text:
# Viewing Inside Faces of a Cube

1. Sep 23, 2011

Hi,

I have no knowledge of Calculus. I am actually studying Computer Science. One of the exercises in the textbook I am using ("Python Programming" by John Zelle) suggests that the answer to the problem can be solved either through a simple simulation program or through calculus. I am still struggling with this, but even if I do find a solution, t I have no way of checking my answer, so I was wondering what solutions using calculus might bring.

The problem (Chapter 9, Exercise 15) is as follows:

You are at the exact center of a cube. You can see all six faces of the cube equally, so you can say that each face of the cube occupies 1/6 of your vision.

Say you move exactly half of the distance from the center of the cube to the center point of one of its faces.

What fraction (or percentage) of your vision does the closest face now occupy?

I would love to know what solution using calculus produces - it would give me a point of reference from which to work through the simulation.

I am having trouble even doing this as a "simple" simulation, because my linear algebra is not up to snuff. Granted it's important for me to get it up to snuff, I am not studying it right now, and this book did not say that knowing it was a requirement.

Thanks, everyone.

2. Sep 24, 2011

### JeffKoch

This is not a calculus problem, it's a simple trigonometry problem. At the center, you know the angles to the edges of a side, so if you move halfway to the face you know the new angles and therefore the new solid angle.

3. Sep 25, 2011

### awkward

It is possible that you may do some calculus, because you have to find the surface area of a region on a sphere-- that's how the solid angle is defined.

4. Dec 4, 2011

Hi everyone,

It's been a while, but I'm still interested in finding an answer to the above problem.

Since I wrote the original post, I have figured out a CS method of getting to an approximate answer by producing random results and calculating a rough average(this was the objective of the exercise), but I still have no certain mathematical answer.

I came up with 7/20 or 35% (remember, this is an approximate answer). I am of neutral confidence on this: it could be correct, or it could be wrong. I would like to know for sure, if someone can solve this using other means.

For those wondering, my CS method of solving this problem was this: using a premade library of graphical objects, first I created a square object with side length L, then I placed a point object in the square L/2 from the top and bottom faces, L/4 from one of the horizontal faces.

Next I created a loop, which generates a line emanating from the point to the edge of the field using a random angle. To check for vertical tilt, the loop then generates a line at a random angle from the point, but only where the x of the end of the line is less than the x of the original point. The loop counts how many times both of the lines succeed in intersecting the face of the square closest to the point. (So, in effect, I'm reusing the same objects I used to find the line's place in two dimensions to find its place in the third).

Running hundreds of thousands of simulations got me an average of 35 hits out of 100.

I understand this might not be interesting to many (or anyone), but I thought I'd put it out there.
Thanks.

5. Dec 7, 2011

### awkward

I calculate that 0.2952 of your vision should be covered by the nearest face when youare half the distance of the cube from the center to the face.

This is based on the formula for the solid angle of a four-sided right rectangular pyramid found here: http://en.wikipedia.org/wiki/Solid_angle#Pyramid
You can find a derivation of the formula here: http://www.slac.stanford.edu/~bgerke/notes/solid_angle.pdf
Caveat: the writer of the derivation reverses the roles of theta and phi in spherical coordinates from the usage I am familiar with.

The apex angle of the pyramid is 2 * arctan(2) = 126.9 degrees, approximately, yielding a solid angle of 3.709 steradians using the Wikipedia formula. Dividing by 4 pi steradians, we find that is 0.2952 of the surface area of a sphere.

If your Monte Carlo approach does not agree with my calculated answer, one possibility is that you are not generating points uniformly distributed on the surface of the sphere. This is a bit tricky. It is incorrect to simply generate the inclination (or polar angle) uniformly distributed in the range from 0 to 180 degrees (0 to pi radians). This link explains why: http://mathworld.wolfram.com/SpherePointPicking.html

6. Dec 7, 2011

Thanks for the excellent exposition, awkward.

I tried plugging in a new $\phi$ function as your link provided, and it did not seem to affect very much. I noticed a problem with the way my lines were being generated. When I fixed that problem, the average went down to 15%. Don't know what's going on there. Other problems, no doubt - certainly due to more kinks in the program.

Thanks for all the insight and for providing a goal for me to shoot towards. Hopefully this thread will be helpful for other people using this textbook.

7. Dec 8, 2011

### Sina

The solid angle can be understood as follows. Say you are at the center of a point. First of all note that the solid angle of any sphere around you with any radius is 4*pi. Moreover if any surface around you completely encloses you it is again 4*pi. This value can be calculated from the formula given for solid angle calculating. It is infinitesimally proportional to Area/r^2. In the case of a sphere this is global and you get for instance (4pi r^2)/r^2 = 4*pi but in the case of less symmetric surfaces you have to calculate it by integration. So if a for instance you are at the center of a half sphere then the solid angle of the half sphere will be 2*pi and just as expected you will have 2*pi/4*pi = 1/2 of your "vision" blocked by the sphere.

Now coming back to the cube, the geometric picture of the solid angle is as follows: from the point you are standing draws rays to the corners of closest face of the cube. This is the outline of a pyramid (assume that it is a full pyramid with interior points included) as explained above. Now draw a unit sphere around you so (that r=1) and look at the intersection between the sphere and the full pyramid you have. This will give you a surface on the sphere. Then (area of that surface) / 4*pi*r^2 (but r=1) is how much of your vision that surface or equivalently, the side of the cube covers. So infact this outlines another numeric method for you to calculate what you want.

Analytically you do as above and find the solid angle occupied by the pyramid descirbed above. This requires trigonometry ofcourse to find the angles and side lengths etc then using the formula given above calculate the solid angle and divide it by 4*pi.


text:
2x^2+3x-3=0

x=(-3 plus/minus $\sqrt {33}$)/4

x=-2.185 and x=0.685

Is this correct? Is there a better way to solve it?

Thank you

2. Originally Posted by oceanmd
2x^2+3x-3=0

x=(-3 plus/minus $\sqrt {33}$)/4

x=-2.185 and x=0.685

Is this correct? Is there a better way to solve it?

Thank you
it is x = -2.186 and x = 0.686

3. ## Slope of Secant Line

How to find the slope of the secant line for f(x) =2x^2-3x. Please solve it, as I have no idea how to do it.

Thank you

4. Originally Posted by oceanmd
How to find the slope of the secant line for f(x) =2x^2-3x. Please solve it, as I have no idea how to do it.

Thank you
we need an interval here. the slope of the secant line between which points?

5. I am doing the chapter review: a printout given by the teacher. It does not say between what points. Can you come up with the points and explain to me how to solve it.

Thank you

6. Originally Posted by oceanmd
I am doing the chapter review: a printout given by the teacher. It does not say between what points. Can you come up with the points and explain to me how to solve it.

Thank you
ok, the slope of the secant line of a function $f(x)$ between the points $x = a$ and $x = b$ is given by:

$\frac {f(b) - f(a)}{b - a}$

(this is just the formula for the slope of the straight line connecting the two points. note its similarity to $\frac {y_2 - y_1}{x_2 - x_1}$)

7. Ok, I will try
The two points A(0.75, -9/8) which is the vertex of the parabola
B (1.5,0), which is the x-intercept.
The slope of the line is 3/2

Is this correct?

8. Originally Posted by oceanmd
Ok, I will try
The two points A(0.75, -9/8) which is the vertex of the parabola
B (1.5,0), which is the x-intercept.
The slope of the line is 3/2

Is this correct?
yes

format: plain


text:
Assuming max(nA, nB) %% min(nA, nB) == 0: compute the distribution of differences of weighted sums between max(1, nB/nA)*sum(a) and max(1, nA/nB)*sum(b) where a is a 0/1 vector of length nA with each item 1 with independent probability (kA+kB)/(nA+nB), and b is a 0/1 vector of length nB with each item 1 with independent probability (kA+kB)/(nA+nB). Then return the significance of a direct two-sided test that the absolute value of this difference is at least as large as the test_rate_difference (if supplied) or the empirically observed rate difference abs(nB*kA - nA*kB)/(nA*nB). The idea is: under this scaling differences in success rates between the two processes are easily observed as differences in counts returned by the scaled processes. The method can be used to get the exact probability of a given difference under the null hypothesis that both the A and B processes have the same success rate (kA+kB)/(nA+nB). When nA and nB don't divide evenly into to each other two calculations are run with the larger process is alternately padded and truncated to look like a larger or smaller experiment that meets the above conditions. This gives us a good range of significances.

Bernoulli_diff_stat(kA, nA, kB, nB, test_rate_difference, common_rate)

## Arguments

kA number of A successes observed. number of A experiments. number of B successes observed. number of B experiments. numeric, difference in rate of A-B to test. Note: it is best to specify this prior to looking at the data. rate numeric, assumed null-rate.

## Value

Bernoulli difference test statistic.

## Details

Note the intent is that we are measuring the results of an A/B test with max(nA, nB) %% min(nA, nB) == 0 (no padding needed), or max(nA,nB) >> min(nA,nB) (padding is small effect).

The idea of converting a rate problem into a counting problem follows from reading Wald's Sequential Analysis.

For very small p-values the calculation is sensitive to rounding in the observed ratio-difference, as an arbitrarily small change in test-rate can move an entire set of observed differences in or out of the significance calculation.

## Examples

Bernoulli_diff_stat(2000, 5000, 100, 200)
#> [1] "Bernoulli difference test: (A=2000/5000=0.4, B=100/200=0.5, =0.4038, post 0.1 two sided; p=0.004677)."Bernoulli_diff_stat(2000, 5000, 100, 200, 0.1)
#> [1] "Bernoulli difference test: (A=2000/5000=0.4, B=100/200=0.5, =0.4038, prior 0.1 two sided; p=0.004677)."Bernoulli_diff_stat(2000, 5000, 100, 199)
#> [1] "Bernoulli difference test: (A=2000/5000=0.4, B=100/199=0.5025, =0.4039, post 0.1025 two sided; pL=0.003753, pH=0.00382)."Bernoulli_diff_stat(2000, 5000, 100, 199, 0.1)
#> [1] "Bernoulli difference test: (A=2000/5000=0.4, B=100/199=0.5025, =0.4039, prior 0.1 two sided; pL=0.004701, pH=0.00474)."Bernoulli_diff_stat(100, 200, 2000, 5000)
#> [1] "Bernoulli difference test: (A=100/200=0.5, B=2000/5000=0.4, =0.4038, post 0.1 two sided; p=0.004677)."
# sigr adjusts experiment sizes when lengths
# don't divide into each other.
Bernoulli_diff_stat(100, 199, 2000, 5000)
#> [1] "Bernoulli difference test: (A=100/199=0.5025, B=2000/5000=0.4, =0.4039, post 0.1025 two sided; pL=0.003753, pH=0.00382)."Bernoulli_diff_stat(100, 199, 2000, 5000)\$pValue
#> [1] 0.003786713


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 63 11 8 33.333%

## 문제

Casting spells is the least understood technique of dealing with real life. Actually, people find it quite hard to distinguish between a real spells like “abrahellehhelleh” (used in the battles and taught at the mage universities) and screams like “rachelhellabracadabra” (used by uneducated witches for shouting at cats).

Finally, the research conducted at the Unheard University showed how one can measure the power of a word (be it a real spell or a scream). It appeared that it is connected with the mages’ ability to pronounce words backwards. (Actually, some singers were burned at the stake for exactly the same ability, as it was perceived as demonic possession.) Namely, the power of a word is the length of the maximum subword of the form wwRwwR (where w is an arbitrary sequence of characters and wR is w written backwards). If no such subword exists, then the power of the word is 0. For example, the power of abrahellehhelleh is 12 as it contains hellehhelleh and the power of rachelhellabracadabra is 0. Note that the power of a word is always a multiple of 4.

The input contains several test cases. The first line of the input contains a positive integer Z ≤ 40, denoting the number of test cases. Then Z test cases follow, each conforming to the format described in section Input. For each test case, your program has to write an output conforming to the format described in section Output.

## 입력

The input is one line containing a word of length at most 3 · 105, consisting of (large or small) letters of the English alphabet.

## 출력

You should output one integer k being the power of the word.

## 예제 입력

2
abrahellehhelleh

12


text: 2 added 545 characters in body edited May 21 '18 at 17:01 Big6 3,27911 gold badge77 silver badges1515 bronze badges One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function. The transfer function: $$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100}$$ This could be re-written as: $$(s^2+12s+100)Y(s)=(6s+100)r(s)$$ Back to the time domain: $$y''+12y'+100y=6r'+100r$$ Here, $$\r(t)\$$ is the unit step function. The derivative of the step function is the dirac delta function, so the differential equation becomes: $$y''+12y'+100y=6\delta(t)+100, \text{for } t>0$$ Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at t=0$$\t=0\$$. Then for t>0$$\t>0\$$: $$y''+12y'+100y=100, \text{for } t>0$$ After solving the differential equation (homogeneous and particular solutions added together), $$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t)$$$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) \tag1$$ NowAnd the constants are found by using the initial conditions. Here is the trick, you have to somehow account for the impulse (dirac delta function), right at t=0 (we have disregarded it so far). Laplace transform are usually defined from $$\t=0^-\$$ in orderin order to include the impulses (such as the dirac delta function). So we need to include this impulse if we want to get the same result one would with the Laplace transform. This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includesincludes a dirac delta, you need to find the solution to the case when t>0$$\t>0\$$ (just what we did before in (1)), with the initial conditions $$\y(0)=0\$$, $$\y'(0)=a/m\$$ (for this case, $$\a=6\$$ and $$\m=1\$$). This is what brings in the fact the you have a dirac delta as part of the forcing function. $$\a\$$ is just the coefficient multiplying the dirac delta function, 6function—6 in this problem. $$\m\$$ is the coefficient for the highest order derivative ($$\y''\$$ is highest order and its coefficient is 1). With those initial conditions ($$\y(0)=0\$$, $$\y'(0)=6\$$), you can find the values of the constants, $$\K_1\$$ and $$\K_2\$$. These turn out to be $$\K_1=-1\$$ and $$\K_2=0\$$. Therefore: $$y(t)=1-e^{-6t}\cos(8t)$$ A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis. Also, the nature of the dirac delta function is what makes this problem a bit trickier. One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function. The transfer function: $$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100}$$ This could be re-written as: $$(s^2+12s+100)Y(s)=(6s+100)r(s)$$ Back to the time domain: $$y''+12y'+100y=6r'+100r$$ The derivative of the step function is the dirac delta function, so: $$y''+12y'+100y=6\delta(t)+100, \text{for } t>0$$ Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at t=0. Then for t>0: $$y''+12y'+100y=100, \text{for } t>0$$ After solving the differential equation, $$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t)$$ Now, you have to somehow account for the impulse (dirac delta function), right at t=0. Laplace transform are usually defined from $$\t=0^-\$$ in order to include the impulses. So we need this to get the same result one would with the Laplace transform. This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when t>0, with the initial conditions $$\y(0)=0\$$, $$\y'(0)=a/m\$$ (for this case, $$\a=6\$$ and $$\m=1\$$). $$\a\$$ is just the coefficient multiplying the dirac delta function, 6 in this problem. $$\m\$$ is the coefficient for the highest order derivative ($$\y''\$$ is highest order and coefficient is 1). With those initial conditions ($$\y(0)=0\$$, $$\y'(0)=6\$$), you can find the values of the constants, $$\K_1\$$ and $$\K_2\$$. These turn out to be $$\K_1=-1\$$ and $$\K_2=0\$$. Therefore: $$y(t)=1-e^{-6t}\cos(8t)$$ A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis. One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function. The transfer function: $$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100}$$ This could be re-written as: $$(s^2+12s+100)Y(s)=(6s+100)r(s)$$ Back to the time domain: $$y''+12y'+100y=6r'+100r$$ Here, $$\r(t)\$$ is the unit step function. The derivative of the step function is the dirac delta function, so the differential equation becomes: $$y''+12y'+100y=6\delta(t)+100, \text{for } t>0$$ Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at $$\t=0\$$. Then for $$\t>0\$$: $$y''+12y'+100y=100, \text{for } t>0$$ After solving the differential equation (homogeneous and particular solutions added together), $$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) \tag1$$ And the constants are found by using the initial conditions. Here is the trick, you have to somehow account for the impulse (dirac delta function), right at t=0 (we have disregarded it so far). Laplace transform are usually defined from $$\t=0^-\$$ in order to include the impulses (such as the dirac delta function). So we need to include this impulse if we want to get the same result one would with the Laplace transform. This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when $$\t>0\$$ (just what we did before in (1)), with the initial conditions $$\y(0)=0\$$, $$\y'(0)=a/m\$$ (for this case, $$\a=6\$$ and $$\m=1\$$). This is what brings in the fact the you have a dirac delta as part of the forcing function. $$\a\$$ is just the coefficient multiplying the dirac delta function—6 in this problem. $$\m\$$ is the coefficient for the highest order derivative ($$\y''\$$ is highest order and its coefficient is 1). With those initial conditions ($$\y(0)=0\$$, $$\y'(0)=6\$$), you can find the values of the constants, $$\K_1\$$ and $$\K_2\$$. These turn out to be $$\K_1=-1\$$ and $$\K_2=0\$$. Therefore: $$y(t)=1-e^{-6t}\cos(8t)$$ A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis. Also, the nature of the dirac delta function is what makes this problem a bit trickier. 1 answered May 21 '18 at 16:46 Big6 3,27911 gold badge77 silver badges1515 bronze badges One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function. The transfer function: $$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100}$$ This could be re-written as: $$(s^2+12s+100)Y(s)=(6s+100)r(s)$$ Back to the time domain: $$y''+12y'+100y=6r'+100r$$ The derivative of the step function is the dirac delta function, so: $$y''+12y'+100y=6\delta(t)+100, \text{for } t>0$$ Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at t=0. Then for t>0: $$y''+12y'+100y=100, \text{for } t>0$$ After solving the differential equation, $$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t)$$ Now, you have to somehow account for the impulse (dirac delta function), right at t=0. Laplace transform are usually defined from $$\t=0^-\$$ in order to include the impulses. So we need this to get the same result one would with the Laplace transform. This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when t>0, with the initial conditions $$\y(0)=0\$$, $$\y'(0)=a/m\$$ (for this case, $$\a=6\$$ and $$\m=1\$$). $$\a\$$ is just the coefficient multiplying the dirac delta function, 6 in this problem. $$\m\$$ is the coefficient for the highest order derivative ($$\y''\$$ is highest order and coefficient is 1). With those initial conditions ($$\y(0)=0\$$, $$\y'(0)=6\$$), you can find the values of the constants, $$\K_1\$$ and $$\K_2\$$. These turn out to be $$\K_1=-1\$$ and $$\K_2=0\$$. Therefore: $$y(t)=1-e^{-6t}\cos(8t)$$ A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis.


text:
Numbers = letters 2.0

SIX + SEVEN + SEVEN = TWENTY

This is a standard alphametic, no hidden rules.

(An alphametic is a puzzle in which every letter corresponds to exactly one number, and only one letter can correspond to one specific number. The goal of the puzzle is to find the numbers for each letter.)

Please provide an explanation/reasoning and don't use brute force as reasoning. Brute force is still allowed, only to support the reasoning.

ONE + NINE + TWENTY + FIFTY = EIGHTY
TEN + TEN + FORTY = SIXTY

• It seems like not a lot of people know what "alphametics" are. tkcs-collins.com/truman/alphamet/alphamet.shtml – Paul L May 6 '16 at 20:08
• By the way, this is a doubly true alphametic because the addends and sum are numbers words and the equation, when expressed as numbers, is true, i.e., 6 + 7 + 7 = 20. – Reti43 May 6 '16 at 21:07
• @Reti43 Yes, that is true as well! – anna328p May 6 '16 at 21:59
• Not sure why you chose Bobby's answer, when mine was a) first and b) more rigorous and doesn't involve guesswork? – astralfenix May 6 '16 at 23:37
• @astralfenix Woops, selected the wrong one. – anna328p May 7 '16 at 1:06

650 + 68782 + 68782 = 138214

because:

Let $C_1$, ... $C_6$ be the carryovers into each column, counting from the right.  So $C_1=0$, $X+N+N=C_2Y \text{ (i.e., }C_2\times 10+Y\text{)}$ (from the least significant column), and $C_6=T$.
Clearly each carryover $<3$.

$C_6$ is either $1$ or $2$.  If $C_6=2$, that implies $S=9$ and $C_5=2$, which is possible only if $E=9$ and $C_4=2$.  That’s a contradiction.  So $C_6=1$ and $T=1$.

$C_5$ is at most $2$, so $S>3$, since $C_6$ is $1$.  If $S=4$, then $C_5=2$, which is possible only if $E=9$ and $C_4=2$.  But then $9+9+2 = 20$, meaning $E=0$, a contradiction.  So $S>4$.

This also means $C_5=1$ and $E<9$.  Also, $E>7$, since otherwise $E+E+C_4$ cannot be a number ending with $E$.  This means $E=8$ and $C_4=2$.  Now, if $S=5$, then $W=1$ (contradiction), so $S>5$.  $S$ cannot be $9$ since then $W=9$. Therefore $S$ is either $6$ or $7$, implying $W$ is either $3$ or $5$.

$C_3$ must be $2$, since we already have two $8$s in column 2 and the sum must end in $1$. Therefore $I+C_2=5$.

If $S=7$, then $V>5$, and this is the most $S$ can be.  So $V>5$.  If $V=6$, then $S=7$, and this leads to $N=1$ (contradiction).  So $V>6$.  If $V=9$, then $N=S$ (contradiction).  So $V=7$.  Therefore $S=6$, $W=3$ and $N=2$.

Now the only values remaining are $4$, $5$, $9$ and $0$.  One of these won't be used.
If $X=4$, then $Y=8$ (contradiction).  If $X=5$, then $Y=9$ and $I=5$ (contradiction).  If $X=9$, then $Y=3$ (contradiction).  So $X=0$.  Therefore $Y=4$ and $I=5$.

S = 6
E = 8
V = 7
N = 2
I = 5
X = 0
T = 1
W = 3
Y = 4

Which results in:

650 + 68782 + 68782 = 138214

Explanation:

I started with the idea that T = 1 since it was unlikely that the result of S + S + carry would be greater than 19.

...SEVEN
...SEVEN
..........SIX
-------------
1WEN1Y

I then noticed that E + E + carry = E. This lead me to initially believe E = 9, as 9+9+1(carry) = 19. After hitting a dead end though, I changed that to E = 8, which meant that V + V + S > 20. With E = 8, I could determine I = 5, as E(8) + E(8)+ I = T(1).

...S8V8N
...S8V8N
+......S5X
-------------
1W8N1Y

From here, I took a guess as S = 6, which gave me W = 3.

..68V8N
..68V8N
+.....65X
-------------
138N1Y

The only digits that would give me a carry of 2 with V + V + 6 are 9 and 7. Again I tried 9, but hit another dead end. With V = 7, N = 2.

..68782
..68782
+....65X
------------
13821Y

Finally, I tried plugging the remaining digits (0, 4, 9) into X and Y and come up with X = 0 and Y = 4.

..68782
..68782
+....650
------------
138214

• Note that $E + E + \text{carry} = E$ could also be solved with $E = 0$ — but then $I + E + E + \text{carry} = 1$ would reduce to $I + \text{carry} = 1$, which would mean $I$ is either $0$ or $1$, which is a contradiction. – Peregrine Rook May 6 '16 at 22:19


text:
## Calculus with Applications (10th Edition)

$$4b^2$$
Factor the terms inside the radical $$\sqrt[3]{64b^6}=\sqrt[3]{2^6b^6}=\sqrt[3]{2^32^3b^3b^3}$$ The cube root of a cubed expression is the expression $$=2\times 2 \times b\times b=4b^2$$


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# Can this matrix equation be solved for X?

$(A^TA+aI)^{-1}A^T = A^TX$

where A and X are real-matrices and a is a positive non-zero real number. A is non-zero and you may assume that A is such that there is a unique X that satisfies the above equation.

-

Using the inverse of the inverse so to speak:

$$A^T=(A^TA+aI)A^TX$$ $$A^T=A^T(AA^T+aI)X$$

and now if $X=(AA^T+aI)^{-1}$ it is solved.

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$(A^TA+aI)A^T=(A^TAA^T+aIA^T)=A^T(AA^T+aI)$ –  adam W Oct 4 '12 at 0:52
Sorry, i forgot to put -1 in the power –  Karan Oct 4 '12 at 0:54
I have changed the question now –  Karan Oct 4 '12 at 0:54
Thanx, this is perfect –  Karan Oct 4 '12 at 1:16
Glad I could help! –  adam W Oct 4 '12 at 1:19


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# Comparing Data in Different Formats

This post about comparing data in different formats is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Liam’s Pay

This graph shows how much Liam gets paid as a function of the number of hours he works. Sarah, on the other hand, is paid according to the formula S=(75/4)h, where S represents Sarah’s pay and h represents the number of hours she has worked.

Is Liam’s rate of pay more or less than Sarah’s, and by how much?

## Solution

You’re being asked to compare data that measure similar things but are given to you in different formats.

The trick is to convert one of those expressions into the same format as the other. Or, if you can’t do that, then convert both to some third format.

P=(75/4)h looks . . . almost straightforward. That fraction may be inconvenient. 75/4 = 18.75, so let’s say S = 18.75h represents Sarah’s rate of pay.

Now look at the graph. Can you represent Liam’s pay as something like L=kh, where k is Liam’s rate of pay? Since the graph goes through the origin, its equation will be of the form L=something*and “something” is the slope. To find it from the graph, pick two points on the line that are easy to read, and calculate. The origin — with coordinates (0, 0) — looks easy. And the point (2, 40) doesn’t look bad. The slope is the change in y divided by the change in for those two points:

$\fn_jvn&space;\frac{40-0}{2-0}=\frac{40}{2}=20$

So Liam’s pay is L=20h.

You’re asked for the difference between Liam’s rate of pay and Sarah’s. That is

$\fn_jvn&space;\20.00/hr&space;-&space;\18.75/hr&space;=&space;\1.25/hr$

Answer: Liam gets paid \$1.25 per hour more than Sarah.

For more about relating a line’s equation to its graph, see “Finding a Line’s Equation from the Graph of Its Perpendicular.

This question is similar to question number 5 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

0 replies


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# American Institute of Mathematical Sciences

January  2018, 14(1): 283-308. doi: 10.3934/jimo.2017047

## Some robust improved geometric aggregation operators under interval-valued intuitionistic fuzzy environment for multi-criteria decision-making process

 School of Mathematics, Thapar University Patiala-147004, Punjab, India

Received  July 2016 Revised  October 2016 Published  June 2017

Fund Project: The author would like to thank the Editor-in-Chief and referees for providing very helpful comments and suggestions

The objective of this manuscript is to present some new interactive geometric aggregation operators for the interval-valued intuitionistic fuzzy numbers (IVIFNs). In order to achieve it, firstly the shortcomings of the existing operators have been highlighted and then resolved it by defining new operational laws based on the pairs of hesitation degree between the membership functions. By using these improved laws, some geometric aggregation operators, namely interval-valued intuitionistic fuzzy Hamacher interactive weighted and hybrid geometric labeled as IIFHIWG and IIFHIHWG operators, respectively have been proposed. Furthermore, desirable properties corresponding to these operators have been stated. Finally, a decision-making method based on the proposed operator has been illustrated to demonstrate the approach. A computed result is compared with the existing results.

Citation: Harish Garg. Some robust improved geometric aggregation operators under interval-valued intuitionistic fuzzy environment for multi-criteria decision-making process. Journal of Industrial & Management Optimization, 2018, 14 (1) : 283-308. doi: 10.3934/jimo.2017047
##### References:

show all references

##### References:
Information about each alternative in the form of the IVIFNs
 $C_1$ $C_2$ $C_3$ $C_4$ $C_5$ $C_6$ $X_1$ $\langle[0.2, 0.3], [0.4, 0.5]\rangle$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.4, 0.5], [0.2, 0.4]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $\langle[0.1, 0.3], [0.5, 0.6]\rangle$ $\langle[0.5, 0.7], [0.2, 0.3]\rangle$ $X_2$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.5, 0.6], [0.1, 0.3]\rangle$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.6, 0.7], [0.1, 0.2]\rangle$ $\langle[0.3, 0.4], [0.5, 0.6]\rangle$ $\langle[0.4, 0.7], [0.1, 0.2]\rangle$ $X_3$ $\langle[0.4, 0.5], [0.3, 0.4]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $\langle[0.5, 0.6], [0.3, 0.4]\rangle$ $\langle[0.6, 0.7], [0.1, 0.3]\rangle$ $\langle[0.4, 0.5], [0.3, 0.4]\rangle$ $\langle[0.3, 0.5], [0.1, 0.3]\rangle$ $X_4$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.5, 0.6], [0.1, 0.3]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $\langle[0.3, 0.4], [0.1, 0.2]\rangle$ $\langle[0.5, 0.6], [0.1, 0.3]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $X_5$ $\langle[0.5, 0.6], [0.3, 0.4]\rangle$ $\langle[0.3, 0.4], [0.3, 0.5]\rangle$ $\langle[0.6, 0.7], [0.1, 0.3]\rangle$ $\langle[0.6, 0.8], [0.1, 0.2]\rangle$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.5, 0.6], [0.2, 0.4]\rangle$
 $C_1$ $C_2$ $C_3$ $C_4$ $C_5$ $C_6$ $X_1$ $\langle[0.2, 0.3], [0.4, 0.5]\rangle$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.4, 0.5], [0.2, 0.4]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $\langle[0.1, 0.3], [0.5, 0.6]\rangle$ $\langle[0.5, 0.7], [0.2, 0.3]\rangle$ $X_2$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.5, 0.6], [0.1, 0.3]\rangle$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.6, 0.7], [0.1, 0.2]\rangle$ $\langle[0.3, 0.4], [0.5, 0.6]\rangle$ $\langle[0.4, 0.7], [0.1, 0.2]\rangle$ $X_3$ $\langle[0.4, 0.5], [0.3, 0.4]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $\langle[0.5, 0.6], [0.3, 0.4]\rangle$ $\langle[0.6, 0.7], [0.1, 0.3]\rangle$ $\langle[0.4, 0.5], [0.3, 0.4]\rangle$ $\langle[0.3, 0.5], [0.1, 0.3]\rangle$ $X_4$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.5, 0.6], [0.1, 0.3]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $\langle[0.3, 0.4], [0.1, 0.2]\rangle$ $\langle[0.5, 0.6], [0.1, 0.3]\rangle$ $\langle[0.7, 0.8], [0.1, 0.2]\rangle$ $X_5$ $\langle[0.5, 0.6], [0.3, 0.4]\rangle$ $\langle[0.3, 0.4], [0.3, 0.5]\rangle$ $\langle[0.6, 0.7], [0.1, 0.3]\rangle$ $\langle[0.6, 0.8], [0.1, 0.2]\rangle$ $\langle[0.6, 0.7], [0.2, 0.3]\rangle$ $\langle[0.5, 0.6], [0.2, 0.4]\rangle$
Effect of the parameter $\gamma$ on the ranking of the alternatives by IIFHIWG and the existing operators
 $\gamma=1$ $\gamma=2$ $\gamma=3$ Wei and Wang [26] Proposed Wang and Liu [24] Proposed Liu [20] Proposed Score value Score value Score value $X_1$ 0.0548 0.1346 0.0727 0.1454 0.0822 0.1517 $X_2$ 0.2874 0.3174 0.2998 0.3310 0.3065 0.3388 $X_3$ 0.2139 0.2713 0.2205 0.2760 0.2245 0.2793 $X_4$ 0.4463 0.4997 0.4535 0.5013 0.4576 0.5024 $X_5$ 0.2985 0.3119 0.3047 0.3166 0.3083 0.3197 ranking $X_4 \succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$
 $\gamma=1$ $\gamma=2$ $\gamma=3$ Wei and Wang [26] Proposed Wang and Liu [24] Proposed Liu [20] Proposed Score value Score value Score value $X_1$ 0.0548 0.1346 0.0727 0.1454 0.0822 0.1517 $X_2$ 0.2874 0.3174 0.2998 0.3310 0.3065 0.3388 $X_3$ 0.2139 0.2713 0.2205 0.2760 0.2245 0.2793 $X_4$ 0.4463 0.4997 0.4535 0.5013 0.4576 0.5024 $X_5$ 0.2985 0.3119 0.3047 0.3166 0.3083 0.3197 ranking $X_4 \succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$
Effect of the parameter $\gamma$ on the ranking of the alternatives by using IIFHIHWG and the existing operators
 $\gamma=1$ $\gamma=2$ $\gamma=3$ Wei and Wang [26] Proposed Wang and Liu [24] Proposed Liu [20] Proposed Score value Score value Score value $X_1$ 0.1221 0.2080 0.1434 0.2163 0.1558 0.2151 $X_2$ 0.3304 0.3674 0.3443 0.3734 0.3522 0.3795 $X_3$ 0.2535 0.3019 0.2630 0.3068 0.2692 0.3113 $X_4$ 0.3705 0.4853 0.3815 0.4815 0.3880 0.4828 $X_5$ 0.3141 0.3414 0.3164 0.3510 0.3203 0.3500 ranking $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$
 $\gamma=1$ $\gamma=2$ $\gamma=3$ Wei and Wang [26] Proposed Wang and Liu [24] Proposed Liu [20] Proposed Score value Score value Score value $X_1$ 0.1221 0.2080 0.1434 0.2163 0.1558 0.2151 $X_2$ 0.3304 0.3674 0.3443 0.3734 0.3522 0.3795 $X_3$ 0.2535 0.3019 0.2630 0.3068 0.2692 0.3113 $X_4$ 0.3705 0.4853 0.3815 0.4815 0.3880 0.4828 $X_5$ 0.3141 0.3414 0.3164 0.3510 0.3203 0.3500 ranking $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4 \succ X_2 \succ X_5 \succ X_3 \succ X_1$
Ordering of the attributes for different $\gamma$
 $\gamma$ By IIFHIWG By IIFHIHWG Aggregated IVIFN Score values Aggregated IVIFN Score values 0.1 $X_1$ $\big\langle[0.3771, 0.5753], [0.2996, 0.4247]\big\rangle$ 0.1140 $\big\langle[0.4562, 0.6029], [0.2818, 0.3971]\big\rangle$ 0.1901 $X_2$ $\big\langle[0.5042, 0.6545], [0.2324, 0.3455]\big\rangle$ 0.2904 $\big\langle[0.5016, 0.6889], [0.1971, 0.3111]\big\rangle$ 0.3411 $X_3$ $\big\langle[0.4719, 0.6441], [0.2310, 0.3559]\big\rangle$ 0.2646 $\big\langle[0.4808, 0.6605], [0.2187, 0.3395]\big\rangle$ 0.2916 $X_4$ $\big\langle[0.6130, 0.7519], [0.1218, 0.2481]\big\rangle$ 0.4975 $\big\langle[0.5404, 0.7670], [0.1097, 0.2330]\big\rangle$ 0.4824 $X_5$ $\big\langle[0.5306, 0.6405], [0.2027, 0.3595]\big\rangle$ 0.3045 $\big\langle[0.5338, 0.6603], [0.1876, 0.3397]\big\rangle$ 0.3334 Ranking $X_4\succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 0.5 $X_1$ $\big\langle[0.3805, 0.5819], [0.2933, 0.4181]\big\rangle$ 0.1255 $\big\langle[0.4597 0.6086], [0.2764 0.3914]\big\rangle$ 0.2003 $X_2$ $\big\langle[0.5092, 0.6634], [0.2249, 0.3366]\big\rangle$ 0.3056 $\big\langle[0.5062 0.6977], [0.1897 0.3023]\big\rangle$ 0.3560 $X_3$ $\big\langle[0.4734, 0.6455], [0.2285, 0.3545]\big\rangle$ 0.2679 $\big\langle[0.4826 0.6633], [0.2157 0.3367]\big\rangle$ 0.2967 $X_4$ $\big\langle[0.6133, 0.7526], [0.1214, 0.2474]\big\rangle$ 0.4986 $\big\langle[0.5406 0.7681], [0.1093 0.2319]\big\rangle$ 0.4838 $X_5$ $\big\langle[0.5318, 0.6429], [0.2009, 0.3571]\big\rangle$ 0.3084 $\big\langle[0.5352 0.6639], [0.1855 0.3361]\big\rangle$ 0.3388 Ranking $X_4\succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 1 $X_1$ $\big\langle[0.3834, 0.5868], [0.2878, 0.4132]\big\rangle$ 0.1346 $\big\langle[0.4626, 0.6125], [0.2717, 0.3875]\big\rangle$ 0.2080 $X_2$ $\big\langle[0.5134, 0.6699], [0.2185, 0.3301]\big\rangle$ 0.3174 $\big\langle[0.5100, 0.7041], [0.1835, 0.2959]\big\rangle$ 0.3674 $X_3$ $\big\langle[0.4750, 0.6467], [0.2260, 0.3533]\big\rangle$ 0.2713 $\big\langle[0.4845, 0.6659], [0.2126, 0.3341]\big\rangle$ 0.3019 $X_4$ $\big\langle[0.6136, 0.7533], [0.1210, 0.2467]\big\rangle$ 0.4997 $\big\langle[0.5409, 0.7693], [0.1089, 0.2307]\big\rangle$ 0.4853 $X_5$ $\big\langle[0.5330, 0.6449], [0.1991, 0.3551]\big\rangle$ 0.3119 $\big\langle[0.5141, 0.6751], [0.1813, 0.3249]\big\rangle$ 0.3414 Ranking $X_4\succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 2 $X_1$ $\big\langle[0.3872, 0.5923], [0.2809, 0.4077]\big\rangle$ 0.1454 $\big\langle[0.4663, 0.6161], [0.2659, 0.3839]\big\rangle$ 0.2163 $X_2$ $\big\langle[0.5186, 0.6770], [0.2106, 0.3230]\big\rangle$ 0.3310 $\big\langle[0.5268, 0.7023], [0.1847, 0.2977]\big\rangle$ 0.3734 $X_3$ $\big\langle[0.4774, 0.6484], [0.2221, 0.3516]\big\rangle$ 0.2760 $\big\langle[0.4890, 0.6646], [0.2047, 0.3354]\big\rangle$ 0.3068 $X_4$ $\big\langle[0.6141, 0.7543], [0.1202, 0.2457]\big\rangle$ 0.5013 $\big\langle[0.5556, 0.7623], [0.1172, 0.2377]\big\rangle$ 0.4815 $X_5$ $\big\langle[0.5348, 0.6473], [0.1963, 0.3527]\big\rangle$ 0.3166 $\big\langle[0.5166, 0.6813], [0.1772, 0.3187]\big\rangle$ 0.3510 Ranking $X_4\succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 5 $X_1$ $\big\langle[0.3922, 0.5987], [0.2716 0.4013]\big\rangle$ 0.1590 $\big\langle[0.4324, 0.6341], [0.2537, 0.3659]\big\rangle$ 0.2235 $X_2$ $\big\langle[0.5256, 0.6849], [0.1999 0.3151]\big\rangle$ 0.3478 $\big\langle[0.5195, 0.7130], [0.1776, 0.2870]\big\rangle$ 0.3840 $X_3$ $\big\langle[0.4815, 0.6506], [0.2153 0.3494]\big\rangle$ 0.2837 $\big\langle[0.4939, 0.6691], [0.1969, 0.3309]\big\rangle$ 0.3176 $X_4$ $\big\langle[0.6150, 0.7558], [0.1189 0.2442]\big\rangle$ 0.5039 $\big\langle[0.5565, 0.7642], [0.1158, 0.2358]\big\rangle$ 0.4846 $X_5$ $\big\langle[0.5379, 0.6504], [0.1917 0.3496]\big\rangle$ 0.3235 $\big\langle[0.5521, 0.6726], [0.1906, 0.3274]\big\rangle$ 0.3534 Ranking $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 10 $X_1$ $\big\langle[0.3952, 0.6020], [0.2659, 0.3980]\big\rangle$ 0.1667 $\big\langle[0.4360, 0.6387], [0.2474, 0.3613]\big\rangle$ 0.2330 $X_2$ $\big\langle[0.5300, 0.6889], [0.1932, 0.3111]\big\rangle$ 0.3573 $\big\langle[0.5220, 0.7368], [0.1572, 0.2632]\big\rangle$ 0.4192 $X_3$ $\big\langle[0.4847, 0.6519], [0.2102, 0.3481]\big\rangle$ 0.2891 $\big\langle[0.4767, 0.6968], [0.1744, 0.3032]\big\rangle$ 0.3479 $X_4$ $\big\langle[0.6158, 0.7568], [0.1178, 0.2432]\big\rangle$ 0.5058 $\big\langle[0.5573, 0.7657], [0.1147, 0.2343]\big\rangle$ 0.4870 $X_5$ $\big\langle[0.5403, 0.6522], [0.1882, 0.3478]\big\rangle$ 0.3282 $\big\langle[0.5608, 0.6594], [0.1933, 0.3406]\big\rangle$ 0.3432 Ranking $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_3 \succ X_5 \succ X_1$ 25 $X_1$ $\big\langle[0.3979, 0.6046], [0.2610, 0.3954]\big\rangle$ 0.1730 $\big\langle[0.4217, 0.6566], [0.2234, 0.3434]\big\rangle$ 0.2558 $X_2$ $\big\langle[0.5338, 0.6920], [0.1874, 0.3080]\big\rangle$ 0.3652 $\big\langle[0.5570, 0.7047], [0.1689, 0.2953]\big\rangle$ 0.3988 $X_3$ $\big\langle[0.4878, 0.6530], [0.2051, 0.3470]\big\rangle$ 0.2943 $\big\langle[0.5176, 0.6687], [0.1871, 0.3313]\big\rangle$ 0.3339 $X_4$ $\big\langle[0.6165, 0.7576], [0.1167, 0.2424]\big\rangle$ 0.5075 $\big\langle[0.6057, 0.7394], [0.1297, 0.2606]\big\rangle$ 0.4774 $X_5$ $\big\langle[0.5426, 0.6536], [0.1847, 0.3464]\big\rangle$ 0.3325 $\big\langle[0.5600, 0.6525], [0.1944, 0.3475]\big\rangle$ 0.3353 Ranking $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$
 $\gamma$ By IIFHIWG By IIFHIHWG Aggregated IVIFN Score values Aggregated IVIFN Score values 0.1 $X_1$ $\big\langle[0.3771, 0.5753], [0.2996, 0.4247]\big\rangle$ 0.1140 $\big\langle[0.4562, 0.6029], [0.2818, 0.3971]\big\rangle$ 0.1901 $X_2$ $\big\langle[0.5042, 0.6545], [0.2324, 0.3455]\big\rangle$ 0.2904 $\big\langle[0.5016, 0.6889], [0.1971, 0.3111]\big\rangle$ 0.3411 $X_3$ $\big\langle[0.4719, 0.6441], [0.2310, 0.3559]\big\rangle$ 0.2646 $\big\langle[0.4808, 0.6605], [0.2187, 0.3395]\big\rangle$ 0.2916 $X_4$ $\big\langle[0.6130, 0.7519], [0.1218, 0.2481]\big\rangle$ 0.4975 $\big\langle[0.5404, 0.7670], [0.1097, 0.2330]\big\rangle$ 0.4824 $X_5$ $\big\langle[0.5306, 0.6405], [0.2027, 0.3595]\big\rangle$ 0.3045 $\big\langle[0.5338, 0.6603], [0.1876, 0.3397]\big\rangle$ 0.3334 Ranking $X_4\succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 0.5 $X_1$ $\big\langle[0.3805, 0.5819], [0.2933, 0.4181]\big\rangle$ 0.1255 $\big\langle[0.4597 0.6086], [0.2764 0.3914]\big\rangle$ 0.2003 $X_2$ $\big\langle[0.5092, 0.6634], [0.2249, 0.3366]\big\rangle$ 0.3056 $\big\langle[0.5062 0.6977], [0.1897 0.3023]\big\rangle$ 0.3560 $X_3$ $\big\langle[0.4734, 0.6455], [0.2285, 0.3545]\big\rangle$ 0.2679 $\big\langle[0.4826 0.6633], [0.2157 0.3367]\big\rangle$ 0.2967 $X_4$ $\big\langle[0.6133, 0.7526], [0.1214, 0.2474]\big\rangle$ 0.4986 $\big\langle[0.5406 0.7681], [0.1093 0.2319]\big\rangle$ 0.4838 $X_5$ $\big\langle[0.5318, 0.6429], [0.2009, 0.3571]\big\rangle$ 0.3084 $\big\langle[0.5352 0.6639], [0.1855 0.3361]\big\rangle$ 0.3388 Ranking $X_4\succ X_5 \succ X_2 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 1 $X_1$ $\big\langle[0.3834, 0.5868], [0.2878, 0.4132]\big\rangle$ 0.1346 $\big\langle[0.4626, 0.6125], [0.2717, 0.3875]\big\rangle$ 0.2080 $X_2$ $\big\langle[0.5134, 0.6699], [0.2185, 0.3301]\big\rangle$ 0.3174 $\big\langle[0.5100, 0.7041], [0.1835, 0.2959]\big\rangle$ 0.3674 $X_3$ $\big\langle[0.4750, 0.6467], [0.2260, 0.3533]\big\rangle$ 0.2713 $\big\langle[0.4845, 0.6659], [0.2126, 0.3341]\big\rangle$ 0.3019 $X_4$ $\big\langle[0.6136, 0.7533], [0.1210, 0.2467]\big\rangle$ 0.4997 $\big\langle[0.5409, 0.7693], [0.1089, 0.2307]\big\rangle$ 0.4853 $X_5$ $\big\langle[0.5330, 0.6449], [0.1991, 0.3551]\big\rangle$ 0.3119 $\big\langle[0.5141, 0.6751], [0.1813, 0.3249]\big\rangle$ 0.3414 Ranking $X_4\succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 2 $X_1$ $\big\langle[0.3872, 0.5923], [0.2809, 0.4077]\big\rangle$ 0.1454 $\big\langle[0.4663, 0.6161], [0.2659, 0.3839]\big\rangle$ 0.2163 $X_2$ $\big\langle[0.5186, 0.6770], [0.2106, 0.3230]\big\rangle$ 0.3310 $\big\langle[0.5268, 0.7023], [0.1847, 0.2977]\big\rangle$ 0.3734 $X_3$ $\big\langle[0.4774, 0.6484], [0.2221, 0.3516]\big\rangle$ 0.2760 $\big\langle[0.4890, 0.6646], [0.2047, 0.3354]\big\rangle$ 0.3068 $X_4$ $\big\langle[0.6141, 0.7543], [0.1202, 0.2457]\big\rangle$ 0.5013 $\big\langle[0.5556, 0.7623], [0.1172, 0.2377]\big\rangle$ 0.4815 $X_5$ $\big\langle[0.5348, 0.6473], [0.1963, 0.3527]\big\rangle$ 0.3166 $\big\langle[0.5166, 0.6813], [0.1772, 0.3187]\big\rangle$ 0.3510 Ranking $X_4\succ X_2 \succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 5 $X_1$ $\big\langle[0.3922, 0.5987], [0.2716 0.4013]\big\rangle$ 0.1590 $\big\langle[0.4324, 0.6341], [0.2537, 0.3659]\big\rangle$ 0.2235 $X_2$ $\big\langle[0.5256, 0.6849], [0.1999 0.3151]\big\rangle$ 0.3478 $\big\langle[0.5195, 0.7130], [0.1776, 0.2870]\big\rangle$ 0.3840 $X_3$ $\big\langle[0.4815, 0.6506], [0.2153 0.3494]\big\rangle$ 0.2837 $\big\langle[0.4939, 0.6691], [0.1969, 0.3309]\big\rangle$ 0.3176 $X_4$ $\big\langle[0.6150, 0.7558], [0.1189 0.2442]\big\rangle$ 0.5039 $\big\langle[0.5565, 0.7642], [0.1158, 0.2358]\big\rangle$ 0.4846 $X_5$ $\big\langle[0.5379, 0.6504], [0.1917 0.3496]\big\rangle$ 0.3235 $\big\langle[0.5521, 0.6726], [0.1906, 0.3274]\big\rangle$ 0.3534 Ranking $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ 10 $X_1$ $\big\langle[0.3952, 0.6020], [0.2659, 0.3980]\big\rangle$ 0.1667 $\big\langle[0.4360, 0.6387], [0.2474, 0.3613]\big\rangle$ 0.2330 $X_2$ $\big\langle[0.5300, 0.6889], [0.1932, 0.3111]\big\rangle$ 0.3573 $\big\langle[0.5220, 0.7368], [0.1572, 0.2632]\big\rangle$ 0.4192 $X_3$ $\big\langle[0.4847, 0.6519], [0.2102, 0.3481]\big\rangle$ 0.2891 $\big\langle[0.4767, 0.6968], [0.1744, 0.3032]\big\rangle$ 0.3479 $X_4$ $\big\langle[0.6158, 0.7568], [0.1178, 0.2432]\big\rangle$ 0.5058 $\big\langle[0.5573, 0.7657], [0.1147, 0.2343]\big\rangle$ 0.4870 $X_5$ $\big\langle[0.5403, 0.6522], [0.1882, 0.3478]\big\rangle$ 0.3282 $\big\langle[0.5608, 0.6594], [0.1933, 0.3406]\big\rangle$ 0.3432 Ranking $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_3 \succ X_5 \succ X_1$ 25 $X_1$ $\big\langle[0.3979, 0.6046], [0.2610, 0.3954]\big\rangle$ 0.1730 $\big\langle[0.4217, 0.6566], [0.2234, 0.3434]\big\rangle$ 0.2558 $X_2$ $\big\langle[0.5338, 0.6920], [0.1874, 0.3080]\big\rangle$ 0.3652 $\big\langle[0.5570, 0.7047], [0.1689, 0.2953]\big\rangle$ 0.3988 $X_3$ $\big\langle[0.4878, 0.6530], [0.2051, 0.3470]\big\rangle$ 0.2943 $\big\langle[0.5176, 0.6687], [0.1871, 0.3313]\big\rangle$ 0.3339 $X_4$ $\big\langle[0.6165, 0.7576], [0.1167, 0.2424]\big\rangle$ 0.5075 $\big\langle[0.6057, 0.7394], [0.1297, 0.2606]\big\rangle$ 0.4774 $X_5$ $\big\langle[0.5426, 0.6536], [0.1847, 0.3464]\big\rangle$ 0.3325 $\big\langle[0.5600, 0.6525], [0.1944, 0.3475]\big\rangle$ 0.3353 Ranking $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$ $X_4\succ X_2\succ X_5 \succ X_3 \succ X_1$
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# Problem Set 6

Due by 11:59pm on Wednesday, March 24, 2021
Submission Instructions | PDF | Rmd

## Problem Set Instructions

This problem set is due on March 24, 11:59 pm Eastern time. Please upload a PDF of your solutions to gradescope. We will accept hand-written solutions but we strongly advise you to typeset your answers in Rmarkdown. Please list the names of other students you worked with on this problem set.

## Question 1

All probability distributions have moments, which are standard expressions that define its shape in ways you’ve already heard of and other more nuanced ways (the variance, the skew, kurtosis, etc.). Describing a population distribution (or empirical sample distribution) in terms of its moments is really useful in social science (e.g. the skew of income in the U.S. population is positive) Specifically, the $$n$$th central moment of a random variable $$X$$ is defined as $$E[(X-E[X])^n]$$, but it is more common to work with the $$n$$th moment defined as as $$E[X^n]$$ (getting rid of the $$E[X]^n$$ part).

Suppose the random variable $$X$$ for your population has the the following first four moments: $$E[X]=1/2$$, $$E[X^2]=1/2$$, $$E[X^3]=3/4$$, $$E[X^4]=3/2$$. Suppose you took an i.i.d. sample $$\{X_1,\ldots,X_{20}\}$$ of size 20 from this distribution. Let $$T=(X^2_1 +...+ X^2_{20})/20 = \overline{X^2}$$, an estimator of the second moment.

1. What are $$E[T]$$ and $$V(T)$$? Be sure to explain why.

2. Use the central limit theorem to approximate the probability (in R) that $$T$$ is less than or equal to 1.

## Question 2

Suppose sample $$n$$ i.i.d draws, whre $$X_i \sim \text{Unif}(0, 1)$$, for $$i \in 1...n$$. Denote the sample mean by $$\bar{X}_n$$.

1. What is the approximate asymptotic distribution of $$\bar{X}_n$$?

2. Now define a new random variables $$Z_n = \bar{X}_n^{1/2}$$. What is the approximate asymptotic distribution of $$Z_n$$?

3. Set $$n = \{3, 25, 1000\}$$ to see how well your approximation in part (b) works as a function of $$n$$ by running $$10000$$ simulations for each sample size. Report the ratio of the analytical variance estimate you found in part b over the simulated variance you find. Explain your results.

## Question 3

In class we learned that if a the variance of a sequence of random variables with finite mean goes to zero as $$n\to\infty$$, then the sequence will converge in probability to some value. But this is a sufficient condition, not a necessary one. To see this, consider the sequence of random variables $$X_n$$ be with probability distribution:

$X_n = \begin{cases} 0 & \text{with probability } 1 - 1/n \\ n & \text{with probability } 1/n \\ \end{cases}$

1. Find $$\mathbb{E}[X_n]$$.

2. Find $$\text{Var}(X_n)$$. Does the variance of the sequence grow or shrink as $$n$$ grows?

3. Use the definition of convergence in probability to show that $$X_n \xrightarrow{p} 0$$.

## Question 4

This problem will use the subprime data from last week’s problem set to walk you through a very common inference problem - testing whether the difference between two population values is non-zero. To begin this problem first download subprime.csv and load it into R.

We are going to be interested primarily in the loan.amount variable - the amount that each loan recipient received. Suppose a lawsuit has been filed in U.S. District Court by a group of Fort Myers women who claim that women in the area were loaned less money than men. The defendants – a group of local mortgage lenders – are vigorously denying these claims, and the case is now advancing to trial. Having heard about your expertise in this area, the federal judge hearing the case has brought you in to provide expert testimony. Your task in this problem is to assist the judge in her determination.

1. Suppose you were only able to interview $$100$$ male and $$100$$ female loan recipients at random, making them iid. To simulate this in R, set the seed to 02138 and draw $$100$$ observations randomly from the male subset of the subprime data and $$100$$ observations randomly from the female subset of the data. These $$200$$ observations constitute your sample. Calculate (1) the average loan amount (loan.amount) for women in your sample, (2) the average loan amount for men, and (3) and (4), the sample standard deviation for each. Report those results in a nicely formatted table.

2. Let $$\mu_{m}$$ and $$\mu_{w}$$ be the population average loan amount for men and women respectively. Let $$\sigma^2_{m}$$ and $$\sigma^2_{w}$$ be the population variances in loan amount for men and women respectively. Denote the sample average loan amounts for men by $$\bar{X}_{m}$$ and for women by $$\bar{X}_{w}$$.

What is the expected value of the sampling distribution of $$\bar{X}_{m} - \bar{X}_{w}$$? What is the variance of the sampling distribution of $$\bar{X}_{m} - \bar{X}_{w}$$?

1. Compute and report your sample difference in average loan amount for men and women. Recall that for large samples, the sampling distribution of a mean or difference-in-means is approximately normal. Suppose that we know that the true population $$\sigma^2_{m} = 32381.57$$ and $$\sigma^2_{w} = 19097.95$$. Using the normal approximation, what is the probability that we would observe a difference-in-means at least as extreme as the one in our sample if the true population difference-in-means $$\mu_{m} - \mu_{w}$$ equals 0? Note that by “at least as extreme,” we mean a value that is further away from $$0$$ than the value we observe - that is, $$P(|\bar{X}_{m} - \bar{X}_{w}| \ge \alpha)$$ where $$\alpha$$ is our observed value and $$||$$ is the absolute value operator.

Hint: In R you can get the probability that a normally distributed random variable takes on a value less than or equal to some value q using the command pnorm(q, mean, sd) where mean is the mean of the normal distribution and sd is the standard deviation.

1. Comment on your result in (c). Given what we observe in our sample, is it likely that there is no difference in loan amounts for men and women? A common threshold for “rejecting” our assumed hypothesis that $$\mu_{m} - \mu_{w} = 0$$ is observing a sample that would occur with probability $$.05$$ or less if that hypothesis were true (that is, a very unlikely sample). Would we reject the hypothesis that there is no difference in average loan amounts between men and women?


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# Tag Info

## New answers tagged optics

0

@Shimanshu Gupta There is more to it than just the wavelength. First of all I assume you mean the path difference at the detection screen So you need to say how far that distance is, A typical slit experiment could be 1 meter (1,000,000,000nm) from the slit to the detection screen. Also you need to know where on the detection screen you are measuring to. It ...

0

I think you did the Fourier Transform (FT) correctly, the first plot is just shifted incorrectly. I'm not familiar with Mathematica's FT but in MATLAB this would be computed as fftshift(fft(data)). What you have shown is what you would see if you only plotted fft(data). That should address the first three questions. The answer to question 4 is yes, your ...

1

At the risk of giving an answer which is to complete, I'm going to offer interpretations for the indicated terms. Let me start by pointing out that the mathematical expression you are using describe the rotation of vectors in the complex number plane. This is convenient for dealing with phase differences, but we normally assume that only the real ...

1

A balanced beam splitter provides a 50% probability for the light from each path to end in each of the two output paths. An asymmetric beam splitter provides different probabilities. I do not know of a distinction between balanced and symmetric beam splitters, but I can envision some authors caring about the relative phases that the light picks up by ...

3

Moonlight is quite reddened compared to Sunlight. The lunar albedo is roughly 0.07 at 400 nm and double that or 0.14 at 700 nm. It doesn't look terribly red because our blue and red receptors aren't pinned at those extremes, but for any given plant you would have to look at the specific receptor molecules at play and what their sensitivities are. See for ...

6

In addition to the other answers, an important physical difference is the perceivable inconsistency between the apparent intensity of the light source (moon vs. sun), and the spectral (color) distribution. Of course, this requires the (obviously wrong) conjecture, that both are thermal light sources, which behave approximately according to the black body ...

15

By far the primary physical difference is intensity- moonlight is much less bright than sunlight. Such flowers bloom when the light intensity is low- it does not matter if they are in moonlight or a greenhouse.

0

The physical difference between sunlight and the light scattered off of the moon aka moonlight is that the scattered light is polarized in nature. However, I am not sure if night blooming cereus uses this.

0

Actually, both the other answer and Veritasium himself are wrong. The shape shown is not due to the pinhole effect (the holes are way too large for that). In fact, the resulting image isn't even a circle! If the sun were a point source - an infinitesimally small source of light - the triangle hole would indeed result in a triangular image. (Projection of a ...

1

If the beam splitter is a non-polarising device and the setup is ideal, then statistically 50% of the photons go in one direction and 50% in the other. But this does not mean that at any point in time the ratio is 2:2. All other scenarios - with lower probability - are possible: 4:0, 3:1, 1:3 and 0:4.

1

Two things you might want to do: your data aren't perfect, in particular, there are two vertical jumps in them, one at approx -2.6 and another at approx 1.5 (not sure if this is what you refer to as "dead time"). These vertical jumps are in fact phase jumps of your oscillations. You won't be able to fit the entire data set with a $\cos^2$ function ...

1

You only see the color that aims at you. From the higher drops that is red. From the lower drops that is violet.

1

To do this efficiently, you want an optical device known as an X-cube. This is a cube that has three input faces: for red, green, blue. The output face combines all of those. Look up "X-cube optics" on the web, (so you don't get X-cube toys). Since you also want a fourth channel, that is a complication. People would typically add a dichroic plate ...

0

In a medium there in general is dispersion, so that the group velocity depends on the frequency. Note that the phase velocity can even exceed $c$, namely when $n<1$ such as occurs for x-rays. A pulse of light contains many frequencies, each with their own amplitude and phase. These components all travel at their own velocity and with their own extinction. ...

0

It sounds like you want a projector. Normally you use it to project a multicolored image on a screen. You want to project 4 different uniformly colored images on a small screen, one at a time. Or perhaps you just want to put 4 different LEDs some distance from the screen and turn on one at a time.

5

This is going to be really hard to do with LEDs since they're not collimated. If they were instead lasers then you would have some options at your disposal. If you have control over the polarization then you can combine the beams using polarizing beam splitters with lower power loss (but you would have differently polarized light at the output, may or may ...

4

Suppose an EM wave is emitted by some source in a step fashion. That is, ideally, prior to some time $t_0$ there is no emitted wave, and after that time there is an emitted wave with some constant non-zero amplitude. An initial wave will leave the emitter at the speed of light in a vacuum, even if the light is passing through matter. This initial "front&...

2

If your system is equivalent to a single lens with focal lenght $f$, then you can proceed like if you had one only lens with that $f$, so $M_L=\frac{s'}{s}$, with $s$ the distance to the object

1

How can an experiment built on the premise that light actually slows down in matter work, when light actually travels through the apparatus at the speed of light in a vacuum and only the phase-velocity of the superposed wave is lowered? Indeed. The correct functioning of the experiment is certainly good evidence supporting the premise upon which the ...

2

I don't have Feynman's lectures available, but I am not sure if you have interpreted him correctly. The fact that the speed of light is different in matter (and varies with wavelength) is the foundation for such profound effects as refraction and dispersion. Just take any prism and see the effects, and you'll know that the speed of light in matter is real.

0

This Numerical simulation by GONDRAN Alexandre is for $R=10\lambda$ The figure on p.5 here goes out to 30R. At $4000R$ you are out of the Fresnel regime. (See the condition on the Fresnel number in Wikipedia). I can't find an image of Fraunhofer diffraction from a disk, but by Babinet's principle it looks (roughly) like the negative of the Airy disk from ...

0

To locate the light circle, look at reflection angles from various parts of the mirror.

1

A metal is reflective because it is conductive. Conductivity will limit reflectivity as it gets thin and narrow. There are considerations beyond this of course if you want to make a solar sail. A metal wire mesh as thin as possible would be weak and fragile. Strength matters. Manufacturability of giant a sail matters. Ability to survive Earth's atmosphere ...

2

An ideal lens takes a point source that is located on its focal plane, say $\mathcal {F}$ emitting monochromatic homocentric rays/spherical waves and transforms them into parallel rays/plane waves whose direction (propagation phase) depends on the location of the point source in the focal plane relative to the symmetry axis. By reciprocity it also goes the ...

2

Assuming everything is correct, and this red light is not a reflection of some other light source or something: one possible way of achieving this is that the medium is absorbing green light and emitting the red light through photoluminescence. So in this case it's technically not light changing frequency in the medium; it's "new" light produced by ...

1

Notation in Figure-01 : \begin{align} n_1,n_2 & \boldsymbol{=} \texttt{indices of refraction} \nonumber\\ \theta_1,\theta_2 & \boldsymbol{=} \texttt{angles of incidence and transmission} \nonumber\\ \mathbf i & \boldsymbol{=} \texttt{unit vector on incident ray} \nonumber\\ \mathbf t & \boldsymbol{=} \texttt{unit vector ...

2

The inverse Fourier transform differs from the Fourier transform only by a flip operator, so for many purposes they can be treated as the same thing. Hence you can apply an inverse transform simply by using a second identical lens.

1

Yes. If you put your thumb in the direction of propagation of light, then as time passes, the electric field vector can move clockwise or anticlockwise only with the help of either your left hand or your right hand. Whichever hand it is gives the sense of polarisation. This tells you the direction in which field vector moves as time passes.

1

There is actually a very intuitive physical difference between left circularly polarized light and right circularly polarized light: left circularly polarized light gets transmitted through a left circular polarisation filter, while right circularly polarized light does not right circularly polarized light gets transmitted through a right circular ...

1

This is a deep question that can be treated at different levels of sophistication. I won't attempt to provide a complete answer, but I can chime in with some information that I hope is insightful. The most basic level of understanding comes from the classical picture of light as a wave in the electromagnetic field. (1) electrons (and all charged particles) ...

1

First off, you must be careful to avoid too sticking too much to a classical picture where the atom is mostly "empty" and the photon is a point particle that you imagine should "pass through it easily". Texts that state that the photon needs exactly the right energy to excite the electron may be trying to keep things simple, but I ...

1

Some substances do absorb photons of just the right frequency. When you shine a white light on them, you see sharp absorption lines in the spectrum of the reflected or transmitted light. The excited atoms may decay and emit the light again. If they do, they emit in all directions. Sometimes they decay in more complex ways. For example, a fluorescent atom may ...

1

First, you confuse permittivity with refractive index. The refractive index of silver for the visible wavelengths is almost 0. Secondly, all laws still apply (ie Maxwell eqs). However, I think its easier to think of it this way: metals have conducting electrons, and they oscillate mostly freely. They are not bound in a potential which affects their mobility ...

0

Thanks for all good comments and answers! After thinking again about the problem I'm posting an additional answer because of one major reason that wasn't clear to me and which is explaining why you see a faint "visual-like" image in thermal images: Diffuse illumination, pretty much like in normal visual images. In normal images you see things ...

0

The suns rays here on earth are only approximately parallel due to the angular extent of the sun. Light coming from opposite limbs of the sun will be about half a degree from parallel. For some purposes, that's close enough and we consider them parallel. For other purposes it isn't. The atmospheric distortion isn't going to add much more than the half ...

2

Such fibers are waveguides which work as photonic crystals, that is, periodic structures with alternating layers of dielectrics. Photonic crystals have stopbands, i.e. spectral regions where the reflectance is high (and transmittance is low). These stopbands allow the waveguide to keep the light inside and actually conduct it (not lose it due to scattering). ...

1

Dispersion, $D$ is defined as: $$D=\frac{\partial^2k}{\partial \omega^2}$$ for the whole system, which is length independent (answering second question). For the first question, then you can see that for highly multimode fibres, waveguinding will not affect propagation much, and the dispersion relation is that of the core's medium....

0

There are quite a few questions here so I will only answer briefly. No, the emerging photons are not entangled. In fact, they do not emergy in pairs. Since the input states can both be described by canonical coherent states, so too can the outputs. 2,3. Still not entangled. SPDC does indeed generate entanglement while [beam splitters with coherent-state ...

0

Wire grid polarizers almost perfectly reflects the orthogonal polarization that does not pass thru the polarizer. There is a small amount of Joule heating in the wires.

1

You are correct, the angle of the transmitted beam depends on the order of diffraction and the frequency shifted. The effect can be significant, for example, we use AOMs in our lab to shift the frequency of a beam to image $^{87}$Rb atoms, we see a sudden drop in intensity after some change in frequency due to the shifted beam not coupling into the fiber ...

-1

The "duality" only comes in when you make a measurement. So the confusion about this duality is really just confusion about the measurement problem in Quantum Mechanics. The fact that light is a wave is reflected in the description of the electromagnetic field as an element of a complex linear vector space. The particle aspect is simply reflected ...

-2

While the classical electrodynamics studies light as a wave, the quantum theory takes account for wave-particle duality, i.e., it pretty much already takes a single prospective. The previous answers used term quantum electrodynamics, which makes sense in order to differentiate it from quantum mechanics, which does not quantize the light. This designation ...

-4

I interpret the question as follows: is there a field that studies electromagnetism either exclusively from the wave concept or exclusively from the particle perspective. The answer to this question is no. There are fields of applied physics that use only the wave perspective, but this is a practical choice. Ray optics and wave optics are tools for optical ...

0

$v=f\lambda$ refers to the relationship between the speed of a wave that is travelling through a medium to its frequency and wavelength. Now, forget the waves and come to oscillations for example like the simple pendulum. The position of the Bob is $x=x_0 sin(\omega t+ \phi)$ . If you differentiate this with respect to time, we get $\dot x = \omega x_0 cos( \... 0 A telescope is a system of two lenses Diagram of a simple telescope. Parallel light rays enter from the left, pass through the objective lens, come to a focus at the focal plane, and exit through the eyepiece lens. The focal length of the objective is F, and the focal length of the eyepiece is f. The magnification happens from the first lens on the left ... 0 If I understand the setup correctly, then the "fringes" you'll see would not be due to interference but because of different reflected orders passing through the lens at different angle producing different spots on the screen. The first beam enters the lens at an angle$\theta_1\$ given according to Snell's law by $$\sin(\theta_1)=n\sin(\theta) .$$...

0

As José Andrade pointed out, Amplitude is just how strong the light is. However, there is a problem with José's assumption that: the color does not change. They look the same. And @Not_Einstein is right when he writes: as the intensity is one component of color, the color will change. Maybe not the hue, but the color will. Bright red is not the same ...

20

Yes, that theory is called quantum electrodynamics, or QED. It predicts all of the “particle” behaviors and all of the “wave” behaviors of light from a single unified mathematical model. QED was the first successful quantum field theory. Since its development other similar field theories have been developed for the strong and weak nuclear forces. So ...

Top 50 recent answers are included


text:
BREAKING NEWS
Twisted cubic

## Summary

In mathematics, a twisted cubic is a smooth, rational curve C of degree three in projective 3-space P3. It is a fundamental example of a skew curve. It is essentially unique, up to projective transformation (the twisted cubic, therefore). In algebraic geometry, the twisted cubic is a simple example of a projective variety that is not linear or a hypersurface, in fact not a complete intersection. It is the three-dimensional case of the rational normal curve, and is the image of a Veronese map of degree three on the projective line.

## Definition

The twisted cubic is most easily given parametrically as the image of the map

${\displaystyle \nu :\mathbf {P} ^{1}\to \mathbf {P} ^{3}}$

which assigns to the homogeneous coordinate ${\displaystyle [S:T]}$  the value

${\displaystyle \nu :[S:T]\mapsto [S^{3}:S^{2}T:ST^{2}:T^{3}].}$

In one coordinate patch of projective space, the map is simply the moment curve

${\displaystyle \nu :x\mapsto (x,x^{2},x^{3})}$

That is, it is the closure by a single point at infinity of the affine curve ${\displaystyle (x,x^{2},x^{3})}$ .

The twisted cubic is a projective variety, defined as the intersection of three quadrics. In homogeneous coordinates ${\displaystyle [X:Y:Z:W]}$  on P3, the twisted cubic is the closed subscheme defined by the vanishing of the three homogeneous polynomials

${\displaystyle F_{0}=XZ-Y^{2}}$
${\displaystyle F_{1}=YW-Z^{2}}$
${\displaystyle F_{2}=XW-YZ.}$

It may be checked that these three quadratic forms vanish identically when using the explicit parameterization above; that is, substitute x3 for X, and so on.

More strongly, the homogeneous ideal of the twisted cubic C is generated by these three homogeneous polynomials of degree 2.

## Properties

The twisted cubic has the following properties:

• It is the set-theoretic complete intersection of ${\displaystyle XZ-Y^{2}}$  and ${\displaystyle Z(YW-Z^{2})-W(XW-YZ)}$ , but not a scheme-theoretic or ideal-theoretic complete intersection; meaning to say that the ideal of the variety cannot be generated by only 2 polynomials; a minimum of 3 are needed. (An attempt to use only two polynomials make the resulting ideal not radical, since ${\displaystyle (YW-Z^{2})^{2}}$  is in it, but ${\displaystyle YW-Z^{2}}$  is not).
• Any four points on C span P3.
• Given six points in P3 with no four coplanar, there is a unique twisted cubic passing through them.
• The union of the tangent and secant lines (the secant variety) of a twisted cubic C fill up P3 and the lines are pairwise disjoint, except at points of the curve itself. In fact, the union of the tangent and secant lines of any non-planar smooth algebraic curve is three-dimensional. Further, any smooth algebraic variety with the property that every length four subscheme spans P3 has the property that the tangent and secant lines are pairwise disjoint, except at points of the variety itself.
• The projection of C onto a plane from a point on a tangent line of C yields a cuspidal cubic.
• The projection from a point on a secant line of C yields a nodal cubic.
• The projection from a point on C yields a conic section.

## References

• Harris, Joe (1992), Algebraic Geometry, A First Course, New York: Springer-Verlag, ISBN 0-387-97716-3.


text:
# Express the edge length of a cube as a function of the cubes diagonal

1. Jan 24, 2006

### jacy

Hello,
Here is my what am trying to solve.
Express the edge length of a cube as a function of the cubes diagonal. Then express the area as a function of diagonal length if the side is x.

This is what i know. The area of a cube is 6x^2 where x is the length. But in the problem i have to express the side length as a function of cubes diagonal. Can someone suggest me the relation between the side length and the diagonal, thanks.

Last edited by a moderator: Jan 7, 2014
2. Jan 24, 2006

### d_leet

Think abut the pythagorean theorem and what is true about cubes but not about all rectangular prisms.

And I'm not entirely sure from your post whether you want a formula for surface area or volume.

3. Jan 25, 2006

### jacy

Okay on using the pythagorean theorem am getting
y=2x^2 where y is the diagonal lenght and x is the side length.

Now the area of the cube = 12x^4
am i right. Plz let me know

4. Jan 25, 2006

### StatusX

You gave the corret area formula in your first post. And remember that the pythagorean theorem is a2+b2=c2. You forgot to square c in your formula. And are you interested in the diagonal of a face or of the whole cube? Because they have different lengths.

5. Jan 26, 2006

### jacy

Thanks for correcting me. Now i have y=sqrt 2x

the area of the cube will be 24 x2
The question says "express the area as a function of diagonal length", so am not sure whether to consider the diagonal of a face or the whole cube. 24x2 will be the answer right.

6. Jan 26, 2006

### HallsofIvy

Staff Emeritus
Be careful- the "diagonal" of a cube is NOT the diagonal across a face. It is the line from one vertex of the cube to the opposite vertex. Yes, if y is the length of a side, then $x= \sqrt{2}y$ is the length of the diagonal of a face of the cube. Now, the diagonal from one vertex of a cube to its opposite vertex is the hypotenuse of a right triangle, having one side of length y and the other $\sqrt{2}y$. If you let z be the length of that diagonal, z2= y2+ 2y2.

7. Jan 26, 2006

### jacy

Thanks, so the length of the diagonal will be z2=3y2
Then the area will be 18y2 square units. Please correct me if am wrong, thank you.


text:
# Calculating the signal-to-noise ratio for CW (Morse Code) signals?

How is the signal-to-noise ratio for received CW and QRSS CW (and/or other ultra narrow-band DSP digital modes) typically computed? e.g. What noise bandwidth should be used to measure the spectrum power against the narrow CW "on" signal power?

Should one use the DSP detection bandwidth (roughly 1X the WPM), the human decode bandwidth (approximately 3X to 5X the WPM according to an ISO or ITU document), the actual bandwidth used by the transmitter rise time, or some other standard radio bandwidth (SSB LSB filter width?)?

How should the CW S/N be computed to compare its data rate (at a given WPM) against the theoretical Shannon limit for that S/N?

Is the same S/N number meaningful depending on whether the noise is mostly Gaussian noise, white noise, or 1/f RF noise?

• I don't think there's a "typical" calculation, but only application-specific ones: Why would you care about some arbitrary SNR definition when it doesn't describe the problem your receiver is facing? May I ask the reason for your question? – Marcus Müller Jan 7 at 15:30
• What do academic papers use for comparison of actual (ultra narrow band?) RF signals versus the Shannon channel limit for S/N? Are any numbers for CW comparable? – hotpaw2 Jan 7 at 16:51
• I'd assume the most sensible papers, they'd use the effective bandwidth of the receiver when describing an S/N from a receiver's perspective. Since ultra-narrow band kind of suggests the channel is flat, it's fair to assume that every practical transceiver system would use matched filtering (since that maximizes the SNR on the flat, AWN channel), so that'd be the equivalent noise bandwidth of the matched filter. – Marcus Müller Jan 7 at 17:46
• So one answer for CW might be to use the ENB of a typical (averaged across the specs of many of the most popular HF rigs) CW filter? – hotpaw2 Jan 7 at 18:52
• hm, yes that's certainly something that would be representative. But I think you could honestly also argue very much on the actual bandwidth/WPM basis and define that a realistically fast keyed channel needs so and so many Hz bandwidth and no more. – Marcus Müller Jan 7 at 19:13

The answer can be fond in The ARRL Handbook 2019, Vol 3, although it's spreaded across different chapters. In short, SNR is typically calculated for the noise floor of 2500 Hz SSB signal. Particularly this is how WSJT-X calculates negative SNRs for FT8 and other modes.

Now the trick is that by deviding the bandwidth in half you decrease the noise floor by 3 dB assuming the noise is distributed evenly. When I use 500 Hz wide DSP filter in my Yaesu FT-891 I typically see S2-S3 noise floor on 20m, and S4 when 2500 Hz bandwidth is used. Now keeping in mind that one S-unit is approximately 6 dB we get:

>>> from math import log2
>>> 3*log2(3000/500)/6
1.292481250360578

And we really see 1-2 less S-units, as expected.

Now back to CW. The ARRL Handbook claims with the reference to "The ITU Classification of Emission Standards" that the bandwidth of CW signal can be estimated as:

BW = WPM * 0.8 * 5

For instance, a 17WPM signal has a bandwidth about 68Hz. This means that the signal can be received:

>>> 3*log2(2500/68)
15.600748614897329

... 15.6 dB "under the noise floor".


text:
## A random sample of 30 patties that were inspected over the course of the last week revealed that the average weight was 95.0 grams. The stan

Question

A random sample of 30 patties that were inspected over the course of the last week revealed that the average weight was 95.0 grams. The standard deviation was 0.25 grams. What percentage of the deliveries is likely to be outside the specification limits (outside the interval of [94.5, 95.5])

in progress 0
2 weeks 2021-09-04T20:51:25+00:00 1 Answers 0 views 0

4.56% of the deliveries are likely to be outside the specification limits.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The average weight was 95.0 grams. The standard deviation was 0.25 grams.

This means that

What percentage of the deliveries is likely to be outside the specification limits (outside the interval of [94.5, 95.5])?

Less than 94.5, or more than 95.5. Since the normal distribution is symmetric, these probabilities are the same, so we can find one of them and multiply by two.

The probability that it is less than 94.5 is the p-value of Z when X = 94.5. So

has a p-value of 0.0228

2*0.0228 = 0.0456

0.0456*100% = 4.56%

4.56% of the deliveries are likely to be outside the specification limits.


text:
# Insertion sort (C)

(Redirected from Insertion sort (C, simple))
Other implementations: ACL2 | C | C, simple | Erlang | Forth | Haskell | Java | OCaml | Python, arrays | Standard ML | Visual Basic .NET

This is a simple example of the insertion sort sorting algorithm, written in C. Since this implementation is intended to be as simple and easy to under as possible, rather than particularly useful in practice, we will focus on sorting a simple array of ints of a given length.

##  Main algorithm

When sorting an array with insertion sort, we conceptually separate it into two parts:

• The list of elements already inserted, which is always in sorted order and is found at the beginning of the array;
• The list of elements we have yet to insert, following.

In outline, our primary function looks like this:

<<insertion_sort>>=
/* Sort an array of integers */
void insertion_sort(int a[], int length) {
int i;
for (i=0; i < length; i++) {
insert a[i] into sorted sublist
}
}

To insert each element, we need to create a hole in the array at the place where it belongs, then place it in it. We can combine the creation of the hole with the searching for the place by starting at the end and shifting each element up by one until we find the place where it belongs. This overwrites the element we're inserting, so we have to save it in a variable first:

<<insert a[i] into sorted sublist>>=
/* Insert a[i] into the sorted sublist */
int j, value = a[i];
for (j = i - 1; j >= 0; j--) {
if (a[j] <= value) break;
a[j + 1] = a[j];
}
a[j + 1] = value;

On average, we move about half the elements of the already-sorted sublist to insert each element. Consequently, the total number of assignments on average to sort n elements is:

$\sum_{i=0}^{n-1} i/2 = \frac{n(n+1)}{4} = O(n^2)$

In the worst case, the list is in reverse sorted order, so that each element must push up all other elements encountered so far. The total number of assignments in this case is twice the average case:

$\sum_{i=0}^{n-1} i = \frac{n(n+1)}{2}$

##  Test driver

To try out the sort, we can write a simple test driver file with the following structure:

<<insertion_sort.c>>=

insertion_sort

test main

Our test main will create a small array, sort it, then print out the result. We use the traditional "sizeof division" trick to get the array length at compile time:

<<test main>>=
int main(void) {
test main variable declarations
int a[] = {5,1,9,3,2};
insertion_sort(a, sizeof(a)/sizeof(*a));
print out a
return 0;
}

A better test main might ask for numbers from the user. We print out the array by just invoking printf on each element in turn:

<<test main variable declarations>>=
unsigned int i;
<<print out a>>=
/* Print out a */
for(i=0; i<sizeof(a)/sizeof(*a); i++) {
printf("%d\n", a[i]);
}

We use unsigned because the sizeof division result is unsigned. We'll need to include stdio.h for printf:

<<header files>>=
#include <stdio.h>

This completes our simple exposition of iterative insertion sort in C. On UNIX, you can build and run this sample using the following script:

<<build_and_run.sh>>=
#!/bin/bash
gcc -Wall -O2 insertion_sort.c -o insertion_sort
./insertion_sort

hijacker
hijacker
hijacker
hijacker


text:
## Momentum polarization: an entanglement measure of topological spin and chiral central charge

Date:
2012-12-31
Author(s):

Hong-Hao Tu, Yi Zhang, Xiao-Liang Qi

Reference:

arXiv:1212.6951v1

Topologically ordered states are quantum states of matter with topological ground state degeneracy and quasi-particles carrying fractional quantum numbers and fractional statistics. The topological spin $\theta_a=2\pi h_a$ is an important property of a topological quasi-particle, which is the Berry phase obtained in the adiabatic self-rotation of the quasi-particle by $2\pi$. For chiral topological states with robust chiral edge states, another fundamental topological property is the edge state chiral central charge $c$. In this paper we propose a new approach to compute the topological spin and chiral central charge in lattice models by defining a new quantity named as the momentum polarization. Momentum polarization is defined on the cylinder geometry as a universal subleading term in the average value of a "partial translation operator". We show that the momentum polarization is a quantum entanglement property which can be computed from the reduced density matrix, and our analytic derivation based on edge conformal field theory shows that the momentum polarization measures the combination $h_a-\frac{c}{24}$ of topological spin and central charge. Numerical results are obtained for two example systems, the non-Abelian phase of the honeycomb lattice Kitaev model, and the $\nu=1/2$ Laughlin state of a fractional Chern insulator described by a variational Monte Carlo wavefunction. The numerical results verifies the analytic formula with high accuracy, and further suggests that this result remains robust even when the edge states cannot be described by a conformal field theory. Our result provides a new efficient approach to characterize and identify topological states of matter from finite size numerics.


text:
# Model Comparison in bvartools

#### 2022-01-21

bvartools comes with the functionality to set up and produce posterior draws for multiple models in an effort to reduce the time required for this potentially laborious process. This vignette illustrates how the package can be used to set up multiple models, produce prior specifications, obtain posterior draws and select the model with the best fit in a few steps.

## Data

For this illustrations the data set E1 from Lütkepohl (2006) is used. It contains data on West German fixed investment, disposable income and consumption expenditures in billions of DM from 1960Q1 to 1982Q4. Like in the textbook only the log-differenced series up to 1978Q4 are used.

library(bvartools)

set.seed(123456) # Set seed for reproducibility

data <- diff(log(e1)) * 100 # Obtain log-differences

# Use date up to 1978Q4
data <- window(data, end = c(1978, 4))

# Plot
plot(data)

## Set up models

Functions gen_var can be used to obtain a list of different model specifications. In the following example five models with an intercept and increasing lag orders are generated.

object <- gen_var(data, p = 0:4,
deterministic = "const",
iterations = 5000, burnin = 1000)

All objects use the same amounts of available observations to ensure consistency for the calculation of information criteria for model selection.

## Priors

Function add_priors can be used to produce priors for each of the models in object models.

coef = list(v_i = 0, v_i_det = 0),
sigma = list(df = "k", scale = 0.0001))

## Estimation

Posterior draws can be obtained using function draw_posterior. The function allows to specify the number of CPUs, which are available for parallel computing.

object <- draw_posterior(object, mc.cores = 3)

If multiple models are estimated the function produces an object of class bvarlist, which is a list of objects of class bvar. Thus, each element of the list can be used for further analysis.

## Model selection

If function summary is applied to an object of class bvarlist, it produces a table of information criteria for each specification. The information criteria are calculated based on the posterior draws of the respective model and calculated in the following way:

• Log-likelihood: $$LL = \frac{1}{R} \sum_{i = 1}^{R} \left( \sum_{t = 1}^{T} -\frac{K}{2} \ln 2\pi - \frac{1}{2} \ln |\Sigma_t^{(i)}| -\frac{1}{2} (u_t^{{(i)}\prime} (\Sigma_t^{(i)})^{-1} u_t^{(i)} \right)$$ for each draw $$i$$ and $$u_t = y_t - \mu_t$$;
• Akaika information criterion: $$AIC = 2 (Kp + M (s + 1) + N) - 2 LL$$;
• Bayesian information criterion: $$BIC = ln(T) (Kp + M (s + 1) + N) - 2 LL$$;
• Hannan-Quinn information criterion: $$HQ = 2 ln(ln(T)) (Kp + M (s + 1) + N) - 2 LL$$.

$$K$$ is the number of endogenous variables and $$p$$ the lag order of the model. If exogenous variables were used $$M$$ is the number of stochastic exogenous regressors and $$s$$ is the lag order for those variables. $$N$$ is the number of deterministic terms.

summary(object)
#>   p s        LL      AIC      BIC       HQ
#> 1 0 0 -420.5955 843.1909 845.4536 844.0907
#> 2 1 0 -413.5161 835.0323 844.0830 838.6314
#> 3 2 0 -405.9250 825.8500 841.6887 832.1485
#> 4 3 0 -408.6642 837.3284 859.9552 846.3264
#> 5 4 0 -406.7913 839.5825 868.9974 851.2799

Since all information criteria have the lowest value for the model with $$p = 2$$, the third element of object is used for further analyis.

plot(irf(object[[3]], impulse = "income", response = "cons", n.ahead = 10))

## Literature

Chan, J., Koop, G., Poirier, D. J., & Tobias, J. L. (2019). Bayesian Econometric Methods (2nd ed.). Cambridge: University Press.

Lütkepohl, H. (2006). New introduction to multiple time series analysis (2nd ed.). Berlin: Springer.


text:
# Angle marker

Martin McBride, 2020-08-23
Tags geometry angle
Categories generativepy generative art

generativepy has moved to the pythoninformer.com website, and this page might not be up to date. Please visit the new location.

The angle_marker function adds an angle marker to the path, that can then be stroked. Angle markers look like this:

An angle marker can have 1, 2 or 3 arcs. The 2 and 3 arc forms are often used to indicate that two angles are equal. You can also draw a right angle marker, as shown.

## angle_marker function

angle_marker(ctx, a, b, c, count=1, radius=8, gap=2, right_angle=False)

Parameter Type Description
ctx Context The Pycairo Context to draw to
a (number, number) Tuple (x, y) for point a
b (number, number) Tuple (x, y) for point b
c (number, number) Tuple (x, y) for point c
count int Number of arcs on angle, 1, 2 or 3.
gap number Gap between arcs in user units.
right_angle boolean True for a right angle.

Draws an angle arc for the angle abc. The arc is centred on point b, and shows the angle between lines ab and cb in the clockwise direction.

If you are using a flipped coordinate system (see the setup function in the drawing module), the angle is drawn in the counterclockwise direction.

count allows you to draw double or triple arcs to indicate different pairs of equal angles. If count is not 1, gap sets the distance between the arcs - typically a value that is about twice the line thickness usually looks best.

right_angle can be set to true to draw a right angle marker rather than an arc. In that case, count is ignored.

Note that it is up to your code do ensure that the angle actually is a right angle before setting this. If you set this flag when the angle isn't a right angle (or very close), you will get strange looking result. If you don't set this flag when the angle is a right angle, a normal angle arc will be drawn. This is to avoid spurious effects in animations if a changing angle happens to be exactly 90 degrees in one frame/

## Example

Here is an example that shows the angle_marker, tick and paratick functions in use:

from generativepy.drawing import make_image, setup
from generativepy.color import Color
from generativepy.geometry import line, polygon, angle_marker, tick, paratick

'''
Create rectangles using the geometry module.
'''

def draw(ctx, width, height, frame_no, frame_count):
setup(ctx, width, height, background=Color(0.8))

ctx.set_source_rgba(*Color(0, 0, 0.5))
ctx.set_line_width(3)

## Draw lines with ticks and paraticks
a = (50, 50)
b = (50, 150)
line(ctx, a, b)
ctx.stroke()
tick(ctx, a, b, length=12, gap=6)
ctx.stroke()

a = (100, 50)
b = (150, 150)
line(ctx, a, b)
ctx.stroke()
tick(ctx, a, b, 2, length=12, gap=6)
ctx.stroke()

a = (250, 50)
b = (200, 150)
line(ctx, a, b)
ctx.stroke()
tick(ctx, a, b, 3, length=12, gap=6)
ctx.stroke()

a = (350, 50)
b = (350, 150)
line(ctx, a, b)
ctx.stroke()
paratick(ctx, a, b, length=12, gap=6)
ctx.stroke()

a = (400, 50)
b = (450, 150)
line(ctx, a, b)
ctx.stroke()
paratick(ctx, a, b, 2, length=12, gap=6)
ctx.stroke()

a = (550, 150)
b = (500, 50)
line(ctx, a, b)
ctx.stroke()
paratick(ctx, a, b, 3, length=12, gap=6)
ctx.stroke()

## Draw lines with angles
a = (50, 250)
b = (50, 450)
c = (150, 450)
polygon(ctx, (a, b, c), closed=False)
ctx.stroke()
angle_marker(ctx, a, b, c, radius=24, gap=6, right_angle=True)
ctx.stroke()

a = (250, 250)
b = (200, 450)
c = (300, 450)
polygon(ctx, (a, b, c), closed=False)
ctx.stroke()
angle_marker(ctx, a, b, c, 3, radius=24, gap=6)
ctx.stroke()

a = (300, 250)
b = (400, 300)
c = (500, 300)
polygon(ctx, (a, b, c), closed=False)
ctx.stroke()
angle_marker(ctx, c, b, a, radius=24, gap=6)
ctx.stroke()

a = (300, 350)
b = (400, 400)
c = (500, 400)
polygon(ctx, (a, b, c), closed=False)
ctx.stroke()
angle_marker(ctx, a, b, c, 2, radius=24, gap=6)
ctx.stroke()

make_image("/tmp/geometry-markers.png", draw, 600, 500)


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

The graphs are different in that $y = |x| - 4$ is translated or shifted vertically four units down on the y-axis and $y = |x - 4|$ is translated or shifted $4$ units horizontally to the right on the x-axis. The graphs are the same because they both represent absolute value functions and, therefore, have $V$ shapes.