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# How to solve this recurrence relation?

There's a frog who could climb either 1 stair or 3 stairs in one shot. In how many ways he could reach at 100th stair?

I came up with the solution $g(n) = g(n-3) + g(n-1)$, where $g(0)=g(1)=g(2)=1$ and $g(3)=2$. But to solve $g(100)$ is quite difficult if I use substitution method so I just want to know if there is a feasible method to solve for $g(100)$ ?

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Small error in your initial conditions: $g(0)=1$. – Brian M. Scott Apr 8 '13 at 3:12
In the case of g(0) the frog remains at its original position only and so it should not contribute to the no. of ways to reach nth floor where n>3. – Maverick Snyder Apr 8 '13 at 3:24
But there is $1$ way for the frog to reach the original position, not $0$. Moreover, you want $g(3)=g(2)+g(0)$, and for that you need $g(0)=1$. – Brian M. Scott Apr 8 '13 at 3:25
No, I am not. I am counting the number of ways of reaching the $n$-th position. There is $1$ way to reach the starting position: do nothing. This is the standard approach to such problems. – Brian M. Scott Apr 8 '13 at 3:35
Yes my mistake you are right. – Maverick Snyder Apr 8 '13 at 3:36

This is OEIS A000930, sometimes known as Narayana’s cows sequence. The link does give a closed form:

$$g(n)=\left\lfloor dc^n+\frac12\right\rfloor\;,$$

where $c$ is the real root of $x^3-x^2-1$, and $d$ is the real root of $31x^3-31x^2+9x-1$. It also says that $d=0.611491991950812\dots$ and

$$c=\frac23\cos\left(\frac13\arccos\left(\frac{29}2\right)\right)+\frac13=1.46557123187676\dots\;,$$

but you’ll need moderately high-precision arithmetic to get an exact value for $g(100)$ from that.

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Just a little something about Narayana's constant $c$. We have $$e^{\pi\sqrt{31}} \approx 2^{12}c^{24}-24.0000069\dots$$ It's not coincidence but has to do with a certain Dedekind eta quotient. :) – Tito Piezas III Feb 12 at 5:54

Essentially, you need to find the number of natural number pairs $(a,b)$ such that $$a+3b = 100$$ More importantly, you need to take into account the order in which $1$'s and $3$'s appear. For instance, $a=1, b=33$ is a solution to the above. And we can choose $1$ one step and $33$ three steps in $34$ ways. If the solution is $(100-3b,b)$, the number of ways we can have $100-3b$ one steps and $b$ three steps is $$\dfrac{(100-2b)!}{b! (100-3b)!}$$ Hence, the total number of ways is $$\sum_{b=0}^{33} \dfrac{(100-2b)!}{b! (100-3b)!}$$ In general, $$g(n) = \sum_{n=0}^{\lfloor n/3 \rfloor} \dfrac{(n-2b)!}{b! (n-3b)!}$$ From this, we get $g(100) = 24382819596721629$.

EDIT

This also agrees with OEIS A000930, Narayana’s cows sequence.

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As a possibly computationally more efficient way of getting the answer I proffer the following (only requires integer arithmetic, and AFAICT at least asymptotically less operations than Marvis' sum of binomial coefficients). We can turn the recurrence formula into a matrix equation. Define the vector $\vec{g}(n)=(g(n),g(n+1),g(n+2))$. Then $\vec{g}(n+1)=\vec{g}(n)A$, where $A$ is the $3\times3$ matrix $$A=\left(\begin{array}{ccc}0&0&1\\1&0&0\\0&1&1\end{array}\right).$$ An obvious induction shows that $\vec{g}(n)=A^n\vec{g}(0)$. You already calculated that $\vec{g}(0)=(1,1,1)$, so `all' we need to do is to compute $A^{100}$ (actually $A^{98}$ would suffice, but that doesn'r really matter this time). There you should use the ever useful square-and-multiply-algorithm: Repeated squaring gives you $A^2$, $A^4=(A^2)^2$, $A^8$, $\ldots$, $A^{64}$ after six matrix multiplications. Three more matrix multiplications (a total of nine) then gives you $$A^{100}=A^{64+32+4}=A^{64}\cdot A^{32}\cdot A^4.$$

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The ordinary generating function for the sequence $g(n)$ is $$G(x) = \sum_{n=0}^\infty g(n)\,x^n = 1 + x + x^2 + 2x^3 + \cdots$$

The relation $g(n) = g(n - 1) + g(n - 3)$ leads to the equation $$G(x) - 1 - x - x^2 = x \big( G(x) - 1 - x \big) + x^3 G(x),$$ which simplifies to $$G(x) = \frac{1}{1 - x - x^3}.$$ Using technology to expand the series to find the $100$th term, $$g(100) = \frac{G^{(100)}(0)}{100!} = 24\,382\,819\,596\,721\,629 \approx 2.4 \times 10^{16}.$$

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I find that waving my hands vigorously is much more fun than "using technology". Besides, taking that 100th derivative seems to me to be much more work than just evaluating the recurrence. – marty cohen Apr 8 '13 at 4:34


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# The determinant of a block Matrix of Order $2n$

Let us consider the $2n \times 2n$ block matrix $$\mathbf{X} = \pmatrix{\mathbf{P} & \mathbf{Q}\\ \mathbf{R} & \mathbf{S} }$$ where $\mathbf{P} , \mathbf{Q} , \mathbf{R} , \mathbf{S}$ are square matrices of order $n$ which commute pairwise. Show that $$\det \mathbf{X} = \det (\mathbf{PS} – \mathbf{QR}).$$

The problem in this case I am having is when none of the matrices $\mathbf{P} , \mathbf{Q} , \mathbf{R} , \mathbf{S}$ are invertible. Any help will be greatly appreciated.

#### Solutions Collecting From Web of "The determinant of a block Matrix of Order $2n$"

It’s enough for $P$ and $R$ to commute. When $P$ is invertible,

$\det(X) = \det(PS – PRP^{-1}Q)$. If $P$ and $R$ commute and $P$ is not invertible, consider

$$f(t) = \det \pmatrix{P+tI & Q\cr R & S\cr}$$

For all but finitely many $t$, $P+tI$ is invertible, and it commutes with $R$, so
$$f(t) = \det((P+tI)S – (P+tI) R (P+tI)^{-1} Q) = \det((P+tI)S – RQ)$$
Now both $f(t)$ and $\det((P+tI)S – RQ)$ are polynomials in $t$, so if they are equal for all but finitely many $t$ they are equal for all $t$.

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Regular version of the site

Modality of representations

arxiv.org. math. Cornell University, 2017. No. arXiv:1707.07720v1 [math.RT] 24 Jul 2017.
We first establish several general properties of modality of al- gebraic group actions. In particular, we introduce the notion of a modali- ty-regular action and prove that every visible action is modality-regular. Then, using these results, we classify irreducible linear representations of connected simple algebraic groups of every fixed modality < 3. Next, ex- ploring a finer geometric structure of linear actions, we generalize to the case of any cyclically graded semisimple Lie algebra the notion of a packet (or a Jordan/decomposition class) and establish the properties of packets.


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# [Tex/LaTex] Option clash with xcolor and TikZ

colorincompatibilitypackage-optionspackagestikz-pgf

I want to draw some coloured circles, so I loaded tikz in my preamble. However, now xcolor is telling me that it has an "option clash". How should I fix this?

\documentclass{article}
\usepackage{tikz}
\usepackage[table]{xcolor}

\begin{document}
Hello world.
\end{document}

The error message is this:

! LaTeX Error: Option clash for package xcolor.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
...

l.4

[]
There has now been an attempt to load it with options
[table]
,table
to your \documentclass declaration may fix this.
Try typing  <return>  to proceed.

Load xcolor before tikz:

\usepackage[table]{xcolor}
\usepackage{tikz}

The problem is that pgf/tikz already loads xcolor so loading it again with a new package option triggers the error message.

Another option is to pass the table option to the class and not to the xcolor package. Then the packages can be loaded in any order:

\documentclass[table]{article}
\usepackage{xcolor}
\usepackage{tikz}

or

\documentclass[table]{article}
\usepackage{tikz}
\usepackage{xcolor}


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# Finding the Inverse Function of a Quadratic Function

What we want here is to find the inverse function – which implies that the inverse MUST be a function itself. Otherwise, we got an inverse that is not a function.

Not all functions are naturally “lucky” to have inverse functions. This happens in the case of quadratics because they all fail the Horizontal Line Test. However, if I restrict their domain to where the $x$ values produce a graph that would pass the horizontal line test, then I will have an inverse function.

But first, let’s talk about the test which guarantees that the inverse is a function.

## Horizontal Line Test

Given a function $f(x)$, it has an inverse denoted by the symbol $\color{red}{f^{ - 1}}\left( x \right)$, if no horizontal line intersects its graph more than one time.

• Example of a graph with an inverse
• Example of a graph without an inverse

The following are the main strategies to algebraically solve for the inverse function.

## Key Steps in Finding the Inverse Function of a Quadratic Function

1. Replace $f(x)$ by $y$.
2. Switch the roles of $\color{red}x$ and $\color{blue}y$. In other words, interchange $x$ and $y$ in the equation.
3. Solve for $y$ in terms of $x$.
4. Replace $y$ by ${f^{ - 1}}\left( x \right)$ to get the inverse function
5. Sometimes, it is helpful to use the domain and range of the original function to identify the correct inverse function out of two possibilities. This happens when you get a “plus or minus” case in the end.

#### Examples of How to Find the Inverse Function of a Quadratic Function

Example 1: Find the inverse function of $f\left( x \right) = {x^2} + 2$, if it exists. State its domain and range.

The first thing I realize is that this quadratic function doesn’t have a restriction on its domain. I am sure that when I graph this, I can draw a horizontal line that will intersect it more than once. Therefore the inverse is not a function. I will not even bother applying the key steps above to find its inverse.

The diagram shows that it fails the Horizontal Line Test, thus the inverse is not a function. I will stop here.

Example 2: Find the inverse function of $f\left( x \right) = {x^2} + 2,\,\,x \ge 0$, if it exists. State its domain and range.

This same quadratic function, as seen in Example 1, has a restriction on its domain which is $x \ge 0$. After plotting the function in $xy-$axis, I can see that the graph is a parabola cut in half for all $x$ values equal to or greater than zero. This should pass the Horizontal Line Test which tells me that I can actually find its inverse function by following the suggested steps.

In its graph below, I clearly defined the domain and range because I will need this information to help me identify the correct inverse function in the end.

Remember that the domain and range of the inverse function come from the range, and domain of the original function, respectively. It’s called the swapping of domain and range.

Even without solving for the inverse function just yet, I can easily identify its domain and range using the information from the graph of the original function: domain is x ≥ 2 and range is y ≥ 0.

Do you see how I interchange the domain and range of the original function to get the domain and range of its inverse?

Now, let’s go ahead and algebraically solve for its inverse.

Graphing the original function with its inverse in the same coordinate axis…

Example 3: Find the inverse function of $f\left( x \right) = - {x^2} - 1,\,\,x \le 0$ , if it exists. State its domain and range.

This problem is very similar to Example 2. The range starts at $\color{red}y=-1$, and it can go down as low as possible.

Now, these are the steps on how to solve for the inverse.

Applying square root operation results in getting two equations because of the positive and negative cases. To pick the correct inverse function out of the two, I suggest that you find the domain and range of each possible answer. Now, the correct inverse function should have a domain coming from the range of the original function; and a range coming from the domain of the same function.

The following are the graphs of the original function and its inverse on the same coordinate axis.

Example 4: Find the inverse of the function below, if it exists. State its domain and range.

I would graph this function first and clearly identify the domain and range. Notice that the restriction in the domain cuts the parabola into two equal halves. I will deal with the left half of this parabola. Clearly, this has an inverse function because it passes the Horizontal Line Test.

Proceed with the steps in solving for the inverse function. In fact, there are two ways how to work this out.

• Solve this by the Quadratic Formula as shown below.

$x = {\Large{{{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}}}$

where $a$, $b$, and $c$ can contain variables.

This is expected since we are solving for a function, not exact values.

The key step here is to pick the appropriate inverse function in the end because we will have the plus (+) and minus (−) cases. We can do that by finding the domain and range of each and compare that to the domain and range of the original function. Remember that we swap the domain and range of the original function to get the domain and range of its inverse.

• The method of completing the squares allows us to isolate the variable in a quadratic trinomial. As you will see in the steps, the quadratic trinomial is converted into linear binomial raised to the power of $2$. Obviously, we can apply the square root operation to get rid of the exponent $2$, therefore leaving us with an easy equation to solve.

If you observe, the graphs of the function and its inverse are actually symmetrical along the line $y = x$ (see dashed line). They are like mirror images of each other.

I hope that you gain some level of appreciation on how to find the inverse of a quadratic function. Although it can be a bit tedious, as you can see, overall it is not that bad. I recommend that you check out the related lessons on how to find inverses of other kinds of functions.

### Practice with Worksheets

You might also be interested in:

Inverse of a 2×2 Matrix

Inverse of Absolute Value Function

Inverse of Constant Function

Inverse of Exponential Function

Inverse of Linear Function

Inverse of Logarithmic Function

Inverse of Rational Function

Inverse of Square Root Function


text:
Question:

A number decreased by $27 \frac{1}{2} \%$ gives 87 . The number is

(a) 58

(b) 110

(c) 120

(d) 135

Solution:

(c) 120

Let the required number be $\mathrm{x}$. Then, we have:

$x-\left(27 \frac{1}{2} \%\right.$ of $\left.x\right)=87$

$\Rightarrow \mathrm{x}-\left(\frac{55}{2} \%\right.$ of $\left.\mathrm{x}\right)=87$

$\Rightarrow \mathrm{x}-\left(\mathrm{x} \times \frac{55}{2} \times \frac{1}{100}\right)=87$

$\Rightarrow \mathrm{x}-\frac{11 \mathrm{x}}{40}=87$

$\Rightarrow \frac{29 \mathrm{x}}{40}=87$

$\Rightarrow \mathrm{x}=\left(87 \times \frac{40}{29}\right)$

$\Rightarrow \mathrm{x}=120$


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# History File Stratification

## History File

In order to stratify a cohort by a time-dependent exposure covariate (aside from age and calendar period), a history file must be created and read in. This file contains one row per person per exposure period. An exposure period is a period of time in which all daily values/levels of an exposure variable are assumed to be constant.

Below are the required variables to be found within the history file:

Variable Description Format
id Unique identifier for each person
begin_dt Beginning date of exposure period character
end_dt End date of exposure period character
<daily exposure variables> Exposure variable(s) numeric

Below is an example layout of a history file with multiple exposures:

id begin_dt end_dt employed exposure_level
1 12/21/1970 6/15/1971 1 71.0
1 6/16/1971 12/31/1980 1 5.0
2 1/19/1972 6/15/1975 1 10.0
3 11/23/1970 12/7/1972 1 41.5

The above example contains 3 persons with 2 exposure variables, employed and exposure_level. Person/id 1 contains 2 non-overlapping exposure periods in which employed is 1 for both but exposure_level drops from 71 to 5 units per day.

## Calculation of Cumulative Exposure From History File

LTASR comes with an example history file, called history_example, that can be used in conjunction with person_example for testing. Below reads in both example files and formats dates appropriately:

person <- person_example %>%
mutate(dob = as.Date(dob, format='%m/%d/%Y'),
pybegin = as.Date(pybegin, format='%m/%d/%Y'),
dlo = as.Date(dlo, format='%m/%d/%Y'))

history <- history_example %>%
mutate(begin_dt = as.Date(begin_dt, format='%m/%d/%Y'),
end_dt = as.Date(end_dt, format='%m/%d/%Y')) %>%
group_by(id)

For the remainder of this section, we will consider Person/id 1 to demonstrate how exposure is calculated over time. Below is the information found within the person file for person/id 1:

id gender race dob pybegin dlo vs rev code
1 M W 11/20/1945 12/21/1970 7/31/2016

This example person’s follow-up starts on 12/21/1970 and continues through 7/31/2016. Below plots their cumulative exposure for both employed and exposure_level variables:

Both exposures start at 0, then employed increases by 1 unit per day for both periods. This can therefore be thought of as a duration variable (in days) of all periods. The exposure_level increases rapidly (71 units per day) during the first period and then increases slower (5 units per day).

NOTE: Any gaps within the history file and the follow-up times (for example, the period between the last exposure period within the history file and through the end of follow-up) is assumed to be 0. That is, exposure values do not change during these periods.

## Stratifying person time

Once the person file and history file have been read in (see Demo for basic stratification vignette for additional information on how to read in files), information on how to stratify the exposure variables must be defined using the exp_strata() function.

Below specifies which exposure variables to consider, what cut-points to use for stratification and any lag (in years) to apply to the cumulative exposure variable:

exp1 <- exp_strata(var = 'employed',
cutpt = c(-Inf, 365, Inf),
lag = 0)
exp2 <- exp_strata(var = 'exposure_level',
cutpt = c(-Inf, 0, 10000, 20000, Inf),
lag = 10)

The employed variable will contain 2 strata: (-Inf, 365] and (365, Inf]. Or, put alternatively, ≤ 1 year and > 1 year.

The exposure_level will contain 5 strata: (-Inf, 0], (0, 10000], (10000, 20000] and (20000, Inf). Therefore, the first category defines unexposed person-time. Additionally, a 10 year lag will be applied when defining strata.

Once the exposure strata have been defined, LTASR provides two functions for stratifying the cohort. One is get_table_history whose usage is:

py_table <- get_table_history(persondf = person,
rateobj = us_119ucod_19602020,
historydf = history,
exps = list(exp1, exp2))

This creates the below table:

ageCat CPCat gender race employedCat exposure_levelCat pdays _o55 _o52
[15,20) [1970,1975) F W (-Inf,365] (-Inf,0] 365 0 0
[15,20) [1970,1975) F W (365, Inf] (-Inf,0] 381 1 0
[25,30) [1970,1975) M N (-Inf,365] (-Inf,0] 55 0 0
[25,30) [1970,1975) M W (-Inf,365] (-Inf,0] 177 0 0
[25,30) [1970,1975) M W (365, Inf] (-Inf,0] 1295 0 0
[25,30) [1975,1980) M W (365, Inf] (-Inf,0] 323 0 0

This function is very fast, and replicates how the original LTAS behaved. It also exactly stratifies the person-days into the appropriate strata. However, it may be desired to calculate mean exposure values for each strata to be used in a Poisson regression later. To implement this exactly is very slow.

Therefore, a separate function, get_table_history_est, calculates these mean exposure values and also allows for a step parameter to be specified defining the number of days to calculate the cumulative exposure.

An example usage is:

py_table_est <- get_table_history_est(persondf = person,
rateobj = us_119ucod_19602020,
historydf = history,
exps = list(exp1, exp2),
step = 7)

By specifying step = 7, person time is considered every 7 days when allocating person-time to strata. This results in a significant increase in speed at the cost of a (generally) trivial amount of inaccuracy.

Specifying step = 1 will calculate strata exactly for each individual day, but is significantly slower.

Below is the result of this specification:

ageCat CPCat gender race employedCat exposure_levelCat pdays _o55 _o52 employed exposure_level
[15,20) [1970,1975) F W (-Inf,365] (-Inf,0] 361 0 0 184.0 0
[15,20) [1970,1975) F W (365, Inf] (-Inf,0] 385 1 0 557.0 0
[25,30) [1970,1975) M N (-Inf,365] (-Inf,0] 53 0 0 30.0 0
[25,30) [1970,1975) M W (-Inf,365] (-Inf,0] 376 0 0 136.9 0
[25,30) [1970,1975) M W (365, Inf] (-Inf,0] 1096 0 0 927.5 0
[25,30) [1975,1980) M W (365, Inf] (-Inf,0] 322 0 0 1636.5 0

As can be seen, the pday are slightly different than the previous table. However, the effects on results will generally be trivial.

In addition, two additional variables are available: employed and exposure_level indicating the person-time weighted mean values.
## Step Specifications

When specifying the step parameter, there is a trade-off between computation speed and accuracy. Specifying step = 1 will result in the most accurate stratification, but can be extremely slow.

To investigate this further, below plots the time (in minutes) taken to stratify a cohort of 5,200 people with 2 exposure variables for various specifications of the step parameter:

There are dramatic savings in computation time when increasing the step parameter in the low end. In this example, at about step = 10, improvements in computation time diminishes. It seems a step parameter of about 5-10 is a good compromise.

Exact times will depend upon:

• the size of the cohort,
• the number of exposure variables and
• the the number of strata per exposure variable.

An additional consideration is the level of detail of the exposure variable. That is, if exposure is dramatically changing, relative to its specified strata, the loss of accuracy will be more dramatic for small increases of the step parameter. For example, age is stratified by 5-year increments, therefore, a step value of 1-week (step = 7) will cause a trivial amount of inaccuracy.

One option is to use a crude step value during initial investigations, but when results are to be published/presented, the function can be run again with a smaller step value.


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# Question about Thurston's paper “A norm for the homology of 3-manifolds”

I have a question about the proof of Theorem 5 in Thurston's paper "A norm for the homology of 3-manifolds". It's the theorem that asserts that the fibered faces are, well, fibered. More precisely, fix a compact 3-manifold $M$, and let $B_x$ be the Thurston polytope. Then the theorem asserts that the set of elements of $H^1(M;\mathbb{R})$ which are representable by non-singular closed $1$-forms is a union of cones on open top-dimensional faces of $B_x$ (minus the origin).

Here's the part that I am having trouble with. Assume that $M$ is closed, and let $\alpha$ be any non-singular closed $1$-form. Thurston already proved that $\alpha$ lies in the cone on an open top-dimensional face of $B_x$. If $S$ is a surface in $M$, then denote by $[S] \in H^1(M;\mathbb{R})$ the Poincare dual cohomology class. Consider an incompressible surface $S$ in $M$ such that $[S]$ lies in the cone on the same open top-dimensional face of $B_x$ as $\alpha$. Most of the work of the proof goes into showing that linear combinations $t[S] + u \alpha$ with $t \geq 0$ and $u > 0$ are representable by nonsingular closed $1$-forms. That I have no problem with. However, Thurston then asserts that $[S]$ can be so represented, and the only justification he gives is that you can get it by "iterating this construction". I cannot figure out what he means here.

Can anyone help me? Thanks!

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Sorry for asking, but I have been trying to find this paper. Does anyone have a link for it please? –  user48617 Mar 23 '14 at 15:14
@user48617: this is available (volume 59 Memoirs AMS, Number 339) at the AMS website. ams.org/books/memo/0339 –  Ian Agol Jun 2 '14 at 17:41

Think in terms of projective coordinates (i.e. I won't say "cone" any more). I think the lemma you describe also holds for surfaces $S$ on the boundary of the face in question. (If this is not the case, then never mind, I'll delete this answer.) That is, we can take $[S]$ to be a vertex of the face, and we know that $t[S] + u\alpha$ is representable by a nonsingular 1-form whenever $\alpha$ is, for $t \ge 0$ and $u>0$. In other words, starting at a representable point in the interior of the face, we can move a arbitrary distance toward any of the corners of the boundary of the face, so long as we stay in the interior. Since the interior of the face is the open convex hull of the corners, we can reach any point in the interior this way by letting $S$ run through the corners of the face.
By the way, if you believe the lemma for $S$ in the interior of the face, you don't have to reprove it for $S$ in the boundary. Instead, you just have to continue the segment from $\alpha$ to $S$ a little further, to another rational point $S'$ still in the interior of the face, and apply the lemma to $S'$. Not that it will be much different to prove for points in the boundary. –  Lee Mosher Feb 29 '12 at 20:27


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# pgfmathsetmacro issue in foreach

I am trying to use \pgfmathsetmacro inside a foreach loop to connect nodes which are adjacent and named according to position as follows:

\foreach \src in {0,...,\T}{
\pgfmathsetmacro{\dest}{\src+1};
\path (n-\src-1) edge[<-] (n-\dest-1);
};

This however leads to the following error during compilation:

./nn.tex:54: Package pgf Error: No shape named n-1 is known. [    }]
./nn.tex:54: Package pgf Error: No shape named n-2 is known. [    }]
./nn.tex:54: Package pgf Error: No shape named n-3 is known. [    }]
./nn.tex:54: Package pgf Error: No shape named n-4 is known. [    }]

which makes it seem as though the \dest value is not available. How do I get around this issue?

After the definition via \pgfmathsetmacro, the macro \dest contains a real number (1.0, 2.0, ...), but very likely you need an integer. This can be achieved by using \pgfmathtruncatemacro instead of \pgfmathsetmacro to cut the decimal fraction part:

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\path
\foreach \src in {0, ..., 4} {
(\src, 0) node[circle, draw] (n-\src-1) {}
}
;

\foreach \src in {0, ..., 3} {
\pgfmathtruncatemacro{\dest}{\src+1};
\path (n-\src-1) edge[<-] (n-\dest-1);
};
\end{tikzpicture}
\end{document}

Alternatively an integer addition can be used, e.g. via eTeX's \numexpr:

\edef\dest{\the\numexpr(\src)+1}

Another method uses the evaluate key (instead of \pgfmathsetmacro) and the int function (to truncate your addition):

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\path
\foreach \src in {0, ..., 4} {
(\src, 0) node[circle, draw] (n-\src-1) {}
}
;

\foreach \src [evaluate=\src as \dest using int(\src + 1)]
in {0, ..., 3} {
\path (n-\src-1) edge[<-] (n-\dest-1);
};
\end{tikzpicture}
\end{document}

• Another possibility would be to run over \dest and use the remember method of TikZ's \foreach, as in \foreach \dest [remember=\dest as \src (initially 0)] in {1,...,4} {…}. – Henri Menke Dec 6 '15 at 11:31


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# What is the type number of a discrete time system given $H(z)$?

Given a continuous time impulse response $$h(t)$$, if I take the Laplace transform and count the no. of poles at origin, that gives the type number of the system. For e.g., $$H(s) = \frac{2}{s(s+2)}$$ Here, there is one pole at origin and hence it is type 1 system. This also implies that the number of integrators in the system is 1. I'm having difficulty in determining the system type when it comes to discrete time systems. For e.g., given the $$Z$$ transform of impulse response, $$H(z) = \frac1z+\frac1{z^6}$$ how to deduce the system type? Is it 6 because there are six poles at origin? Also how does it relate to no. of integrators in the system given that it is discrete time. Does it correspond to no. of accumulators in any way? I came across this while searching but it talks about the order of the input polynomial as the type of the system which makes the type number independent of the impulse response. Is this correct?

I couldn't find a definition of system type for discrete-time control systems, but in analogy with continuous-time systems you should count the number of poles at $$z=1$$, not at $$z=0$$, because in discrete time, $$z=1$$ is the DC point, just like $$s=0$$ in continuous time.
A discrete-time integrator (an accumulator), has a pole at $$z=1$$. Its transfer function is given by
$$H(z)=\frac{z}{z-1}\tag{1}$$
• @Navin: What people usually mean by DC is the point $s=0$ and $z=1$. Anyway, what is important is that these two points correspond to each other, unlike the points $s=0$ and $z=0$. – Matt L. Apr 29 '19 at 17:41


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# Which are “big theorems” of descriptive set theory?

Question: If one were to fully understand 10 theorems in DST, or 15,20,25,30 theorems, which ones would be the most important to understand in order to work towards an understanding of descriptive set theory (viewed from the boldface side of things)? Specifically, I mean theorems which are in Kechris, Moschovakis, Srivastava. I know that a lot of knowledge goes into understanding the "big theorems" and this will have to be obtained in the journey.

My point: I'm working through Kechris and would like some sort of guide posts to help me know where I am in the subject. I would like the theorems to be from these books and not any theorems from current literature.

Also, are there any books that can help me in the subject besides the 3 I mentioned above?

-
This question should probably be community wiki, because there is not a unique correct answer. – Trevor Wilson Oct 15 '12 at 3:54
@Trevor, I disagree about this being CW (and the SE 2.0 does not allow the OP to set CW on their own anyway; just for answers). – Asaf Karagila Oct 15 '12 at 8:09

I would like to add on Trevor's list:

• Gale-Stewart theorem (or determinacy of any simple-enough class).
• Borel determinacy.
• $\Sigma^1_1$ determinacy implies $0^\#$.
• Solovay's construction of a model where all sets are Lebesgue measurable (while from a descriptive point of view it is not the most exciting theorem, it is an important proof).

I'm not sure whether or not it's too far, but in a yearly course I'd definitely expect to hear something about it:

• The relations between Woodin cardinals and Projective Determinacy

It is somewhat unclear to me how far do you want to go with this. Do you want to end up knowing? I would consider the projective determinacy equivalence with Woodin cardinals (and the natural extension of AD with infinitely many of those) quite an advanced theorem, but maybe you would like to go beyond that. Maybe you would like to know stationary tower forcing; and theorems about the absoluteness of the theory of $L(\mathbb R)$. That really depends on you.

As for references, I would also suggest Miller's book:

Descriptive Set Theory and Forcing: how to prove theorems about Borel sets the hard way. Lecture Notes in Logic 4(1995), Springer-Verlag.

Which one can find for free on his homepage or on ArXiv.

-

Here are the first few I can think of.

• The Baire category theorem
• Mostowski's absoluteness Theorem
• Shoenfield's absoluteness theorem
• The Mansfield–Solovay theorem
• The Martin–Solovay Theorem
• Martin's proof of analytic determinacy
-


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# Thermodynamic potential

Thermodynamic potential are thermodynamics sizes of their information content fro the behavior of a thermodynamic system in equilibrium fully described. The independent state variables of a thermodynamic potential are called its natural variables if their derivative of the potential is equal to one of the dependent state variables (for example:) . A thermodynamic potential corresponds to the information content of the internal energy , the natural variables of which are all extensive ( fundamental equation ). ${\ displaystyle \ left (\ partial U / \ partial S \ right) _ {V, N} = T}$ ${\ displaystyle U}$${\ displaystyle S, V, N}$

Thermodynamic potentials, which are energies, can be derived from the internal energy by Legendre transformation , but, unlike this, have one or more intense quantities as natural variables . The intensive quantities arise during the coordinate transformation as derivatives of the internal energy according to its extensive variables. ${\ displaystyle U (S, V, N)}$${\ displaystyle (T, p, \ mu)}$

There are also other thermodynamic potentials that are not energies, for example entropy . ${\ displaystyle S (U, V, N)}$

The term thermodynamic potential was generalized by Gottfried Falk to the term Massieu-Gibbs functions (according to Josiah Willard Gibbs ), which are corresponding state functions in not necessarily thermodynamic systems.

## Physical meaning

An extreme value (not always a minimum) of a thermodynamic potential indicates the thermodynamic equilibrium .

After connecting one closed system to another, a thermodynamic equilibrium has been established as soon as the entropy of the entire system is at its maximum. In this case, all intensive parameters of the two systems are also the same:

{\ displaystyle {\ begin {aligned} T_ {1} & = T_ {2} \\ p_ {1} & = p_ {2} \\\ mu _ {1} & = \ mu _ {2} \ end { aligned}}}

In addition, thermodynamic potentials summarize the equations of state of the system, since these are accessible by differentiating a thermodynamic potential according to its dependent variables.

## description

The internal energy and the functions resulting from it by means of the Legendre transformation are thermodynamic potentials (with one exception , see below). The Legendre transformation transforms the potentials ${\ displaystyle U (S, V, N)}$${\ displaystyle K}$

• from entropy to temperature , there${\ displaystyle S}$ ${\ displaystyle T}$${\ displaystyle {\ frac {\ partial U} {\ partial S}} = T}$
• from volume to pressure , there${\ displaystyle V}$ ${\ displaystyle p}$${\ displaystyle {\ frac {\ partial U} {\ partial V}} = - p}$
• from particle number to chemical potential , there${\ displaystyle N}$ ${\ displaystyle \ mu}$${\ displaystyle {\ frac {\ partial U} {\ partial N}} \, = \ mu}$

Because of these 3 pairs of variables there are possible thermodynamic potentials: ${\ displaystyle 2 ^ {3} = 8}$

Thermodynamic Potentials
Name (alternative name) Formula symbol natural variables
(intense bold)
Characteristic function f Connections
Inner energy ${\ displaystyle U}$ ${\ displaystyle S, V, N}$ ${\ displaystyle \ mathrm {d} U = TdS-pdV + \ mu {d} N}$ ${\ displaystyle U (S, V, N) = TS-pV + {\ sum _ {i = 1} ^ {K}} \, \ mu _ {i} N_ {i}.}$
( Euler equation for the internal energy )
Free energy
(Helmholtz potential)
${\ displaystyle F}$ ${\ displaystyle {\ boldsymbol {T}}, V, N}$ ${\ displaystyle \ mathrm {d} F = -SdT-pdV + \ mu {d} N}$ ${\ displaystyle F = U (S, V, N) -TS}$
Enthalpy ${\ displaystyle H}$ ${\ displaystyle S, {\ boldsymbol {p}}, N}$ ${\ displaystyle \ mathrm {d} H = TdS + Vdp + \ mu {d} N}$ ${\ displaystyle H = U (S, V, N) \ qquad \ + pV}$
Gibbs energy
(free enthalpy)
${\ displaystyle G}$ ${\ displaystyle {\ varvec {T}}, {\ varvec {p}}, N}$ ${\ displaystyle \ mathrm {d} G = -SdT + Vdp + \ mu {d} N}$ {\ displaystyle {\ begin {aligned} G & = U (S, V, N) -TS + pV \\ & = H (S, p, N) -TS \\ & = F (T, V, N) \ qquad \ quad + pV \ end {aligned}}}
- ${\ displaystyle R}$ ${\ displaystyle S, V, {\ boldsymbol {\ mu}}}$ ${\ displaystyle \ mathrm {d} R = TdS-pdV-N {d} \ mu}$ ${\ displaystyle R = U (S, V, N) \ qquad \ qquad \ quad - \ mu N}$
Grand canonical potential ${\ displaystyle \ Omega}$ ${\ displaystyle {\ varvec {T}}, V, {\ varvec {\ mu}}}$ ${\ displaystyle \ mathrm {d} \ Omega = -S {d} Tp {d} VN {d} \ mu}$ {\ displaystyle {\ begin {aligned} \ Omega & = U (S, V, N) -TS \ qquad \ - \ mu N \\ & = R (S, V, \ mu) -TS \\ & = F (T, V, N) \ qquad \ qquad \ quad - \ mu N \ end {aligned}}}
- ${\ displaystyle J}$ ${\ displaystyle S, {\ boldsymbol {p}}, {\ boldsymbol {\ mu}}}$ ${\ displaystyle \ mathrm {d} J = TdS + Vdp-N {d} \ mu}$ {\ displaystyle {\ begin {aligned} J & = U (S, V, N) \ qquad \ quad + pV- \ mu N \\ & = R (S, V, \ mu) \ qquad \ quad + pV \\ & = H (S, p, N) \ qquad \ quad \ qquad - \ mu N \ end {aligned}}}
- ${\ displaystyle (K)}$ ${\ displaystyle {\ varvec {T}}, {\ varvec {p}}, {\ varvec {\ mu}}}$
Not useful because the Gibbs-Duhem relation forbids the independent specification of the variables . ${\ displaystyle T, p, \ mu}$
${\ displaystyle \ mathrm {d} K = -SdT + Vdp-N {d} \ mu}$ {\ displaystyle {\ begin {aligned} K & = U (S, V, N) -TS + pV- \ mu N = 0 \\ & = J (S, p, \ mu) -TS \\ & = \ Omega (T, V, \ mu) \ qquad \ quad + pV \\ & = G (T, p, N) \ qquad \ quad \ qquad - \ mu N \ end {aligned}}}
Guggenheim Square

All meaningful thermodynamic potentials, which result from U (S, V, N) by Legendre transformation, provide the same, complete information about a system. The simplest description of the system, however, only provides one of the potentials depending on the ensemble ; this is extremal in equilibrium.

One way of memorizing thermodynamic potentials with their natural variables is the Guggenheim square .

## Example of transformation

As an example, the change from one potential to the other is to be carried out using characteristic functions ( total differentials ).

The aim is to transform from the internal energy  U with the natural variables to a new potential with the natural variables . We start with the inner energy: ${\ displaystyle S, V, N}$${\ displaystyle T, V, N}$

{\ displaystyle {\ begin {aligned} \ mathrm {d} U (S, V, N) & = {\ frac {\ partial U} {\ partial S}} \, \ mathrm {d} S + {\ frac { \ partial U} {\ partial V}} \, \ mathrm {d} V + {\ frac {\ partial U} {\ partial N}} \, \ mathrm {d} N \\ & = T \, \ mathrm { d} Sp \, \ mathrm {d} V + \ mu \, \ mathrm {d} N \ end {aligned}}}

With

{\ displaystyle {\ begin {aligned} \ mathrm {d} (TS) & = T \, \ mathrm {d} S + S \, \ mathrm {d} T \\\ Leftrightarrow T \, \ mathrm {d} S & = \ mathrm {d} (TS) -S \, \ mathrm {d} T \ end {aligned}}}

follows

${\ displaystyle \ Rightarrow \ mathrm {d} U = \ mathrm {d} (TS) -S \, \ mathrm {d} Tp \, \ mathrm {d} V + \ mu \, \ mathrm {d} N.}$

Now bring to the left: ${\ displaystyle \ mathrm {d} (TS)}$

{\ displaystyle {\ begin {aligned} \ Leftrightarrow \ mathrm {d} (U-TS) & = - S \, \ mathrm {d} Tp \, \ mathrm {d} V + \ mu \, \ mathrm {d} N =: \ mathrm {d} F \\ & = {\ frac {\ partial F} {\ partial T}} \, \ mathrm {d} T + {\ frac {\ partial F} {\ partial V}} \ , \ mathrm {d} V + {\ frac {\ partial F} {\ partial N}} \, \ mathrm {d} N \ end {aligned}}}

The new potential now depends on the natural variables and is called "free energy  ". It is the Legendre transform of U (S, V, N) with respect to the variable S. ${\ displaystyle U-TS}$${\ displaystyle T, V, N}$${\ displaystyle F}$

## Equations of state

Let be any thermodynamic potential, then its total differential can be written as ${\ displaystyle \ Phi}$

${\ displaystyle \ mathrm {d} \ Phi = \ sum _ {i} x_ {i} \ cdot \ mathrm {d} y_ {i}}$

in which

• ${\ displaystyle y_ {i}}$the natural variables of are and${\ displaystyle \ Phi}$
• ${\ displaystyle x_ {i}}$ the respective conjugated variables.

From the chain rule follows:

${\ displaystyle x_ {j} = {\ frac {\ partial \ Phi} {\ partial y_ {j}}} {\ Bigg |} _ {\ {y_ {i \ neq j} \}}}$

Here is the set of all natural variables of none . They are expressions of thermodynamic parameters as a function of derivatives of the thermodynamic potential, which depends on its natural variables. The so is precisely the state functions of the system. ${\ displaystyle \ {y_ {i \ neq j} \}}$${\ displaystyle \ Phi}$${\ displaystyle y_ {i}}$${\ displaystyle x_ {j}}$${\ displaystyle x_ {j}}$

## Individual evidence

1. a b c Wolfgang Nolting : Basic course in theoretical physics. Volume 6: Statistical Physics. 6th edition. Springer, Berlin et al. 2007, ISBN 978-3-540-68870-9 , p. 373 ff., ( Google Books ).
2. Wolfgang Nolting: Basic course in theoretical physics. Volume 4: Special Theory of Relativity, Thermodynamics. 6th, updated edition. Springer, Berlin et al. 2005, ISBN 3-540-24119-1 , p. 206.
3. Gottfried Falk : Theoretical Physics. On the basis of a general dynamic (= Heidelberg pocket books. ). 2 volumes (Vol. 1: Elementary Point Mechanics . Vol. 2: General Dynamics, Thermodynamics. 7 and 27, ). Springer, Berlin et al. 1966–1968.
4. Gottfried Falk, Wolfgang Ruppel : Energy and Entropy. An introduction to thermodynamics. Springer, Berlin et al. 1976, ISBN 3-540-07814-2 .
5. Christoph Strunk: Modern Thermodynamics. 2015, pp. 144 ff., ( Google Books ).
6. Walter Greiner , Ludwig Neise, Horst Stöcker : Thermodynamics and statistical mechanics (= Theoretical Physics. Vol. 9). 2nd, completely revised and enlarged edition. Deutsch, Thun et al. 1993, ISBN 3-8171-1262-9 , p. 150.
7. ^ Callen: Thermodynamics and an Introduction to Thermostatistics. 2nd edition. 1985, p. 37.


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# Shouldn't the matrix of eigen vectors of a defective matrix be non-invertible?

Is it true that a defective (non-diagonalizable) square matrix has a set of eigen vectors that don't span the whole space? In that case, if we do its singular value decomposition, shouldn't the matrix containing the eigen vectors be less than full rank and hence non-invertible?

If this is true, I'm getting a contradiction. Based on this page: What kind of matrices are non-diagonalizable?, the matrix:

$$A = \pmatrix{ 2 & 1\\ 0 & 2 }$$

is non-diagonalizable.

Now, I get its eigen matrix in python and try to invert it and nothing explodes.

import numpy as np
a=np.array([[2,1],[0,2]])
np.linalg.inv(np.linalg.svd(a)[0])

What am I missing?

EDIT: Here are the $$U$$ matrix, diagonal eigen values and $$V$$ matrix returned by Python. Are they incorrect? Should we even expect the $$U$$ and $$V$$ matrices to be non-invertible in this case?

u=[[ 0.78820544, -0.61541221],
[ 0.61541221,  0.78820544]]
lambda=[2.56155281, 1.56155281],
v=[[ 0.61541221,  0.78820544],
[-0.78820544,  0.61541221]]

• For your matrix, eigenvectors are just multiples of $[1,0]^T$ with eigenvalue $2$. So of course the resulting $2 \times 1$ matrix is "non-invertible". Is this, then, a question about Python or something? Jan 20 at 19:20
• When you say nothing explodes, does the returned inverse matrix actually function as an inverse? Or is it rather a Moore-Penrose pseudoinverse, or perhaps just some junk? Jan 20 at 19:33

svd doesn't calculate the eigenvector matrix; it calculates the Singular Value Decomposition. For singular values, it doesn't matter whether $$A$$ itself is diagonalizable, but that self-adjoint operator $$A^*A = \begin{pmatrix} 4 & 2 \\ 2 & 5 \end{pmatrix}$$ is diagonalizable, and it is.

The answer by @Dan is spot on. I should have been looking for the eigen decomposition, not the singular value decomposition. And if I do that with python, I get an Eigen matrix whose inverse explodes as expected.

(code continues from the question)

np.linalg.eig(a)

produces:

(array([2., 2.]),
array([[ 1.0000000e+00, -1.0000000e+00],
[ 0.0000000e+00,  4.4408921e-16]]))


text:
• Category:
• matlab
• Platform:
• matlab
• File Size:
• 2048
• Update:
• 2018-06-09
• 0
• 大古-1
Description: optimization algorithm.Genetic algorithm is a search algorithm used to solve optimization in computational mathematics. It is a kind of evolutionary algorithm. Evolutionary algorithms originally developed from some phenomena in evolutionary biology, such as heredity, mutation, natural selection and hybridization. Genetic algorithm is usually implemented in a computer simulation mode. For an optimization problem, a number of candidate solutions (called individuals) are represented by abstract representations (called chromosomes) to better solutions. Traditionally, binary representations (i.e. 0 and 1 strings) are used, but other representations can also be used. Evolution starts from the population of completely random individuals, and then occurs from generation to generation. In each generation, the fitness of the whole population is evaluated.
文件名大小更新时间遗传算法代码\best.m 320  2018-05-31


text:
# For what value of x is x/0 = 2/1 ?

Dec 29, 2016

The left hand side of this equation is undefined, regardless of the value of $x$, so there are no solutions.
In the context of algebra, division should be the same as multiplication by the multiplicative inverse, but $0$ has no multiplicative inverse. So division by $0$ is undefined.
If ${0}^{- 1}$ were defined then:
$1 = 0 \cdot {0}^{- 1} = \frac{0}{0} = \frac{0 + 0}{0} = \frac{0}{0} + \frac{0}{0} = 0 \cdot {0}^{- 1} + 0 \cdot {0}^{- 1} = 1 + 1 = 2$


text:
# Class 10 Maths Worksheet for Similar triangles

1) ABC is an isosceles triangle, right – angled at C. Prove that AB2 = 2BC2.
2) The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.
3) The foot of a ladder is 6 m away from a wall and its op reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?
4) Determine whether the triangle having sides (b – 1) cm, 2√b  cm and (b + 1) cm is a right angled triangle.
5) The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1cm, find the corresponding median of the other.
6) In  ABC, AD is perpendicular to BC. Prove that:
a)AB2 + CD2 = AC2 + BD2
b) AB2 – BD2 = AC2 – CD2
7) In a quadrilateral ABCD, given that <A + <D = 90o. Prove that AC2 + BD2 = AD2 + BC2
8)Prove that a line drawn through the mid- point of one side of a triangle parallel to another side bisects the third side.
9) Prove that the line joining the mid – points of any two sides of a triangle is parallel to the third side

10) Sides of triangles are given below. Determine which of them are right triangles.In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 4 cm, 5 cm
(iii) 40 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
11) DEF is an equilateral triangle of side 2b. Find each of its altitudes.
12) Triangle ABC is right – angled at B and D is the mid – point of BC.
Prove that: AC2 = 4AD2 – 3AB2
13) The sides of a  ABC are in the ratio AB : BC : CA = 1 : √2 : 1. Show that ABC is a right triangle, right – angled at A.
14) A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower

Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

You can use above books for extra knowledge and practicing different questions.

Note to our visitors :-

Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.
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### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20


text:
# Including both transformed and original data (untransformed) in a multivariable linear regression.

This may have a quick response (i.e. don't do it).

Just attended a lecture on multivariable linear regression where the outcome is forced expiratory volume (the amount of air you can push out of your lungs in one go). The explanatory variables are age, height, gender and smoking status. In this example, the lecturer finds that height is skewed and therefore transforms the variable to Height^2.

After performing forward selection he ends up with 5 significant explanatory variables in his model: age, height, gender, smoking status and height^2.

My question is: How can you end up with both height^2 and height in your model. Wouldn't they interact (be highly related)? More broadly, when we transform a variable, should the untransformed variable ever be in the model with the transformed?

• There are two ways to think about it. One is to think of it as an interaction with itself. When we use interactions we generally do include the main effects and the interaction term: e.g. stats.idre.ucla.edu/stata/faq/…. Another is to think of it as a polynomial. There is no reason why the relationship should be linear. If a polynomial fits better then why not use it? The Data Generating Process might be non-linear. – Toby May 3 '17 at 11:47
• BTW in this case think of your lungs as two cylinders that are a function of someone's size. The volume of a cylinder is $\pi r^2 h$. People who are taller probably are also larger in other dimensions, they tend to be right?, so the $r$ will be larger. And since the cylinder is a nonlinear function of $r$ it will be a nonlinear function of size. It seems to make sense to me to include height squared. Write it in logs and you have log($\pi r^2 h$) = log(\pi) + log($r^2$) + log($h$). If $h$ and $r$ are both a function of size for which height is a reasonable proxy then this is your model. – Toby May 3 '17 at 11:51
• Using height and height$^2$ is fitting a quadratic in height, other predictors aside. It could be just a way of catching some curvature in the data. There are even cases, e.g. trajectory of a projectile, in which a quadratic is exactly the shape you expect. – Nick Cox May 3 '17 at 15:00
• That's very helpful. I can't see a biological reason to have height#height^2 as an interaction term? I do, however, see why you'd square height in this specific example. My take on his question is, more generally when we transform non-normal covariates, whether we should include both the transformed and original variables in the model? – bobmcpop May 3 '17 at 15:02
• Let say he needed to trasform x, and the best transformation was log(x), would he add both x and log(x) into a model? – bobmcpop May 3 '17 at 15:06

BTW in this case think of your lungs as two cylinders that are a function of someone's size. The volume of a cylinder is $$πr^2h$$. People who are taller probably are also larger in other dimensions, they tend to be right?, so the $$r$$ will be larger. And since the cylinder is a nonlinear function of $$r$$ it will be a nonlinear function of size. It seems to make sense to me to include height squared. Write it in logs and you have $$\log(πr^2 h) = \log(\pi) + \log(r^2) + \log(h)$$. If $$h$$ and $$r$$ are both a function of size for which height is a reasonable proxy then this is your model. – Toby
Toby's argument in comments can be extended this way: often $$r$$ wil be only approximately proportional to $$h$$, $$r$$ wil grow slower (and slower and slower) as $$h$$ grows, so maybe $$r\propto h^\alpha$$ for some $$0<\alpha<1$$ is a better model. That leads to $$y (\text{volume})\propto h^{1+2\alpha}$$ and taking logarithms on both sides leads to a model in only $$\log h$$, where we in addition have an theoretical expectation for the value of slope: $$1+2\alpha$$, so we can see from the fit if this geometrical speculations holds for this case. And no, in this model we do not need $$h$$ untransformed as an additional term.


text:
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text:
1. Intersection between two lines

Hi there. I got this exercise, where I must find the intersection between these two lines:

$L_1=2x-y=2$

$L_2=\begin{Bmatrix} x=-1+2\lambda & \mbox{ }& \\y=-1 & \mbox{}& \end{matrix}$

I've parametrized $L_1$

$P_0(1,0)\in{L_1}$ y $P_1(0,-2)\in{L_1}$

From these points of $L_1$ I found a vector parallel to L_1: $u=(-1,-2)$

And then: $L_1=\begin{Bmatrix} x=1-\lambda & \mbox{ }& \\y=-2\lambda & \mbox{}& \end{matrix}$

$L_1\cap{L_2}=\begin{Bmatrix} -1+2\lambda=1-\lambda & \mbox{ }& \\-1=-2\lambda & \mbox{}& \end{matrix}\Longrightarrow{\begin{Bmatrix} \lambda=\displaystyle\frac{3}{2} & \mbox{ }& \\\displaystyle\frac{1}{2}=\lambda & \mbox{}& \end{matrix}}$

Lambda should be in both equations $\displaystyle\frac{1}{2}$, but for x it gives me $\displaystyle\frac{3}{2}$, and I don't know where the error is.

Any help would be appreciated.

Well, I don't know where is the error on my latex code neither

2. Originally Posted by Ulysses
Hi there. I got this exercise, where I must find the intersection between these two lines:

$L_1=2x-y=2$

$L_2=\begin{Bmatrix} x=-1+2\lambda & \mbox{ }& \\y=-1 & \mbox{}& \end{matrix}$

I've parametrized $L_1$

$P_0(1,0)\in{L_1}$ y $P_1(0,-2)\in{L_1}$

From these points of $L_1$ I found a vector parallel to L_1: $u=(-1,-2)$

And then: $L_1=\begin{Bmatrix} x=1-\lambda & \mbox{ }& \\y=-2\lambda & \mbox{}& \end{matrix}$

$L_1\cap{L_2}=\begin{Bmatrix} -1+2\lambda=1-\lambda & \mbox{ }& \\-1=-2\lambda & \mbox{}& \end{matrix}\Longrightarrow{\begin{Bmatrix} \lambda=\displaystyle\frac{3}{2} & \mbox{ }& \\\displaystyle\frac{1}{2}=\lambda & \mbox{}& \end{matrix}}$

Lambda should be in both equations $\displaystyle\frac{1}{2}$, but for x it gives me $\displaystyle\frac{3}{2}$, and I don't know where the error is.

Any help would be appreciated.

Well, I don't know where is the error on my latex code neither

L1: 2x1-y1=2 or y1=2x1-2

L2: x=-1+2t
y=-1

To find the intersection point you need to do this: y1=y or 2x1-2=-1 and x1=1/2, so the intersection point is (1/2,-1).

Regards.


text:
# Why is a capacitor to ground a low frequency pole, instead of a high frequency zero?

In building my intuition about how analog circuits relate to the s-plane, I am wondering why a pole emerges from a capacitor to ground, for example in a simple lowpass filter consisting of only a resistor followed by a capacitor to ground.

I have done the math and it's pretty clear that a pole emerges, but I am struggling with any intuition that leads there. Since we effectively "shunt high frequencies away", I intuitively expect the capacitor to cause a zero for those high frequencies, not a pole.

I think I actually understand this in the z-plane instead of the s-plane, where the z transform relates to difference equations. There, the poles relate to what is fed back from the output to the input, which intuitively means that there can be some complex exponential where the signal is amplified to infinity.

But in a simple analog lowpass filter, there is no feedback at all, or at least no feedback that I can readily identify. Am I wrong in that point, or is there another reason for the emergence of the pole instead of a zero even without opamp or other active element?

EDIT: There seems to be some confusion about what my question was about. The thing is, I do understand what zeroes and a poles in the s-plane are, and how they influence the frequency magnitude/bode plot. What I was lacking was the same intuition about how the poles and zeroes in the plane emerge that I got from the z-plane, where it was easier for me to accept why a feedback path would cause a pole, and a feedforward path would cause a zero. I was not seeing the same for electronic components because I was not taking the current's feedback action into account.

I intuitively expect the capacitor to cause a zero for those high frequencies, not a pole.

It does create a zero but only when the frequency is infinite.

For more normal frequencies (such as the 3 dB point of an RC low-pass filter), there is a pole present but, it's not obvious when looking at a bode-plot. The pole is hidden behind the bode-plot but, nevertheless it is still present.

Consider a 2nd order system (mainly because I have nice pictures of one). The bode plot can have a moderate to high peak in the spectrum or, it may not exhibit any peak at all. This is dictated by something that doesn't exist in a first order system, namely Q-factor or, its close reciprocal damping factor. However, these next diagrams should convince you that a pole does exist (as it does for a first order low-pass filter): -

And, if we examined the mathematics of the transfer function, we would see a pole like this: -

Hopefully, it doesn't take too much of a leap to realize that this applies to a simple RC low-pass filter; the pole lies behind the bode-plot (as if hidden inside the page or the screen). Pictures from my boring website here.

But, remember that poles are a concept to help join together the real-world spectral image (the bode-plot) and the mathematics. So, going back to a low-pass filter, it has a transfer-function like this: -

$$H(s) = \dfrac{1}{1+sRC}$$

The transfer function rises to infinity when $$\sRC = -1\$$ or $$\s = \frac{-1}{RC}\$$ and, given that a bode plot x-axis is totally fixated on the complex part of $$\s\$$, i.e. the $$\j\omega\$$ part, we have to accept that when $$\s\$$ is not a complex value, it cannot exist along the $$\j\omega\$$ axis and, must only be found beneath the page or behind the bode-plot.

• Thanks! I think I knew the above, but my trouble was with why that pole "somewhere" in the s-plane (elsewhere than on the imaginary axis since that would be infinite amplification) is a pole and not a zero, and not just "because the math works out that way". Especially because in the z-plane, it seems obvious to me: Only the coefficients that feed back the output form poles. Seems clear: they could add up to infinity! The (past) input coefficients can only form zeroes: The most they can do is cancel input out. But I get it now I think, I did not notice the feedback action of the current! May 23 at 0:46
• Nevertheless, this is a really clear explanation of the concept that helps reinforce how things fit together. May 23 at 0:47
• Poles and zeroes are mathematical concepts to help explain the bode plot. May 23 at 9:27
• There seems to be some confusion about what my question was about. The thing is, I do understand what zeroes and a poles in the s-plane are, and how they influence the frequency magnitude/bode plot. What I was lacking was the same intuition about how the poles and zeroes in the plane emerge that I got from the z-plane, where it was easier for me to accept why a feedback path would cause a pole, and a feedforward path would cause a zero. I was not seeing the same for electronic components because I was not taking the current's feedback action into account. May 23 at 21:19
• This appears to be your other question: is there another reason for the emergence of the pole instead of a zero even without opamp or other active element? and the answer has nothing to do with feedback; it's just plain ordinary analogue circuit analysis and not feedback. May 23 at 21:54

You are probably mixing some meanings for the term "zero". Think of it like this: in a transfer function it represents the root in the numerator, so it "zeroes out" higher frequencies. That's because the meaning the magnitude of a transfer function drops when a zero is encountered,whereas a pole causes the magnitude to peak like a pole in a tent.

But in a simple analog lowpass filter, there is no feedback at all

The feedback is due to the current. If you add a load at the output of an RC lowpass, the input will be influenced by it. The current is the hidden feedback in analog circuits.

• Thanks. I did understand what you wrote in your first paragraph, but the rest of your answer seems to contain the key to further intuition. How exactly does the current provide the feedback here? May 22 at 22:20
• I think I get it now. By adding that capacitor, I'm forming a parallel path for some of the current, the current that is allowed to go through that capacitor. That obviously adds current, and forces the source to react. Hence a feedback path exists. And assuming no resistance at all, that capacitor's impedance could go all the way to zero (as the frequency goes to infinity), forcing the current to go infinite, zo it's not too hard for me to accept that that is a pole. Does that make sense? May 23 at 0:36
• @anyfoo is the current the output? or is the voltage across the capacitor the output? May 23 at 8:26
• @anyfoo If you talk about an ideal capacitor then it's capacitor's reactance, (minor mishap). The pole would be whatever causes the magnitude to peak, or be kept high. In a lowpass that happens at $\omega_c$. And since you are talking about Vout/Vin, then you have to look at how the output voltage is influenced by the diminishing reactance: when it reaches (a theoretical) zero, the output is zero. May 23 at 8:32

Any cap can have a series R (ESR) , a series L (ESL), shunt leakage resistance, Rp and a small (pF) stray capacitance, Cp across Rp.

But mainly for 1st order realism, you are totally right on the pole created by the ESR.

Here's a clear example of Rs / ESR ratio = 100:1 the pole results in only 40 dB attenuation. The ESL causes a notch down to ESR when Zc=ZL at one resonant frequency and cancel each other out then rise in impedance.

Here's a bit of intuition that I picked up from "Linear Circuit Transfer Functions: An Introduction to Fast Analytical Techniques".

A zero is manifested when the output is nulled from a specific input frequency.

## RC Lowpass

Let's look at the RC lowpass filter. What conditions causes the output to be 0V?

• One is that R = infinite Ohms (open). Ideal Resistors don't have frequency dependance, so that won't happen here.
• Another is if C = 0 Ohms. Let's look at the impedance vs frequency of an ideal cap to see if there is a frequency that that happens.

$$\Zc(\omega) = \frac{1}{j\omega C}\$$

From the above equation, Zc (capacitor impedance) approaches 0 as omega (frequency in rad/s) approaches infinity. But note, this is not a real frequency. There is no infinite frequency where Zc = 0. Zc only approaches zero, meaning no null, meaning that there is no zero.

Let's look at the transfer function plot for the simple RC case. The output voltage drops off at a rate of -20 dB/dec forever (ideally of course).

If we were to keep increasing frequency, the output keeps dropping off. But there is always an output. The capacitor impedance never gets to 0, so the output voltage will decrease forever.

.

## RC Highpass

Now let's try an RC highpass filter. This filter should have a pole and zero.

Now the zero would manifest if the C became infinite impedance. Does this ever happen?

$$\Zc(\omega) = \frac{1}{j\omega C}\$$

If omega = 0 rad/s (a valid frequency mind you), Zc is 1/0 which for our purposes = infinity, or an open circuit. This means the input can not effect the output, so we have a zero. We even know we have a zero at 0 rad/s (commonly called a differentiator).

Although on the frequency must be > 0 for LTSpices AC analysis, the fact remains that there is a real frequency (DC) that cannot pass to the output. The RC Lowpass filter didn't have an absolute frequency where the output is zero. If you named a high frequency and said it was close enough to zero, I could simply name a higher frequency to get even closer.

The RC highpass filter is commonly used as a "DC Block".

.

## RC Lowpass with Extra Resistor

As a final check of your understanding, does the following circuit have a zero?


text:
Lemma 66.38.7. If $f$ is unramified at $x$ then $f$ is quasi-finite at $x$. In particular, an unramified morphism is locally quasi-finite.

Proof. Via a diagram as in Lemma 66.22.1 this translates into Morphisms, Lemma 29.35.10. $\square$

Comment #3021 by Brian Lawrence on

Suggested slogan: An unramified morphism is locally quasi-finite.

Comment #3139 by on

Going to ignore this slogan because it is identical to part of the statement of the lemma.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
In this work a numerical investigation of a four stage heavy-duty gas turbine is presented. Fully three-dimensional, multistage, Navier-Stokes analyses are carried out to predict the overall turbine performance. Coolant injections, cavity purge flows, and leakage flows are included in the turbine modeling by means of suitable wall boundary conditions. The main objective is the evaluation of the impact of gas modeling on the prediction of the stage and turbine performance parameters. To this end, four different gas models were used: three models are based on the perfect gas assumption with different values of constant $cp$, and the fourth is a real gas model which accounts for thermodynamic gas properties variations with temperature and mean fuel∕air ratio distribution in the through-flow direction. For the real gas computations, a numerical model is used which is based on the use of gas property tables, and exploits a local fitting of gas data to compute thermodynamic properties. Experimental measurements are available for comparison purposes in terms of static pressure values at the inlet∕outlet of each row and total temperature at the turbine exit.

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text:
Linearization of Remez algorithm rational case

In the rational case, we are interested to find polynomials $$P(x)$$ and $$Q(x)$$ s.t. $$f(x_k)-P(x_k)/Q(x_k)=(-1)^kE$$ for $$k=1,2,\ldots, N$$ where $$N=deg(P)+deg(Q)+2$$

This can be rewritten as $$(1)~~~~~~(f(x_k)-(-1)^kE)Q(x_k)-P(x_k)=0$$

The equation (1) is non-linear. The notes in this document suggest using the following variation of the system of equations for linearizing (1). $$(2)~~~~~~((-1)^k E_0 - f(x_k))\sum_{i=1}^{q}b_i x_k^i+\sum_{i=1}^{p}a_i x_k^i + (-1)^kE=f(x_k)$$

It seems quite different from (1). Is (2) above is the correct interpretation of (1)? How this was derived from the definition? Where the additional $$f(x_k)$$ and $$(-1)^kE$$ come from?

I have found that this is Fraser and Hart's variant of the Remez algorithm.

Let us fix the constant term of $$Q(x_k)$$ to $$1$$:

$$f(x_k) - \frac{\sum_{i=0}^{p}a_i x_k^i}{\sum_{i=1}^{q}b_i x_k^i + 1}=(-1)^kE$$

from which we obtain $$f(x_k) + f(x_k)\sum_{i=1}^{q}b_i x_k^i - \sum_{i=0}^{p}a_i x_k^i=(-1)^kE\sum_{i=1}^{q}b_i x_k^i + (-1)^kE$$

regrouping, we get

$$(*)~~~~~~~\left((-1)^kE- f(x_k)\right)\sum_{i=1}^{q}b_i x_k^i + \sum_{i=0}^{p}a_i x_k^i + (-1)^kE = f(x_k)$$

Since we don't know $$E$$, $$(*)$$ is not linear. The approach of Fraser and Hart is to make an initial guess $$E_0$$ so that $$(*)$$ becomes $$(2)$$ which is a linear system.


text:
# Subspaces of linear transformation problem

I have a linear transformation $T : V \to W$, and $Z$ is a subspace of $W$. We also have $U = \{v \in V\}$ such that $T(v) \in Z$, and we want to show that U is a subspace of V.

So basically they're telling me that T maps V into W, and U is a subset of V that maps into Z, which is a subset of W.

For $U$ to be a subspace of $V$, it must be closed under addition and scalar multiplication. For any two vectors $u,v \epsilon U$, it's clear that $T(u + v)$ maps into Z because we know T is a linear transformation, so it's also closed under addition. The same argument holds for scalar multiplication.

Is that sufficient to show that U is a subspace of V? Is there a simpler way to do it? I also thought of maybe using a proof by contradiction, but I don't think that's possible due to insufficient information. Are there other ways to show that U is a subspace of V?

Thanks,

-

Your method works just fine (the only missing argument is that $Z$ itself is closed under addition and scalar multiplication).