The Official Math/Physics/Whatever Homework questions thread - Page 351

I'm taking my first physics course. I have no prior background in physics.

The topic is special relativity. Specifically, it is about (inertial) reference frames.

Which of the following are (at least nearly) intertial reference frames (some could be considered both inertial and non-intertial)?

a) A nonrotating frame floating in deep space.
b) A rotating frame floating in deep space.
c) A nonrotating frame attached to the sun.
d) A frame attached to the surface of the earth.
e) A frame attached to a car moving at a constant velocity.
f) A frame attached to a roller-coaster car.

My thoughts:

a) is technically not mentioning constant velocity, so this is non-inertial.
b) rotation implies acceleration, so non-inertial.
c) technically the Sun is rotating, so it is not inertial. However, due to gravity, we can consider it inertial for practical purposes.
d) technically the Earth is rotating, so it is not inertial. However, due to gravity, we can consider it inertial for practical purposes.
e) inertial frame, because constant velocity.
f) a roller-coaster car experiences acceleration, so non-inertial.

Appreciate all input.

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11-14-2015 , 06:52 PM

#8752

masque de Z

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Quote:

Originally Posted by Enough Is Enough

a) A nonrotating frame floating in deep space.
b) A rotating frame floating in deep space.
c) A nonrotating frame attached to the sun.
d) A frame attached to the surface of the earth.
e) A frame attached to a car moving at a constant velocity.
f) A frame attached to a roller-coaster car.
.

a) is likely inertial depending what they mean by floating (of course nothing really is inertial as long as you have gravity even from distant galaxies but the accelerations can be locally very small to ignore them. If you are not accelerating in deep space the distant objects cancel each other more or less and the end result is a very tiny acceleration (ie rotation around the galactic center) and we may ignore it and call it inertial.

b) is not inertial ie centripetal force etc.

c) Well the surface of the sun is not inertial, it has rather big gravitational acceleration (much more than earth's). The center is close to 0 gravity but under tremendous random fluctuation pressure and the asymmetries in the surrounding system and the distant planets affect how you perceive acceleration even at the center so i dont know what they want here. I would still like to say no in general.

d) is not inertial as earth is rotating and also has gravitational acceleration at the surface.

e) (same as earth above ) the car is an inertial frame only if you ignore gravity (also rotation of earth, moon etc) ie if you leave an object to fall inside the car it will be accelerated to the floor of the car. So the car at constant speed is not an inertial frame but will appear as one in situations you ignore the vertical forces (and the earth rotation and distant celestial bodies) that cancel each other say when in contact. The space station (any object in orbit say) is closer to a real inertial frame (although there is microgravity really there too)

f) is also not inertial.

Quote

11-14-2015 , 09:20 PM

#8753

MXdotCH

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Join Date: Jan 2009 Posts: 924

I am doing some statistical analysis in R for a class and am confused about how to assess whether one predictor inside my model is significant using partial F-test

Here is my R ouput:

Call:
lm(formula = w ~ a + b + I(b^2) + c + d, data = W)

Coefficients:
(Intercept) a b I(b^2) c
-2.120092 0.530091 0.204842 -0.004127 0.133669
d
-1.790226

> anova(fit.Complete)
Analysis of Variance Table

Response: w
Df Sum Sq Mean Sq F value Pr(>F)
a 1 1179.7 1179.73 142.470 < 2.2e-16 ***
b 1 432.5 432.52 52.233 1.768e-12 ***
I(b^2) 1 315.6 315.62 38.116 1.344e-09 ***
c 1 527.0 526.99 63.641 9.579e-15 ***
d 1 399.6 399.65 48.263 1.119e-11 ***
Residuals 520 4305.9 8.28
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

I need to asses whether b^2 is significant if all the other predictors are also included in the model.

Am I able to confirm it is significant just by the F value of 38.116 from the Anova table? But one thing which confuses me is that these F values are sequential right? So the F value of b^2 is when a and b are in the model, and c and d arent?

Am I reading this right?

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11-15-2015 , 09:27 AM

#8754

reno expat

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Quote:

Originally Posted by MXdotCH

I need to asses whether b^2 is significant if all the other predictors are also included in the model.

Am I able to confirm it is significant just by the F value of 38.116 from the Anova table? But one thing which confuses me is that these F values are sequential right? So the F value of b^2 is when a and b are in the model, and c and d arent?

Am I reading this right?

I believe you are reading this right. To do partial F-tests in R, You want to have some control of which restricted model you are using. So instead of just doing anova(model) you want to do anova(model1, model2). I forget what the order is but if you get a negative test statistic, just flip them. The code would look something like

full <- lm(formula = w ~ a + b + I(b^2) + c + d, data = W)
restricted <- lm(formula = w ~ a + b + c + d, data = W)
anova(full, restricted)

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11-15-2015 , 05:04 PM

#8755

Ryanb9

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Trying to do Tn of integral from 0 to pi/2 of sqrt(sin(x)) dx when N = 6

Formula is Tn = 1/2 * delta x * ( f(x)1 + 2*f(x)2 + 2*f(x)3 + f(x)n)

I thought this should work but its not... I'm sure I'm making a stupid mistake that I'm not seeing somewhere...

edit: never mind, found the stupid mistake. I needed to pull that 2 out from inside the function b/c its 2 * f(x) not f(2*x)

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11-15-2015 , 05:40 PM

#8756

Ryanb9

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In another question, they say find Tn and Mn (Tn is trapezoidal rule and m is midpoints) to do integral of ln(x)dx from 1 to 2, where N = 5.

The answer they give in the back of the book says T4 = 0.3846... and M5 = 0.3871....

why did they do T4 instead of T5? It says N = 5.

edit: i think it might just be a typo in back of book, I think they meant T5 = and M5 =

because wolphram said their T4 answer was really the answer to T5 : http://www.wolframalpha.com/input/?i...ule%2C+5+areas

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11-17-2015 , 03:49 PM

#8757

Enough Is Enough

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Let A and B be n x n matrices and let C = AB. Prove that if B is singular then C must be singular by using this theorem:

Let A be an n x n matrix. The following are equivalent:

(a) A is nonsingular.
(b) Ax = 0 has only the trivial solution 0.
(c) A is row equivalent to I.

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11-17-2015 , 04:45 PM

#8758

Cueballmania

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I'd try a proof by contradiction.

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11-17-2015 , 04:52 PM

#8759

PairTheBoard

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Quote:

Originally Posted by Enough Is Enough

Should be easy to show using (b) and associativity of matrix multiplication. (A vector x is a column matrix)

PairTheBoard

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11-17-2015 , 06:51 PM

#8760

Enough Is Enough

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I don't see it.

Do you mean I should start with Ax = 0 and assume X is a solution?

X = I*X = (A^-1*A)*X = A^-1*(A*X) = A^-1*0 = 0.

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11-17-2015 , 07:02 PM

#8761

masque de Z

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All you need to show is that if B has eg a nontrivial solution Bx=0
(with x not 0 column)then C has it too Cx=0 also so C becomes a singular matrix as well.

By the way row equivalent to identity is equivalent to invertible too (ie A^-1 exists for some A) although that was not given in the OP. But you do not necessarily need to use that here.

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11-17-2015 , 07:29 PM

#8762

Cueballmania

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Maybe you're getting bogged down in the notation. A way to show that any matrix is singular is by finding a non-zero vector such that the matrix times this vector is the zero vector.

Here's an important question that helps with your proof:
What happens when you multiply a matrix by the zero vector?

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11-17-2015 , 08:24 PM

#8763

PairTheBoard

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You need the associative property of matrix multiplication to realize that:

(AB)x = A(Bx)

PairTheBoard

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11-18-2015 , 11:02 AM

#8764

Enough Is Enough

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Think I have it.

B is singular, so Bx = 0 with x not being the zero vector.
C = AB, postmultiplying by x gives Cx = (AB)x = A(Bx)
Cx = A*0 = 0
Cx = 0 and we know that x is not the zero vector, which means C is singular.

Thanks for the help

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11-18-2015 , 04:35 PM

#8765

Enough Is Enough

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Prove that B is row equivalent to A if and only if there exists a nonsingular matrix M such that B = M*A.

My thoughts:

If B is row equivalent to A, then we can write B = E_(k)*E_(k-1)...E_(1)*A or B = M*A with M = E_(k)*E_(k-1)...E_(1).

But how do I go from B = M*A to B = E_(k)*E_(k-1)...E_(1)*A without being circular and is this even the right way to approach this?

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11-18-2015 , 10:46 PM

#8766

PairTheBoard

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Quote:

Originally Posted by Enough Is Enough

From Wiki on Row Equivalence
https://en.wikipedia.org/wiki/Row_equivalence
----------------------------------------
Strictly speaking, in linear algebra, two matrices are row equivalent if one can be changed to the other by a sequence of elementary row operations. Alternatively, two m × n matrices are row equivalent if and only if they have the same row space. The concept is most commonly applied to matrices that represent systems of linear equations, in which case two matrices of the same size are row equivalent if and only if the corresponding homogeneous systems have the same set of solutions, or equivalently the matrices have the same null space. (my bold)
----------------------------------------------

I would use the bolded, if available to you, along with your last proof.

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11-18-2015 , 11:34 PM

#8767

masque de Z

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Use these links ;
(for nonsingular matrix)
https://en.wikipedia.org/wiki/Invertible_matrix

https://en.wikipedia.org/wiki/Elementary_matrix (pay attention to the inverses of these matrices too)

and of course the row equivalence definition

https://en.wikipedia.org/wiki/Row_equivalence

Do you know that a matrix is invertible/nosingular if it is row equivalent to the identity? Also do you know that when you perform a row equivalence operation its like multiplying with an elementary matrix and that each one of them has an inverse such that the operation is undone if you apply them afterwards you have applied the original elementary matrix?

Spoiler:

Last edited by masque de Z; 11-19-2015 at 12:03 AM.

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11-25-2015 , 06:12 AM

#8768

Enough Is Enough

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For question a) I get x1 = 5, x2 = 4, z* = 49. Could somebody give me a hint on how to approach question b)?

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11-25-2015 , 03:20 PM

#8769

masque de Z

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Think what form the expression z=c1*x1+c2*x2 has (line with slope passing from certain points etc). Now what must happen for 2,6 to be maximum? Same as the other point you found, it must be that the line cannot go higher than that corner (to find a better value for the linear combination sum or higher constant for the linear equation z=c1*x1+c2*x2). Every line c1*x1+c2*x2=constant1 is parallel to another line x1*x1+c2*x2=constant2.

So the line must be passing at the maximum from 2,6 when it is parallel transported and it must have slope that is bounded by the 2 other lines in that corner for this to be the case, no ? It can be of course any other line that passes from there in the limit of maximization consistent with the form z=c1*x1+3*x2. Consider all lines (they are of the form a*x1+b*x2=constant) that can eventually pass from 2,6 for example when you change their constant , ie the c in the c= c1*x1+c2*x2 expression (here a=c1 and b=3). Plot a few test cases to see what i mean. Eg consider c1=0, c1=1, c1=2, c1=2.9999 etc to see what i mean (check if all those have the same maximum point , then try other values and see that they fail to get a maximum there). The unknown c1 is the one that determines the slope of the line c1*x1+c2*x2 ie the line x2=c/c2-c1/c2*x1 wher c is some constantand c1 is the parameter you are changing. How about considering all the lines that have slopes that vary from the one side to the other side (not the x2 axis side, the other 2 ) of the left triangle in your chart (the triangle that has one side from x2 2 to 7.2 or so ).

Last edited by masque de Z; 11-25-2015 at 03:29 PM.

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11-25-2015 , 05:32 PM

#8770

Enough Is Enough

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Thanks. I think I have it.

-2/3 <= c1/-3 <= 2
or
-6 <= c1 <= 2

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12-01-2015 , 10:38 PM

#8771

CyberShark93

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Join Date: Dec 2013 Posts: 3,325

Hi friends,
quick question,

suppose I was doing some multivariate regression, and one of the independent variables I chose have a correlation with the dependent variable, however the relationship is not linear. what is the best practice in this situation?? should I leave out this variable and just do regression on the rest of the variables or should I pre-process it, and use say x^2 of the independent variable as part of my regression?

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12-01-2015 , 11:00 PM

#8772

Wyman

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Transform the variable into something with a linear relationship w the dependent variable, eg by squaring it.

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12-02-2015 , 12:02 AM

#8773

BrianTheMick2

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Join Date: May 2012 Posts: 19,410

Quote:

Originally Posted by Wyman

Transform the variable into something with a linear relationship w the dependent variable, eg by squaring it.

That is most likely correct, but that whole overfitting thing comes to mind.

You need more info before offering a cure.

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12-02-2015 , 12:33 AM

#8774

reno expat

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Quote:

Originally Posted by Wyman

Transform the variable into something with a linear relationship w the dependent variable, eg by squaring it.

If you add a squared term to your regression, you need to also have the linear term in the model too

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12-02-2015 , 01:24 AM

#8775

Wyman

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Not sure why you need the linear term though the model can figure out that the coeff is 0 if it is.

Also if you have a real reason to believe that y ~ x^2 not sure why we're worried about overfitting. I'm not suggesting we throw a billion terms in and see what sticks...

But this isn't really my area of expertise, so I'll defer to the experts here

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