y$$, it is $$\mathcal{O} (y/x)$$. Hence, $$T(x)=\mathcal{O}(1/x)$$.
Just looking at the sum, it is less than $$\ln(x)+O(1)$$ and, by looking at successive terms of $$\sin^2(n)$$, I'm pretty sure that I could show the s is greater than $$c\ln(x)$$ where $$c > 1/2$$ or so.
The key would be showing that if $$\sin^2(n)$$ is small then $$\sin^2(n+1)$$ is not small.
• I believe that by the Euler--Maclaurin summation formula and properties of the cosine integral, the sum is $\frac{1}{2}\log x + \mathcal{O}(1) = \log \sqrt x + \mathcal{O}(1)$. – Gary Feb 24 at 7:12


text:
Get Dimension - Maple Help

# Online Help

###### All Products    Maple    MapleSim

Units

 GetDimension
 return dimension as a product of powers of base dimensions

 Calling Sequence GetDimension(dim)

Parameters

 dim - symbol; dimension

Description

 • The function GetDimension(dim) returns the argument for a base dimension, and the product of powers of base dimensions for any other dimension.

Examples

 > $\mathrm{with}\left(\mathrm{Units}\right):$
 Automatically loading the Units[Simple] subpackage
 > $\mathrm{GetDimension}\left('\mathrm{length}'\right)$
 ${\mathrm{length}}$ (1)
 > $\mathrm{GetDimension}\left('\mathrm{speed}'\right)$
 $\frac{{\mathrm{length}}}{{\mathrm{time}}}$ (2)
 > $\mathrm{GetDimension}\left('\mathrm{acceleration}'\right)$
 $\frac{{\mathrm{length}}}{{{\mathrm{time}}}^{{2}}}$ (3)
 > $\mathrm{GetDimension}\left('\mathrm{surge}'\right)$
 > $\mathrm{AddDimension}\left('\mathrm{surge}'=\frac{'\mathrm{length}'}{{'\mathrm{time}'}^{3}}\right)$
 > $\mathrm{GetDimension}\left('\mathrm{surge}'\right)$
 $\frac{{\mathrm{length}}}{{{\mathrm{time}}}^{{3}}}$ (4)

 See Also


text:
# Revision history [back]

### codomain could not be determined

I created an InfiniteDimensionalFreeAlgebra using CombinatorialFreeModule. It looks something like this:

class InfiniteDimensionalFreeAlgebra(CombinatorialFreeModule):
r""" The algebra generated by a[0], a[1], a[2], ... over the integers. """
def __init__(self,
base_ring=IntegerRing(),
prefix='a',
index_set=NonNegativeIntegerSemiring()):
self._base_ring = base_ring
self._basis_monoid = FreeMonoid(index_set=index_set, commutative=True, prefix=prefix)
# category
category = Algebras(self._base_ring.category()).WithBasis().Commutative()
category = category.or_subcategory(category)
# init
CombinatorialFreeModule.__init__(
self,
self._base_ring,
self._basis_monoid,
category=category,
prefix='',
bracket=False)

Now I want to view it as a ring and take symmetric polynomials over the ring. For example, in 2 variables, the following is one such symmetric polynomial:

2a[7] * x[1] + 2a[7] * x[2] + x[1]x[2] + 5

In my code, I attempt to initialize some Schur symmetric functions over this ring with something like:

a = InfiniteDimensionalFreeAlgebra()
sym = SymmetricFunctions(a)
s = sym.s()
one = s.one()
one_poly = one.expand(2)

at which point I run into the error codomain could not be determined when it attempts to apply some sort of morphism to the Schur function one. The full stacktrace is

Traceback (most recent call last):
File "./test_all.py", line 1162, in <module>
h = double_homogeneous(1, 1)
File "/Users/Matthew/programming/sage/morse-code/k_combinat_for_sage/all.py", line 664, in double_homogeneous
one_poly = one.expand(n)
File "/Users/Matthew/programming/sage/official-git-repo/local/lib/python2.7/site-packages/sage/combinat/sf/schur.py", line 537, in expand
return self._expand(condition, n, alphabet)
File "/Users/Matthew/programming/sage/official-git-repo/local/lib/python2.7/site-packages/sage/combinat/sf/sfa.py", line 4986, in _expand
return parent._apply_module_morphism(self, f)
File "/Users/Matthew/programming/sage/official-git-repo/local/lib/python2.7/site-packages/sage/categories/modules_with_basis.py", line 1096, in _apply_module_morphism
raise ValueError('codomain could not be determined')
ValueError: codomain could not be determined

and the full code can be found at k_combinat_for_sage.

Thank you, your help is greatly appreciated.


text:
# Is always possible say that $\langle df(\frac{\partial\varphi}{\partial u})\wedge df(\frac{\partial\varphi}{\partial v}) ,N_2\rangle>0$

Let $S_2$ be an orientable regular surface and $f:S_1\to S_2$ be a differentiable map which is a local diffeomorphism at every $p\in S_1$. Prove that $S_1$ is orientable.

My attemp: Suppose that $S_2$ is oriented by $N_2$ (since all orientable surface has a vector field of unit vectors N, we can assume that $S_2$ is oriented by $N_2$), and let $\{\varphi_{\lambda}:U_\lambda\to S_1\}_{\lambda\in\Gamma}$ a family of coordinate neighborhood of $S_1$.

Since $f$ is a local diffeomorphism, then $df_p$ is injective, but in particular $T_{f(p)}(S_2)=df_p(T_p S_1)$, and if is necessary we interchange $u$ and $v$, we can assume $$\left\langle df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial u}\right)\wedge df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial v}\right) ,N_2\right\rangle_{f(p)}>0\qquad (*)$$

Now, such $\{\varphi_{\lambda}\}_{\lambda\in\Gamma}$ is an orientation of $S_1$.

Suppose that $\varphi_{\alpha},\varphi_{\beta}\in\{\varphi_{\lambda}\}_{\lambda\in\Gamma}$ and such that $\varphi_{\alpha}(r,s)\cap\varphi_{\alpha}(u,v)\neq0$. Now, for the change of cordinates we have $$(\varphi^{-1}_{\alpha}\circ\varphi_{\beta})(r,s)=(u(r,s),v(r,s))\implies \varphi_{\beta}(r,s)=\varphi_{\alpha}((u(r,s),v(r,s)))$$ Then, $$\dfrac{\partial\varphi_{\beta}}{\partial r}=\dfrac{\partial\varphi_{\alpha}}{\partial u}\dfrac{\partial u}{\partial r}+\dfrac{\partial\varphi_{\alpha}}{\partial v}\dfrac{\partial v}{\partial r}$$

$$\dfrac{\partial\varphi_{\beta}}{\partial s}=\dfrac{\partial\varphi_{\alpha}}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial\varphi_{\alpha}}{\partial v}\dfrac{\partial v}{\partial s}$$

So, $df\left(\dfrac{\partial\varphi_{\beta}}{\partial r}\right)\wedge df\left(\dfrac{\partial\varphi_{\beta}}{\partial s}\right)=\dfrac{\partial(u,v)}{\partial(r,s)}df\left(\dfrac{\partial\varphi_{\alpha}}{\partial u}\right)\wedge df\left(\dfrac{\partial\varphi_{\alpha}}{\partial v}\right)$. Therefore,

$$0<\left\langle df\left(\dfrac{\partial\varphi_{\beta}}{\partial r}\right)\wedge df\left(\dfrac{\partial\varphi_{\beta}}{\partial s}\right) ,N_2\right\rangle=\dfrac{\partial(u,v)}{\partial(r,s)}\left\langle df\left(\dfrac{\partial\varphi_{\alpha}}{\partial u}\right)\wedge df\left(\dfrac{\partial\varphi_{\alpha}}{\partial v}\right),N_2\right\rangle$$ Which implies that $\dfrac{\partial(u,v)}{\partial(r,s)}>0$ and a regular surface is say "orientable" if is possible cover it with a family of coordinates neighborhood such that if a point belongs two differents coordinates family then, the change of coordinates has positive jacobian in such point.

The solution of this problem looks fine to me, but I have the little problem that is, if is always possible say that $\left\langle df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial u}\right)\wedge df_p\left(\dfrac{\partial\varphi_{\lambda}}{\partial v}\right) ,N_2\right\rangle_{f(p)}>0$.

My Attemp: If I have a basis for $T_{p}(S_1)$, $\{\varphi_u,\varphi_v\}$, and $df_p:T_p(S_1)\to T_{f(p)}(S_2)$ is a injective linear map, then $\{df_p(\varphi_u),df_p(\varphi_v)\}$ is linearly independent in $T_{f(p)}(S_2)$, i.e., a basis for $T_{f(p)}(S_2)$. Now, we can define a positive basis if $$\left\langle df_p(\varphi_u)\wedge df_p(\varphi_v), N_2 \right\rangle>0$$ Because a orientation $N_2$ on $S_2$ induces an orientation on each tangent space $T_{f(p)}(S_2)$. This is right?. Thanks!

• I don't know what you mean by "oriented by N," but the fact that $f$ is a local diffeomorphism tells you that $df_p$ maps $T_pS_1$ isomorphically to $T_{f(p)}S_2$. – Ted Shifrin Oct 21 '16 at 2:13
• @TedShifrin Hi! thanks for your answer, this is exactly what I wanna prove. If $f$ is a local diffeomorphism then $df_p$ mas $T_pS_1$ isomorphically to $T_{f(p)}S_2$, thanks – MathUser Oct 21 '16 at 3:43
• A local diffeomorphism $f$ has a local inverse $g$, so write down the chain rule (for both $g\circ f$ and $f\circ g$) and see that $df_p$ is an invertible linear map onto its image. (Clearly, $df_p(T_pS_1\subset T_{f(p)}S_2$, but you'll show they have the same dimension. Indeed, you only need $df_p$ maps onto, so you'll need only one of the equations you get from the chain rule.) – Ted Shifrin Oct 21 '16 at 3:47
• @TedShifrin I wrote that mean "Oriented by $N_2$", You believe that the answer is right? – MathUser Oct 24 '16 at 3:01
• OK. Then it seems fine to me :) – Ted Shifrin Oct 24 '16 at 14:43

Orientability can be defined in various ways. I'll stick to the following (alluded to in your question): An abstract two-dimensional surface $S$ is orientable if we can select an atlas for $S$ such that all transition maps have positive Jacobian wherever defined.
We are given an orientable surface $S_2$ (maybe embedded in ${\mathbb R}^3$). Let $\bigl(V_\beta,\psi_\beta\bigr)_{\beta\in J}$ be an atlas for $S_2$ with patches $V_\beta\subset S_2$ and coordinate functions $\psi_\beta:\>V_\beta\to{\mathbb R}^2$, such that all transition maps $\psi_{\beta'}\circ\psi_\beta^{-1}$ have positive Jacobian wherever defined.
In addition we are given a surface $S_1$ and a differentiable map $f:\>S_1\to S_2$ which is locally a diffeomorphism. We then may choose an atlas $\bigl(U_\alpha,\phi_\alpha\bigr)_{\alpha\in I}$ for $S_1$, such that for each $\alpha\in I$ the restriction $f\restriction U_\alpha$ maps $U_\alpha$ diffeomorphically into $V_\beta$ for a suitably chosen $\beta=\beta_\alpha$. The map $$f_{\alpha\beta}:=\psi_\beta\circ( f\restriction U_\alpha)\circ \phi_\alpha^{-1}: \qquad(x,y)\mapsto(u,v)$$ has nonzero Jacobian throughout, and after renumbering the output variables $(x,y)$ of $\phi_\alpha$ (if necessary) we may assume that this Jacobian is positive. Note that our assumption on the $\psi_\beta$ guarantees that the resulting numbering is independent of the $\beta$ we have chosen for this particular $\alpha$.
When the necessary renumberings have been performed all transition maps $\phi_{\alpha'}\circ\phi_\alpha^{-1}$, being compositions of ${\mathbb R}^2\to{\mathbb R}^2$ maps with positive Jacobian, have positive Jacobian as well.
• Thanks for your answer, but I think, if I prove that $(*)$ is always true, then my answer will be correct, can give me some hint to show this, thanks! – MathUser Oct 25 '16 at 19:30


text:
# Which of the following statements is true? a. Systematic reviews and meta-analyses are the same. b. Census studies determine casual relationships. c. Qualitative and quantitative data collection methods produce the same level of evidence. d. none of the above

Which of the following statements is true?
a. Systematic reviews and meta-analyses are the same.
b. Census studies determine casual relationships.
c. Qualitative and quantitative data collection methods produce the same level of evidence.
d. none of the above
You can still ask an expert for help

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Cynthia Lester
$\cdot$ First is True because A systematic review attempts to gather all available empirical research by using clearly defined, systematic methods to obtain answers to a specific question. A meta-analysis is the statistical process of analyzing and combining results from several similar studies. So both are same;
$\cdot$ Second is wrong because sample studies determines casual relationship;
$\cdot$ Third is wrong because Qualitative data and quantitative data collection methods produce the different level of evidence. Even we can say that Quantitative data is more useful and have good estimates.)

Systematic reviews and meta-analyses are the same.


text: Golden Archive RussianEnglish About the article MAGNETO-RESISTANCE OF TWO-DIMENSIONAL SYSTEMS A.I. LARKIN JETP LETTERS , 31, 219 (1980) 2015-05-22 15:22:02 D.E.Khmelnitskii Cavendish Laboratory, University of Cambridge J.J.Thomson ave, Cambridge, CB3 0HE, UK Sent to publication in January of 1980, this short paper finalised the stormish development in 1979 of what was later called the weak localisation. The year begun with publication of the famous "gang of four" paper [1], in which its authors proposed the scaling theory of localisation and claimed that all states in a 2D conductor with an arbitrary disorder are localised. A microscopic derivation of the scaling equation and the arguments in support of the conjecture of renormalisation group (RG) were given in the paper [2]. One month later, F. Wegner [3] proposed - phenomenologically - a model of the Field Theory, which possessed the RG and, in the limit of a weak disorder, gave the scaling equation conjectured by the authors of Ref [1]. The further development of the microscopic theory indicated the strong dependence of the quantum correction to conductivity on magnetic field [4], which led to suppression of the negative corrrection to conductivity and contributed to a negative mag-netirestance. This finding shifted attention to dependence of the resistance on magnetic field, which proved to be an easier way to conduct an experimental study of the whole phenomenon. It also brought up the necessity to discuss the influence of various factors on magneto-resistance. One month later, Hikami, Larkin and Nagaoka investigated effect of spin-orbit interaction on magnetoresistance [5] and found out that the latter changes its sign. Now, in this paper, Larkin attracted attention to the contribution to magneto-resistance of the Maki-Thompson correction to conductivity [6], discovered ten years earlier. The Maki-Thompson correction is connected to the scattering of the electrons on the Cooper pairs, which arose due to fluctuations. Originally, this correction was studied at the temperature close to the critical temperature of superconducting transition because it results in a smoothing out the sharp resistance drop. Still, the Cooper pairs could arise due to fluctuations far from superconducting transition and even for the repulsive sign of electron-electron interaction. Larkin pointed out that the sensitivity of the wave function of the Cooper pair to external magnetic field results in the contribution to magneto-resistance. Dependence of this contribution upon the magnetic field turned out to be identical to that due to weak localisation and its magnitude being dependent on the magnitude of the strength g of electron-electron interaction in the "Cooper channel". As the result, the magnitude of mag-netoresistance due to weak localisation must be corrected by the account of new contribution.  The factor a (a = 1 for potential scattering and a = -0.5 for the case of strong spin-orbit interaction) at the front of the standard expression for magneto-resistance must be replaced by the combination $a - \beta$. After exposition of the argument in favor of the new effect and the analytical derivation of the formulae, Larkin presents -in the forme of a tableaux - the results of numerical calculation of the factor $\beta$ for different values of the coupling constant g. For all the values of the coupling constant, the factor $\beta$ remains positive. The results of this paper were quickly adsorbed by all practicing researchers and remain now a part of the uniformorly used formula for magneto-resistance. References [1] E. Abrahams, P.W. Anderson, D.C. Licciardello & T.V. Ra-makrishnan, Phys Rev Lett, 42, 673, (1979) [2] L.P. Gorkov, A.I. Larkin & D.E. Khmelnitskii, JETP Letters, 30, 248, (1979) [3] F. Wegner, Z.f Phys., B 35, 207 (1979) [4] B.L. Altshuler, D.E. Khmelnitskii, A.I. Larkin, P.A.Lee, Phys Rev , B22, 5142 (1980) [5] S. Hikami, A.I. Larkin & Y. Nagaoka, Progr. Theor. Phys., 63 , 707 (1980) [6] R. Thompson, Phys. Rev., B1, 327 (1970)   Download as PDF


text:
# A projectile is shot from the ground at a velocity of 84 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?

Aug 5, 2017

$t = 16.5$ $\text{s}$

#### Explanation:

We're asked to find the time it takes for a projectile to land, given its initial velocity.

To do this, we can use the kinematics equation

ul(Deltay = v_0sinalpha_0t - 1/2g t^2

where

• $\Delta y$ is the change in height (which is $0$ when it lands)

• ${v}_{0}$ is the initial speed (given as $84$ $\text{m/s}$)

• ${\alpha}_{0}$ is the launch angle (given as $\frac{7 \pi}{12}$)

• $t$ is the time (what we're trying to find)

• $g = 9.81 \textcolor{w h i t e}{l} {\text{m/s}}^{2}$

Plugging in known values, we have

$0 = \left(84 \textcolor{w h i t e}{l} {\text{m/s")sin((7pi)/12)t - 1/2(9.81color(white)(l)"m/s}}^{2}\right) {t}^{2}$

$\left(4.905 \textcolor{w h i t e}{l} \text{m/s"^2)t^2 = (81.1color(white)(l)"m/s}\right) t$

(4.905color(white)(l)"m/s"^2)t = 81.1color(white)(l)"m/s"

t = color(red)(ulbar(|stackrel(" ")(" "16.5color(white)(l)"s"" ")|)


text:
×

# [closed] Data Structure Tutorial : Array

 4 `If you find any error please comment. I will try to update this post. Array is a collection of homogeneous data elements. It is a very simple data structure. The elements of an array are stored in successive memory location. Array is refered by a name and index number. Array is nice because of their simplicity and well suited for situations where the number is known. Array operation : Traverse Insert Delete Sort Search There are two types of array. One dimensional array and multi dimensional array.One dimensional array This type of array of array represent and strore in linear form. Array index start with zero. Declaration : datatype arrayname[size]; int arr[10]; Input array : for(int i=0; i<10; i++) cin>>arr[i]; We can use store integer type of data to the array arr using above segment. ---------- Traverse : Traversing can easy in linera array. Algorithm: C++ implement : void traverse(int arr[]) { for(int i=0; i<10; i++) cout<=k; i--) { arr[i]=arr[i+1]; } arr[k] = item; n++; } ---------- Deletion : Deletion is very easy on linear array. Algorithm : Deletion(arr, n, k) Here arr is a linear array with n number of items. K is position of elememt which can be deleted. Step 1:Start Step 2: Repeat for i=k upto n [Move the element upword] arr[i]=arr[i+1]; [End of the loop] Step 3: n--; Step 4 : Exit. C++ implementation : void deletion(int arr[], int n, int k) { for(int i=k; iarr[j+1]) then interchange arr[j] and arr[j+1] [End of inner loop] [End of outer loop] Step 4: Exit. C++ implement : void BubbleSort(int arr, int n) { for(int i=0; iarr[j+1]) swap(arr[j],arr[j+1]); } } } Quick Sort: Quick sort is a divide and conquer paradism. In this paradism one element is to be chosen as partitining element . We divide the whole list array into two parts with respect to the partitiong elemnt . The data which are similar than or equal to the partitining element remain in the first part and data data which are greater than the partitioning element remain in the second part. If we find any data which is greater than the partitioning value that will be transfered to the second part., If we find any data whichis smaller than the partitioning element that will be transferred to first part. Transferring the data have been done by exchanging the position of the the data found in first and second part. By repeating this process , we can sort the whole list of data. Algorithm: QUICKSORT(arr, l, h) if l arr[i]) { swap(arr[pivotPosition+1], arr[i]); swap(arr[pivotPosition] , arr[pivotPosition+1]); pivotPosition++; } } return pivotPosition; } void quickSort(int arr[], int low, int high) { if (low < high) { int pi = partition(arr, low, high); quickSort(arr, low, pi - 1); quickSort(arr, pi + 1, high); } } C++ example for simple sorting program with stl function:  #include using namespace std; int main() { int n, arr[100]; cin >> n; for(int i=0; i>arr[i]; } sort(arr, arr + n); for (int i=0; i

### The question has been closed for the following reason "Other" by rashedcs 01 Jan, 11:45

 2 For the deletion algo, wouldn't k-1 be the position of element to be deleted? answered 21 Nov '16, 11:00 0★byteword 26 accept rate: 100% I modified this post. Tnq for finding error. (21 Nov '16, 11:35) rashedcs1★
 2 Nice tutorial. Upvoted. answered 21 Nov '16, 11:53 2.6k●3●20 accept rate: 8%
 0 How do we do a binary search in a matrix or a 2-D array whose columns and rows are sorted? And how to sort a 2-D array? answered 21 Nov '16, 11:52 2.6k●3●20 accept rate: 8% 2 https://www.quora.com/How-do-I-search-in-a-row-wise-and-column-wise-sorted-matrix (21 Nov '16, 11:56) rashedcs1★ Thanks and how do we sort a 2-D array? (21 Nov '16, 11:58)
 0 @rashedcs types of array: one-dimensional and two-dimensional, shouldn't it be multi-dimensional array instead of two-dimensional array, as in dynamic programming sometimes 3D and 4D arrays are also used. answered 21 Nov '16, 11:59 5★srd091 1.5k●1●11 accept rate: 42%
 0 In quick sort, you have written "here's the link". The link is not working. answered 21 Nov '16, 12:01 2.6k●3●20 accept rate: 8% where you have written 'details about quick sort[link]', the link is still not working. (21 Nov '16, 13:25) 2 (22 Nov '16, 13:29) rashedcs1★
 0 @rashedcs Small typo its bubble sort not buble sort or bible sort ... Some suggestions to improve this tutorial , try to explain each function in sorting separately , like provide quicksort(parameters) and partition(parameters) example separately with examples as raw theory is hard to understand for someone who is coming across these algo for first time. Also when you use any built in function try to include its other overloaded versions if any. answered 21 Nov '16, 12:02 3★smsubham 673●2●14 accept rate: 15% Yeap. I always try to improve my tutorial. Tnq for suggestion @smsubham (21 Nov '16, 12:09) rashedcs1★ 2 It is typing mistake. I modified that. (21 Nov '16, 13:20) rashedcs1★
 0 How do we sort a 2-D array by rows and columns both? answered 22 Nov '16, 13:05 2.6k●3●20 accept rate: 8% (22 Nov '16, 13:21) rashedcs1★
 0 You should add the tag algorithm as this tutorial is more about algo's I think. answered 23 Nov '16, 15:06 2.6k●3●20 accept rate: 8% yeap. You are right. (23 Nov '16, 15:09) rashedcs1★

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# Math Help - Distribution to density to expected

1. ## Distribution to density to expected

A random variable X has the cumulative distribution function
$F(x) = 0$ for $x<1$
$F(x) = \frac{x^2 - 2x + 2}{2}$ for $1 \leq x < 2$
$F(x) = 1$ for $x \geq 2$

Calculate the variance of X.

The answer specifies that the density function is
$f(x) = 0.5$ if x=1
$f(x) = x-1$ if 1<x<2
$f(x) = 0$ otherwise

Then
$E(X) = 0.5 + \int_1^2 x(x-1) dx$
$E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx$

I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

My questions are:
Where do we get $f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.

2. Originally Posted by Boris B
A random variable X has the cumulative distribution function
$F(x) = 0$ for $x<1$
$F(x) = \frac{x^2 - 2x + 2}{2}$ for $1 \leq x < 2$
$F(x) = 1$ for $x \geq 2$

Calculate the variance of X.

The answer specifies that the density fucntion is
$f(x) = 0.5$ if x=1
$f(x) = x-1$ if 1<x<2
$f(x) = 0$ otherwise

Then
$E(X) = 0.5 + \int_1^2 x(x-1) dx$
$E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx$

I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

My questions are:
Where do we get $f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
$F(x) = 0$ for $x<1$ implies that $\Pr(X < 1) = 0$.

$F(x) = \frac{x^2 - 2x + 2}{2}$ for $1 \leq x < 2$ implies that $\Pr(X \leq 1) = \frac{1^2 - 2(1) + 2}{2} = \frac{1}{2}$.

It follows that $\Pr(X = 1) = \Pr(X \leq 1) - \Pr(X < 1) = \frac{1}{2}$.

3. Originally Posted by Boris B
A random variable X has the cumulative distribution function
$F(x) = 0$ for $x<1$
$F(x) = \frac{x^2 - 2x + 2}{2}$ for $1 \leq x < 2$
$F(x) = 1$ for $x \geq 2$

Calculate the variance of X.

The answer specifies that the density fucntion is
$f(x) = 0.5$ if x=1
$f(x) = x-1$ if 1<x<2
$f(x) = 0$ otherwise

Then
$E(X) = 0.5 + \int_1^2 x(x-1) dx$
$E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx$

I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

My questions are:
Where do we get $f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.
The $1/2$ comes from the jump discontinuity in the cumulative distribution $F(x)$ at $x=1$, indicating a probability mass of $1/2$ at that point.

Another way of thinking about this is to think of the density as a generalised function, then we may represent the density of a piecewise continuous cumulative distribution as the sum of a continuous function and delta functionals at the discontiuities of amplitude equal to the size of the jumps.

RonL

4. Originally Posted by Boris B
A random variable X has the cumulative distribution function
$F(x) = 0$ for $x<1$
$F(x) = \frac{x^2 - 2x + 2}{2}$ for $1 \leq x < 2$
$F(x) = 1$ for $x \geq 2$

Calculate the variance of X.

The answer specifies that the density fucntion is
$f(x) = 0.5$ if x=1
$f(x) = x-1$ if 1<x<2
$f(x) = 0$ otherwise
If this last is the given answer for the "density" then it is wrong, as as given it is not a density since:

$\int_{-\infty}^{\infty} f(x)~dx=1/2$

The density is the generalised function:

$f(x)=g(x)+(1/2)\delta(x-1)$

where

$g(x) = x-1, \ \ 1
$g(x) = 0, \ \ \mbox{otherwise}.$

RonL

5. Originally Posted by Boris B

What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
If you use the generalised function approach you have a perfectly normal looking equation for the expected value. The problem arrises if you represent the distribution as mixed discrete/continuous, when you have to work with the two components seperately.

RonL