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time-series consisting of a nonlinear function of the true but unknown parameter corrupted by noise. This example shows how to implement an online recursive least squares estimator. Kernel Recursive Least-Squares (KRLS) algorithm with approximate linear dependency criterion, as proposed in Y. Engel, S. Mannor, and R. Meir. What are synonyms for Recursive least squares algorithm? Although there are several approaches and algorithms set up for the nonlinear stochastical system, the strong consistency of the parameter … Abstract | PDF (2479 KB) PubMed. 64, no. recursive-least-squares algorithms based on the filtered-x and the adjoint gradient approaches. The engine model includes nonlinear elements for the throttle and manifold system, and the combustion system. Based on a set of independent variables, we try to estimate the magnitude of a dependent variable which is the outcome variable. filters digital-communications adaptive-filters least-squares lms. The Recursive Least Squares Estimator estimates the parameters of a system using a model that is linear in those parameters. While simple models (such as linear functions) may not be able to capture the underlying relationship among Synonyms for Recursive least squares algorithm in Free Thesaurus. SIAM Journal on Scientific and Statistical Computing 13 :3, 771-793. 2, pp. In addition, a nonlinear least -squares fitting is utilized to predict the future values of infected populations. 1 synonym for least squares: method of least squares. What is the difference between least mean square and recursive least squares adaptive linear filters? The EX-KRLS is preferable to 1) a standard kernel recursive least squares (KRLS) in applications that require tracking the state-vector of general linear state-space models in the kernel space, or 2) an EX-RLS when the application requires a nonlinear observation and state models. 2nd edition, Springer Vieweg, 2016, ISBN 978-3-658-11455-8. Linear Model Estimation of Nonlinear Systems Using Least-Squares Algo-rithm Dr. Alireza Rahrooh, Daytona State College Alireza Rahrooh received B.S., M.S., and Ph.D. degrees in electrical engineering from the University of Akron, Ohio in 1979, 1986, and 1990, respectively. A very important “real-life” problem that is inherent in the continuous use of adaptive algorithms is their behavior in ﬁnite-precision environments. Nearly all physical systems are nonlinear at some level, but may appear linear over restricted operating ranges. The linear compounds of indicators that PLS substitutes for the latent variables do not obey the equations that the latter satisfy. Via, and I. Santamaria. Partial Least Squares as applied to models with latent variables, measured indirectly by indicators, is well-known to be inconsistent. In this paper, a new nonlinear system identification method using Partial Least-Squares (PLS) method is proposed, which is called a local linear PLS method because it is based on local models. SysIdentPy is an open source package for System Identification using NARMAX models (Nonlinear Autoregressive models with Moving Average and Exogenous Input) developed by me with the collaboration of my friends Luan Pascoal, Samuel Oliveira, and Samir Martins. Antonyms for Recursive least squares algorithm. 8, pages 2275-2285, 2004. The usual reason is: too many equations. Although KRLS can be applied to non-linear tracking prob-lems, it is not as effective as particle ﬁltering, particula rly in noisy environments. 31 May 1996 Nonlinear adaptive recursive least squares (NRLS) algorithm for target detection in infrared imagery. I am searching for a recursive or online non linear least squares algorithm. [38] Wold, H. (1973a) Non-linear iterative partial least squares (NIPALS) modelling. From the other side, convenience interface is somewhat slower than original algorithm because of additional level of abstraction it provides. of adaptive FIR recursive least squares (RLS) ﬁlters is well developed and provides the user, at each time instant, a set of parameters optimal in the least-squares sense [1], [2]. NARMAX models are a generalization of ARMAX family. The block can provide both infinite-history and finite-history (also known as sliding-window), estimates for θ. This chapter discusses extensions of basic linear least ‐ squares techniques, including constrained least ‐ squares estimation, recursive least squares, nonlinear least squares, robust estimation, and measurement preprocessing. F or the usual Euclidean inner pro duct in an m-dimensional space, where < a i; b j > = a 0 i b j, w e simply ha v A; B = A. F or the inner pro duct de ned y < i; > S for a p ositiv e de nite, Hermitian matrix S, w ha v A; B = A 0. This paper focuses on recursive nonlinear least squares parameter estimation in multi-agent networks, where the individual agents observe sequentially over time an independent and identically distributed (i.i.d.) y and H are known quantities that you provide to the block to estimate θ. Recursive multiple least squares Multicategory discrimination abstract In nonlinear regression choosing an adequate model structure is often a challenging problem. Google … Such a system has the following form: y (t) = H (t) θ (t). Linear Regression is a statistical analysis for predicting the value of a quantitative variable. Some current developments . Working with specialized interface is more convenient that using underlying optimization algorithm directly. In this paper we design a recursive bifurcation model for analyzing COVID-19 virus spread in different countries. Derivation of a Weighted Recursive Linear Least Squares Estimator $$\let\vec\mathbf \def\myT{\mathsf{T}} \def\mydelta{\boldsymbol{\delta}} \def\matr#1{\mathbf #1}$$ In this post we derive an incremental version of the weighted least squares estimator, described in a previous blog post. Sliding-Window Kernel Recursive Least-Squares (SW-KRLS), as proposed in S. Van Vaerenbergh, J. There are more equations than unknowns (m is greater than n). The n columns span a small part of m-dimensional space. First step is to identify local linear models by using the conventional Recursive Least-Squares (RLS) method. The parameter estimates by using the weight recursive least squares (W-RLS) algorithm are listed as follows: ... C. L. P. Chen, and H. X. Li, “A regularized variable projection algorithm for separable nonlinear least-squares problems,” IEEE Transactions on Automatic Control, vol. "The kernel recursive least-squares algorithm", IEEE Transactions on Signal Processing, volume 52, no. Extended recursive least squares algorithm for nonlinear stochastic systems Abstract: The strong consistency of parameter estimation has always been one of the main problems in system identification theory especially for the nonlinear systems. 'AN ALGORITHM FOR NONLINEAR LEAST SQUARES vscht cz May 9th, 2018 - AN ALGORITHM FOR NONLINEAR LEAST SQUARES M Balda the method transforms into fast Newton A code of the function has been later recasted into MATLAB with' 'working of Recursive least square method with an example Engine Model . share | improve this question | follow | edited Aug 13 '18 at 8:16. This leads to the development of new recursive-least-squares algorithms for the training of the controller neural network in the two networks structure. Recursive Least Squares Parameter Estimation for Linear Steady State and Dynamic Models Thomas F. Edgar Department of Chemical Engineering University of Texas Austin, TX 78712 1. 383 – 407 . I want to spread the computation out as new data is sampled like in the linear Recursive Least Squares or the LMS. We start with the original closed form formulation of the weighted least squares estimator: … 4.3 Least Squares Approximations It often happens that Ax Db has no solution. Ideally a recursive Levenberg–Marquardt algorithim would exist as Levenberg–Marquardt works great on my non linear problem but need to reuse all the samples to calculate a new estimate. 526–537, 2019. Recursive Least Squares Identification Algorithms for Multiple-Input Nonlinear Box–Jenkins Systems Using the Maximum Likelihood Principle Feiyan Chen, Feiyan Chen Key Laboratory of Advanced Process Control for Light Industry (Ministry of Education), Jiangnan University, Wuxi 214122, China e-mail: fychen12@126.com. Nonlinear least squares solver described here is actually a convenience wrapper around Levenberg-Marquardt optimizer. Moreover, standard implementations of KRLS do not account for unobserved state parameters. RLS (Recursive Least Squares), can be used for a system where the current state can be solved using A*x=b using least squares. (1992) A Parallel Nonlinear Least-Squares Solver: Theoretical Analysis and Numerical Results. Unless all measurements are perfect, b is outside that column space. You estimate a nonlinear model of an internal combustion engine and use recursive least squares to detect changes in engine inertia. Mohammad Abu-Tahnat, Michael W. Thompson. Matt L. 67.4k 3 3 gold badges 53 53 silver badges 126 126 bronze badges. The matrix has more rows than columns. Although there are several approaches and algorithms set … Search for other works by this author on: This Site. asked Aug 12 '18 at 22:16. user24907 user24907. He worked as an Electronic Engineer in Kurdistan from 1979 to 1984. Extended Recursive Least Squares Algorithm for Nonlinear Stochastic Systems Huixin Chen School of Computing & Technology University of Sunderland Sunderland SR6 0DD, UK Abstract The strong consistency of parameter estimation has always been one of the main problems in system iden-ti cation theory especially for the nonlinear systems. In Multivariate Analysis III, pp. that unfolds the iterations of the well-known recursive least squares (RLS) algorithm into the layers of a deep neural net-work in order to perform nonlinear PCA. K. Madsen, H. B. Nielsen, O. Tingleff, Methods for Non-Linear Least Squares Problems (nonlinear least-squares tutorial; L-M code: analytic Jacobian secant) T. Strutz: Data Fitting and Uncertainty (A practical introduction to weighted least squares and beyond) . least squares estimation: of zero-mean r andom variables, with the exp ected v alue E (ab) serving as inner pro duct < a; b >.) 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# Identifying the forces in a difficult circular / non-circular motion hybrid problem.

by greenskyy
Tags: circular, difficult, forces, hybrid, identifying, motion, noncircular
 P: 17 1. The problem statement, all variables and given/known data The problem is to create an equation to find the acceleration of the cart based upon theta. All that is given is the ball hanging is known as mass m. 2. Relevant equations $$f=ma$$ $$f=\frac{mv^{2}}{r}$$ 3. The attempt at a solution I am assuming that the car is moving to the right, since the hanging mass is tilted backwards. I am really at a loss of even just identifying all the forces exactly. What confuses me the most is that we are assuming the cart is accelerating forward, but there is no force causing it to move forward. So far, this is what I have: m being the mass, c being the cart $$F_{mx}=Tsin\theta$$ $$F_{my}=Tcos\theta - mg$$ $$F_{cx}=m_{c}a - Tsin\theta$$ $$F_{cy}=-Tcos\theta = 0$$ I know that these are probably wrong, but can someone please tell me why?
 HW Helper P: 3,394 I'm a little confused by all the subscripts! And can't make thetas here. Vertical: Tcos(A) - mg = 0 Horizontal: Tsin(A) = ma If you solve one for T and sub in the other, you'll get something that makes sense. How do you make a theta? PS it is accelerating to the right because theta > 0. If it wasn't accelerating, it would hang straight. So no need to think about what is making it accelerate.
 P: 17 Haha, well you make a theta using latex code. For the theta symbol, it would be [ t e x ] \ t h e t a [ \ t e x ] (minus the spaces)
HW Helper
P: 3,394

## Identifying the forces in a difficult circular / non-circular motion hybrid problem.

Thank you greenskyy. I fear that is too awkward for me.
I have seen someone with a whole bunch of symbols (theta, pi, etc.) in his signature. I wonder if that is so he can copy a theta whenever he wants it? Do you know how I can have a signature here?
 P: 17 Nope, I looked around and couldn't find it anywhere in the user control panel =\
 P: 672 What is the net force on the ball-cart system? What quantity is conserved, then?

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Why treat complex scalar field and its complex conjugate as two different fields?

I am new to QFT, so I may have some of the terminology incorrect.

Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field: $$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$ The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT in this book by Robert D. Klauber, "Student Friendly Quantum Field Theory".)

What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$.

Is it simply a shortcut to obtaining the equations of motion $$(\square +m^2)\phi=0\\ (\square + m^2)\phi^*=0~?$$

I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, as is done here; perhaps this addresses my question in a way that I don't understand.

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Yes, it is just a short-cut. The main point is that the complexified map

$$\tag{A} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix} ~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix} \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix}$$

is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2$.

Notation in this answer: In this answer, let $\phi,\phi^{*}\in \mathbb{C}$ denote two independent complex fields. Let $\overline{\phi}$ denote the complex conjugate of $\phi$.

I) Let us start with the beginning. Imagine that we consider a field theory of a complex scalar field $\phi$. We are given a Lagrangian density

$$\tag{B} {\cal L}~=~{\cal L}(\phi,\overline{\phi},\partial\phi, \partial\overline{\phi})$$

that is a polynomial in $\phi$, $\overline{\phi}$, and spacetime derivatives thereof. We can always decompose a complex field in real and imaginary parts

$$\tag{C} \phi~\equiv~\phi_1+ i \phi_2 ,$$

where $\phi_1,\phi_2 \in \mathbb{R}$. Hence we can rewrite the Lagrangian density (B) as a theory of two real fields

$$\tag{D}{\cal L}~=~{\cal L}(\phi_1,\phi_2,\partial\phi_1, \partial\phi_2).$$

II) We can continue in at least three ways:

1. Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.

2. Originally $\phi_1,\phi_2 \in \mathbb{R}$ are of course two real fields. But we can complexify them, vary the action wrt. the two independent complex variables $\phi_1,\phi_2 \in \mathbb{C}$, if we at the end of the calculation impose the two real conditions $$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2).$$

3. Or equivalently, we can replace the complex conjugate field $\overline{\phi}\to \phi^{*}$ in the Lagrangian density (B) with an independent new complex variable $\phi^{*}$, i.e. treat $\phi$ and $\phi^{*}$ as two independent complex variables, vary the action wrt. the two independent complex variables $\phi,\phi^{*} \in \mathbb{C}$, if we at the end of the calculation impose the complex condition $$\tag{F} \phi^{*} ~=~ \overline{\phi}.$$

III) The Euler-Lagrange equations that we derive via the two methods (1) and (2) will obviously be exactly the same. The Euler-Lagrange equations that we derive via the two methods (2) and (3) will be just linear combinations of each other with coefficients given by the constant matrix from eq. (A).

IV) We mention for completeness that the complexified theory [i.e. the theory we would get if we do not impose condition (E), or equivalently, condition (F)] is typically not unitary, and therefore ill-defined as a QFT.

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Although I agree that in the context described by the OP, its usually used as a mathematical trick, it's also that case that one could consider an extended field theory of independent, complex scalar fields if one so desired. Also is it not also important to make some remark about how, complexification + EL equations + imposing conjugate condition will yield the same equations of motion as not having used the trick in the first place? –  joshphysics Dec 5 '13 at 0:02


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What are velocity vs. time graphs?

How to analyze graphs that relate velocity and time to acceleration and displacement.

What does the vertical axis represent on a velocity graph?

The vertical axis represents the velocity of the object. This probably sounds obvious, but be forewarned—velocity graphs are notoriously difficult to interpret. People get so used to finding velocity by determining the slope—as would be done with a position graph—they forget that for velocity graphs the value of the vertical axis is giving the velocity.
Try sliding the dot horizontally on the example graph below to choose different times and see how the velocity changes.

Concept check: What is the velocity of the object at time $t=4\text{ seconds}$, according to the graph above?
The velocity is $+3\text{ m/s}$ at time $t=4\text{ s}$. We can see this by moving the dot to the time $t=4\text{ s}$ and seeing that the value of the graph is $+3\text{ m/s}$ on the vertical axis.

What does the slope represent on a velocity graph?

The slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.
The slope of a velocity graph will be given by the following formula:
$\text{slope}=\dfrac{\text{rise}}{\text{run}}=\dfrac{v_2-v_1}{t_2-t_1}=\dfrac{\Delta v}{\Delta t}$
Since $\dfrac{\Delta v}{\Delta t}$ is the definition of acceleration, the slope of a velocity graph must equal the acceleration of the object.
$\text{slope}=\text{acceleration}$
This means that when the slope is steep, the object will be changing velocity rapidly. When the slope is shallow, the object will not be changing its velocity as rapidly. This also means that if the slope is negative—directed downwards—the acceleration will be negative, and if the slope is positive—directed upwards—the acceleration will be positive.
Try sliding the dot horizontally on the example velocity graph below to see what the slope looks like for particular moments in time.

The slope of the curve is positive between the times $t=0\text{ s}$ and $t=2 \text{ s}$ since the slope is directed upward. This means the acceleration is positive.
The slope of the curve is negative between $t=2 \text{ s}$ and $t=8 \text{ s}$ since the slope is directed downward. This means the acceleration is negative.
At $t=2\text{ s}$, the slope is zero since the tangent line is horizontal. This means the acceleration is zero at that moment.
Concept check: Is the velocity of the object whose motion is described by the graph above increasing, decreasing, or constant at time $t=4\text{ s}$?
The object's velocity is decreasing at time $t=4\text{ s}$. One way to see this is by dragging the slider to $t=4\text{ s}$. Notice that the slope of the graph is negative, which means the acceleration is negative. The velocity at $t=4\text{ s}$ is positive and has a value of $+3\text{ m/s}$. Since the acceleration and the velocity have opposite signs, the object must be slowing down.
Another way to see this is by recognizing that at $t=4\text{ s}$ the curve is approaching the horizontal axis. Since the horizontal axis represents $v=0\text{ m/s}$, a curve that is heading toward the horizontal axis represents an object that is slowing down. Similarly, a curve that is heading away from the horizontal axis represents an object that is speeding up. So at $t=7\text{ s}$ in the graph above, the curve is heading away from the horizontal axis and the object's velocity is increasing in the negative direction.

What does the area under a velocity graph represent?

The area under a velocity graph represents the displacement of the object. To see why, consider the following graph of motion that shows an object maintaining a constant velocity of 6 meters per second for a time of 5 seconds.
To find the displacement during this time interval, we could use this formula
$\Delta x=v\Delta t=(6\text{ m/s})(5\text{ s})=30\text{ m}$
which gives a displacement of $30\text{ m}$.
Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of area made by the graph as seen below.
The area of this rectangle can be found by multiplying height of the rectangle, 6 m/s, times its width, 5 s, which would give
$\text{ area}=\text{height} \times \text{width} = 6\text{ m/s} \times 5\text{ s}=30\text{ m}$
This is the same answer we got before for the displacement. The area under a velocity curve, regardless of the shape, will equal the displacement during that time interval.
For an arbitrary shape, you can consider breaking the area into rectangles of very small width as seen in the graph below. The displacement during each small time interval would be given by $\Delta x=v\Delta t$ since the velocity would be roughly constant if the time intervals are small enough. If you were to make the rectangles infinitesimally small, they would exactly represent the area under the curve. Adding up all the small areas would equal the total area and also the displacement of the object. When doing an actual problem, you never have to actually break a curve into an infinite amount of rectangles. Once you have proven to yourself that the area represents the displacement, you are free to determine the area in whichever way is most convenient.
$\text{area}=\text{displacement}$

What do solved examples involving velocity vs. time graphs look like?

Example 1: Windsurfing speed change

A windsurfer is traveling along a straight line, and her motion is given by the velocity graph below.
Select all of the following statements that are true about the speed and acceleration of the windsurfer.
(A) Speed is increasing.
(B) Acceleration is increasing.
(C) Speed is decreasing.
(D) Acceleration is decreasing.
Options A, speed increasing, and D, acceleration decreasing, are both true.
The slope of a velocity graph is the acceleration. Since the slope of the curve is decreasing and becoming less steep this means that the acceleration is also decreasing.
It might seem counterintuitive, but the windsurfer is speeding up for this entire graph. The value of the graph, which represents the velocity, is increasing for the entire motion shown, but the amount of increase per second is getting smaller. For the first 4.5 seconds, the speed increased from 0 m/s to about 5 m/s, but for the second 4.5 seconds, the speed increased from 5 m/s to only about 7 m/s.

Example 2: Go-kart acceleration

The motion of a go-kart is shown by the velocity vs. time graph below.
A. What was the acceleration of the go-kart at time $t=4\text{ s}$?
B. What was the displacement of the go-kart between $t=0\text{ s}$ and $t=7\text{ s}$?

A. Finding the acceleration of the go-kart at $t=4\text{ s}$

We can find the acceleration at $t=4\text{ s}$ by finding the slope of the velocity graph at $t=4\text{ s}$.
$\text{slope}=\dfrac{\text{rise}}{\text{run}}$
For our two points, we'll choose the start—$3\text{ s}, 6\text{ m/s}$—and end—$7\text{ s}, 0\text{ m/s}$—of the diagonal line as points one and two respectively. Plugging these points into the formula for slope we get
$\text{slope}=\dfrac{v_2-v_1}{t_2-t_1}=\dfrac{0\text{ m/s}-6\text{ m/s}}{7\text{ s}-3\text{ s}}=\dfrac{-6\text{ m/s}}{4\text{ s}}=-1.5\dfrac{\text{m}}{\text{s}^2}$
$\text{acceleration}=-1.5\dfrac{\text{m}}{\text{s}^2}$

B. Finding the displacement of the go-kart between $t=0\text{ s}$ and $t=7\text{ s}$

We can find the displacement of the go-kart by finding the area under the velocity graph. The graph can be thought of as being a rectangle (between $t=0\text{ s}$ and $t=3\text{ s}$) and a triangle (between $t=3\text{ s}$ and $t=7\text{ s}$). Once we find the area of these shapes and add them, we will get the total displacement.
The area of the rectangle is found by
$\text{area}=h\times w=6\text{ m/s} \times 3\text{ s}=18\text{ m}$
The area of the triangle is found by
$\text{area}=\dfrac{1}{2} bh=\dfrac{1}{2} (4\text{ s})(6\text{ m/s})=12\text{ m}$
Adding these two areas together gives the total displacement.
$\text{total~area}=18\text{ m}+12\text{ m}=30\text{ m}$
$\text{total~displacement}=30\text{ m}$


text:
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• 12-05-2017, 03:47 PM
@tkhunny great, I've got it now, thanks for clearing that up. It feels good to get that annoying bit of confusion out of the way.
10 replies | 320 view(s)
• 12-03-2017, 11:19 PM
Actually, having re-read that link to Paul's Online Maths Notes, it is indeed that a convergent series has a sequence that tends to 0. I can see the...
10 replies | 320 view(s)
• 12-03-2017, 10:56 PM
My understanding is that the divergence test only tells you that if the limit of the sequence a_n is not zero, then the series diverges, and that it...
10 replies | 320 view(s)
• 12-03-2017, 09:02 PM
So the sequence would look like a_0+a_1(-4)+a_2(16)+a_3(-64)+... I still don't see how the limit of a_n, \lim_\limits{n\to\infty}a_n, is 0?...
10 replies | 320 view(s)
• 12-03-2017, 12:10 AM
@tkhunny so I guess that would be a_0+a_1(-4)+a_2(16)+a_3(-64)... I still don't see how the limit of the sequence a_n, \lim\limits_{n\to\infty}a_n...
10 replies | 320 view(s)
• 12-02-2017, 06:35 PM
So the power series \sum_{n=0}^{\infty}a_n (x-4)^n converges at 0 and diverges at 9. How do we know that the limit of the sequence a_n is zero?...
10 replies | 320 view(s)
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# Is 1 degree K/C equivalent at all temperatures?

## Main Question or Discussion Point

This is an odd question. I was talking with a social scientist the other day and he seemed to think that the amount of energy it takes to raise the temperature 1 degree Celsius or Kelvin was minutely different depending on the temperature (ie near absolute zero the energy it takes to increase T by 1 is different than the energy it takes to increase T by 1 at 300K). I tried explaining to him that kinetic energy is linearly proportional to temperature, but he insisted that he was right and that the scale was very slightly nonlinear (no proof or source I'm afraid, so it's been driving me nuts :/ ). I have done some searching to no avail, I turn to you for ideas. Are there any nonlinear effects that cause the energy separation between degrees K/C to be unequal?

Related Classical Physics News on Phys.org
Pythagorean
Gold Member
Maybe in his mind, he wasn't thinking of an isolated system, but a system on Earth, in which you're generally fighting strong gradients to keep any particular region cooled, and the colder you get, the bigger a gradient you generate, and the more the heat wants to get into your cooled region.

It just occurred to me that near enough to absolute zero only translational modes are occupied (sorry I know this is the classical forum) but at higher temperatures non-translational degrees of freedom get populated; not all of it goes into kinetic energy. Is this right?

D H
Staff Emeritus
This is an odd question. I was talking with a social scientist the other day and he seemed to think that the amount of energy it takes to raise the temperature 1 degree Celsius or Kelvin was minutely different depending on the temperature (ie near absolute zero the energy it takes to increase T by 1 is different than the energy it takes to increase T by 1 at 300K). I tried explaining to him that kinetic energy is linearly proportional to temperature, but he insisted that he was right and that the scale was very slightly nonlinear (no proof or source I'm afraid, so it's been driving me nuts :/ ). I have done some searching to no avail, I turn to you for ideas. Are there any nonlinear effects that cause the energy separation between degrees K/C to be unequal?
The Kelvin and Celsius scales are defined so that a change of 1K is exactly the same as a change of 1 °C. The two scales differ by a constant offset.

However, the amount of energy needed to raise the temperature of some real substance by 1K or by 1 °C does vary with temperature. The heat capacity of a real substance is not constant. It is a function of temperature. An ideal gas -- that's an ideal. Some gases do come close to this ideal over some finite temperature range, but there is no such thing as an ideal gas. Nor an ideal solid.

Pythagorean
Gold Member
Sorry, I'm a classical dude. I have no idea what translational modes are..

Substances undergo phase-change, in which a substance absorbs energy but the temperature doesn't rise, at the critical points of a substance (where it freezes/sublimates/melts/evaporates/etc). Once the substance has received enough energy to change phase, it starts raising in temperature again with additional energy absorbed.

Pythagorean
Gold Member
The Kelvin and Celsius scales are defined so that a change of 1K is exactly the same as a change of 1 °C. The two scales differ by a constant offset.

However, the amount of energy needed to raise the temperature of some real substance by 1K or by 1 °C does vary with temperature. The heat capacity of a real substance is not constant. It is a function of temperature. An ideal gas -- that's an ideal. Some gases do come close to this ideal over some finite temperature range, but there is no such thing as an ideal gas. Nor an ideal solid.
hrmm... does that mean that

$$\frac{1}{2}mv^2 = \frac{3}{2}kT$$

is not true for real substances..? or that I'm misinterpreting it?

D H
Staff Emeritus
hrmm... does that mean that

$$\frac{1}{2}mv^2 = \frac{3}{2}kT$$

is not true for real substances..? or that I'm misinterpreting it?
Correct. That's for a monatomic ideal gas. It's not even true for a diatomic or multiatomic ideal gas.

There is no such thing as an ideal gas. Some gases come close to that ideal, some very, very close. But even the inert gases condense into liquids, and the atoms have a non-zero size.

Chestermiller
Mentor
Correct. That's for a monatomic ideal gas. It's not even true for a diatomic or multiatomic ideal gas.

There is no such thing as an ideal gas. Some gases come close to that ideal, some very, very close. But even the inert gases condense into liquids, and the atoms have a non-zero size.
Even in the limit of ideal gas behavior, the heat capacity is a function of temperature.

D H
Staff Emeritus
Even in the limit of ideal gas behavior, the heat capacity is a function of temperature.
No, Chester. Heat capacity is constant an ideal gas. For a monatomic ideal gas, cv=3/2 R. Look at post #6. It's right there.

Chestermiller
Mentor
No, Chester. Heat capacity is constant an ideal gas. For a monatomic ideal gas, cv=3/2 R. Look at post #6. It's right there.
Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics:

"Ideal-gas heat capacities (designated $C_p^{ig}$ and $C_v^{ig}$) are different for different gases; although functions of temperature, they are independent of pressure.
The temperature dependence may be shown graphically, as illustrated in Fig. 4.1, where $C_p^{ig}/R$ is plotted vs. temperature for argon, nitrogen, water, and carbon dioxide.

SteamKing
Staff Emeritus
Homework Helper
Don't worry about what a social scientist thinks he knows about temperature. The most dangerous thing about social scientists is what they know that just ain't so. He was just trying to get inside your head and you let him succeed.

The temperature scale is linear by definition. There are 100 divisions in the amount of heat required to bring water from zero degrees C (after it has just fused) to 100 degrees C (where it boils at one atmosphere). All of this was decided before anyone knew about translational modes or statistical mechanics.

D H
Staff Emeritus
Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics:

"Ideal-gas heat capacities (designated $C_p^{ig}$ and $C_v^{ig}$) are different for different gases; although functions of temperature, they are independent of pressure.
The temperature dependence may be shown graphically, as illustrated in Fig. 4.1, where $C_p^{ig}/R$ is plotted vs. temperature for argon, nitrogen, water, and carbon dioxide.
That first sentence makes no sense. Look at the post #6. It's right there, plain as day. Specific heat is constant for an ideal gas. As far as the second sentence is concerned, argon, nitrogen, water, and carbon dioxide are not ideal gases. There is no such thing. It's an ideal.

UltrafastPED
Gold Member
The thermochemical calorie was originally defined to be the amount of heat required to raise the temperature of a gram of water by one degree Celsius. However this definition is inadequate - which is something that your friend the sociologist knows:

http://en.wikipedia.org/wiki/Calorie

Andrew Mason
Homework Helper
It just occurred to me that near enough to absolute zero only translational modes are occupied (sorry I know this is the classical forum) but at higher temperatures non-translational degrees of freedom get populated; not all of it goes into kinetic energy. Is this right?
It depends. If the substance is monatomic, there are only translational modes (no rotational or vibrational modes). If the substance is diatomic or polyatomic, the heat capacity will begin to increase when it reaches temperatures where the rotational and vibrational modes start to become active. As temperature decreases rotational and vibrational modes cease to be active heat capacity decreases. When any substance approaches absolute 0, even the translational modes start to get frozen out.

Regarding the point that your sociologist friend was trying to make, was he referring to a particular substance or all matter? If he was simply saying that the heat capacity of water varies with temperature, he would be right. That has to do more with the intermolecular forces in water (ie. potential energy of molecules increases with temperature).

AM

AlephZero
Homework Helper
I thought sociologists were mostly concerned with the properties of hot air

Chestermiller
Mentor
That first sentence makes no sense. Look at the post #6. It's right there, plain as day. Specific heat is constant for an ideal gas. As far as the second sentence is concerned, argon, nitrogen, water, and carbon dioxide are not ideal gases. There is no such thing. It's an ideal.
This entire quote makes perfect sense to me. Smith and Van Ness has been used as a text in thermo courses for over fifty years, with several new editions, and the quote I gave is in complete agreement with the corresponding sections in numerous other highly regarded treatises, including Hougan and Watson, Perry's Chemical Engineers' Handbook, Transport Phenomena by Bird, Stewart, and Lightfoot.

I think what we may be dealing here is a difference in the definition used for an "ideal gas" by physicists vs engineers. In engineering we define an ideal gas as the limiting behavior of a real gas at very low pressures. Thus, an ideal gas is one whose
(a) PVT behavior is PV=nRT
(b) Cv and Cp are functions only of temperature.
(c) H and U are functions only of temperature.

Apparently, in physics, this is not the definition that is used. In order to clarify this issue, I'm going to start a new thread in which I solicit responses from both physicists and engineers as to their understanding of what an ideal gas is. I will be titling this thread "Definition of an Ideal Gas."

Chet

D H
Staff Emeritus
Apparently, in physics, this is not the definition that is used.
Briefly, what physicists call an ideal gas, engineers call a perfect gas. I provided a more detailed answer in your thread on this side topic.

With that nomenclature discrepancy cleared away, I'll back to the original post.

I tried explaining to him that kinetic energy is linearly proportional to temperature ...
That's only true for ideal (perfect) substances. It's not true for real substances. I'll give a couple of examples.

Example 1: A mix of ice and water.

Add energy and the temperature does not increase. Instead, some of the ice melts but the mix remains at 0 °C. The temperature doesn't start rising until all the ice has melted. In an ideal gas, the only form of energy is kinetic energy. That's not the case here, or in any real substance. There are also various forms of potential energy. The phase transition represents a change in potential energy and also in entropy.

Example 2: A diatomic gas.

Even kinetic energy is a bit problematic when dealing with real gases. An ideal gas has its kinetic energy equally partitioned amongst all possible outlets for that energy. There are only three degrees of freedom for a low temperature monatomic gas: Motion in three dimensional space. Helium is about as close to ideal (perfect) as a gas can get. Except at very low temperatures, specific heat is pretty much independent of temperature, and it's just what kinetic theory suggests, $c_v = \frac 3 2 R$. Diatomic gases at low temperature behave similar to monatomic gases. Increase the temperature and specific heat eventually starts increasing toward $\frac 5 2 R$. It's as if a couple of new degrees of freedom have been added -- and they have. The diatomic gas starts exhibiting two rotational degrees of freedom. Add even more energy and specific heat eventually starts increasing even further. Bottom line: The simple linear relation between energy and temperature seen in an ideal gas is not linear and not so simple even for a near-ideal diatomic gas.

D H
Staff Emeritus
Chestermiller: I moved your post on the engineer's distinction between an ideal gas and a perfect gas to your own thread. That discussion is off-topic here. The topic of this thread is ultimately "what is temperature?"

Chestermiller
Mentor
Never mind.

Andrew Mason
Homework Helper
Chestermiller: I moved your post on the engineer's distinction between an ideal gas and a perfect gas to your own thread. That discussion is off-topic here. The topic of this thread is ultimately "what is temperature?"
Actually, the title is misleading. It seems to me that this thread is really about whether heat capacity depends on temperature. The answer is that it depends...

AM

Chestermiller
Mentor
Actually, the title is misleading. It seems to me that this thread is really about whether heat capacity depends on temperature. The answer is that it depends...

AM
It is about whether the heat capacity of what one calls an ideal gas depends on temperature. If you are a physicist, you might have one answer to this question, while, if you are from some other discipline, you might have a different answer.

Chet

D H
Staff Emeritus
It is about whether the heat capacity of what one calls an ideal gas depends on temperature. If you are a physicist, you might have one answer to this question, while, if you are from some other discipline, you might have a different answer.
No, it's not, Chet. There's nothing in the OP that talks about ideal gases. And please, stop with the comments about how different groups have different names for things. It's off topic in this thread.

The original post, which is admittedly a bit confused, asks three different questions, one implied:
• Is a temperature change of 1K always the same as a temperature change of 1 °C?
The answer to this question is yes. That's how the Kelvin and Celsius scales are defined.

• Is energy (always) linear with respect to change in temperature?
The answer to this question is no. That implied always is a killer. It's not just ideal (or perfect) gases versus real ones where that linear relation falls apart. The thermodynamic behaviors of a crystalline solid can be rather funky at very low temperatures thanks to quantum mechanics. And liquid helium? Superfunky.

• Is kinetic energy all that one needs to be concerned with when it comes to temperature?
The answer to this question is an emphatic no. Ideal substances don't go through phase transitions, undergo chemical reactions, dissociate, undergo radioactive decay, etc. Real substances do, and those changes involve some form of energy other than kinetic energy.

SteamKing
Staff Emeritus
Homework Helper
See what troublemakers social scientists are.

Chestermiller
Mentor
No, it's not, Chet. There's nothing in the OP that talks about ideal gases. And please, stop with the comments about how different groups have different names for things. It's off topic in this thread.

The original post, which is admittedly a bit confused, asks three different questions, one implied:
• Is a temperature change of 1K always the same as a temperature change of 1 °C?
The answer to this question is yes. That's how the Kelvin and Celsius scales are defined.

• Is energy (always) linear with respect to change in temperature?
The answer to this question is no. That implied always is a killer. It's not just ideal (or perfect) gases versus real ones where that linear relation falls apart. The thermodynamic behaviors of a crystalline solid can be rather funky at very low temperatures thanks to quantum mechanics. And liquid helium? Superfunky.

• Is kinetic energy all that one needs to be concerned with when it comes to temperature?
The answer to this question is an emphatic no. Ideal substances don't go through phase transitions, undergo chemical reactions, dissociate, undergo radioactive decay, etc. Real substances do, and those changes involve some form of energy other than kinetic energy.
Sorry DH. I apologize again.

Chet

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text:
# How does a computer compute negative(-) and positive(+) Infinity?

If we divide (1.0/0.0) we will get +Infinity and if we divide (-1.0/0.0) we will get -Infinity.

How does a computer calculate this value internally?

The simple answer is that a computer that follows a floating point standard such as IEEE 754 does not really compute +/- infinity at all. Instead it treats it as an exception - a special case. Whenever the processor detects an attempt to divide a non-zero floating point number by zero, it sets the result to a special value that is then interpreted as +/-infinity, where the sign depends on the sign of the original number.

In IEEE 754 I believe the special values of +/- infinity are represented internally as a floating point number with a sign bit, an exponent set to all $$1$$s, and a significand set to all $$0$$s. This is a value that cannot result from normal (non-exceptional) floating point operations.

Technically speaking, not all computers evaluate 1.0 / 0.0 to +Inf. But most of recent computers follows a specification named IEEE 754, which especially defines the division on floating-point numbers 1.0 / 0.0 is +Inf.

Division by zero: an operation on finite operands gives an exact infinite result, e.g., 1/0 or log(0). By default, returns ±infinity.

Also note that an evaluation of a division on integers 1 / 0 often throws "Division By Zero" error in many programming languages.

Division is a rather complex operation.

If the computer is implementing IEEE 754 arithmetic then most likely it categorises its inputs as NaN (not a number), Infinity, Normalised numbers, denormalise numbers, and zeroes. That is done by looking at the exponent and mantissa. The rules are:

1. Largest possible exponent with zero mantissa is Infinity.
2. Largest possible exponent with nonzero mantissa is NaN.
3. Smallest possible exponent with zero mantissa is zero.
4. Smallest possible exponent with nonzero mantissa is denormalised.
5. Everything else is normalised.

Denormalised numbers are made normalised. Then if the operands are not both normalised, the hardware has defined values for the 15 other combinations. If both operands are normalised, the division is performed, and the result can be overflow, normalised, denormalised, or so small it is rounded to zero.

It just checks if the denominator is zero and, if so, does whatever it needs to do, such as returning a value representing infinity, an error, or whatever else is required in the context.


text:
# How do I find the limit as x approaches infinity of xsin(1/x)?

Sep 26, 2014

By l'Hopital's Rule,

${\lim}_{x \to \infty} x \sin \left(\frac{1}{x}\right) = 1$

Let us look at some details.

${\lim}_{x \to \infty} x \sin \left(\frac{1}{x}\right)$

by rewriting a bit,

$= {\lim}_{x \to \infty} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}$

by l'Hopital's Rule,

$= {\lim}_{x \to \infty} \frac{\cos \left(\frac{1}{x}\right) \cdot \left(- \frac{1}{x} ^ 2\right)}{- \frac{1}{x} ^ 2}$

by cancelling out $- \frac{1}{x} ^ 2$,

$= {\lim}_{x \to \infty} \cos \left(\frac{1}{x}\right) = \cos \left(0\right) = 1$


text:
Build a party-line Intercom

cbenjes

Member
Here’s a great success story that may help someone out with an intercom requirement and minimal funds.

Recently, I have become the technical advisor for my son’s high school theater. First duties were to get the equipment fully functional. In the process, I realized how much we truly needed a technician intercom system for communication between the catwalks, backstage and the booth. Walkie-talkies simply were not desirable. How can you work a follow spot and still communicate with a hand-held walkie-talkie?

I was fortunate to discover that the theater, built in 1978, had the makings of an old intercom system that was probably never used. There were a total of 7 intercom stations, three in the catwalks, one in the booth, one immediately outside the booth, and one in each of the wings backstage. Each intercom station consisted of a 1/4" TRS (Tip-Ring-Sleeve) jack (stereo headphones jack) for plugging in a headset. The stations were connected using a 2-wire shielded cable with the red lead connecting the tips and the black lead connecting the rings. After a little troubleshooting and some very basic repairs, I managed to get the stations solidly connected with no short circuits. I could loop back a signal from one end of the theater to the other proving continuity of the circuits. Now I had the basis of an intercom.

After some research, I devised and implemented the following set-up:

First things first, we needed a power supply to drive the voice signal. I acquired and connected a filtered, low-voltage DC power supply to a centrally-located intercom station. Positive went to the ring, negative went to tip. The power supply provided approximately 11 volts (with no load) at 70mA of clean DC power filtered through a 1000mF capacitor inside the power supply. (The power supply needs to be carefully selected to match the headsets and to provide clean power. Voltage anywhere from 6-14 v DC should be OK.) Filtered (minimal ripple), regulated power is the key. Without the filtering effect of the power supply's capacitor, the system would have had too much noise to be usable. Communications are full duplex utilizing the "party-line" style of simple telephone circuitry.

Extension cords were run from the intercom station in the first catwalk to each of the follow spot operator positions.

Headsets in use are Western Electric WE-52 vintage telephone operator's headsets fitted with a 1/4" TRS plug instead of the traditional 2-prong telephone switchboard plug. Most of the ones I have were manufactured back in the 1950's and 1960's and still work great. Just make sure they are wired correctly. They just don't build them like this anymore....

Interestingly enough, a traditional Western Electric (or AT&T) G3 telephone handset can also be used as a cheap communications device (I bought one on e-bay for 99 cents, you can probably find one at a yard-sale cheap) and is a good choice for backstage areas since these handsets are virtually indestructible. The coiled cord can be refitted with a TRS plug which will connect right into the intercom station. In order to make a traditional telephone handset fully compatible, the resistor on the earpiece speaker must be removed, otherwise users of the phone are heard clearly, but the volume in the earpiece for phone users is very low.

So now we have a basic full-duplex party-line intercom system that works great. You can build your own cheap! The tough part is running the cable to the stations, but if you are lucky like we were, it may already be done, or can be done easily. Parts needed are a DC regulated filtered power supply, some headsets (refurbished WE-52 headsets are about $40, new ones are about$85), some TRS-jacks, 1000’ of shielded 2-wire cable (about $55), and misc TRS jacks, plugs and extension cords (nominal). Hope this info can help someone out. It was a lot of fun to build, and even more fun to use. Last edited: Tex Active Member This is just like the system used at the theatre where I was a resident company member for several years. It was decidedly low-tech, but worked great. Thanks for this. I love a good hack. I've been in a couple of theatres that had no com system, but did have unused aux speaker lines. This system would be great for that application. Studio Active Member Sounds like the system in my schools old theater except you could "call" (flash a light) on the other boxes. Les Well-Known Member cbenjes Member The intercom system in my old high school (circa 1974-1978) had indicator lamps on each of the intercom wall plates that were wired in series. Press a button on the wall plate and all of the indicator lights flashed, and you hoped the right person would pick up a headset to see what you wanted. This feature requires a separate power supply and cable that were totally separate from the voice circuit. For the system I just installed, I didn't have the luxury or ability to run a separate circuit and install indicator lights and push-buttons. Wish I did, but heck, I'll take what we have over crappy walkie-talkies ANYDAY. Studio Active Member The intercom system in my old high school (circa 1974-1978) had indicator lamps on each of the intercom wall plates that were wired in series. Press a button on the wall plate and all of the indicator lights flashed, and you hoped the right person would pick up a headset to see what you wanted. This feature requires a separate power supply and cable that were totally separate from the voice circuit. For the system I just installed, I didn't have the luxury or ability to run a separate circuit and install indicator lights and push-buttons. Wish I did, but heck, I'll take what we have over crappy walkie-talkies ANYDAY. Thats what I ment. Pics: erosing The Royal Renaissance Man Looks fun, good to have in the back of my head for emergencies/no funding. museav CBMod CB Mods Departed Member I moved this thread to the Sound forum since that is where most other Clear-Com and production communication system discussions live. A very creative solution to your situation and an interesting story, I just want people to realize that it may not be a practical approach for all applications. Did you consider adding push-to-talk at the headsets and handsets? Having all sets live all the time would seem to potentially make communication difficult and be undesirable in many applications. And not having volume control for each unit also seems a potential issue in many applications. cbenjes Member Thanks for the move. This topic fits better here. Believe it or not, we really haven't had problems with people talking over each other, but push-to-talk might be a good solution to eliminate chatty stage managers that don't think about moving the mic away from their mouth when they are screaming at someone NOT on the headset system. I did modify one of the telephone handsets over to push-to-talk because one particular intercom jack is very close to our ETC Sensor+ racks, and their cooling fans create a lot of background noise that gets picked up by an idle handset. It was simple enough to install a red push-button switch from Radio Shack on the handset and wire it to the mouthpiece. The trick is getting novice users to push (AND HOLD) the button to talk. Chris15 CBMod CB Mods Premium Member Departed Member Sounds like a solution to a problem has been found. However the use of TRS connectors for the connections is a cause of slight concern. As they are connected, a momentary short is created across the contacts. This will eventually kill the power supply and if someone if particularly slow about plugging in, it may kill it then and there... The risk will be lessened if a current limiting resistor is installed in series with the power supply connection... cbenjes Member Very true, you do get a distinct click when you plug-in and plug-out, so there may be a momentary short, but I think a good power supply will be able to compensate for it, though. The design for this intercom system came from my high school's theater. Their original intercom system was installed in the late 1960's and was used through the 2002 school year. Through those years, the system only ever had 1 power supply - a Bogen PS-1 power supply that was manufactured in the mid-1960's and was powered continuously for over 30 years. In 2002, the school moved to a wireless system and disabled the original wired intercom, but left the power supply unplugged in the catwalks until I picked it up in October 2009. I used it as a test power supply during the install phase of my current project. I now have that power supply in my shop and it still works great for testing and repairing headsets and handsets. They just don't build stuff like they used to. rervin Member Hi have a similar system and am trying to replace it with a newer version. Have you found any source for this party line type of system? I just can't see spending 3-5K for the off-the-shelf systems that usual supplier are trying to get me to purchase. Was wondering also is you have tried to update the headsets. We have lite headsets made by eartech from a wireless system that no longer works( never did work that well but dropping the master in a mop bucket during a show finished them off :-( ) We have tried to hook them up in place of the old style telephone headsets but seem to need an amplifier???? rob cbenjes Member Rervin, this is the only type of intercom system I am familiar with, so I really can't speak to any type of modern system. From doing research, I do know that if your theater is wired with 4-wire cable, you should be able to install some type of modern system that incorporates beltpacks, but you're still talking lots of dollars per station once you get the headsets, beltpacks, master stations, power supplies, etc. The beauty of this simple party-line set is that it is: simple - plug them in and they work. durable - I have headsets that are 50 years old that still work! cheap - headsets are$85 new at Audiosears , a power supply is \$35, and all you need is a 2-wire cable strung throughout the system terminated with TRS connectors.

Yeah, the old telephone operators headsets look like pictures from a Life magazine article from the 1960's, but hey, I don't have any budget to speak of, but my kids have a working intercom. Believe me, that's better then the cheap walkie-talkies they had.

I think only these headsets work on this type of system because of the make-up of the elements in the mic/ear piece, and that you can't use newer types of headsets without amplification. These are carbon elements and it's a design that's been around since Alexander called Watson and told him that he needed him quickly.

cbenjes

Member
Update: The school just acquired a real Clearcom headset/beltpack system. No more 1950's technology! So nice to have modern technology. I think I'll donate the WE52 headsets to the Smithsonian.

macsound

Well-Known Member
Dredging up an old thread, but thinking about the viability of making a wired telephone party line using regular phones and possibly headsets.
Is it as simple as just having a power supply and people can pickup phones anywhere and they would just have an open line?
Thanks

Fight Leukemia

TimMc

Well-Known Member
Yes, that was pretty common. A DC supply and some kind of twisted pair wire to each station. Wire the stations in parallel. Experiment with putting a pair of 300 ohm, 1 Watt resistors between the line and the power supply.
This is where we need Granny's crank magneto wall phone... {evil laugh}

FMEng

Well-Known Member
Fight Leukemia
This is where we need Granny's crank magneto wall phone... {evil laugh}
And like most things built by Western Electric, that phone would still work perfectly.


text:
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text:
CCSC Mid-South Programming Contest

#### Start

2018-04-06 10:00 AKDT

## CCSC Mid-South Programming Contest

#### End

2018-04-06 14:00 AKDT
The end is near!
Contest is over.
Not yet started.
Contest is starting in -901 days 3:22:17

4:00:00

0:00:00

# Problem AClimbing Sloths

The Cool Climbing Sloths Consortium (CCSC) is holding their annual sloth climbing contest. Sloths from far and wide have come to test their prowess in ascending trees of various designs. May the “fastest” (perhaps “least slow” would be more accurate) sloth win!

Since there are so many sloths in the contest, the contest organizing committee has decided to produce the trees computationally. They will generate a number of binary search trees, which will then be produced into physical structures for the sloths to climb. However, to ensure the contest is fresh and exciting, they don’t want to use the same tree more than once. They need a way of determining whether any pair of trees are structurally identical.

Two trees are considered structurally identical if they share the same node layout, regardless of the values stored in the nodes. Furthermore, since these trees will be produced into physical structures, two trees are also considered structurally identical if mirroring one about its root produces a tree with the same node layout as the other.

For example, the two trees below are structurally identical:

    6           7
/ \         / \
1   8       2   9
\           \
3           5

The two trees below are also structurally identical, since mirroring one about its root produces the same node layout as the other:

    6           7
/ \         / \
1   8       2   9
\             /
3           8

## Input

Input consists of two lines. Each line contains between $1$ and $10^4$ space-separated, distinct integers, where each integer is between $1$ and $10^5$. Each line represents the binary search tree that is formed by inserting those integers in that order into an initially empty tree.

## Output

Output YES if the input represents structurally identical trees, or NO if not.

Sample Input 1 Sample Output 1
6 8 1 3
7 2 9 8

YES

Sample Input 2 Sample Output 2
6 8 1 3
7 2 8 9

NO


text:
## Section20.7Sage

Many computations, in seemingly very different areas of mathematics, can be translated into questions about linear combinations, or other areas of linear algebra. So Sage has extensive and thorough support for topics such as vector spaces.

### SubsectionVector Spaces

The simplest way to create a vector space is to begin with a field and use an exponent to indicate the number of entries in the vectors of the space.

V = QQ^4; V

F.<a> = FiniteField(3^4)
W = F^5; W

Elements can be built with the vector constructor.

v = vector(QQ, [1, 1/2, 1/3, 1/4]); v

v in V

w = vector(F, [1, a^2, a^4, a^6, a^8]); w

w in W

Notice that vectors are printed with parentheses, which helps distinguish them from lists (though they alos look like tuples). Vectors print horizontally, but in Sage there is no such thing as a “row vector” or a “column vector,” though once matrices get involved we need to address this distinction. Finally, notice how the elements of the finite field have been converted to an alternate representation.

Once we have vector spaces full of vectors, we can perform computations with them. Ultimately, all the action in a vector space comes back to vector addition and scalar multiplication, which together create linear combinations.

u = vector(QQ, [ 1, 2,  3, 4,   5,  6])
v = vector(QQ, [-1, 2, -4, 8, -16, 32])
3*u - 2*v

w = vector(F, [1, a^2, a^4, a^6,  a^8])
x = vector(F, [1,   a, 2*a,   a,    1])
y = vector(F, [1, a^3, a^6, a^9, a^12])
a^25*w + a^43*x + a^66*y

### SubsectionSubspaces

Sage can create subspaces in a variety of ways, such as in the creation of row or column spaces of matrices. However, the most direct way is to begin with a set of vectors to use as a spanning set.

u = vector(QQ, [1, -1, 3])
v = vector(QQ, [2, 1, -1])
w = vector(QQ, [3, 0, 2])
S = (QQ^3).subspace([u, v, w]); S

3*u - 6*v + (1/2)*w in S

vector(QQ, [4, -1, -2]) in S

Notice that the information printed about S includes a “basis matrix.” The rows of this matrix are a basis for the vector space. We can get the basis, as a list of vectors (not rows of a matrix), with the .basis() method.

S.basis()

Notice that Sage has converted the spanning set of three vectors into a basis with two vectors. This is partially due to the fact that the original set of three vectors is linearly dependent, but a more substantial change has occurred.

This is a good place to discuss some of the mathematics behind what makes Sage work. A vector space over an infinite field, like the rationals or the reals, is an infinite set. No matter how expansive computer memory may seem, it is still finite. How does Sage fit an infinite set into our finite machines? The main idea is that a finite-dimensional vector space has a finite set of generators, which we know as a basis. So Sage really only needs the elements of a basis (two vectors in the previous example) to be able to work with the infinitely many possibilities for elements of the subspace.

Furthermore, for every basis associated with a vector space, Sage performs linear combinations to convert the given basis into another “standard” basis. This new basis has the property that as the rows of a matrix, the matrix is in reduced row-echelon form. You can see this in the basis matrix above. The reduced row-echelon form of a matrix is unique, so this standard basis allows Sage to recognize when two vector spaces are equal. Here is an example.

u = vector(QQ, [1, -1,  3])
v = vector(QQ, [2,  1, -1])
w = vector(QQ, [3,  0,  2])
u + v == w

S1 = (QQ^3).subspace([u, v, w])
S2 = (QQ^3).subspace([u-v, v-w, w-u])
S1 == S2

As you might expect, it is easy to determine the dimension of a vector space.

u = vector(QQ, [1, -1,  3,  4])
v = vector(QQ, [2,  1, -1, -2])
S = (QQ^4).subspace([u, v, 2*u + 3*v, -u + 2*v])
S.dimension()

### SubsectionLinear Independence

There are a variety of ways in Sage to determine if a set of vectors is linearly independent or not, and to find relations of linear dependence if they exist. The technique we will show here is a simple test to see if a set of vectors is linearly independent or not. Simply use the vectors as a spanning set for a subspace, and check the dimension of the subspace. The dimension equals the number of vectors in the spanning set if and only if the spanning set is linearly independent.

F.<a> = FiniteField(3^4)
u = vector(F, [a^i for i in range(0,  7, 1)])
v = vector(F, [a^i for i in range(0, 14, 2)])
w = vector(F, [a^i for i in range(0, 21, 3)])
S = (F^7).subspace([u, v, w])
S.dimension()

S = (F^7).subspace([u, v, a^3*u + a^11*v])
S.dimension()

So the first set of vectors, [u, v, w], is linearly independent, while the second set, [u, v, a^3*u + a^11*v], is not.

### SubsectionAbstract Vector Spaces

Sage does not implement many abstract vector spaces directly, such as $$P_n\text{,}$$ the vector space of polynomials of degree $$n$$ or less. This is due in part to the fact that a finite-dimensional vector space over a field $$F$$ is isomorphic to the vector space $$F^n\text{.}$$ So Sage captures all the functionality of finite-dimensional vector spaces, and it is left to the user to perform the conversions according to the isomorphism (which is often trivial with the choice of an obvious basis).

However, there are instances where rings behave naturally as vector spaces and we can exploit this extra structure. We will see much more of this in the chapters on fields and Galois theory. As an example, finite fields have a single generator, and the first few powers of the generator form a basis. Consider creating a vector space from the elements of a finite field of order $$7^6=117\,649\text{.}$$ As elements of a field we know they can be added, so we will define this to be the addition in our vector space. For any element of the integers mod 7, we can multiply an element of the field by the integer, so we define this to be our scalar multiplication. Later, we will be certain that these two definitions lead to a vector space, but take that for granted now. So here are some operations in our new vector space.

F.<a> = FiniteField(7^6)
u = 2*a^5 + 6*a^4 + 2*a^3 + 3*a^2 + 2*a + 3
v = 4*a^5 + 4*a^4 + 4*a^3 + 6*a^2 + 5*a + 6
u + v

4*u

2*u + 5*v

You might recognize that this looks very familiar to how we add polynomials, and multiply polynomials by scalars. You would be correct. However, notice that in this vector space construction, we are totally ignoring the possibility of multiplying two field elements together. As a vector space with scalars from $${\mathbb Z}_7\text{,}$$ a basis is the first six powers of the generator, $$\{1,\,a,\,a^2,\,a^3,\,a^4,\,a^5\}\text{.}$$ (Notice how counting from zero is natural here.) You may have noticed how Sage consistently rewrites elements of fields as linear combinations — now you have a good explanation.

Here is what Sage knows about a finite field as a vector space. First, it knows that the finite field is a vector space, and what the field of scalars is. We suppress additional output with isomorphisms between the finite field structure and the vector space structure.

V = F.vector_space(map=False); V

R = V.base_ring(); R

R == FiniteField(7)

V.dimension()

So the finite field (as a vector space) is isomorphic to the vector space $$({\mathbb Z}_7)^6\text{.}$$ Notice this is not a ring or field isomorphism, as it does not fully address multiplication of elements, even though that is possible in the field.

Second, elements of the field can be converted to elements of the vector space easily.

x = V(u); x

y = V(v); y

Notice that Sage writes field elements with high powers of the generator first, while the basis in use is ordered with low powers first. The computations below illustrate the isomorphism preserving the structure between the finite field itself and its interpretation as a vector space, $$({\mathbb Z}_7)^6\text{.}$$

V(u + v) == V(u) + V(v)

two = R(2)
V(two*u) == two*V(u)

### SubsectionLinear Algebra

Sage has extensive support for linear algebra, well beyond what we have described here, or what we will need for the remaining chapters. Create vector spaces and vectors (with different fields of scalars), and then use tab-completion on these objects to explore the large sets of available commands.