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# The length of arc AXB is twice the length of arc BZC, and

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The length of arc AXB is twice the length of arc BZC, and [#permalink]

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03 Dec 2010, 20:08
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The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

A. 20
B. 40
C. 60
D. 80
E. 120

[Reveal] Spoiler:
My reasoning was:

AXB=2BCZ
AYC=3AXB
AYC=6BCZ

2BCZ+BCZ+3BCZ=360 Degrees

BCZ=40 degrees

AXB=80 degrees

AYC=240 degrees

Since AXB=80 and is opposite of angle BCA will be half of that (40)
At the same time, since BCZ is 40 degrees, BAC will be 20, and finally, since AYC is 240, ABC will be 120. I am not sure if this rule exists, it just made sense to me, and my answer of 40 degrees was correct. Am i right here?
[Reveal] Spoiler: OA

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Re: Can this MGMAT Geometry question be solved this way? [#permalink]

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03 Dec 2010, 21:47
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Since AXB=80 and is opposite of angle BCA will be half of that (40)

It's not quite because of the fact that it's opposite to BCA. Any arc will subtend an angle at the center - that's the one you've found, where AXB = 80. If the same arc subtends any angle on the circumference, then that angle is half of the angle subtended at the center, and hence BCA is 40 (Central Angle Theorem)

For more information on the properties of the circle, check out the Math Book's section on Circles

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Re: Can this MGMAT Geometry question be solved this way? [#permalink]

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03 Dec 2010, 23:39
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It should be solved like following
length of AXB + BZC + AYC = Perimeter of the circle

also, length of AXB = 2 BZC, therefore ½ AXB = BZC
also, length of AYC = 3 AXB

Hence, length AXB + ½ AXB + 3 AXB = 2πr
Θ/360 2πr + ½ Θ/360 2πr + 3 Θ/360 2πr = 2πr
2πr Θ/360 ( 1 + ½ + 3 ) = 2πr
Θ = (2πr*360*2)/(2πr*9)
= 80
And ∠BCA = ½ m (arc AXB)
= ½ * 80
= 40

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Re: Can this MGMAT Geometry question be solved this way? [#permalink]

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04 Dec 2010, 03:53
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MisterEko wrote:
Following is from MGMAT CAT:

The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

1. 20
2. 40
3. 60
4. 80
5. 120

My reasoning was:

AXB=2BCZ
AYC=3AXB
AYC=6BCZ

2BCZ+BCZ+3BCZ=360 Degrees

BCZ=40 degrees

AXB=80 degrees

AYC=240 degrees

Since AXB=80 and is opposite of angle BCA will be half of that (40)
At the same time, since BCZ is 40 degrees, BAC will be 20, and finally, since AYC is 240, ABC will be 120. I am not sure if this rule exists, it just made sense to me, and my answer of 40 degrees was correct. Am i right here?

The red part is not correct.

When we are told that "the length of arc AXB is twice the length of arc BZC" means that $$\angle{AOB}=2*\angle{BOC}$$, where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means that $$\angle{YOC}=3*\angle{AOB}$$

So if $$\angle{BOC}=x$$ then $$x+2x+3*2x=360$$ --> $$x=40$$ --> $$\angle{AOB}=2*\angle{BOC}=2*40=80$$. Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle so $$\angle{AOB}=80=2*\angle{BCA}$$ --> $$\angle{BCA}=40$$.

To elaborate more on angles and arcs:
You should know that when you measure the length of an arc in degrees then it's the measure of corresponding central angle. For example $$\angle{BOC}=40$$ means that arcBZC is 40/360=1/9 th of the circumference. Next, $$\angle{BAC}$$, according to the central angle theorem, will be half the measure of the central angle, so $$\angle{BAC}=\frac{\angle{BOC}}{2}=20$$. As for $$\angle{BZC}$$: it'll be supplementary angle to $$\angle{BAC}$$ (supplementary angles are two angles that add up to 180°), so it equals to $$\angle{BZC}=180-\angle{BAC}=160$$.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it helps.
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Re: Can this MGMAT Geometry question be solved this way? [#permalink]

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04 Dec 2010, 05:02
Hey friend Bunuel
as u have mention

Quote:
When we are told that "the length of arc AXB is twice the length of arc BZC" means that , where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means that

So if then --> --> . Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle so --> .

You are taking Length of an arc and measure angle of an arc as same
Length of an arc is a part of the circumference and it is calculated with formula Θ/360 2πr
where Θ = the measure angle of the arc

Measure angle of the arc is the central angle, which arc makes with the center.

What mentioned here in the question is not the measure angle, which will total to 360; But rather, what mentioned is length of the arcs and that will be equal to circumference i.e. 2πr.

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Re: Can this MGMAT Geometry question be solved this way? [#permalink]

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04 Dec 2010, 05:18
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vyassaptarashi wrote:
Hey friend Bunuel
as u have mention

Quote:
When we are told that "the length of arc AXB is twice the length of arc BZC" means that , where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means that

So if then --> --> . Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle so --> .

You are taking Length of an arc and measure angle of an arc as same
Length of an arc is a part of the circumference and it is calculated with formula Θ/360 2πr
where Θ = the measure angle of the arc

Measure angle of the arc is the central angle, which arc makes with the center.

What mentioned here in the question is not the measure angle, which will total to 360; But rather, what mentioned is length of the arcs and that will be equal to circumference i.e. 2πr.

I had a little typo, which I corrected (initially I calculated x to be equal to 60 not to 40). But else is correct in my solution.

You can transform the arc length into the central angle measure. So $$arcAXB+arcBZC+arcAYC=2\pi{r}$$ can be written as $$\angle{AOB}+\angle{BOC}+\angle{AOC}=360$$ which I did.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it's clear.
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Re: Can this MGMAT Geometry question be solved this way? [#permalink]

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04 Dec 2010, 06:22
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hey Bunuel

Quote:
You can transform the arc length into the central angle measure. So can be written as which I did.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it's clear.

You are right, it can be converted into angle form. I actually solve it and then found u r right.
Thank you for making my concepts clearer to this next limit.

+1 for you

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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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04 Mar 2013, 12:07
simply tooooooo gooodd Bunuel

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Re: Angle Measure from Length of Arcs [#permalink]

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04 May 2013, 12:25
bharatdasaka wrote:
The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

a) 20
b) 40
c) 60
d) 80
e) 120

Can Some one Please Explain this

Total length of the arcs = 2pi*r = AXB+BZC+AYC =$$\frac{9}{2}*AXB$$.Thus, as for arc length of 2pi*r the central angle measure is 360 degrees, thus for arc AXB, the angle measure = $$\frac{2}{9}*360$$ = 80 degrees. Thus, from the central angle theorem, the required angle measure is half of this = 40 degrees.

B.
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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09 May 2013, 21:04
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Hi Buneul, Is my approach correct?
As question said Arc AXB is twice Arc BZC and Arc AYC is three times AXB.
Let,
BAC=x
ABC=6x
BCA=2x

Now, x+2x+6x=180 ie x=20.

Hence, Required angle BCA=2x ie 2*20= 40 degree.

Sorry for the poor writing....
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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10 May 2013, 01:07
atalpanditgmat wrote:
Hi Buneul, Is my approach correct?
As question said Arc AXB is twice Arc BZC and Arc AYC is three times AXB.
Let,
BAC=x
ABC=6x
BCA=2x

Now, x+2x+6x=180 ie x=20.

Hence, Required angle BCA=2x ie 2*20= 40 degree.

Sorry for the poor writing....

Yes, that's correct.
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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10 Jun 2013, 08:09
Hi Bunuel,

I would like you to clarify one below doubt.

It has been deduced above that :
$$\angle {AOB} = 80 \angle{BOC}= 40 and \angle {AOC}=240$$
Now, here I'm getting confused, how can angle {AOC} be 240, If I join, AO and OC, then in this case, it becomes a triangle, and the sum of all the angles of triangle is equal to 180, which is contradictory to the above finding.

Regards,
imhimanshu

Quote:

You can transform the arc length into the central angle measure. So $$arcAXB+arcBZC+arcAYC=2\pi{r}$$ can be written as $$\angle{AOB}+\angle{BOC}+\angle{AOC}=360$$ which I did.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it's clear.

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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 05:18
Bump. Can someone please respond to the above doubt.
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 05:27
imhimanshu wrote:
Bump. Can someone please respond to the above doubt.

O is the center of the circle.

It has been deduced above that :
$$\angle {AOB} = 80\angle{BOC}= 40and \angle {AOC}=240$$
Now, here I'm getting confused, how can angle {AOC} be 240, If I join, AO and OC, then in this case, it becomes a triangle, and the sum of all the angles of triangle is equal to 180, which is contradictory to the above finding.

I do not understand the red part: the sum of the angles of a triangle will always be 180, and this does not contradict the above finding, which refers to the angles that are made with the center O. I attached an image, I think that you are probably not considering the correct drawing of the thing...
Attachments

1479_TriangleinCircleQ(edit).gif [ 6.68 KiB | Viewed 9079 times ]

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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 05:43
However, I feel, I still have the same doubt. Let me try to explain again.

It has been deduced above that Angle AOC is 240 degrees.
Now, If I join AO and OC and AC; then it will become a Triangle with Angle AOC is 240, which is against the rules of Triangles.

Can you please tell me where I'm going wrong. I have no idea..

Regards,
imhimanshu
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 06:14
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imhimanshu wrote:
However, I feel, I still have the same doubt. Let me try to explain again.

It has been deduced above that Angle AOC is 240 degrees.
Now, If I join AO and OC and AC; then it will become a Triangle with Angle AOC is 240, which is against the rules of Triangles.

Can you please tell me where I'm going wrong. I have no idea..

Regards,
imhimanshu

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.
Attachments

Untitled.png [ 5.61 KiB | Viewed 9063 times ]

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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 07:48
Zarrolou wrote:

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.

Hi Zarrolou,
Thanks for the explanation.
I have another follow up question, so does that mean that we should not trust the diagrams in PS as well. I know that trusting a figure in DS is a sin, but Is the same goes for PS as well.
In addition, can we do such reshuffle of the points as well, as we did in the this problem, as long as logic is permitting us to do so.

Regards,
imhimanshu
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 10:06
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imhimanshu wrote:
Zarrolou wrote:

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.

Hi Zarrolou,
Thanks for the explanation.
I have another follow up question, so does that mean that we should not trust the diagrams in PS as well. I know that trusting a figure in DS is a sin, but Is the same goes for PS as well.
In addition, can we do such reshuffle of the points as well, as we did in the this problem, as long as logic is permitting us to do so.

Regards,
imhimanshu

Actual figure looks as shown below:
Attachment:

Arcs.png [ 5.75 KiB | Viewed 9028 times ]

OG13, page150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

So, the question is not designed according to the GMAT rules, it should have been mentioned that the figure is not drawn accurately.

Hope it helps.
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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12 Jun 2013, 23:17
imhimanshu wrote:
Zarrolou wrote:

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.

Hi Zarrolou,
Thanks for the explanation.
I have another follow up question, so does that mean that we should not trust the diagrams in PS as well. I know that trusting a figure in DS is a sin, but Is the same goes for PS as well.
In addition, can we do such reshuffle of the points as well, as we did in the this problem, as long as logic is permitting us to do so.

Regards,
imhimanshu

Responding to a pm:

In addition, note that when you read the question, you can make out that the diagram is not to scale. AXB certainly doesn't look like twice of BZC. The moment you notice this, realize that you need to make your own diagram. As I have pointed out before, if they are mutilating the actual diagram, it means the actual diagram can make the question easier for you. So make the diagram on your own according to scale. That said, the actual GMAT will probably have more dependable diagrams.
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Re: The length of arc AXB is twice the length of arc BZC, and [#permalink]

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# Newton's method

Newton's method uses the derivative of a differentiable function to approximate its real or complex roots. For a function $f(x)$, the approximations are defined recursively by $$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)}.$$ To begin the recursion, an initial guess $x_0$ must be chosen. Often the choice of $x_0$ determines which of several possible roots is found, and in some cases the initial guess can cause the recursion to cycle or diverge instead of converging to a root.

## Derivation

At each step of the recursion, we have $x_i$ and seek a root of $f(x)$. Since all nonconstant linear functions have exactly one root, as long as $f'(x_i) \neq 0$ we can construct a tangent-line approximation to $f(x)$ and find its root as an approximation. The tangent-line approximation of $f(x)$ about $x_i$ is $$f(x_i) + (x - x_i)f'(x_i).$$ We seek the value $x = x_{i+1}$ such that the above expression equals $0$; hence, $$f(x_i) + (x_{i+1} - x_i)f'(x_i) = 0.$$ Dividing by $f'(x_i)$ (as long as $f'(x_i) \neq 0$), $$\frac{f(x_i)}{f'(x_i)} + x_{i+1} - x_i = 0.$$ Therefore, the desired value of $x_{i+1}$ is $$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_{i})}.$$

## Worked example

Problem: Find the values of the roots of $f(x) = x^2 - x - 1$.

Solution: We could use the quadratic formula to find that the roots are $\frac{1 \pm \sqrt 5}{2}$, but this approach does not immediately yield a decimal value.

To form a guess, we note that $\sqrt 5$ is slightly larger than $2$, so a suitable guess for the greater root is $\frac{1 + 2}{2} = 1.5$.

The derivative of $x^2 - x - 1$ is $2x - 1$ by the power rule for derivatives. Therefore, the recursive formula is

$$x_{i+1} = x_i - \frac{x_i^2 - x_i - 1}{2x_i - 1} = \frac{x_i^2 + 1}{2x_i - 1}.$$

Starting at $x_0 = 1.5$, we apply this formula repeatedly: \begin{align*} x_1 &= \frac{1.5^2 + 1}{2 * 1.5 - 1} &= 1.625, \\ x_2 &= \frac{1.625^2 + 1}{2 * 1.625 - 1} &= 1.6180556, \\ x_3 &= \frac{1.6180556^2 + 1}{2 * 1.6180556 - 1} &= 1.6180340, \\ x_4 &= \frac{1.6180340^2 + 1}{2 * 1.6180340 - 1} &= 1.6180340. \\ \end{align*}

Because $x_4 = x_3$ as calculated, the ratio $\frac{f(x_3)}{f'(x_3)}$ must be very close to $0$. Since $f'(x_3) = 2.2360680$ is not too large, $f(x_3)$ must be quite close to $0$, so $x_3 = 1.6180340$ is a very good estimate of the greater root.

The sum of the roots is $1$ by Vieta's formulas, so the lesser root is simply $1 - 1.6180340 = -0.6180340$.

## Failure cases

Although powerful, Newton's method is delicate and can be very sensitive to the initial guess and type of function, even when the function is differentiable everywhere.

### Zero derivative

Suppose in the above example of finding a root of $x^2 - x - 1$, we had started with $x_0 = 0.5$. In the process of calculating $x_1$, then, we would have to divide by $f'(x_0) = 2x_0 - 1 = 0$, creating an undefined result.

### Wrong root

Suppose we try to estimate the greater root as before, but start with $x_0 = -0.5$. Applying Newton's method to $x^2 - x - 1$ yields the following sequence of estimates:

\begin{align*} x_1 &= \frac{-0.5^2 + 1}{2 * (-0.5) - 1} &= -0.625, \\ x_2 &= \frac{(-0.625)^2 + 1}{2 * (-0.625) - 1} &= -0.6180556, \\ x_3 &= \frac{(-0.6180556)^2 + 1}{2 * (-0.6180556) - 1} &= -0.6180340, \\ x_4 &= \frac{(-0.6180340)^2 + 1}{2 * (-0.6180340) - 1} &= -0.61803395, \\ x_5 &= \frac{(-0.61803395)^2 + 1}{2 * (-0.61803395) - 1} &= -0.61803406, \\ x_6 &= \frac{(-0.61803406)^2 + 1}{2 * (-0.61803406) - 1} &= -0.61803395. \\ \end{align*}

These estimates converge to the lesser root, in this case because the initial estimate was closer to the lesser root than the greater root. In general, predicting which root Newton's method will finally converge to is difficult, and Newton's method may converge to a farther root even if a closer root is available within the interval between the initial estimate and the root.

### Periodic behavior without finding a root

When finding roots of $x^2 - x - 1$ as above, the estimates eventually stabilize into a predictable pattern: in the $x_0 = 1.5$ case, $x_4 = x_3$, so every subsequent estimate will equal $x_3$; in the $x_0 = -0.5$ case, $x_6 = x_4$, so the subsequent estimates will oscillate, with every even estimate equal to $x_4$ and every odd estimate equal to $x_5$. These behaviors are due to the limited precision of the calculations; with indefinite precision, the estimates would continue to approach the root without becoming periodic.

However, for some functions, Newton's method can cycle without coming close to a root.

### Divergence

The function $f(x) = e^x$ is always positive, so it has no roots. Since $f(x) = e^x$ satisfies the differential equation $f(x) = f'(x)$ for all $x$, the ratio $\frac{f(x)}{f'(x)}$ is always $1$, so the sequence of estimates produced by Newton's method becomes $x_{i+1} = x_i - 1$, diverging in the negative direction.

## Applications in Analytic Number Theory (Advanced)

The applications of Newton's Method actually apply to Hensel's Lemma, specifically in the $p$ adic integers $\mathbb{Z}_p$. In particular, take some polynomial $p\in\mathbb{Z}[x]$ and an integer $x$ such that $p(x)\equiv 0\pmod{p}$. Our goal is to approximate some $\hat{x}$ such that $p(\hat{x})\equiv 0\pmod{p^2}$. However, in the $p$ adics, we can't just lift" arbitrarily. We actually need to approximate such a solution in the next $p$ power level, and then refine it iteratively.

Here's what that looks like realistically. WLOG we assume that $x$ is some $a_0$ between $0$ and $p-1$. We search for $\hat{x}=a_0+a_1p$ such that $p(\hat{x})\equiv 0\pmod{p^2}$. Note the following critical step:

$$p(\hat{x})=p(a_0+a_1p)=p(a_0)+p'(a_0)a_1p+(a_1p)^2b$$

for some $b\in\mathbb{Z}$. By assumption we know that $p(a_0)\equiv 0\pmod{p}$ so let $p(a_0)=pt$ for some $t$. The desired congruence holds modulo $p^2$ when $t+p'(a_0)a_1\equiv 0\pmod{p}$. When $p'(a_0)\not\equiv 0\pmod{p}$, we see that $a_1\equiv -t/p'(a_0)\pmod{p}$ which means that

$$\hat{x}=a_0+a_1p=a_0-\frac{pt}{p'(a_0)}=x-\frac{p(a_0)}{p'(a_0)}.$$

This is exactly the statement of the Newtonian Algorithm! Our root is approximated, so we found some $\hat{x}$ such that $p(\hat{x})\equiv 0\pmod{p^2}$. But this isn't the whole story.

Theorem: Let $p(x)\in\mathbb{Z}_p[x]$ and $x\in\mathbb{Z}_p$ such that $p(x)\equiv 0\pmod{p^n}$. If $k=v(P'(x)) then there exists an $\hat{x}=x-p(x)/p'(x)$ such that (1) $p(\hat{x})\equiv 0\pmod{p^{n+1}}$, (2) $\hat{x}\equiv x\pmod{p^{n-k}}$ and (3) $k=v(p'(x))=v(p'(\hat{x}))$.

Proof: Let $p(x)=p^ny$ for $y\in\mathbb{Z}_p$ and $p'(x)=p^ku$ where $u\in\mathbb{Z}_p^{\times}$. By definition we see that

$$\hat{x}-x=-\frac{p(x)}{p'(x)}=-\frac{p^ny}{p^ku}=-p^{n-k}yu^{-1}\in p^{n-k}\mathbb{Z}_p.$$

This proves (2). Notice, however, that in the Taylor expansion of $p$ about $x$ we see that

$$p(\hat{x})=p(x)-\frac{p(x)p'(x)}{p'(x)}+(\hat{x}-x)^2t=(\hat{x}-x)^2t$$

where $t\in\mathbb{Z}_p$. Also notice that

$$p(\hat{x})=(\hat{x}-x)^2t\in p^{2n-2k}\mathbb{Z}_p$$

but we finish (1) since $p^{2n-2k}\mathbb{Z}_p=p^n\cdot p^{n-2k}\mathbb{Z}_p\subset p^{n-1}\mathbb{Z}_p$. To prove (3) we just compute the order of $p'(\hat{x})$. Taking the Taylor expansion of $p'$ at point $x$ gives

$$p'(\hat{x})=p'(x+(\hat{x}-x))=p'(x)+(\hat{x}-x)s=p^ku+p^{n-k}zs=p^k(u+p^{n-2k}zs)=p^kv.$$

But since $k and $u\in\mathbb{Z}_p^{\times}$ we have

$$v=u+p^{n-2k}zs\in u+p\mathbb{Z}_p\subset \mathbb{Z}_p^{\times}$$

which proves that $k=v(p'(x))=v(p'(\hat{x}))$ $\square$

This sets up the entire motivation behind Hensel's Lemma in a more formal way. In addition, the proof of the lemma is just an extension of what we previously noted, which actually makes all of our previous work worth it in the long run.

Hensel's Lemma: Assume $p\in\mathbb{Z}_p[x]$ and $x\in\mathbb{Z}_p$ satisfies $p(x)\equiv 0\pmod{p^n}$. If $k=v(p'(x)), then there exists a unique root $\xi$ of $p\in\mathbb{Z}_p[x]$ such that $\xi\equiv x\pmod{p^{n-k}}$ and $k=v(p'(x))=v(p'(\xi))$.

Proof: We first prove the existence of $\xi$. Let $x_0=x$ be given. We can construct $x_1\in\mathbb{Z}$ such that $$x_1\equiv x\pmod{p^{n-k}}~\text{and}~p(x_1)\equiv 0\pmod{p^{n+1}},~v(p'(x_0))=v(p'(x_1))=k.$$ We can do this for $x_2$ and $x_n$ such that we have a sequence $(x_n)_{n\ge 0}$ where $x_n\to \xi$ as $n\to \infty$ where $p(\xi)=0$ and $\xi\equiv x\pmod{p^{n-k}}$. Now we prove that $\xi$ is unique. For the sake of contradiction, suppose that there are two roots $\xi$ and $\omega$ satisfying

$$\xi\equiv \omega\pmod{p^{n-k}}.$$

Since it was given that $k we have $2k. This means that $n-k\ge k+1$ so we have at the very least

$$\xi\equiv \omega\pmod{p^{k+1}}.$$

Now we have,

$$p(\omega)=p(\xi)+p'(\xi)(\omega-\xi)+(\omega-\xi)^2a$$

for some $a\in\mathbb{Z}_p$. Since $\xi$ and $\omega$ were assumed to be roots, we factor to get

$$(\omega-\xi)(p'(\xi)+(\omega-\xi)a)=0$$

but since $v(p'(\xi))=k$ and $v((\omega-\xi)a)\ge k+1$

$$p'(\xi)+(\omega-\xi)a\ne 0$$

which means that $\omega-\xi=0\implies \omega=\xi$, so uniqueness holds $\square$

format: plain


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# Non-convex constraint

Hi, I want to formulate a maximization problem in CVX. The objective function is concave but there exists a constraint in the following form:
\sum_{i=1}^K \dfrac{a_i}{1+e^{-b_i x_i}} \geq c_i
where x_i, i=1,..., K are variables to be optimized and we also have the upper and lower limit of each x_i as
l_i \leq x_i \leq u_i.
Can I formulate this problem in CVX?

Presuming a_i, b_i > 0, the LHS of the constraint is convex for x <= 0 and concave for x >= 0. Therefore, the constraint is only convex for x >= 0. Therefore, this is only a convex optimization problem if l_i >= 0. If that is the case, I am not optimistic (other than for K = 1) this can be formulated in CVX, because the LHS is not concave over the entirety of its natural domain. Your only hope is if a (Mosek) wizard comes along and proves my pessimism to be misplaced.

1 Like

Thanks for your answer. Yes, a_i, b_i > 0 and l_i \geq 0.
So, if CVX cannot be used to solve this problem, is there any other tool which can be used to do so? Because the problem is not straightforward to be solved analytically.

You can use a tool such as YALMIP, which with a nonlinear solver, should handle it, even if not in conic form.

1 Like


text:
ISSN 0439-755X
CN 11-1911/B

Acta Psychologica Sinica ›› 2018, Vol. 50 ›› Issue (2): 243-252.

### Social class and social perception: Is warmth or competence more important?

WEI Qingwang; LI Muzi; CHEN Xiaochen

1.  (Department of Psychology, Renmin University of China, Beijing 100872, China)
• Received:2017-02-20 Published:2018-02-25 Online:2017-12-26
• Contact: CHEN Xiaochen, E-mail: xiaochenchen@ruc.edu.cn
• Supported by:

Abstract:  Warmth and competence are the two fundamental dimensions (i.e. Big Two) in social cognition. According to the Dual Perspective Model (DPM), warmth is the primacy of the Big Two and the Big Two are differentially linked to the actor (self) vs. observer (other) perspectives. In the observer perspective, warmth is more relevant and more important; whereas in the actor perspective, competence is more relevant and more important. Another domain of literature on social class psychology suggested that lower-class individuals were more sensitive to external environment and valued interdependent self; whereas upper-class individuals were more self-focused and valued independent self. The current study combined these two domains of literature and examined possible moderating role of social class on the link between the Big Two and the actor vs. observer perspectives. Specifically, we hypothesized that both lower-class individuals and upper-class individuals would value warmth more than competence in evaluating others, and this primacy of warmth would be more evident for lower-class individuals (H1). In contrast, lower-class individuals would also value warmth more than competence whereas upper-class individuals would value competence more than warmth when evaluating themselves (H2). Two studies were carried out to test these hypotheses. In Study 1, 122 undergraduate participants were presented with a list of 8 sentences each describing a behavior of a stranger. The behavioral acts were deliberately chosen to be amenable to both warmth and competence traits. Participants were asked to use a single word to describe the character of the subject in each sentence. Information on participants’ objective socioeconomic status (SES, family income and highest parental education level) was also collected. In Study 2, 137 community participants were asked to rate the importance of 12 traits (6 on warmth dimension, 6 on competence dimension) in evaluating themselves. The MacArthur scale was used to assess participants’ subjective social class rank. In Study 1, more warm words (as compared to words on the competence dimension) were chosen to describe the character of the subjects in the sentences. This pattern was consistent among both lower- and upper-class individuals. In addition, lower-class individuals used significantly more warm words than upper-class participants. In Study 2, lower-class individuals scored significantly higher on warmth than competence. Contrary, upper-class individuals scored significantly higher on competence than warmth. Taken together, findings from these two studies provided evidences to support our hypotheses. The current study contributes to the social cognition literature by integrating the DPM and social class psychology. Social class does influence the primacy of warmth vs. competence as demonstrated previously in DPM. The primacy of warmth in evaluating others is more typical for lower-class individuals and the primacy of competence in evaluating themselves is more typical for upper-class individuals. Moreover, DPM further develops the ideas of social class psychology. Lower-class individuals showing contextual social cognitive tendencies and upper-class individuals showing solipsistic social cognitive tendencies are reflected very well in the primacy of warmth vs. competence in evaluating others vs. themselves. Uncovering the nuances in social cognition between lower and upper classes also provides important practical implications to promote healthy inter-class communications.

CLC Number:


text:
## Classical logic, storage operators and second-order lambda-calculus.(English)Zbl 0814.03009

Storage operators, introduced by the author [Arch. Math. Logic 30, 241- 267 (1990; Zbl 0712.03009)], are a device to simulate call-by-value in a lambda-calculus with head-reduction. In this paper this concept is used to characterize the lambda-terms encoding integers in classical logic. Let $$\text{Nat} [x]$$ be the formula expressing in second-order lambda- calculus (also with first-order variables and quantifiers) the data type of natural numbers. An important consequence of the normalization theorem is that if $$\vdash M: \text{Nat} [x]$$, then $$M$$ is beta-equivalent to $$\underline{n}$$, the Church numeral for $$n$$. When one extends the calculus with axioms (i.e., when the typing context is extended with typed constants $$c_ i: \sigma_ i$$), this in general is no longer true. The paper considers the especially relevant extension obtained with the axiom $${\mathbf c}$$: $$\forall X(\neg\neg X\to X)$$. This system, in view of Gödel’s double-negation translation, axiomatizes classical logic. M. Parigot [Lect. Notes Comput Sci. 592, 361-380 (1991)] has shown that closed elements of type NAT in this extended system can be characterized in terms of storage operators: If $${\mathbf c} \vdash M: \text{Nat} [n]$$ and if $$T$$ is a storage operator for integers, then, for any variable $$f$$, $$TfM$$ reduces by head reduction to $$fN$$, with $$N$$ beta- equivalent to $$\underline {n}$$, depending only on $$n$$ (and not on $$M$$). The nontrivial proof uses a realizability argument. All the preliminaries, including those on storage operators and the proof of Parigot’s result, are clearly explained in the first part (pp. 53-70) of the paper. The result, besides its direct technical interest, is of potential relevance also for the ongoing research on the extraction of programs from classical proofs.
Reviewer: S.Martini (Pisa)

### MSC:

 03B40 Combinatory logic and lambda calculus 68N15 Theory of programming languages

Zbl 0712.03009
Full Text:

### References:

 [1] Barendregt, H., The lambda calculus, (1984), North-Holland Amsterdam · Zbl 0549.03012 [2] Felleisen, M., The calculi of λ_{v}-CS conversion: a syntactic theory of control and state in imperative higher order programming, Ph.D. thesis, (1987), Indiana Univ [3] Girard, J.Y., Une extension de l’interprétation de Gödel á l’analyse, (), 63-92 [4] Girard, J.Y., A new constructive logic: classical logic, (1991), preprint · Zbl 0752.03027 [5] Girard, J.Y., On the unity of logic, Ann. pure appl. logic, 59, 201-217, (1993) · Zbl 0781.03044 [6] Griffin, T., A formulae-as-type notion of control, Conf. record of the 17th ACM symp. on principles of programming languages, (1990) [7] Krivine, J.L., Lambda-calcul, types et modéles, (1990), Masson Paris [8] Krivine, J.L., Opérateurs de mise en mémoire et traduction de Gödel, Arch. math. logic, 30, 241-267, (1990) · Zbl 0712.03009 [9] Krivine, J.L.; Parigot, M., Programming with proofs, J. inform. process. cybernet. EIK, 26, 3, 149-167, (1990) · Zbl 0699.68020 [10] Leivant, D., Reasoning about functional programs and complexity classes associated with type disciplines, 24th ann. symp. on found. of comp. sci., 460-469, (1983) [11] Murthy, C., Extracting constructive content from classical proofs, Ph.D. thesis, (1990), Cornell Univ [12] Parigot, M., Free deduction: an analysis of computations in classical logic, (), 361-380, St. Petersbourg, Lecture Notes in Computer Science · Zbl 0925.03203 [13] Parigot, M., Λμ-calculus: an algorithmic interpretation of classical natural deduction, (), 190-201, St Petersbourg, Lecture Notes in Computer Science · Zbl 0925.03092 [14] Parigot, M., Séminaire de logique, Université Paris, VII, (1992) [15] Plotkin, G., Call-by-name, call-by-value, and the λ-calculus, Theoret. comp. sci., 1, 125-159, (1975) · Zbl 0325.68006
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Search…
Contractions & DBond

## 1. "When can I swap $DIBS for$DBOND?"

$DBOND will only become available in the bonds section following epochs in which the Time Weighted Average Price (TWAP) of$DIBS is under peg. This means that $DIBS's price will have had to have been under 1$BNB per 1000 DIBS for the majority of the previous epoch in order to trigger the bonds section to "open".
The bonds section will always open at the very beginning of a new epoch, and remain open for the entire epoch — the bonds section can not and will never open mid-epoch — and during epochs in which the bonds section is open, $DIBS will not be printed in the Piggybank. ## 2. "What is the formula to calculate the redemption bonus for$DBOND?"

To encourage redemption of $DBOND for$DIBS when $DIBS's TWAP > 1.1, and in order to incentivize users to redeem at a higher price,$DBOND redemption will be more profitable with a higher $DIBS TWAP value. The$DBOND to $DIBS ratio will be 1:R, where R can be calculated in the formula as shown below: $R=1+[(DIBStwapprice)-1)*coeff)]$ $coeff = 0.7$ To further illustrate why the longer you hold$DBOND the more profitable it is, let's take an initial $1000 investment into consideration. In this example, say this$1000 is used to buy $DIBS when$DIBS TWAP is 0.95 and then swapped for $DBOND. If these$DBOND are redeemed when: -$DIBS TWAP is 1.5, your investment would now be worth$1421. -$DIBS TWAP is 2, your investment would now be worth$1789. -$DIBS TWAP is 3, your investment would now be worth$2526. -$DIBS TWAP is 5, your investment would now be worth$4000.

## 4. "When can I swap $DBOND back to$DIBS?"

You can swap it back again when the following two criteria are met:
1: $DIBS TWAP is above peg and 2. There is enough in the treasury to cover it the redemption. ## 5. "Is$DBOND right for me?"

Like anything else in crypto, obtaining $DBOND is not risk-free. Just like in the real world, you are purchasing debt from the protocol with the expectation that you will be redeemed at a premium in the future. To date, this has occurred after all contractions, but past performance does not guarantee the same future outcomes.$DBOND is ideal for those with a medium to long-term time preference, as it incentivizes hodling in exchange for potentially extremely lucrative rewards. If you are looking for a quick flip or have short-term time preference, \$DBOND may not be the right investment option for you.


text:
# Cmaes¶

The Boulder Opal Toolkits are currently in beta phase of development. Breaking changes may be introduced.

class Cmaes(initial_mean=None, initial_step_size=None, initial_covariance=None, seed=None)

The covariance matrix adaptation evolution strategy (CMA-ES) optimizer.

Parameters
• initial_mean (np.ndarray, optional) – The array of the initial means of the parameters for the multivariate normal distribution. Defaults to an array of ones.

• initial_step_size (float, optional) – The initial step size for the multivariate normal distribution from which new test points are sampled. Defaults to one.

• initial_covariance (np.ndarray, optional) – The initial covariance matrix between the different parameters for the multivariate normal distribution, as a 2D square array. Defaults to the identity matrix.

• seed (int, optional) – Seed for the random number generator. Use this option to generate deterministic results from the optimizer.

Notes

The CMA-ES optimizer uses a multivariate normal distribution to generate new test points. The mean, step size, and covariance of this distribution are determined from their previous values and the current set of results.

Note that this optimizer requires at least $$4 + \lfloor 3 \times \log N \rfloor$$ initial test parameters, where $$N$$ is the number of optimizable parameters. Otherwise, it will throw an error.

For more detail on CMA-ES see CMA-ES on Wikipedia.

Methods

 create_optimizer Return the initialized optimizer.


text:
Notes On Types of Media - CBSE Class 6 Science
Luminous Objects Objects that emit light on their own are called luminous objects. The light emitted by luminous objects enables us to see things around us. Examples of luminous objects are a tubelight, the sun, a lit candle, a glowing bulb, a bonfire and a lit torch. Non-luminous Objects Objects that do not emit light on their own are called non-luminous objects. The light emitted by luminous objects falls on non-luminous objects, and then bounces back to our eyes, which enables us to see non-luminous objects. Examples of non-luminous objects are the moon, a book, a pen, a wooden box, a cupboard and a chair. Light travels in a straight line, and its rays represent the path of light. The material that light passes through is called a medium.  Opaque Objects Objects through which we cannot see anything are called opaque objects. A medium that does not allow light to pass through it is called an opaque medium. Examples of opaque objects are a pencil box, a wooden screen, a book, a towel, a ceramic plate and chart paper. Most objects in our surroundings, like buildings and trees, are opaque objects. Transparent Objects Objects through which we can see clearly are called transparent objects. A medium that allows all the light incident on it to pass through it is called a transparent medium. Examples of transparent objects are a plain glass plate, a clear plastic scale, windowpanes, a soap bubble, and pure water. Translucent Objects Objects through which we cannot see the objects on the other side clearly but can see some light are called translucent objects. A medium that allows only a part of the light incident on it to pass through it is called a translucent medium.  Examples of translucent objects  are ground glass, frosted glass, smoked glass, sun glasses and butter paper.

#### Summary

Luminous Objects Objects that emit light on their own are called luminous objects. The light emitted by luminous objects enables us to see things around us. Examples of luminous objects are a tubelight, the sun, a lit candle, a glowing bulb, a bonfire and a lit torch. Non-luminous Objects Objects that do not emit light on their own are called non-luminous objects. The light emitted by luminous objects falls on non-luminous objects, and then bounces back to our eyes, which enables us to see non-luminous objects. Examples of non-luminous objects are the moon, a book, a pen, a wooden box, a cupboard and a chair. Light travels in a straight line, and its rays represent the path of light. The material that light passes through is called a medium.  Opaque Objects Objects through which we cannot see anything are called opaque objects. A medium that does not allow light to pass through it is called an opaque medium. Examples of opaque objects are a pencil box, a wooden screen, a book, a towel, a ceramic plate and chart paper. Most objects in our surroundings, like buildings and trees, are opaque objects. Transparent Objects Objects through which we can see clearly are called transparent objects. A medium that allows all the light incident on it to pass through it is called a transparent medium. Examples of transparent objects are a plain glass plate, a clear plastic scale, windowpanes, a soap bubble, and pure water. Translucent Objects Objects through which we cannot see the objects on the other side clearly but can see some light are called translucent objects. A medium that allows only a part of the light incident on it to pass through it is called a translucent medium.  Examples of translucent objects  are ground glass, frosted glass, smoked glass, sun glasses and butter paper.

#### Activities

Activity 1 Learnthings.co.za has created a good animation that simulates that transparent and translucent materials let light through, whereas opaque materials do not.  If an opaque material is put in front of a light source, it blocks the light and a shadow forms. A shadow made by the sun will change in length and position throughout the day.   We can explore that the size of the shadow at different times of the day changes by changing the height of the stick by dragging it up and down.  Go to Actitivity http://www.bootslearningstore.com/ks2/sunfun2.html


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# Finding the vector equation of a plane

i_love_science
Homework Statement:
Find a vector equation of the plane that passes through the point (6, 0, 0) and contains the line 𝑥 = 4 - 2𝑡, 𝑦 = 2 + 3𝑡, 𝑧 = 3 + 𝑡.
Relevant Equations:
vector equation
Solution:
u = [-2,3,1]
Po = (6,0,0) & P = (4,2,3)
PoP = v = [-2,2,3]
Therefore, the answer is [6,0,0] + r[-2,3,1] + q[-2,2,3]; r, q are real numbers

I don't understand why (6,0,0) is used as the point in the vector equation, since it only lies on the [-2,2,3] vector, not the u = [-2,3,1] vector.

My answer is r = [4,2,3] + r[2,-2,-3] + q[-2,3,1], r, q are real numbers

Could anyone explain the solution, and is my answer correct? Thanks.

$$[4,2,3]+q[-2,3,1]-r[-2,2,3]=[6,0,0]+q[-2,3,1]-(r-1)[-2,2,3]$$


text:
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# Ross (Michigan) 2013 - Calling All Waitlisted Applicants

Author Message
Intern
Joined: 01 Mar 2013
Posts: 7
Concentration: Operations, Strategy
GPA: 3.6
WE: Project Management (Consulting)
Followers: 0

Kudos [?]: 7 [0], given: 4

Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

17 Mar 2013, 17:00
Waitlisted after interview
Intern
Joined: 05 Feb 2013
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

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17 Mar 2013, 21:49
do we have any number on how many people are on waitlist?

what was the conversion percentage for last year (and may be the year before that)?
Intern
Joined: 06 Nov 2012
Posts: 46
GMAT 1: 640 Q V
GPA: 3.82
WE: Operations (Non-Profit and Government)
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

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18 Mar 2013, 10:04
Waitlisted without interview. Any others in this situation? For those that were invited to interview after being waitlisted w/o interview, how long did that invitation come after the initial WL decision?
Manager
Joined: 17 Nov 2011
Posts: 75
Followers: 1

Kudos [?]: 14 [1] , given: 19

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18 Mar 2013, 11:59
1
KUDOS
PDX105 wrote:
Waitlisted without interview. Any others in this situation? For those that were invited to interview after being waitlisted w/o interview, how long did that invitation come after the initial WL decision?

I was waitlisted in R1 without interview, and was subsequently invited to interview when the first wave of R2 interview invitations went out (around Jan 10th). I know, from this board, quite a few WL applicants were invited to interview in that wave, and in the subsequent waves. You can probably find the dates of these waves on Soojin's blog.
Manager
Joined: 15 Oct 2010
Posts: 69
Followers: 0

Kudos [?]: 9 [0], given: 1

Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

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19 Mar 2013, 03:56
nullvoid325 wrote:
PDX105 wrote:
Waitlisted without interview. Any others in this situation? For those that were invited to interview after being waitlisted w/o interview, how long did that invitation come after the initial WL decision?

I was waitlisted in R1 without interview, and was subsequently invited to interview when the first wave of R2 interview invitations went out (around Jan 10th). I know, from this board, quite a few WL applicants were invited to interview in that wave, and in the subsequent waves. You can probably find the dates of these waves on Soojin's blog.

I am one of those too. Waitlisted in R1, and interviewed in R2. But still find myself on waitlist after round 2.
Current Student
Joined: 10 Jan 2012
Posts: 7
Concentration: Strategy, Healthcare
Schools: Ross '16 (M)
GMAT 1: 730 Q48 V41
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### Show Tags

24 Mar 2013, 20:19
I'm not sure if I have anything for an update to my application. They seem pretty adamant that updates should be substantial, and I'm not sure if I have anything that would meet that threshold. Would an assignment to train a new analyst be substantial enough for an update? I don't have new test scores or anything like that to update. I don't want to look disinterested or stagnant at work, but I also don't want to send in something that would be detrimental to my application. Any suggestions?
Current Student
Joined: 10 Jan 2012
Posts: 7
Concentration: Strategy, Healthcare
Schools: Ross '16 (M)
GMAT 1: 730 Q48 V41
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

26 Mar 2013, 12:18
So it looks like there are about 435 people in the waitlist google group. Is this the total number of people that have accepted a waitlist offer? If so does anyone know how many people in a given year drop off of the waitlist because they accept an offer from a different school or may just not be able to stay on the waitlist for financial aid reasons?
Manager
Joined: 14 Dec 2009
Posts: 57
Location: United States
Concentration: Finance, Accounting
GMAT 1: 760 Q49 V44
GPA: 3.38
WE: Other (Other)
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

27 Mar 2013, 21:47
2
KUDOS
Hi All,

I was admitted with $$to Ross in R1. Just found out I was admitted to Booth today (off the WL), so will be accepting and opening up a spot. Here's hoping it goes to one of you. Cheers, Kcw984 Manager Joined: 04 Feb 2012 Posts: 84 Followers: 0 Kudos [?]: 13 [0], given: 17 Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink] ### Show Tags 29 Mar 2013, 17:16 kcw984 wrote: Hi All, I was admitted with$$$to Ross in R1. Just found out I was admitted to Booth today (off the WL), so will be accepting and opening up a spot. Here's hoping it goes to one of you. Cheers, Kcw984 Congrats and thanks!!!! Any tips on how you got off the WL at Booth? Maybe it can help me with Ross. Thanks. Manager Joined: 14 Dec 2009 Posts: 57 Location: United States Concentration: Finance, Accounting GMAT 1: 760 Q49 V44 GPA: 3.38 WE: Other (Other) Followers: 0 Kudos [?]: 17 [4] , given: 8 Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink] ### Show Tags 29 Mar 2013, 21:04 4 This post received KUDOS Hi eyjose, I can't be sure that Ross is exactly the same as Booth, but I had someone who is currently on the Booth WL pm me earlier asking for advice. Here is a slightly edited (just regarding the Booth-specific references) version of my response: The best way I can sum up my strategy is with the phrase, "the diplomatic squeaky wheel gets the grease". I applied to Booth two years ago and was waitlisted from R1 until the end of August, when I was finally rejected. I spent much of the two years between then and my most recent application attempt improving my profile in every way possible (went from a 730 to 760 gmat, took additional classes and earned As, progressed at work, took on more responsibility in my extracurricular activities, cultivated Booth relationships, etc). When I applied most recently in R1, I was disappointed to learn I had been waitlisted again. However, I was lucky enough to be admitted (with $$) to another top program (Ross), but I had my heart set on Booth. Knowing that this would be my last time applying to MBA programs, I took every measure I could to get in. I reached out to alumni and others connected to Booth, about 20 people in total, taking in-person meetings, phone calls, etc, and was able to secure multiple (five) letters of support, including three from very prominent alumni, including a partner at the firm I work for. I visited campus again (I visited two years ago for my original application) and met with a member of the admissions committee to personally present my story, emphasizing my passion for the program and my very specific reasons for wanting to attend. I sent in two long update letters, submitted a recent professional performance review, and provided 3-4 additional shorter updates that shed light on certain aspects of my profile I wanted to highlight. I also timed everything so that I was essentially in constant contact with the admissions committee, with new material sent every 1-2 weeks. I did my best to ensure that I was on their minds without being an annoyance. Luckily, everything finally paid off. My experience shows that programs looks favorably upon people who really go the extra mile to show how badly they want to be there, and have a deep and nuanced knowledge of the program beyond the usual headlines (i.e. you have very specific reasons for wanting to go, and well thought-out plans for maximizing your time there). It will also help if you can secure alumni or other strong community support (i.e. professors, prominent professionals connected to the program, or even current students). Ultimately programs want to admit students who will definitely attend, and who want to be at the program as their top choice. If you are able to convey that sentiment effectively and keep yourself on the minds of the adcom, you will have a solid shot. I hope this helps. Best of luck, kcw984 Manager Joined: 04 Feb 2012 Posts: 84 Followers: 0 Kudos [?]: 13 [0], given: 17 Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink] ### Show Tags 29 Mar 2013, 22:00 kcw984 wrote: Hi eyjose, I can't be sure that Ross is exactly the same as Booth, but I had someone who is currently on the Booth WL pm me earlier asking for advice. Here is a slightly edited (just regarding the Booth-specific references) version of my response: The best way I can sum up my strategy is with the phrase, "the diplomatic squeaky wheel gets the grease". I applied to Booth two years ago and was waitlisted from R1 until the end of August, when I was finally rejected. I spent much of the two years between then and my most recent application attempt improving my profile in every way possible (went from a 730 to 760 gmat, took additional classes and earned As, progressed at work, took on more responsibility in my extracurricular activities, cultivated Booth relationships, etc). When I applied most recently in R1, I was disappointed to learn I had been waitlisted again. However, I was lucky enough to be admitted (with$$$) to another top program (Ross), but I had my heart set on Booth. Knowing that this would be my last time applying to MBA programs, I took every measure I could to get in. I reached out to alumni and others connected to Booth, about 20 people in total, taking in-person meetings, phone calls, etc, and was able to secure multiple (five) letters of support, including three from very prominent alumni, including a partner at the firm I work for. I visited campus again (I visited two years ago for my original application) and met with a member of the admissions committee to personally present my story, emphasizing my passion for the program and my very specific reasons for wanting to attend. I sent in two long update letters, submitted a recent professional performance review, and provided 3-4 additional shorter updates that shed light on certain aspects of my profile I wanted to highlight. I also timed everything so that I was essentially in constant contact with the admissions committee, with new material sent every 1-2 weeks. I did my best to ensure that I was on their minds without being an annoyance.

Luckily, everything finally paid off. My experience shows that programs looks favorably upon people who really go the extra mile to show how badly they want to be there, and have a deep and nuanced knowledge of the program beyond the usual headlines (i.e. you have very specific reasons for wanting to go, and well thought-out plans for maximizing your time there). It will also help if you can secure alumni or other strong community support (i.e. professors, prominent professionals connected to the program, or even current students). Ultimately programs want to admit students who will definitely attend, and who want to be at the program as their top choice. If you are able to convey that sentiment effectively and keep yourself on the minds of the adcom, you will have a solid shot.

I hope this helps.

Best of luck,
kcw984

Thanks for the advice. I def have not done as much networking. That may be an area of opportunity.
Intern
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

01 Apr 2013, 09:58
Hopefully someone can answer this question, but it appears that the online form has some contradictory information (or rather, omitted information) when compared to the waitlist instructions page. The instructions page seems to imply that they only want you to write an update if you have anything SINGIFICANT to update the committee on, including a promotion, higher test scores, etc. But when you open up the online form it contains the following: "You may also describe what you have done to get to know Ross better since you submitted your application." Does that mean if you made contact with Ross students or alumni, visited campus, etc this would be the place to inform them of that? I'm not sure if this is a good idea because they seem pretty adamant about only hearing about significant updates.
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

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16 Apr 2013, 13:39
Intern
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

08 May 2013, 13:08
Getting the thread restarted... I'm a R2 w/ interview waitlister.

Has anyone heard anything yet? Is it safe to assume that everyone will be hearing on the 15th? Or does the adcom start contacting before then?
Manager
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

09 May 2013, 11:18
Hey Guys. I wanted to see how you all had interpreted this excerpt from the blog entry on 4/15:

"We'll need to see how many deposits are received by May 1 to have a sense of the number of admission offers we can make for Round 3 and waitlisted applicants. Round 3 decisions will be released on May 15. The waitlist will continue to be reviewed throughout the summer as space becomes available in the class."

It makes me think that they will only be notifying round 3 people and not waitlist. Thoughts?
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GMAT 2: 700 Q48 V37
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

### Show Tags

10 May 2013, 12:37
1
KUDOS
watersorbet wrote:
Hey Guys. I wanted to see how you all had interpreted this excerpt from the blog entry on 4/15:

"We'll need to see how many deposits are received by May 1 to have a sense of the number of admission offers we can make for Round 3 and waitlisted applicants. Round 3 decisions will be released on May 15. The waitlist will continue to be reviewed throughout the summer as space becomes available in the class."

It makes me think that they will only be notifying round 3 people and not waitlist. Thoughts?

Hi watersorbet

if you look at the post from 2/26, it says:

"Included in our Round 2 review will be a re-review of Round 1 waitlisted candidates.". If they follow the same pattern, people WL'ed in R2 will be considered among R3, so good luck for you
Intern
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Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants [#permalink]

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02 Jan 2014, 07:03
I'm Wait listed w/o interview as well
Re: Ross (Michigan) 2013 - Calling All Waitlisted Applicants   [#permalink] 02 Jan 2014, 07:03

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text:
## How To Make Archive Tapes of Stokes Polarimeter Data

### Mees to Manoa

Presently, the Haleakala Stokes Polarimeter data is FTP'ed from Mees Solar Observatory to Manoa in the early morning the day after the data was taken, at approximately 4 AM. Data from each day is stored in a separate directory where the name is given by the day number of the year. The magnetograms, calibration files, and preliminary reductions are stored in the directory. For a "normal" day of observing, with three to four magnetograms, the amount of data for a single day is 5-15 Mb of data. This of course strongly depends on the number of active regions present and the occurence of special campaign observations. Even after the data is transferred to Manoa, copies are kept on-line at Mees. These on-line files are retained until the data is sucessfully copied onto exabyte in Manoa. In addition, the Stokes data is also stored on system backup tapes of the computers at Mees. Therefore, there should not be any single point failure allowing data to be lost.

### Archiving the Data in Manoa

Once the amount of data transferred to Mees reaches a critical mass, it can be tarred off to exabyte and stored. This should generally be done when the amount of data reaches 200-300 Mb, or when Tom needs disk space, whichever comes first. It generally takes about two months for this much data to accumulate.

The data is stored on milo on /solarm/FTP/pub/stokes in individual directories as detailed above. The data is archived using the tar command to write all the files/directories into a single compressed file on an exabyte tape. However, milo does not have its own exabyte drive, so the tar file must be transferred over the network to a machine that does have an exabyte. The basic command sequence for tarring the data and transferring it to the exabyte drive on akala is as follows (these commands are explained in the UNIX man pages for the tar command):

akala>102% allocate st0
milo>101% cd /solarm/FTP/pub/stokes
milo>102% tar -cvfb - 20 * | rsh akala dd of=/dev/nrst0 obs=20b

This will take an hour or so. After the tar is complete, always rewind the tape and check it again with tar to make sure it can be read properly using the following commands:
akala>104% mt -f /dev/nrst0 rew
akala>105% tar -tvf /dev/nrst0 > /solar/Stokes/tape_logs/stokes.tar.list.DATE

where DATE is the current date in the DDmmmYY format (e.g. 06apr95).
For the truly paranoid, this check can be done on another exabyte drive on a separate machine in order to guarantee it is generally readable. The output from this tar listing is saved into a new file. This file is kept on-line in /solar/Stokes/tape_logs (e.g. /solar/Stokes/tape_logs/stokes.tar.list.11oct95) in order to allow quick searches of the tape contents, see what data is available, etc.

In order to also assure that all the data in the directories was actually stored on the tape, one can compare the out from the "tar -t" command above with a general directory listing, produced with the following command:

milo>103% ls -lR > /solar/Stokes/tape_logs/stokes.ls.list.DATE

The numbers of files, their names, and their sizes can be compared between the two files to assure that no data was lost. Comparing the entire listing can be quite time consuming, so generally a spot check of several directories is all that is done. The above command will also store the output of the "ls -lR" command in the Stokes tape log directory (e.g. /solar/Stokes/tape_logs/stokes.ls.list.11oct95). This is simply done for the sake of completeness and to allow one to check in the future the actual contents of the directory with what was archived.

When the integrity of the tar file on exabyte has been checked, a listing of all the archived directories is made and sent to Elaine Kiernan (kiernan@koa), using a command such as:

milo>104% ls -l * > ~/stokes.dir.list

This lets Elaine knows that the data has been sucessfully transferred to exabyte and can also be deleted from the machine there.

After, the data has been archived, it can be deleted from /solarm. I actually like to leave the data for the past week or so on /solarm so that there is some recent data there to look at, should the need arise. To delete the data from /solarm, the following command can be used:

milo>105% \rm -r NN*

where NN is the beginning two numbers of the directories you want to delete (e.g. 21*).

The exabyte tape is then labelled and put upstairs in the data archive. The command used to archive the data ("tar -xvf ... | dd ...") and what machine was used should be written on the card in the tape case. In addition, the days of the year included on that tape, and the range of dates that covers, should also be written on the tape.

### Miscellaneous

There is a automatic email message from polsyn@koa every time data is transferred from Mees to Manoa sent to the current Stokes archiver and the observers. I have been keeping a file with all these messages (/solar/Stokes/tape_logs/stokes.data.199N) since they may be useful in the future if there is some discrepancy between the taken and archived data (i.e. where did the link break down?). But, as you can tell, I just like to be overly cautious.

If there is a problem in the transfer, or if /solarm fills up, then polsyn@koa will automatically send a message detailing any discrepencies between the directories for the data on koa and the directories on milo. If you clear space on /solarm, the missing files will automatically be re-transferred the following night.

In the past, a listing of the data that was archived off to tape was stored in a master file that showed the contents of each tape. This was useful and appropriate when the data was stored on nine-track tape (which only held two or three days) and the notification and backup procedures were more complex. However, the files in /solarm/Stokes/tape_logs (described above) take over most of this functionality. I mostly kept the list up to date, but I don't think it is necessary anymore. The file, which lists tapes back to May, 1991, is in /solar/daily/Mees/tape_logs/stokes.tapes.logs.

And that's all there is to it. One thing to watch out for is that new data may come in between the time you made the tar archive and when you go to delete the archived data. This could happen if you start the tar job in the evening, go home, and then delete the archived files the next day. In this case, be careful not to delete files that haven't been archived yet.


text:
2013
12-21

# Bob and Alice’s Meet

Bob and Alice are quite closely friends. They are both intersted on Researching ciper. As everybody knows , they ,of course, are experienced computer users. So they always communicate their new reaserches on ciyper by Internet. But now, unfortunetly, their country is suffering from a cruel war caused by nucleus. However the enemy state is so powerful much more than their motherland. So after several months Bob and Alice’s country is controled by enemy on many areas, including network. Any communications on the network are highly suspiciously. If anybody uncareful,he may lost his life! This situation is so dangerious to computer users. So Bob’s mother and Alice’s father make a decision together for their childern — forbid them to use computer again. Bob and Alice can’t disobey this dicesion ,but they also don’t want to pause their new research. So they also make a decison together — to meet at somewhere personally ,and , of coure,don’t let their parents konw.

Now , Bob and Alice fasten the last chance on network communication to confirm the place they will meet at in the next days. The city they lived , has many roads and hotels. Of course, they can’t talk just on the roads, and they also can’t enter any buildings except hotels. So they decide to meet in a certain hotel in their city. As they know,the city’s building pattern is very regular like a rectangle with many griddings in it. And all hotels and houses are built in lines and columns, so are roads. So they can only walk straightly by lines or columns. While it is a mountainly city, every buliding may has a different height above sea level.If you walk form a hotel with a height x at an initial speed v0 to a hotel with a height y, then the speed on this road is surposed to be v0 × 2(x-y) . We already know that Bob lives in the city’s left-up corner,and Alice lives in the right-down corner. The buliding in the grid among their houses are all hotels. and the length of the road between two neighbor( four directions connected) hotel is alway a same value L, and they set up at their own initial speed at the same time.

Now , they want to pick a path which will take them a fastest time to meet each other. Pitfully for them ,can you help them to commpute the minmum time to meet ? You should always remember that ,they only meet at a certain hotel ,so maybe one of them will reach a hotel and don’t walk again to wait the other one to arrive.

There will be more than a test case. every case data goes like below:

Line 1 : Four space-separated integers: N, Vb, Va, L.
N is the size of the rectangle of Bob and Alice’s city. Va is Alice’s initial speed, while Vb is Bob’s initial speed. L is the length between two neighbor hotels .

Line 2 ~ Line N + 1 : every line has N space-separated integers,each one represents a hotel’s height.

N = 0 indicates the end of input and should not be processed.

For each case , ouput an interger in a separated line which is the time they take to meet each other.The answer must round to three decimal places.

3 1 1 1
1 2 3
3 2 1
1 2 1
0 0 0 0

4.000

1. /*
* =====================================================================================
*
* Filename: 1366.cc
*
* Description:
*
* Version: 1.0
* Created: 2014年01月06日 14时52分14秒
* Revision: none
* Compiler: gcc
*
* Author: Wenxian Ni (Hello World~), [email protected]
* Organization: AMS/ICT
*
* =====================================================================================
*/

#include
#include

using namespace std;

int main()
{
stack st;
int n,i,j;
int test;
int a[100001];
int b[100001];
while(cin>>n)
{
for(i=1;i>a[i];
for(i=1;i>b[i];
//st.clear();
while(!st.empty())
st.pop();
i = 1;
j = 1;

while(in)
break;
}
while(!st.empty()&&st.top()==b[j])
{
st.pop();
j++;
}
}
if(st.empty())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}

2. 代码是给出了，但是解析的也太不清晰了吧！如 13 abejkcfghid jkebfghicda
第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)，为什么要这样拆分，原则是什么？


text:
# Finite normal series operator

This article defines a group property modifier (a unary group property operator) -- viz an operator that takes as input a group property and outputs a group property

## Definition

### Property-theoretic definition

The finite normal series operator is a group property modifier that takes as input a group property $p$ and outputs the Kleene star closure of $p$ with respect to the group extension operator, right-bracketed.

### Definition with symbols

Given a group property $p$, the finite normal series operator applied to $p$ returns the group property $q$ defined as follows: a group $G$ has property $q$ if there is a finite normal series $e = H_0 \le H_1 \le \ldots \le H_r = G$ such that each $H_i$ is normal in $G$ and such that $H_i/H_{i-1}$ satisfies property $p$ for every $i$.

## Metaproperties

### Monotonicity

This group property modifier is monotone, viz if $p \le q$ are group properties and $f$ is the operator, then $f(p) \le f(q)$

If $p \le q$ are group properties, then the image of the finite normal series operator on $p$ implies the image of the finite normal series operator on $q$.

### Ascendance

This group property modifier is ascendant, viz the image of any group property under this modifier is always weaker than the group property we started with

For any group property $p$, any group satisfying $p$ also satisfies the image of $p$ under the finite normal series operator.

### Idempotence

The finite normal series operator need not be idempotent.


text:
# falzoon's folly ....

Number Theory Level 5

Consider the set $$S$$ of all (non-degenerate) triangles that meet the following two conditions:

(i) the side lengths are consecutive positive integers;

(ii) the area is integer valued.

What is the area of the third-smallest triangle in $$S$$?

(Credit goes to the mysterious falzoon for this question.)

×

Problem Loading...

Note Loading...

Set Loading...


text:
# Thread: how many orientations does a simple graph g have?

1. ## how many orientations does a simple graph g have?

hi

I'm stuck on the followinq question

How many orientations does a simple graph g have?

for one vertex there is one, just the isolated vertex

for two vertices there are 3, i think...two isolated vertices u and v, a digraph with arc uv and a digraph with arc vu.

i'm not sure where to go from here

2. ## Check definition of "simple" graph

Your description of your calculation already lets you count how many directed graphs there are. Between each pair of points, there seem to be 3 possibilities - no edge, an edge directed one way, an edge directed the other. You can count how many pairs of points there are - n choose 2 - points. So you've counted
$\displaystyle 3 ^ { \binom{n}{2} }$
graphs.

However there is another definition of simple graph that doesn't allow for directed vertices. If so, your calculation overcounts because there's only 2 possibilities - no edge, or yes, there's an edge.

3. hi, i just want to check i understand

i'm working from the definition of a simple graph which states that g contains no loops or multiple edges, so i don't think i need to worry about overcounting

i understand now that i need to divide into the three cases for each pair of vertices

n number of orientations
1 1
2 3
3 9
4 24
....

are these values correct?

4. ## Definition of simple graph

I'm still not sure if you're working with simple graphs or simple directed graphs. The definition I'm getting from the web for simple graph is: "unweighted, undirected" graph. Please note the "undirected"!

The formula is not being interpreted correctly. The
$\displaystyle \binom {n}{m} = \frac {n!}{m!(n-m)!}$

so the first few values are:
$\displaystyle \binom {2}{2} = 1, \implies 3^{\binom{2}{2}} = 3$
$\displaystyle \binom {3}{2} = 3, \implies 3^{\binom{3}{2}} = 27$
$\displaystyle \binom {4}{2} = 6, \implies 3^{\binom{4}{2}} = 729$

5. I need the graphs to be directed since i'm looking for the number of orientations.

i think that if i have 3 vertices and and a directed edge (arc) between 1 pair i count this once not 6 times for alternate direction and the choice of the two other pairs.

this means that for n=1,2,3,4,5,.... i get 1,2,7,42,582

(this sequence is from wolfram mathworld)

im trying to find the formula for this sequence, does anyone know it?

thank you

,

,

,

,

,

,

# howmany orientation does a simple graph G have

Click on a term to search for related topics.


text:
# Biggest Reuleaux Triangle inscribed within a square which is inscribed within a hexagon?

Here we will see the area of biggest Reuleax triangle inscribed within a square which is inscribed in a regular hexagon. Suppose ‘a’ is the side of the hexagon. The side of the square is x and the height of the Reuleaux triangle is h.

From the formula of each side of inscribed square inside one hexagon is −

𝑥 = 1.268𝑎

The height of the Reuleaux triangle is the same as x. So x = h. So the area of the Reuleaux triangle is −

## Example

#include <iostream>
#include <cmath>
using namespace std;
float areaReuleaux(float a) { //side of hexagon is a
if (a < 0) //if a is negative it is invalid
return -1;
float area = ((3.1415 - sqrt(3)) * (1.268*a) * (1.268*a))/2;
return area;
}
int main() {
float side = 5;
cout << "Area of Reuleaux Triangle: " << areaReuleaux(side);
}

## Output

Area of Reuleaux Triangle: 28.3268
Published on 01-Aug-2019 08:08:16
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text:
# Newton's third law and General relativity

Is Newton's third law valid at the General Relativity?

Newton's second law, the force exerted by body 2 on body 1 is: $$F_{12}$$ The force exerted by body 1 on body 2 is: $$F_{21}$$

According to Newton's third law, the force that body 2 exerts on body 1 is equal and opposite to the force that body 1 exerts on body 2: $$F_{12}=-F_{21}$$

-
Why are you asking this(curious)? –  namehere Nov 29 '12 at 13:21
@namehere During the calculation of the properties of black holes –  Neo Nov 29 '12 at 20:42
Which black hole metric are you using? note that Schwarschild, Kerr, and Newmann metrics are stationary(Schwarschild is also static), so they don't represent physical black holes. I'm also not very sure how Newton's third law affects black holes. Note that objects only under the influence of gravitation in General Relativity are in free fall, so that their 'motion' cannot be described by forces, and instead you should look at the geodesic equation. Also, with the metric being generally different at general points, any apparent 'force' that is exerted should NOT obey Newton's Third Law. –  namehere Nov 30 '12 at 0:27
@namehere: the Kerr model is closer to a physical black hole than almost any exact model is to any physical body. –  Jerry Schirmer Sep 18 '13 at 16:43

First, let's note that newton's third law is really equivalent to conservation of momentum, by example of object one exerting a force on object two, and vice versa, and these two forces being the only forces in the universe:

\begin{align} F_{12} &= -F_{21}\\ m_{2}a_{2} &= -m_{1}a_{1}\\ \int m_{2}a_{2} dt &= -\int m_{1}a_{1} dt\\ m_{2}v_{2f}-m_{2}v_{2i} &= m_{1}v_{1i}-m_{1}v_{1f}\\ m_{1}v_{1f} + m_{2}v_{2f} &= m_{1}v_{1i} + m_{2}v_{2i}\\ \sum p_{f} &= \sum p_{i} \end{align}

Now, we know that we are looking for conservation of momentum, rather than just Newton's third law (and conservation of momentum is a more general concept anyway--Newton's third law will come up false in a variety of electromagnetic applications, but conservation of momentum will still be true). How do we get conservation of momentum? Well, the motion of a particle can be found by looking for the minium of something known as the Lagrangian:

$$L = KE - PE$$

It turns out that there is a result called Noether's theorem that says that if the Lagrangian is doesn't change when you modify your variables in a certain way, then the dynamics defined by that Lagrangian will necessarily have a conserved quantity associated with that transformation. It turns out that conservation of momentum arises when the invariance is a translation of the coordinates: $x^{a^\prime} = x^{a} + \delta^{a}$. Now, let's go back to general relativity. Here, the motion of a particle is the one that maximizes the length of:

$$\int ds^{2} =\int g_{ab}{\dot x^{a}}{\dot x^{b}}$$

If the metric tensor $g_{ab}$ has a translation invariance, this motion will necessarily have a conserved momentum associated with it, and will not otherwise. Note: common solutions, like the Schwarzschild solution of GR are NOT translation invariant--that's because the model assumes that the central black hole does not move. A more general solution that included the back-reaction of the test particle's motion WOULD have a conserved momentum (and would end with a moving black hole after some orbiting was completed).

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"Newton's third law will come up false in a variety of electromagnetic applications" i cant immediately recall any effects\phenomena of this sort(unless you count quantum stuff). enlighten me! –  namehere Nov 29 '12 at 15:51
@namehere: the simplest case of this is if you have two charged particles of different masses exerting a force on each other, they will radiate asymmetrically, and this will produce different radiation back-reaction forces, so you will observe that the force that the first particle exerts on the second particle will be different than vice versa. Looked at using momentum, you will see that the radiation carries momeuntum, and overall momentum is conserved. –  Jerry Schirmer Nov 29 '12 at 16:42
ohhh, thankyou, I couldnt recall that. –  namehere Nov 30 '12 at 0:21
Yes, this is an elegant proof. +1 (a short while ago.). –  Dimensio1n0 Sep 18 '13 at 16:20

Newton's Third Law partially holds in General Relativity.

In General Relativity, spacetime is "curved", and momentum(and hence force) at a point cannot be directly meaningfully corresponded to momentum(and force) at a different point in spacetime. To correspond quantities from different points in spacetime, we need something called "parallel transport" which maps the momentum(or force) space at one point to the momentum(or force) space at another point, given a path. As you can see, the resulting momentum(and force) at the destination is dependent on the path chosen and the particular geometry of the particular spacetime. So forces at a distance cannot really obey things like Newton's Third Law.

However, locally(aka at one spacetime point), Newton's Third Law (sort of) does hold. Basically Newton's Third Law is equivalent to the conservation of momentum. In General Relativiy, this holds in the form of $$\nabla_\beta T^{\alpha\beta} = 0$$ where $T^{\alpha\beta}$ is the energy-momentum tensor and $\nabla_\beta$ denotes the covariant derivative. The energy-momentum tensor which you guessed it, represents the energy and momentum plus some other components irrelevant here at a spacetime point, and covariant derivative is basically a derivative modified slightly say that it works in curved spacetime. So the equation says the derivative of momentum is zero and this can be interpreted as equal to momentum being conserved, which is equal to Newton's Third Law.

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+1 your answer is good too –  Neo Nov 29 '12 at 20:43
Yes this answer is also good because of the mention of the sem tensor. I'll +1 this tomorrow, when my daily vote limit is reset. –  Dimensio1n0 Sep 18 '13 at 16:20


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Electrical signal power and energy calculations by example

In electronics and signal processing, you have to deal with electrical signals. In many cases, you may need to calculate signal power and energy.

Power and energy for DC

In a standard situation, when DC power supply is applied to a known resistor or another device like an LED, motor you can calculate its power very easy by applying Ohms law:

$P = \frac{U^{2}}{R}$

If we rung this device for time T then we can calculate total energy used:

$E = T\cdot \frac{U^{2}}{R}$

In some cases, you may not know the resistance of your circuit. In this case, you can measure the current flow. So your power formula can be transformed by using same Ohms law:

$P = U\cdot I$

So for instance if you connect your circuit to the 3V power supply and measure that it uses 0.5A of current, you can calculate the instant power usage:

$P = 3\cdot 0.5 = 1.5W$

And the energy used during one hour:

$E = 1\cdot 3\cdot 0.5 = 1.5Wh$

Variable signal power and energy applied to load

Things are getting a bit more complicated when signal current isn’t continuous DC. In signal processing applications, the term signal power and signal energy are introduced. The idea is practically the same as in DC. But since current (or voltage) change in time, it is written as v(t) and i(t). for instance, for periodic harmonic signals, we can write

$v(t) = V\cdot cos(\omega \cdot t)$.

To calculate signal power, we can apply Ohms formula from above.

This way we can calculate instantaneous power of signal at any time: $p(t) = \frac{v(t)^{2}}{R}$.

Lets say our voltage is periodical $v(t) = 220cos(2 \pi f t)$ where f = 50Hz. And we apply this voltage across 100Ω resistor. At 20 seconds time we can calculate momentary power usage:

$p(t) = \frac{v(t)^{2}}{R}=\frac{(220cos(10000 \pi ))^{2}}{100} = 484W$

If we want to calculate energy loss during 100s wee need to integrate the power in period from 0 to T:

$E = \int_{0}^{T} p(t)dt = \frac{1}{R}\int_{0}^{T}v^2(t)dt$

I know – everyone hates integrals, but this is how math works. In many cases you don’t need to integrate manually – just place formulas into one of the packages like WolframAlpha and you will get the answer immediately.

Lets apply energy formula to our signal:

$E = \frac{1}{100}\int_{0}^{100}(220cos(100 \pi t ))^{2}(t)dt = \frac{121(200\pi))}{5\pi}=672.22Wh$

!dont apply 220V to 100Ω resistor.

When we know the energy consumed during time T we can calculate average power over period of time:

$P_{avg} = \frac{E}{T} = \frac{1}{RT}\int_{0}^{T}v^2(t)dt$

and with our previous example we calculate average power:

$P_{avg} = \frac{672.22Wh}{0.03h} = 22407W = 22.4kW$

!dont apply 220V to 100Ω resistor.

Hope you get the idea what’s going on here.

Signal power and energy

In signal processing for it is agreed to eliminate the resistance parameter from equations by making it equal to 1 (R = 1). Also, instead of v(t) usually, signals are expressed as s(t). Thus we can express signal energy as follows:

$E = \int_{0}^{T}s^2(t)dt$

signal power:

$p(t) = s^2(t)$

and average power over the time period:

$P_{avg} = \frac{E}{T} = \frac{1}{T}\int_{0}^{T}s^2(t)dt$

Finite and infinite signal energy

There is one more important matter about signals. They can be the finite length and infinite. If we need to find the power of an infinite signal, we need to find its average power overall time:

$P_{avg} = \lim_{T\rightarrow \infty }\frac{1}{T}\int_{-T/2}^{T/2}s^2(t)dt$

Power of infinite signal can be calculated as follows:

$E = \int_{-\infty}^{\infty}s^2(t)dt$

For some signals, power and energy can be found by intuition. For instance, let us take a step signal:

We can see that the energy of the signal grows to infinity when time also goes to infinity. Average power in other hand is 0.5W.

In another case we have a sinusoidal signal:

If we applied the formulas above we would get signal energy = ∞ and average power = 0.5W.

And lastly very interesting case – sinc signal:

$s(t) = sinc(t) = \frac{sin(t)}{t}$

signal energy = π and average power = 0W.

The RMS

If we take square root of average power then we get root mean square (RMS) of the signal:

$RMS = \sigma_{s} = \sqrt{P_{avg}} = \sqrt{\lim_{T\rightarrow \infty }\frac{1}{T}\int_{-T/2}^{T/2}s^2(t)dt}$

For instance if we get sine wave average power equal to 0.5W then RMS:

$RMS = \sigma_{s} = \sqrt{0.5} = 0.707W$

What does RMS mean in practice? RMS is equivalent to DC voltage to get the same heating effect. It is easy to calculate the RMS voltage of the sine wave. You need to multiply peak voltage by 0.707, and you get RMS value.

If you have sine signal amplitude 220V then its RMS = 220·0.707 = 155.54V.


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> AP Calculus Exam Prep | CK-12 Foundation
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# 9.4: AP Calculus Exam Prep

Created by: CK-12

This activity is designed for AP Calculus students as a review for the AB Calculus Exam at the end of the year.

## Part 1 – Calculator Active Exam Tips

The AP* Calculus exam is long. It is a marathon of the mind. Success requires training.

Tip #1 – Know the course

There is a free-response (FR) section and a multiple-choice (MC) section. For both there is a ‘calculator NOT allowed’ section and a ‘graphing calculator is required’ section.

On the two FR sections you have a total 1 hour and 30 minutes to do 6 questions, 3 with your TI-89 handheld and 3 without. In Section 1 Part A, you will have 55 minutes to complete 28 MC questions where a calculator is NOT allowed. Section 1 Part B requires a calculator to complete 17 MC questions in 50 minutes.

1. About how much time should you spend on each question in each part of the exam?

Tip #2 – Know the expectations

On the calculator active questions you should:

• Give answers to 3 decimal places (unless it says otherwise, e.g., give answer to the nearest cent).
• Be able to do the following four things:
• graph on an arbitrary window
• solve equations (find zeros and intersection points)
• numerically calculate the derivative
• numerically calculate the definite integral

2. The position function of a particle is $s(t) = 3t + 6.5\sin(2t)$. What is $v(2)$?

On the HOME screen, enter $d(3t+6.5\sin(2t),t)|t=2$. You can access the derivative command by selecting F3:Calc > 1:d ( differentiate, or by using the shortcut $2^{nd}\ [d]$. The “such that” bar after the derivative command is input by pressing the $[ \ | \ ]$ key.

Notice the mode settings at the bottom of your screen. It should say RAD indicating that you are in radian mode. To change the angle mode, go to MODE > Angle. You will only need to be in radian mode for the entire AP exam.

Using AUTO mode under MODE > Exact/Approx is also nice. You get an exact answer if possible, or if there is a decimal in what you enter, you get a decimal answer. Pressing $[\blacklozenge]$ [ENTER] $(\approx)$ will give an approximation.

$^*$AP, College Board, and SAT are registered trademarks of the College Board, which was not involved in the production of and does not endorse this product.

Tip #3 – Have the right gear

Besides the right settings, bring a couple of pencils. You can actually bring $2$ graphing calculators. Be sure they have reasonably fresh batteries.

3. $h(x) = \int\limits_0^x \left (t \cos(t^3)-0.7 \right ) dt$. Where is $h(x)$ increasing on the interval $-1.8 < x < 1.4$?

Find the solution graphically. Press $[\blacklozenge][Y=]$ and enter the function in $y1(x)$. The integration command can be entered by pressing $2 \ + <$. Change the window to view only the domain by pressing $[2^{nd}]\ [\int\limits]$ and entering XMin: -1.8 and XMax: 1.4. Graph the derivative. Why? What are you looking for?

Find the zeros by selecting F5:Math > 2:Zero. You can trace to input your lower and upper limits to find each root or enter your lower and upper limits on the keypad.

Try solving for the zeros algebraically. Press [HOME] to return to the HOME screen. Then press F2:Algebra > Solve. If you have already entered the function in $y1(x)$ and the derivative of the function in $y2(x)$, enter solve$(y2(x)=0,x)|-1.8.

The ‘such that’ symbol needs to be outside of the parentheses.

4. Find the area of the region bounded by $y = \frac{1}{x}, y = 3\ln(x)$, and the vertical line $x = 3$.

One way to find the intersection point is to graph the functions using a suitable window and then selecting F5:Math > 5:Intersection. Use the ENTER key and the arrow keys to select the two curves and input a lower and upper bound. To store a value, like the $x-$coordinate of the point of intersection, press the [STO$\rightarrow$] key. This will store the $x-$coordinate of the point of intersection as $xc$. Then, use $xc$ in the limits of integration. By doing this, you won’t make a rounding error and you won’t make a mistake in typing the numbers. Next, return to the HOME screen and solve the integral.

## Part 2 – AB Exam Practice Non-Calculator Type Questions

Answer the following AP-type exam questions. These are similar to non-calculator questions you will encounter on the exam. You should spend on average less than 2 minutes for each question. Don’t spend too much time on any one question. On the actual exam, you will have plenty of space so extra paper will not be needed. However, for this practice you may need additional space.

1. If $y = (4x^2 + 5)^3$, then $\frac{dy}{dx} =$

(A) $12(8x)^2$

(B) $4x(4x^2 + 5)^3$

(C) $8x(4x^2 + 5)^2$

(D) $24x(4x^2 + 5)^2$

(E) $12x(4x^2 + 5)^2$

2. $\int\limits^3_0 \left (\frac{1}{3} e^{- \frac{x}{3}} \right )dx$

(A) $\frac{1}{9} \left (1-e^{-1} \right )$

(B) $e^{-1} - 1$

(C) $e^{-1}$

(D) $\frac{1}{3} \left (1-e^{-1} \right )$

(E) $1 - e^{-1}$

3. If $\tan(x) = e^y$, what is $\frac{dy}{dx}$ ?

(A) $\frac{\sec(x)}{\tan(x)}$

(B) $\frac{1}{\sin(x) \cos(x)}$

(C) $\frac{\cos(x)}{\sin(x)}$

(D) $\ln(\tan(x))$

(E) $\tan(x)\sec^2(x)$

4. If $y = \frac{5x+7}{2x+3}$, then $\frac{dy}{dx} =$

(A) $\frac{-10x-14}{(2x+3)^2}$

(B) $\frac{29}{(2x+3)^2}$

(C) $\frac{1}{(2x+3)^2}$

(D) $\frac{22}{(2x+3)^2}$

(E) $\frac{5}{2}$

5. $\int\limits_0^{\frac{\pi}{6}} (2 \cos x)dx$

(A) $\sqrt{3}$

(B) $\sqrt{3} - 2$

(C) -1.5

(D) 1

(E) $\frac{\sqrt{3}}{2}$

6. $\lim_{x \to \infty} \left (\frac{4x^2 - 5x-3}{3x^2+2x+4} \right )$

(A) $- \frac{4}{9}$

(B) $- \frac{3}{4}$

(C) 1

(D) $\frac{4}{3}$

(E) $\infty$

7. If $f(x) = 1 + 3g(x)$ when $2 \le x \le 7$, find $\int\limits_{2}^{7}(f(x)-g(x))dx$.

(A) $4 \int\limits_{2}^{7} (g(x))dx$

(B) $x+2 \int\limits_{2}^{7} (g(x)) dx$

(c) $5+2 \int\limits_{2}^{7}(g(x))dx$

(D) $- \int\limits_{2}^{7}(1+3g(x))dx$

(E) $-10-2 \int\limits_{2}^{7}(g(x))dx$

8. If $y = \sqrt{2x} \cdot \tan(3x)$, then $\frac{dy}{dx} =$

(A) $\frac{1}{2}(2x)^{- \frac{1}{2}} \sec^2 (3x)$

(B) $3(2x)^{- \frac{1}{2}} \sec^2 (3x)$

(C) $\sqrt{2x} \sec^2 (3x)+ \frac{1}{2}(2x)^{- \frac{1}{2}} \tan (3x)$

(D) $3 \sqrt{2x} \sec^2 (3x) + (2x)^{- \frac{1}{2}} \tan (3x)$

(E) $\sqrt{2x} \sec^2 (3x)+ (2x)^{- \frac{3}{2}} \tan (3x)$

9. Find the equation of the tangent line to $y = \sin(2x)$ at the point $(\pi, 0)$.

(A) $y = 2x - 2\pi$

(B) $y = -2x + 2\pi$

(C) $y = 2x$

(D) $y = x - \pi$

(E) $y = x - 2\pi$

## Date Created:

Feb 23, 2012

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# Mechanical index

(Redirected from Mechanical Index)

Mechanical index (MI) is an ultrasound metric. It is defined as [1]

$\text{MI} = \frac{\text{PNP}}{\sqrt{F_c}},$

where

• PNP is the peak negative pressure of the ultrasound wave (MPa), derated by 0.3 dB cm-1 MHz-1 to account for the difference between in-water and in-tissue acoustic attenuation, and
• Fc is the center frequency of the ultrasound wave (MHz).

Mechanical index can be used as an estimate for the degree of bio-effects a given set of ultrasound parameters will induce. A higher mechanical index means a larger bio-effect. Currently the FDA stipulates that diagnostic ultrasound scanners cannot exceed a mechanical index of 1.9.[2]

## References

1. ^ Thomas Szabo (2004). Diagnostic Ultrasound Imaging: Inside Out (Biomedical Engineering). ISBN 978-0-12-680145-3
2. ^ http://www.accessdata.fda.gov/cdrh_docs/pdf9/K091970.pdf (accessed July 2010)