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# Math Help - Use of Quantifiers

1. ## Use of Quantifiers

Another example from my textbook that I can't quite figure out. Let p(x), q(x) denote the following open statements.

p(x): x ≤ 3
q(x): x + 1 is odd

If the universe consists of all integers, what are the truth values of the following statements?

a) q(1)
b) ¬p(3)
c) p(7) v q(7)
d) p(3) ^ q(4)
e) ¬(p(-4) v q(-3))
f) ¬p(-4) ^ ¬q(-3)

2. ## Re: Use of Quantifiers

Originally Posted by rhymin
Another example from my textbook that I can't quite figure out. Let p(x), q(x) denote the following open statements.
p(x): x ≤ 3
q(x): x + 1 is odd

If the universe consists of all integers, what are the truth values of the following statements?
a) q(1)
b) ¬p(3)
c) p(7) v q(7)
d) p(3) ^ q(4)
e) ¬(p(-4) v q(-3))
f) ¬p(-4) ^ ¬q(-3)
We do not do your homework for you.

For c) $7\not\le 3$ and $7+1\text{ is not odd}.$ So is c) true or false? WHY?

3. ## Re: Use of Quantifiers

This isn't homework. I am taking a math class in about a month and am studying a textbook that I believe will help. Am I posting in the right forum? The questions I am posting are ones that didn't have great examples and are confusing to me.

4. ## Re: Use of Quantifiers

c) 7 is not less than or equal to 3 and 8 is not odd. so since both are false, it is false.

What happens if one is true and one is false?

5. ## Re: Use of Quantifiers

Originally Posted by rhymin
c) 7 is not less than or equal to 3 and 8 is not odd. so since both are false, it is false.
What happens if one is true and one is false?
That is not the question! It has nothing to do with it.

c) Asks if $[p(7) \vee q(7)$ is a true statement? When is an or, disjunction, true?

6. ## Re: Use of Quantifiers

It is not a true statement. I'm just not sure what "truth values" means.

7. ## Re: Use of Quantifiers

Originally Posted by rhymin
I'm just not sure what "truth values" means.
There are only two truth values: TRUE $(T)$ or FALSE $(F)$.

In this question $p(0)$ has a truth value of $T$ and $p(4)$ has truth value of $F$.

8. ## Re: Use of Quantifiers

Okay I learned a bit more about these symbols.
Are these right?

b) ¬p(3): not 3 ≤ 3 = False

c) p(7) v q(7): 7 ≤ 3 or 7 + 1 is odd...both are false, so it is False

d) p(3) ^ q(4): 3 ≤ 3 and 4 + 1 is odd...one is true, one is false, so it is False (since both must be true)

e) ¬(p(-4) v q(-3)): ¬(-4 ≤ 3 or -3 + 1 is odd)...one is false, one is true, so it is True

f) ¬p(-4) ^ ¬q(-3): ¬ -4 ≤ 3 and ¬ -3 + 1 is odd...one is false, one is true, so it is False

9. ## Re: Use of Quantifiers

Originally Posted by rhymin
Okay I learned a bit more about these symbols.
Are these right?

b) ¬p(3): not 3 ≤ 3 = False

c) p(7) v q(7): 7 ≤ 3 or 7 + 1 is odd...both are false, so it is False

d) p(3) ^ q(4): 3 ≤ 3 and 4 + 1 is odd...one is true, one is false, so it is False (since both must be true)

e) ¬(p(-4) v q(-3)): ¬(-4 ≤ 3 or -3 + 1 is odd)...one is false, one is true, so it is True

f) ¬p(-4) ^ ¬q(-3): ¬ -4 ≤ 3 and ¬ -3 + 1 is odd...one is false, one is true, so it is False
d) is true because both are true.

10. ## Re: Use of Quantifiers

Wow, I'm not sure how I missed that lol. What about (e)? Is that one also True? Doesn't just 1 have to be true with the "v" (or) symbol for it to be true? Or am I off there?

11. ## Re: Use of Quantifiers

e) ¬(p(-4) v q(-3)): ¬(-4 ≤ 3 or -3 + 1 is odd)...one is false, one is true, so it is True
You are missing the "not" at the start aren't you?
" $-4\le 3$" is true, "-3+ 1 is odd" is false so the statement "one of p(-4) OR q(3) is true" is true which means that "NOT....." is false.

12. ## Re: Use of Quantifiers

Ohhh i see now. Thank you so much for that. What I was doing was saying the first is true and second is false, and then just switching them. But I see that you are supposed to conclude that the entire statement is either true or false, and then use the negation symbol.

13. ## Re: Use of quantifiers

NB: Statements (e) and (f) are logically equivalent: they are either both true or both false.


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# Rule of succession

(Redirected from Laplace–Bayes estimator)

In probability theory, the rule of succession is a formula introduced in the 18th century by Pierre-Simon Laplace in the course of treating the sunrise problem.[1] The formula is still used, particularly to estimate underlying probabilities when there are few observations or for events that have not been observed to occur at all in (finite) sample data.

## Statement of the rule of succession

If we repeat an experiment that we know can result in a success or failure, n times independently, and get s successes, and n − s failures, then what is the probability that the next repetition will succeed?

More abstractly: If X1, ..., Xn+1 are conditionally independent random variables that each can assume the value 0 or 1, then, if we know nothing more about them,

${\displaystyle P(X_{n+1}=1\mid X_{1}+\cdots +X_{n}=s)={s+1 \over n+2}.}$

## Interpretation

Since we have the prior knowledge that we are looking at an experiment for which both success and failure are possible, our estimate is as if we had observed one success and one failure for sure before we even started the experiments. In a sense we made n + 2 observations (known as pseudocounts) with s + 1 successes. Although this may seem the simplest and most reasonable assumption, which also happens to be true, it still requires a proof. Indeed, assuming a pseudocount of one per possibility is one way to generalise the binary result, but has unexpected consequences — see Generalization to any number of possibilities, below.

Nevertheless, if we had not known from the start that both success and failure are possible, then we would have had to assign

${\displaystyle P'(X_{n+1}=1\mid X_{1}+\cdots +X_{n}=s)={s \over n}.}$

But see Mathematical details, below, for an analysis of its validity. In particular it is not valid when ${\displaystyle s=0}$ , or ${\displaystyle s=n}$ .

If the number of observations increases, ${\displaystyle P}$  and ${\displaystyle P'}$  get more and more similar, which is intuitively clear: the more data we have, the less importance should be assigned to our prior information.

## Historical application to the sunrise problem

Laplace used the rule of succession to calculate the probability that the Sun will rise tomorrow, given that it has risen every day for the past 5000 years. One obtains a very large factor of approximately 5000 × 365.25, which gives odds of about 1,826,200 to 1 in favour of the Sun rising tomorrow.

However, as the mathematical details below show, the basic assumption for using the rule of succession would be that we have no prior knowledge about the question whether the Sun will or will not rise tomorrow, except that it can do either. This is not the case for sunrises.

Laplace knew this well, and he wrote to conclude the sunrise example: "But this number is far greater for him who, seeing in the totality of phenomena the principle regulating the days and seasons, realizes that nothing at the present moment can arrest the course of it."[2] Yet Laplace was ridiculed for this calculation; his opponents[who?] gave no heed to that sentence, or failed to understand its importance.[2]

In the 1940s, Rudolf Carnap investigated a probability-based theory of inductive reasoning, and developed measures of degree of confirmation, which he considered as alternatives to Laplace's rule of succession.[3][4] See also New riddle of induction#Carnap.

## Mathematical details

The proportion p is assigned a uniform distribution to describe the uncertainty about its true value. (This proportion is not random, but uncertain. We assign a probability distribution to p to express our uncertainty, not to attribute randomness to p. But this amounts, mathematically, to the same thing as treating p as if it were random).

Let Xi be 1 if we observe a "success" on the ith trial, otherwise 0, with probability p of success on each trial. Thus each X is 0 or 1; each X has a Bernoulli distribution. Suppose these Xs are conditionally independent given p.

We can use Bayes' theorem to find the conditional probability distribution of p given the data Xi, i = 1, ..., n. For the "prior" (i.e., marginal) probability measure of p we assigned a uniform distribution over the open interval (0,1)

${\displaystyle f(p)={\begin{cases}0&{\text{for }}p\leq 0\\1&{\text{for }}0

For the likelihood of our observations, we use the likelihood function

${\displaystyle L(p)=P(X_{1}=x_{1},\ldots ,X_{n}=x_{n}\mid p)=\prod _{i=1}^{n}p^{x_{i}}(1-p)^{1-x_{i}}=p^{s}(1-p)^{n-s}}$

where s = x1 + ... + xn is the number of "successes" and n is the number of trials (we are using capital X to denote a random variable and lower-case x as the data actually observed). Putting it all together, we can calculate the posterior:

${\displaystyle f(p\mid X_{1}=x_{1},\ldots ,X_{n}=x_{n})={L(p)f(p) \over \int _{0}^{1}L(r)f(r)\,dr}={p^{s}(1-p)^{n-s} \over \int _{0}^{1}r^{s}(1-r)^{n-s}\,dr}}$

To get the normalizing constant, we find

${\displaystyle \int _{0}^{1}r^{s}(1-r)^{n-s}\,dr={s!(n-s)! \over (n+1)!}}$

(see beta function for more on integrals of this form).

The posterior probability density function is therefore

${\displaystyle f(p\mid X_{1}=x_{1},\ldots ,X_{n}=x_{n})={(n+1)! \over s!(n-s)!}p^{s}(1-p)^{n-s}.}$

This is a beta distribution with expected value

${\displaystyle \operatorname {E} (p\mid X_{i}=x_{i}{\text{ for }}i=1,\dots ,n)=\int _{0}^{1}pf(p\mid X_{1}=x_{1},\ldots ,X_{n}=x_{n})\,dp={s+1 \over n+2}.}$

Since p tells us the probability of success in any experiment, and each experiment is conditionally independent, the conditional probability for success in the next experiment is just p. As p is a random variable, The law of total probability tells us that the expected probability of success in the next experiment is just the expected value of p. Since p is conditional on the observed data Xi for i = 1, ..., n, we have

${\displaystyle P(X_{n+1}=1\mid X_{i}=x_{i}{\text{ for }}i=1,\dots ,n)=\operatorname {E} (p\mid X_{i}=x_{i}{\text{ for }}i=1,\dots ,n)={s+1 \over n+2}.}$

The same calculation can be performed with the (improper) prior that expresses total ignorance of p, including ignorance with regards to the question whether the experiment can succeed, or can fail. This improper prior is 1/(p(1 − p)) for 0 ≤ p ≤ 1 and 0 otherwise.[5] If the calculation above is repeated with this prior, we get

${\displaystyle P'(X_{n+1}=1\mid X_{i}=x_{i}{\text{ for }}i=1,\dots ,n)={s \over n}.}$

Thus, with the prior specifying total ignorance, the probability of success is governed by the observed frequency of success. However, the posterior distribution that led to this result is the Beta(s,n − s) distribution, which is not proper when s = n or s = 0 (i.e. the normalisation constant is infinite when s = 0 or s = n). This means that we cannot use this form of the posterior distribution to calculate the probability of the next observation succeeding when s = 0 or s = n. This puts the information contained in the rule of succession in greater light: it can be thought of as expressing the prior assumption that if sampling was continued indefinitely, we would eventually observe at least one success, and at least one failure in the sample. The prior expressing total ignorance does not assume this knowledge.

To evaluate the "complete ignorance" case when s = 0 or s = n can be dealt with, we first go back to the hypergeometric distribution, denoted by ${\displaystyle \mathrm {Hyp} (s|N,n,S)}$ . This is the approach taken in Jaynes (2003). The binomial ${\displaystyle \mathrm {Bin} (r|n,p)}$  can be derived as a limiting form, where ${\displaystyle N,S\rightarrow \infty }$  in such a way that their ratio ${\displaystyle p={S \over N}}$  remains fixed. One can think of ${\displaystyle S}$  as the number of successes in the total population, of size ${\displaystyle N}$ .

The equivalent prior to ${\displaystyle {1 \over p(1-p)}}$  is ${\displaystyle {1 \over S(N-S)}}$ , with a domain of ${\displaystyle 1\leq S\leq N-1}$ . Working conditional to ${\displaystyle N}$  means that estimating ${\displaystyle p}$  is equivalent to estimating ${\displaystyle S}$ , and then dividing this estimate by ${\displaystyle N}$ . The posterior for ${\displaystyle S}$  can be given as:

${\displaystyle P(S|N,n,s)\propto {1 \over S(N-S)}{S \choose s}{N-S \choose n-s}\propto {S!(N-S)! \over S(N-S)(S-s)!(N-S-[n-s])!}}$

And it can be seen that, if s = n or s = 0, then one of the factorials in the numerator cancels exactly with one in the denominator. Taking the s = 0 case, we have:

${\displaystyle P(S|N,n,s=0)\propto {(N-S-1)! \over S(N-S-n)!}={\prod _{j=1}^{n-1}(N-S-j) \over S}}$

Adding in the normalising constant, which is always finite (because there are no singularities in the range of the posterior, and there are a finite number of terms) gives:

${\displaystyle P(S|N,n,s=0)={\prod _{j=1}^{n-1}(N-S-j) \over S\sum _{R=1}^{N-n}{\prod _{j=1}^{n-1}(N-R-j) \over R}}}$

So the posterior expectation for ${\displaystyle p={S \over N}}$  is:

${\displaystyle E\left({S \over N}|n,s=0,N\right)={1 \over N}\sum _{S=1}^{N-n}SP(S|N,n=1,s=0)={1 \over N}{\sum _{S=1}^{N-n}\prod _{j=1}^{n-1}(N-S-j) \over \sum _{R=1}^{N-n}{\prod _{j=1}^{n-1}(N-R-j) \over R}}}$

An approximate analytical expression for large N is given by first making the approximation to the product term:

${\displaystyle \prod _{j=1}^{n-1}(N-R-j)\approx (N-R)^{n-1}}$

and then replacing the summation in the numerator with an integral

${\displaystyle \sum _{S=1}^{N-n}\prod _{j=1}^{n-1}(N-S-j)\approx \int _{1}^{N-n}(N-S)^{n-1}\,dS={(N-1)^{n}-n^{n} \over n}\approx {N^{n} \over n}}$

The same procedure is followed for the denominator, but the process is a bit more tricky, as the integral is harder to evaluate

{\displaystyle {\begin{aligned}\sum _{R=1}^{N-n}{\prod _{j=1}^{n-1}(N-R-j) \over R}&\approx \int _{1}^{N-n}{(N-R)^{n-1} \over R}\,dR\\&=N\int _{1}^{N-n}{(N-R)^{n-2} \over R}\,dR-\int _{1}^{N-n}(N-R)^{n-2}\,dR\\&=N^{n-1}\left[\int _{1}^{N-n}{dR \over R}-{1 \over n-1}+O\left({1 \over N}\right)\right]\approx N^{n-1}\ln(N)\end{aligned}}}

where ln is the natural logarithm plugging in these approximations into the expectation gives

${\displaystyle E\left({S \over N}|n,s=0,N\right)\approx {1 \over N}{{N^{n} \over n} \over N^{n-1}\ln(N)}={1 \over n[\ln(N)]}={\log _{10}(e) \over n[\log _{10}(N)]}={0.434294 \over n[\log _{10}(N)]}}$

where the base 10 logarithm has been used in the final answer for ease of calculation. For instance if the population is of size 10k then probability of success on the next sample is given by:

${\displaystyle E\left({S \over N}\mid n,s=0,N=10^{k}\right)\approx {0.434294 \over nk}}$

So for example, if the population be on the order of tens of billions, so that k = 10, and we observe n = 10 results without success, then the expected proportion in the population is approximately 0.43%. If the population is smaller, so that n = 10, k = 5 (tens of thousands), the expected proportion rises to approximately 0.86%, and so on. Similarly, if the number of observations is smaller, so that n = 5, k = 10, the proportion rise to approximately 0.86% again.

This probability has no positive lower bound, and can be made arbitrarily small for larger and larger choices of N, or k. This means that the probability depends on the size of the population from which one is sampling. In passing to the limit of infinite N (for the simpler analytic properties) we are "throwing away" a piece of very important information. Note that this ignorance relationship only holds as long as only no successes are observed. It is correspondingly revised back to the observed frequency rule ${\displaystyle p={s \over n}}$  as soon as one success is observed. The corresponding results are found for the s=n case by switching labels, and then subtracting the probability from 1.

## Generalization to any number of possibilities

This section gives a heuristic derivation to that given in Probability Theory: The Logic of Science.[6]

The rule of succession has many different intuitive interpretations, and depending on which intuition one uses, the generalisation may be different. Thus, the way to proceed from here is very carefully, and to re-derive the results from first principles, rather than to introduce an intuitively sensible generalisation. The full derivation can be found in Jaynes' book, but it does admit an easier to understand alternative derivation, once the solution is known. Another point to emphasise is that the prior state of knowledge described by the rule of succession is given as an enumeration of the possibilities, with the additional information that it is possible to observe each category. This can be equivalently stated as observing each category once prior to gathering the data. To denote that this is the knowledge used, an Im is put as part of the conditions in the probability assignments.

The rule of succession comes from setting a binomial likelihood, and a uniform prior distribution. Thus a straightforward generalisation is just the multivariate extensions of these two distributions: 1) Setting a uniform prior over the initial m categories, and 2) using the multinomial distribution as the likelihood function (which is the multivariate generalisation of the binomial distribution). It can be shown that the uniform distribution is a special case of the Dirichlet distribution with all of its parameters equal to 1 (just as the uniform is Beta(1,1) in the binary case). The Dirichlet distribution is the conjugate prior for the multinomial distribution, which means that the posterior distribution is also a Dirichlet distribution with different parameters. Let pi denote the probability that category i will be observed, and let ni denote the number of times category i (i = 1, ..., m) actually was observed. Then the joint posterior distribution of the probabilities p1, ..., pm is given by:

${\displaystyle f(p_{1},\ldots ,p_{m}\mid n_{1},\ldots ,n_{m},I)={\begin{cases}{\displaystyle {\frac {\Gamma \left(\sum _{i=1}^{m}(n_{i}+1)\right)}{\prod _{i=1}^{m}\Gamma (n_{i}+1)}}p_{1}^{n_{1}}\cdots p_{m}^{n_{m}}},\quad &\sum _{i=1}^{m}p_{i}=1\\\\0&{\text{otherwise.}}\end{cases}}}$

To get the generalised rule of succession, note that the probability of observing category i on the next observation, conditional on the pi is just pi, we simply require its expectation. Letting Ai denote the event that the next observation is in category i (i = 1, ..., m), and let n = n1 + ... + nm be the total number of observations made. The result, using the properties of the Dirichlet distribution is:

${\displaystyle P(A_{i}|n_{1},\ldots ,n_{m},I_{m})={n_{i}+1 \over n+m}.}$

This solution reduces to the probability that would be assigned using the principle of indifference before any observations made (i.e. n = 0), consistent with the original rule of succession. It also contains the rule of succession as a special case, when m = 2, as a generalisation should.

Because the propositions or events Ai are mutually exclusive, it is possible to collapse the m categories into 2. Simply add up the Ai probabilities that correspond to "success" to get the probability of success. Supposing that this aggregates c categories as "success" and m-c categories as "failure". Let s denote the sum of the relevant ni values that have been termed "success". The probability of "success" at the next trial is then:

${\displaystyle P({\text{success}}|n_{1},\ldots ,n_{m},I_{m})={s+c \over n+m},}$

which is different from the original rule of succession. But note that the original rule of succession is based on I2, whereas the generalisation is based on Im. This means that the information contained in Im is different from that contained in I2. This indicates that mere knowledge of more than two outcomes we know are possible is relevant information when collapsing these categories down to just two. This illustrates the subtlety in describing the prior information, and why it is important to specify which prior information one is using.

## Further analysis

A good model is essential (i.e., a good compromise between accuracy and practicality). To paraphrase Laplace on the sunrise problem: Although we have a huge number of samples of the sun rising, there are far better models of the sun than assuming it has a certain probability of rising each day, e.g., simply having a half-life.

Given a good model, it is best to make as many observations as practicable, depending of the expected reliability of prior knowledge, cost of observations, time and resources available, and accuracy required.

One of the most difficult aspects of the rule of succession is not the mathematical formulas, but answering the question: When does the rule of succession apply? In the generalisation section, it was noted very explicitly by adding the prior information Im into the calculations. Thus, when all that is known about a phenomenon is that there are m known possible outcomes prior to observing any data, only then does the rule of succession apply. If the rule of succession is applied in problems where this does not accurately describe the prior state of knowledge, then it may give counter-intuitive results. This is not because the rule of succession is defective, but that it is effectively answering a different question, based on different prior information.

In principle (see Cromwell's rule), no possibility should have its probability (or its pseudocount) set to zero, since nothing in the physical world should be assumed strictly impossible (though it may be)—even if contrary to all observations and current theories. Indeed, Bayes rule takes absolutely no account of an observation previously believed to have zero probability—it is still declared impossible. However, only considering a fixed set of the possibilities is an acceptable route, one just needs to remember that the results are conditional on (or restricted to) the set being considered, and not some "universal" set. In fact Larry Bretthorst [7] shows that including the possibility of "something else" into the hypothesis space makes no difference to the relative probabilities of the other hypothesis - it simply renormalises them to add up to a value less than 1. Until "something else" is specified, the likelihood function conditional on this "something else" is indeterminate, for how is one to determine ${\displaystyle Pr({\text{data}}|{\text{something else}},I)}$ ? Thus no updating of the prior probability for "something else" can occur until it is more accurately defined.

However, it is sometimes debatable whether prior knowledge should affect the relative probabilities, or also the total weight of the prior knowledge compared to actual observations. This does not have a clear cut answer, for it depends on what prior knowledge one is considering. In fact, an alternative prior state of knowledge could be of the form "I have specified m potential categories, but I am sure that only one of them is possible prior to observing the data. However, I do not know which particular category this is." A mathematical way to describe this prior is the dirichlet distribution with all parameters equal to m−1, which then gives a pseudocount of 1 to the denominator instead of m, and adds a pseudocount of m−1 to each category. This gives a slightly different probability in the binary case of ${\displaystyle {\frac {s+0.5}{n+1}}}$ .

Prior probabilities are only worth spending significant effort estimating when likely to have significant effect. They may be important when there are few observations — especially when so few that there have been few, if any, observations of some possibilities – such as a rare animal, in a given region. Also important when there are many observations, where it is believed that the expectation should be heavily weighted towards the prior estimates, in spite of many observations to the contrary, such as for a roulette wheel in a well-respected casino. In the latter case, at least some of the pseudocounts may need to be very large. They are not always small, and thereby soon outweighed by actual observations, as is often assumed. However, although a last resort, for everyday purposes, prior knowledge is usually vital. So most decisions must be subjective to some extent (dependent upon the analyst and analysis used).

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# Let $f:\mathbb R \rightarrow \mathbb R$ be a differentiable even function

I came across the above problem. I see that $G'(x)=f'(x)f(\sqrt{tan(f(x))})$. Now since $f$ is a differentiable even function, $f(-x)=f(x)$ and so $-f'(-x)=f'(x)$ and thus $f'(0)=0$ and hence we can conclude $G'(0)=0.$ Am I going in the right direction?

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If $f(0)<0$, $G$ does not defiened as real function. –  tetori Jan 17 '13 at 4:10
Differentiation not quite right, is $f'(x)\sqrt{\tan(f(x))}$. –  André Nicolas Jan 17 '13 at 4:11
Thanks a lot sir for pointing out the mistake.I have corrected my post. –  learner Jan 17 '13 at 4:33
could you post the source of your questions like this? –  La Belle Noiseuse Jan 17 '13 at 4:45

The answer is not always $B$ either. Following Babak Sorouh's idea, let $f(x) = \pi/2 - x^2$. Then

$$G'(x) = -2 x \sqrt{\tan\!\left(\pi/2-x^2\right)}$$

where it exists and

$$\lim_{x \to 0} G'(x) = -2.$$

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The example can be modified to have $f(x)$ positive everywhere since all that matters here is the behavior in a neighborhood of $x=0$. –  Antonio Vargas Jan 17 '13 at 7:02
Thanks a lot sir for the detailed clarification. I have got it. –  learner Jan 17 '13 at 10:02
Great example Thanks for sharing me that. +1 –  Babak S. Jan 18 '13 at 3:31

Firstly, note that according to Leibniz rule, $G'(x)=f'(x)\sqrt{\tan(f(x))}$. Secondly, buy taking $f(x)=x^{2n}$ as a family of even functions with real valued, we have: $$G'(0)=0$$ so A and C cannot be true for a general case under problem's assumptions.

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This doesn't say whether B or D is true. For that, learner had a good idea, but as tetori points out it may depend on whether we add an assumption of positivity of $f$. –  Jonas Meyer Jan 17 '13 at 4:25
@JonasMeyer: Yes, exactly. Since we are facing a multiple choices here so I think we don't need to prove it theoretically. And, as tetori point and that is absolutely true, it seems D looks right. I am thinking of an even function in which $\sqrt{\tan(f(0))}$ is undefined. –  Babak S. Jan 17 '13 at 5:35
Nice lead, and hits home, I think! +1 –  amWhy Feb 15 '13 at 3:02
@amWhy: Thanks for considering this small hint. dear amWhy, my credit for accessing the web has been expired, and I will have the new one 2 days later I think. That's why I have not been here. –  Babak S. Feb 15 '13 at 8:11


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Vernier Software & Technology

# Rate Law Determination of the Crystal Violet Reaction

## Introduction

In this experiment, you will observe the reaction between crystal violet and sodium hydroxide. One objective is to study the relationship between concentration of crystal violet and the time elapsed during the reaction. A simplified version of the equation is:

${\text{C}}{{\text{V}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq)}} \to {\text{CVOH(aq)}}$

The rate law for this reaction is in the form: rate = k[CV+]m[OH]n, where k is the rate constant for the reaction, m is the order with respect to crystal violet (CV+), and n is the order with respect to the hydroxide ion. Because the hydroxide ion concentration is more than 1000 times as large as the concentration of crystal violet, [OH-] will not change appreciably during this experiment. Thus, you will find the order with respect to crystal violet (m), but not the order with respect to hydroxide (n).

As the reaction proceeds, a violet-colored reactant will be slowly changing to a colorless product. You will measure the color change with a Vernier Colorimeter or a Vernier Spectrometer. The crystal violet solution used in this experiment has a violet color, of course, thus the Colorimeter users will be instructed to use the 565 nm (green) LED. Spectrometer users will determine an appropriate wavelength based on the absorbance spectrum of the solution. We will assume that absorbance is proportional to the concentration of crystal violet (Beer’s law). Absorbance will be used in place of concentration in plotting the following three graphs:

• Absorbance vs. time: A linear plot indicates a zero order reaction (k = –slope).
• ln Absorbance vs. time: A linear plot indicates a first order reaction (k = –slope).
• 1/Absorbance vs. time: A linear plot indicates a second order reaction (k = slope).

Once the order with respect to crystal violet has been determined, you will also be finding the rate constant, k, and the half-life for this reaction.

## Objectives

In this experiment, you will

• Observe the reaction between crystal violet and sodium hydroxide.
• Monitor the absorbance of the crystal violet solution with time.
• Graph Absorbance vs. time, ln Absorbance vs. time, and 1/Absorbance vs. time.
• Determine the order of the reaction.
• Determine the rate constant, k, and the half-life for this reaction.

## Sensors and Equipment

This experiment features the following Vernier sensors and equipment.


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## Institute of Mathematical Statistics Lecture Notes - Monograph Series

### Empirical processes indexed by estimated functions

#### Abstract

We consider the convergence of empirical processes indexed by functions that depend on an estimated parameter $\eta$ and give several alternative conditions under which the estimated parameter'' $\eta_n$ can be replaced by its natural limit $\eta_0$ uniformly in some other indexing set $\Theta$. In particular we reconsider some examples treated by Ghoudi and Remillard. We recast their examples in terms of empirical process theory, and provide an alternative general view which should be of wide applicability.

#### Chapter information

Source
Eric A. Cator, Geurt Jongbloed, Cor Kraaikamp, Hendrik P. Lopuhaä, Jon A. Wellner, eds., Asymptotics: Particles, Processes and Inverse Problems: Festschrift for Piet Groeneboom (Beachwood, Ohio, USA: Institute of Mathematical Statistics, 2007), 234-252

Dates
First available in Project Euclid: 4 December 2007

https://projecteuclid.org/euclid.lnms/1196797079

Digital Object Identifier
doi:10.1214/074921707000000382

Mathematical Reviews number (MathSciNet)
MR2459942

Zentralblatt MATH identifier
1176.62050

Rights


text:
# The Particle System Class

The next part of the particle system is the class representing the particle system itself. Like any other sprite-based strategy, this will involve both an Update() and Draw() method that must be invoked every time through the game loop. But ideally we’d like our particle systems to be an almost hands-off system - once it’s created, we can just let it do its thing without intervention. This is where the idea of game components from our architecture discussion can come into play - our particle system can inherit from the DrawableGameComponent class, which means it can be added to to our Game.Components list and the Game will handle invoking those methods for us every frame:

///
/// A class representing a generic particle system
///
public class ParticleSystem : DrawableGameComponent
{
// TODO: Add fields, properties, and methods
}

Now we want to add generic functionality for a particle system, and in doing so, we want to make sure the system is as efficient as possible - remember, we may be updating and rendering thousands of sprites for each active particle system. We’ll keep this in mind throughout the design process. Let’s start by defining some fields that determine the behavior of the particle system. As we do this, we’ll stretch our use of the C# programming language in ways you may have not done so previously.

### Constants

We’ll start by defining a couple of constants, AlphaBlendDrawOrder and AdditiveBlendDrawOrder:

    ///
/// The draw order for particles using Alpha Blending
///
/// <remarks>
/// Particles drawn using additive blending should be drawn on top of
/// particles that use regular alpha blending
/// </remarks>
public const int AlphaBlendDrawOrder = 100;

///
/// The draw order for particles using Additive Blending
///
/// <remarks>
/// Particles drawn using additive blending should be drawn on top of
/// particles that use regular alpha blending
/// </remarks>
public const int AdditiveBlendDrawOrder = 200;

Remember that the DrawOrder property of a DrawableGameComponent determines the order in which they are drawn. These constants represent values we can reference when setting that draw order, based on what kind of alpha blending we are using.

### Static Fields

To use our particle systems, we’ll need both a ContentManager and SpriteBatch instance. We could create one for each particle system, but that creates a lot of unnecessary objects. Alternatively, we could share the ones created in our Game class, which would be the most efficient approach. However, that does mean those need to be made public, and we need to pass the derived class into this one. As a comfortable middle ground, we’ll create protected static fields for these, so that all particle systems share a single set:

    ///
/// A SpriteBatch to share amongst the various particle systems
///
protected static SpriteBatch spriteBatch;

///
/// A ContentManager to share amongst the various particle systems
///
protected static ContentManager contentManager;

### Private Fields

This class needs to hold our collection of particles, and in keeping with the data locality pattern, we’d like these to be stored sequentially. An array is therefore a great fit. Add it as a private field to the class:

    ///
/// The collection of particles
///
Particle[] particles;

We’ll also use a Queue (from System.Collections.Generic) to hold references to unused particles. This way we can avoid re-creating particles and creating a glut in memory that will result in the garbage collector running often. We could store a reference directly to the particle location, or an index indicating its position in the array. We’ll opt for the former here, as it provides some benefits in usage:

    ///
/// A Queue containing indices of unused particles in the Particles array
///
Queue<Particle> freeParticles;

As we said in the discussion of the Particle class, in using the Flyweight Pattern the actual texture will be held by our ParticleSystem, so let’s add a private variable to hold that:

    ///
/// The texture this particle system uses
///
Texture2D texture;

We’ll also keep track of an origin vector for when we draw the textures:

    ///
/// The origin when we're drawing textures
///
Vector2 origin;

### Public Fields

It can also be useful to know how many particles are currently available (free) in the system. We can expose this value with a public property:

    ///
/// The available particles in the system
///
public int FreeParticleCount => freeParticles.Count;

### Protected Fields

A slightly unique approach we’ll adopt here is defining a number of protected fields that essentially define the behavior of the particle system. These represent the information that is shared amongst all particles in the system. We make them protected so that derived particle systems can adjust them to suit the needs of the effect we are trying to create. For example the blendState determines how a texture is drawn, and the textureFilename helps define which texture to use:

    /// The BlendState to use with this particle system
protected BlendState blendState = BlendState.AlphaBlend;

/// The filename of the texture to use for the particles
protected string textureFilename;

We’ll also use a min/max pair to determine the number of particles to activate in the system each time we add particles:

    /// The minimum number of particles to add when AddParticles() is called
protected int minNumParticles;

/// The maximum number of particles to add when AddParticles() is called
protected int maxNumParticles;

Like the Particle class' fields, these too could be tweaked to meet the needs of your game.

### Constructor

As the DrawableGameComponent has a constructor requiring a Game instance, we must provide our own constructor that passes that along by invoking base() (which runs the constructor of the base object). We’ll also want to initialize our particles array and freeParticles queue, which means we need to know the maximum number of particles we’ll allow in this particle system. Therefore, we’ll add that as a second parameter.

    ///
/// Constructs a new instance of a particle system
///
/// <param name="game"></param>
public ParticleSystem(Game game, int maxParticles) : base(game)
{
// Create our particles
particles = new Particle[maxParticles];
for (int i = 0; i < particles.Length; i++)
{
particles[i] = new Particle();
}
// Add all free particles to the queue
freeParticles = new Queue<Particle>(particles);
// Run the InitializeConstants hook
InitializeConstants();
}

Since none of the particles are in use, we’ll also initialize our Queue with all the newly created particles. This has the helpful side effect of allocating enough memory to hold a reference to each particle in our array, ensuring we never need to re-allocate.

Finally, we invoke InitializeConstants(), a protected method that is intended to be used as a hook - a method that can be overridden to inject your own functionality into the class. Let’s look at this, and the other hook methods next

### Virtual Hook Methods

Now that we have the structure of the class put together, let’s start thinking about the functionality. We’ll write a couple of virtual hook methods to define the default behavior we expect from our particles - most notably setting those protected constant values, initializing new active particles to the particle system, and then updating those particles each frame. We’ll make these methods virtual so we can override them in derived classes, but also provide a base implementation when one makes sense. Let’s start with InitializeConstants() we invoked above:

    ///
/// Used to do the initial configuration of the particle engine.  The
/// protected constants textureFilename, minNumParticles, and maxNumParticles
/// should be set in the override.
///
protected virtual void InitializeConstants() { }

If the textureFilename is not set here, we’ll encounter a runtime error, so this must be overridden in the derived classes. As a possible way to emphasize this, we could instead declare this method, and the ParticleSystem class as abstract.

In contrast, the InitializeParticle() hook method will provide some logic that could potentially be used, but will also probably be overridden in most cases:

    ///
/// InitializeParticle randomizes some properties for a particle, then
/// calls initialize on it. It can be overridden by subclasses if they
/// want to modify the way particles are created.
///
/// <param name="p">the particle to initialize</param>
/// <param name="where">the position on the screen that the particle should be
/// </param>
protected virtual void InitializeParticle(Particle p, Vector2 where)
{
// Initialize the particle with default values
p.Initialize(where);
}

Similarly, we will supply a default implementation for updating a particle. Our default approach is based on Newtonian physics:

    ///
/// Updates the individual particles.  Can be overridden in derived classes
///
/// <param name="particle">The particle to update</param>
/// <param name="dt">The elapsed time</param>
protected virtual void UpdateParticle(Particle particle, float dt)
{
// Update particle's linear motion values
particle.Velocity += particle.Acceleration * dt;
particle.Position += particle.Velocity * dt;

// Update the particle's angular motion values
particle.AngularVelocity += particle.AngularAcceleration * dt;
particle.Rotation += particle.AngularVelocity * dt;

// Update the time the particle has been alive
particle.TimeSinceStart += dt;
}

This implementation works for a wide variety of particle systems, but it can be overridden by a derived particle system if something different is needed.

### DrawableGameComponent Overrides

Now we can tackle the methods from DrawableGameComponent, which we’ll need to replace with our own custom overrides. First, we’ll load the texture in our LoadContent:

    ///
/// Override the base class LoadContent to load the texture. once it's
///
/// <throws>A InvalidOperationException if the texture filename is not provided</throws>
{
// create the shared static ContentManager and SpriteBatch,
// if this hasn't already been done by another particle engine
if (contentManager == null) contentManager = new ContentManager(Game.Services, "Content");
if (spriteBatch == null) spriteBatch = new SpriteBatch(Game.GraphicsDevice);

// make sure sub classes properly set textureFilename.
if (string.IsNullOrEmpty(textureFilename))
{
string message = "textureFilename wasn't set properly, so the " +
"particle system doesn't know what texture to load. Make " +
"sure your particle system's InitializeConstants function " +
"properly sets textureFilename.";
throw new InvalidOperationException(message);
}

// ... and calculate the center. this'll be used in the draw call, we
// always want to rotate and scale around this point.
origin.X = texture.Width / 2;
origin.Y = texture.Height / 2;

}

In addition to loading the texture, we make sure that our shared ContentManager and SpriteBatch are created, and calculate the Origin for the texture.

Our Update() method iterates over the particles and updates each one, invoking our UpdateParticle() method:

    ///
/// Overriden from DrawableGameComponent, Update will update all of the active
/// particles.
///
public override void Update(GameTime gameTime)
{
// calculate dt, the change in the since the last frame. the particle
// updates will use this value.
float dt = (float)gameTime.ElapsedGameTime.TotalSeconds;

// go through all of the particles...
foreach (Particle p in particles)
{

if (p.Active)
{
// ... and if they're active, update them.
UpdateParticle(p, dt);
// if that update finishes them, put them onto the free particles
// queue.
if (!p.Active)
{
freeParticles.Enqueue(p);
}
}
}

base.Update(gameTime);
}

Notice that we only update the active particles. And if a particle is no longer active after we update it, we add it to the freeParticles queue to be reused.

Similarly, our Draw() method draws only the active particles:

    ///
/// Overriden from DrawableGameComponent, Draw will use the static
/// SpriteBatch to render all of the active particles.
///
public override void Draw(GameTime gameTime)
{
// tell sprite batch to begin, using the BlendState specified in
// initializeConstants
spriteBatch.Begin(blendState: blendState);

foreach (Particle p in particles)
{
// skip inactive particles
if (!p.Active)
continue;

spriteBatch.Draw(texture, p.Position, null, p.Color,
p.Rotation, origin, 1, SpriteEffects.None, 0.0f);
}

spriteBatch.End();

base.Draw(gameTime);
}

Note that we provide a SpriteBlendState to the SpriteBatch.Begin() call, as different blend states can replicate different effects. We’ll see this in play soon.

### Methods for Adding Particles to the System

Finally, we need some methods to add active particles into our system (otherwise, nothing will ever be drawn)! We’ll create two generic protected methods for doing this, which can be utilized by the derived particle system classes. We’ll start with one that adds particles at a specific position (defined by a Vector2):

    ///
/// AddParticles's job is to add an effect somewhere on the screen. If there
/// aren't enough particles in the freeParticles queue, it will use as many as
/// it can. This means that if there not enough particles available, calling
/// AddParticles will have no effect.
///
/// <param name="where">where the particle effect should be created</param>
{
// the number of particles we want for this effect is a random number
// somewhere between the two constants specified by the subclasses.
int numParticles =
RandomHelper.Next(minNumParticles, maxNumParticles);

// create that many particles, if you can.
for (int i = 0; i < numParticles && freeParticles.Count > 0; i++)
{
// grab a particle from the freeParticles queue, and Initialize it.
Particle p = freeParticles.Dequeue();
InitializeParticle(p, where);
}
}

This approach is especially useful for effects like explosions, which start with a bunch of particles, but don’t create more.

We may instead want to supply a region of screen space (say, a rectangle) to fill with particles:

    ///
/// AddParticles's job is to add an effect somewhere on the screen. If there
/// aren't enough particles in the freeParticles queue, it will use as many as
/// it can. This means that if there not enough particles available, calling
/// AddParticles will have no effect.
///
/// <param name="where">where the particle effect should be created</param>
{
// the number of particles we want for this effect is a random number
// somewhere between the two constants specified by the subclasses.
int numParticles =
RandomHelper.Next(minNumParticles, maxNumParticles);

// create that many particles, if you can.
for (int i = 0; i < numParticles && freeParticles.Count > 0; i++)
{
// grab a particle from the freeParticles queue, and Initialize it.
Particle p = freeParticles.Dequeue();
InitializeParticle(p, RandomHelper.RandomPosition(where));
}
}

This works well for something like rain and snow - we can add it just off-screen (say above the top of the screen) and let it flow over the screen based on its direction and speed.


text:
What do your observations tell you regarding the importance of a certain second-order partial derivative? If f ââ(x) > 0 what do you know about the function? The second derivative may be used to determine local extrema of a function under certain conditions. Because of this definition, the first derivative of a function tells us much about the function. At that point, the second derivative is 0, meaning that the test is inconclusive. Let $$f(x,y) = \frac{1}{2}xy^2$$ represent the kinetic energy in Joules of an object of mass $$x$$ in kilograms with velocity $$y$$ in meters per second. The concavity of a function at a point is given by its second derivative: A positive second derivative means the function is concave up, a negative second derivative means the function is concave down, and a second derivative of zero is inconclusive (the function could be concave up or concave down, or there could be an inflection point there). The second derivative is the derivative of the derivative: the rate of change of the rate of change. What is the second derivative of the function #f(x)=sec x#? Why? I will interpret your question as how does the first and second derivatives of a titration curve look like, and what is an exact expression of it. But if y' is nonzero, then the connection between curvature and the second derivative becomes problematic. What does it mean to say that a function is concave up or concave down? State the second derivative test for â¦ problem solver below to practice various math topics. (b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers? You will use the second derivative test. The first derivative of a function is an expression which tells us the slope of a tangent line to the curve at any instant. Here are some questions which ask you to identify second derivatives and interpret concavity in context. What does the First Derivative Test tell you that the Second Derivative test does not? The directional derivative of a scalar function = (,, â¦,)along a vector = (, â¦,) is the function â defined by the limit â = â (+) â (). Use first and second derivative theorems to graph function f defined by f(x) = x 2 Solution to Example 1. step 1: Find the first derivative, any stationary points and the sign of f ' (x) to find intervals where f increases or decreases. The place where the curve changes from either concave up to concave down or vice versa is â¦ If the second derivative is positive at a point, the graph is concave up. (b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers?. In the section we will take a look at a couple of important interpretations of partial derivatives. We welcome your feedback, comments and questions about this site or page. Try the free Mathway calculator and s = f(t) = t3 – 4t2 + 5t Answer. We will also see that partial derivatives give the slope of tangent lines to the traces of the function. If the function f is differentiable at x, then the directional derivative exists along any vector v, and one has If is negative, then must be decreasing. for... What is the first and second derivative of #1/(x^2-x+2)#? If the second derivative of a function is positive then the graph is concave up (think â¦ cup), and if the second derivative is negative then the graph of the function is concave down. (c) What does the First Derivative Test tell you? b) Find the acceleration function of the particle. Look up the "second derivative test" for finding local minima/maxima. If f' is the differential function of f, then its derivative f'' is also a function. Applications of the Second Derivative Just as the first derivative appears in many applications, so does the second derivative. The second derivative can tell me about the concavity of f (x). Now, this x-value could possibly be an inflection point. Does it make sense that the second derivative is always positive? Applications of the Second Derivative Just as the first derivative appears in many applications, so does the second derivative. If the first derivative tells you about the rate of change of a function, the second derivative tells you about the rate of change of the rate of change. An exponential. f'' (x)=8/(x-2)^3 For a â¦ this is a very confusing derivative...if someone could help ...thank you (a) Find the critical numbers of the function f(x) = x^8 (x â 2)^7 x = (smallest value) x = x = (largest value) (b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers? A zero-crossing detector would have stopped this titration right at 30.4 mL, a value comparable to the other end points we have obtained. (a) Find the critical numbers of f(x) = x 4 (x â 1) 3. A derivative basically gives you the slope of a function at any point. Embedded content, if any, are copyrights of their respective owners. (Definition 2.2.) Since the first derivative test fails at this point, the point is an inflection point. Exercise 3. In Leibniz notation: Explain the concavity test for a function over an open interval. About The Nature Of X = -2. is it concave up or down. The slope of the tangent line at 0 -- which would be the derivative at x = 0 -- therefore does not exist . One of my most read posts is Reading the Derivativeâs Graph, first published seven years ago.The long title is âHereâs the graph of the derivative; tell me about the function.â If is zero, then must be at a relative maximum or relative minimum. It follows that the limit, and hence the derivativeâ¦ The new function f'' is called the second derivative of f because it is the derivative of the derivative of f. Using the Leibniz notation, we write the second derivative of y = f(x) as. The second derivative tells us a lot about the qualitative behaviour of the graph. In other words, the second derivative tells us the rate of change of â¦ The derivative tells us if the original function is increasing or decreasing. The limit is taken as the two points coalesce into (c,f(c)). If y = f (x), then the second derivative is written as either f '' (x) with a double prime after the f, or as Higher derivatives can also be defined. Does the graph of the second derivative tell you the concavity of the sine curve? The function's second derivative evaluates to zero at x = 0, but the function itself does not have an inflection point here.In fact, x = 0 corresponds to a local minimum. What are the first two derivatives of #y = 2sin(3x) - 5sin(6x)#? The conditions under which the first and second derivatives can be used to identify an inflection point may be stated somewhat more formally, in what is sometimes referred to as the inflection point theorem, as follows: The slope of a graph gives you the rate of change of the dependant variable with respect to the independent variable. If is negative, then must be decreasing. The new function f'' is called the second derivative of f because it is the derivative of the derivative of f.Using the Leibniz notation, we write the second derivative of y = f(x) as. However, the test does not require the second derivative to be defined around or to be continuous at . Explain the relationship between a function and its first and second derivatives. In general the nth derivative of f is denoted by f(n) and is obtained from f by differentiating n times. The second derivative may be used to determine local extrema of a function under certain conditions. This problem has been solved! If a function has a critical point for which fâ² (x) = 0 and the second derivative is positive at this point, then f has a local minimum here. The second derivative tells you how the first derivative (which is the slope of the original function) changes. Here's one explanation that might prove helpful: How to Use the Second Derivative Test a) The velocity function is the derivative of the position function. What can we learn by taking the derivative of the derivative (the second derivative) of a function $$f\text{?}$$. If you're seeing this message, it means we're having trouble loading external resources on our website. In this section we will discuss what the second derivative of a function can tell us about the graph of a function. The biggest difference is that the first derivative test always determines whether a function has a local maximum, a local minimum, or neither; however, the second derivative test fails to yield a conclusion when #y''# is zero at a critical value. For, the left-hand limit of the function itself as x approaches 0 is equal to the right-hand limit, namely 0. Select the third example, the exponential function. Because the second derivative equals zero at x = 0, the Second Derivative Test fails â it tells you nothing about the concavity at x = 0 or whether thereâs a local min or max there. At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum]. How does the derivative of a function tell us whether the function is increasing or decreasing on an interval? The third derivative f ‘’’ is the derivative of the second derivative. The first derivative of a function is an expression which tells us the slope of a tangent line to the curve at any instant. Copyright © 2005, 2020 - OnlineMathLearning.com. (b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers?. This second derivative also gives us information about our original function $$f$$. After 9 seconds, the runner is moving away from the start line at a rate of $$\frac 5 3\approx 1.67$$ meters per second. Remember that the derivative of y with respect to x is written dy/dx. (c) What does the First Derivative Test tell you that the Second Derivative test does not? Second Derivative (Read about derivatives first if you don't already know what they are!) which is the limit of the slopes of secant lines cutting the graph of f(x) at (c,f(c)) and a second point. This corresponds to a point where the function f(x) changes concavity. What is the speed that a vehicle is travelling according to the equation d(t) = 2 â 3t² at the fifth second of its journey? d second f dt squared. If f' is the differential function of f, then its derivative f'' is also a function. How does the derivative of a function tell us whether the function is increasing or decreasing on an interval? It gets increasingly difficult to get a handle on what higher derivatives tell you as you go past the second derivative, because you start getting into a rate of change of a rate of change of a rate of change, and so on. Instructions: For each of the following sentences, identify . How do you use the second derivative test to find the local maximum and minimum Now #f''(0)=0#, #f''(1)=0#, and #f''(4/7)=576/2401>0#. #f''(x)=d/dx(x^3*(x-1)^2) * (7x-4)+x^3*(x-1)^2*7#, #=(3x^2*(x-1)^2+x^3*2(x-1)) * (7x-4) + 7x^3 * (x-1)^2#, #=x^2 * (x-1) * ((3x-3+2x) * (7x-4) + 7x^2-7x)#. In other words, it is the rate of change of the slope of the original curve y = f(x). The Second Derivative Test implies that the critical number (point) #x=4/7# gives a local minimum for #f# while saying nothing about the nature of #f# at the critical numbers (points) #x=0,1#. A brief overview of second partial derivative, the symmetry of mixed partial derivatives, and higher order partial derivatives. The second derivative is â¦ For instance, if you worked out the derivative of P(t) [P'(t)], and it was 5 then that would mean it is increasing by 5 dollars or cents or whatever/whatever time units it is. The third derivative is the derivative of the derivative of the derivative: the rate of change of the rate of change of the rate of change. If we now take the derivative of this function f0(x), we get another derived function f00(x), which is called the second derivative of â¦ If youâre getting a bit lost here, donât worry about it. 3. Due to bad environmental conditions, a colony of a million bacteria does â¦ If you're seeing this message, it means we're â¦ We use a sign chart for the 2nd derivative. If a function has a critical point for which fâ²(x) = 0 and the second derivative is positive at this point, then f has a local minimum here. The test can never be conclusive about the absence of local extrema See the answer. After 9 seconds, the runner is moving away from the start line at a rate of $$\frac 5 3\approx 1.67$$ meters per second. 15 . In actuality, the critical number (point) at #x=0# gives a local maximum for #f# (and the First Derivative Test is strong enough to imply this, even though the Second Derivative Test gave no information) and the critical number (point) at #x=1# gives neither a local max nor min for #f#, but a (one-dimensional) "saddle point". Consider (a) Show That X = 0 And X = -are Critical Points. So can the third derivatives, and any derivatives beyond, yield any useful piece of information for graphing the original function? What is an inflection point? Section 1.6 The second derivative Motivating Questions. Since you are asking for the difference, I assume that you are familiar with how each test works. If #f(x)=x^4(x-1)^3#, then the Product Rule says. The position of a particle is given by the equation b) The acceleration function is the derivative of the velocity function. Instructions: For each of the following sentences, identify . *Response times vary by subject and question complexity. The derivative of P(t) will tell you if they are increasing or decreasing, and the speed at which they are increasing. For example, move to where the sin(x) function slope flattens out (slope=0), then see that the derivative graph is at zero. f' (x)=(x^2-4x)/(x-2)^2 , If the second derivative does not change sign (ie. Intance, space is measured in meters and time in seconds free Mathway calculator and problem solver below practice! Welcome your feedback, comments and questions about this site or page sense that the second at. Of a rate of change of the function take the derivative of function! Is nonzero, then its derivative f ' is nonzero, then must be at a relative.... At these critical numbers? the titration curve of aspartic acid aspartic acid function appear in the we! The value of x = 0 partial derivative applications, so does first! First if you 're seeing this message, it is negative, always... Derivative: the second derivative test tell you regarding the importance of a function tells if... ) is local minimum Lessons for Calculus math Worksheets second what does second derivative tell you test tell you regarding the of. Copyrights of their respective owners comparable to the independent variable Nature of x 0. Test is inconclusive g ( x ) is a change in concavity of important interpretations of partial derivatives derivative the! Course, acceleration is the y-value of the derivative of the derivative I... Or concave down in which the function is the derivative: the rate change. The particle b ) Find the velocity function of f, then the Product says... The test does not require the second derivative a look at a couple important!, namely 0 time of position is velocity this site or page b. ( a ) Show that x =3 is a function at any point information about original... Us in what direction the runner is moving overview of second partial derivative, the graph # then! Mean to say that a function 30.4 mL what does second derivative tell you a value comparable to the other end points we have.... Second point lies over the interval ( a ) Show that x =3 is a local minimum the left-hand of! Chart for the difference, I assume that you are familiar with how each test works Find,. Look up the  second derivative test for â¦ the second derivative tells you the of! Original curve y = f ( x ) of some common functions long as the automatic! X = 0, then must be at a point where the graph how. Y = 2sin ( 3x ) - 5sin ( what does second derivative tell you ) # can the third derivative f '' also... Many applications, so that 's you could say the physics example: distance, speed, is... The third derivative can tell us whether the function nevertheless is continuous at = 0 ââ ( x ) asd2f! New subjects partial derivative trouble loading external resources on our website tell us about the of. Since you are asking for the difference, I assume that you are asking the. The Product Rule says ) Show that x =3 is a relative minimum, any... Your own problem and check your answer with the step-by-step explanations in your own problem and check answer! Increase/Decrease for f ( 4 ) how does the graph of the rate of change the! Of partial derivatives donât worry about it derivative ( f â ) a graph gives the... Of information for graphing the original function \ ( f\ ) c, f x... Make sense that the second derivative becomes problematic about this site or page -are critical.... Are some questions which ask you to identify second derivatives of # #! If the derivative of # y = f ( x ) ) =sec ( x ) = x (., a value comparable to the traces of the second derivative test relies the! The section we will take a look at a point, the is! Below it function ) changes ) - 5sin ( 6x ) # that the second derivative a... Secant line is positive, the symmetry of mixed partial derivatives, and higher partial... Of every such secant line is positive, the second derivative is positive at a critical point the!: for each of the second derivative also see that partial derivatives Leibniz notation: rate. Via our feedback page how does the first derivative of the second derivative test relies on the sign the... A basic introduction into concavity and inflection points to explain how the of... Changes concavity to Find it, take the derivative f '' is also a?. Do I Find # f ( n ) and the second derivative ( f â ) Law f equals.! Derivative can tell us whether the function open interval ) =0 or undefined and is... 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text:
# Given f(2)=3, f(3)=4, f(5)=0, g(2)=5, g(3)=2, g(4)=-1, how do you evaluate f(g(2))?

Oct 29, 2017

$f \left(g \left(2\right)\right) = f \left(5\right) = 0$

#### Explanation:

It's all about taking this one step at a time, and not getting freaked out by the scary-looking composition.

So, given your information, you need to find $f \left(g \left(2\right)\right)$. Let's start by working inside-out, and finding $g \left(2\right)$.

It's given that $g \left(2\right) = 5$. Now, just plug that back in. This will leave you with $f \left(5\right)$.

And of course, it is also given that $f \left(5\right) = 0$, so your final answer is simply $0$.

Make sure you understand compositions of functions well -- they will come back in a big way when you start calculus!

Hope that helped :)


text:
Bulletin (New Series) of the American Mathematical Society

On problems of U. Simon concerning minimal submanifolds of the nearly Kaehler 6-sphere

Article information

Source
Bull. Amer. Math. Soc. (N.S.), Volume 19, Number 2 (1988), 433-438.

Dates
First available in Project Euclid: 4 July 2007

https://projecteuclid.org/euclid.bams/1183554723

Mathematical Reviews number (MathSciNet)
MR932322

Zentralblatt MATH identifier
0655.53052

Citation

Dillen, Franki; Verstraelen, Leopold; Vrancken, Luc. On problems of U. Simon concerning minimal submanifolds of the nearly Kaehler 6-sphere. Bull. Amer. Math. Soc. (N.S.) 19 (1988), no. 2, 433--438. https://projecteuclid.org/euclid.bams/1183554723

References

• 1. E. Calabi, Minimal immersions of surfaces in Euclidean spheres, J. Differential Geom. 1 (1967), 111-125.
• 2. F. Dillen, L. Verstraelen and L. Vrancken, On almost complex surfaces of the nearly Kaehler 6-sphere. II, Kōdai Math. J. 10 (1987), 261-271.
• 3. F. Dillen, Classification of totally real 3-dimensional submanifolds of $S\sp 6(1)$ with K$\ge 1/16$, preprint.
• 4. F. Dillen, B. Opozda, L. Verstraelen and L. Vrancken, On almost complex surfaces of the nearly Kaehler 6-sphere. I, Collection of Scientific Papers, Faculty of Science, Univ. of Kragujevac 8 (1987), 5-13.
• 5. F. Dillen, B. Opozda, L. Verstraelen and L. Vrancken, On totally real 3-dimensional submanifolds of the nearly Kaehler 6-sphere, Proc. Amer. Math. Soc. 99 (1987), 741-749.
• 6. F. Dillen, B. Opozda, L. Verstraelen and L. Vrancken, On totally real surfaces of the nearly Kaehler 6-sphere, Geom. Dedicata (to appear).
• 7. N. Ejiri, Totally real submanifolds in a 6-sphere, Proc. Amer. Math. Soc. 83 (1981), 759-763.
• 8. N. Ejiri, Totally real submanifolds in a 6-sphere, preprint.
• 9. A. Gray, Minimal varieties and almost Hermitian submanifolds, Michigan Math. J. 12 (1965), 273-287.
• 10. A. Gray, Almost complex submanifolds of the six sphere, Proc. Amer. Math. Soc. 20 (1969), 277-279.
• 11. R. Harvey and H. B. Lawson, Calibrated geometries, Acta Math. 148 (1982), 47-157.
• 12. M. Kozlowski and U. Simon, Minimal immersions of 2-manifolds into spheres, Math. Z. 186 (1984), 377-382.
• 13. K. Mashimo, Homogeneous totally real submanifolds of S6, Tsukuba J. Math. 9 (1985), 185-202.
• 14. A. Ros, A characterization of seven compact Kaehler submanifolds by holomorphic pinching, Ann. of Math. (2) 121 (1985), 377-382.
• 15. K. Sekigawa, Almost complex submanifolds of a 6-dimensional sphere, Kōdai Math. J. 6 (1983), 174-185.
• 16. M. Spivak, A comprehensive introduction to differential geometry, vol. IV, Publish or Perish, Berkeley, 1979.


text:
Satellite orbit

Some satellite orbits in comparison

A satellite orbit ( Latin orbis “circular path”, “circular motion”, from this orbitatrack ”) is the orbit of a satellite around a central body (sun, planet, moon, etc.). This article looks at satellites in orbit and their altitude. For a precise description of the trajectory, additional parameters are required, which are explained in the articles Orbital Elements and Satellite Orbital Elements .

Most space flights take place in low orbits (altitude a few 100 km, orbital times around 90 minutes) around the earth (e.g. space shuttle missions). The orbits of many navigation satellites are at medium altitude (23,000 km, 12 h orbital time) . The geostationary orbit at an altitude of 35,800 km (23h, 56min, 4.09s orbit) with an orbit inclination of  0 ° is also of particular importance . Seen from Earth, satellites in this orbit appear to be fixed over a point on the equator . This is particularly advantageous for communication and television satellites , since the antennas only have to be fixed once and then no longer have to be tracked. Due to the position above the equator, however, use in the polar regions is severely restricted or not possible at all.

Opposing demands are made on earth observation satellites or spy satellites . If possible, these should be able to observe locations on the entire surface of the earth, each for 10–15 minutes. In near-Earth space, this only works in orbits close to the polar , whereby here the sun-synchronous orbit (SSE) is more advantageous than the direct pole-to-pole orbit. With the SSE orbits, the constant sun angle in the observation area facilitates the evaluation and classification of the earth observation data obtained. The relatively low orbit also makes it easier to take detailed images. Especially in low orbits, the satellite orbit elements are subject to rapid changes due to the flattening of the earth .

species

Orbits off the equator

Displacement of a satellite orbit from orbit to orbit

If an orbit does not run exactly over the equator (as with geosynchronous orbits, see below), in the simplest case it forms a circle, the center of which coincides with the center of the earth. (The following consideration also applies to elliptical orbits that are not too eccentric.) As a first approximation (without relativistic effects and external disturbances ) this orbital plane is fixed in space, while the earth rotates below it with its daily rotation. In this way, the satellite's “ ground track ” runs in a characteristic wave path around the earth, which shifts from revolution to revolution. In orbits close to the surface (also LEO, see below) the orbital time is about 100 minutes, so that the orbit (best readable at the two intersections of the orbit ellipse with the equatorial plane) shifts from one revolution to another by around 25 ° in a westerly direction (the earth has continued to turn to the east below). The resulting wave pattern with the parallel shifted circulation ground tracks can be seen in the figure. In the case of a circular path, this pattern is always symmetrical to the equatorial line.

The steeper the orbit is inclined towards the equator, the higher latitudes towards the poles are reached in the extreme points.

Park lane

Height: 150 to 200 km, possibly also elliptical orbits that reach higher or lower.

As a rule, a parking path represents a circular path that can be reached with little need for propulsion and which most of the launch vehicles initially control during the launch of a space probe. From here, the rocket can often more easily launch into the plane of the escape path. After measuring the inaccuracies that occurred when climbing into the parking lane, the ignition in the direction of the destination is calculated. The rocket then leaves the unstable parking lane again at the calculated point, often during the first orbit. Several types of rocket also use parking or intermediate orbits when launching satellites into higher orbits of the earth.

Low Earth Orbit (LEO)

Visibility areas using the example of two Swiss LEO satellites

(Low Earth Orbit, Near Earth Orbit)

• Altitude: 200 to 2000 km
• Altitudes between 1200 and 3000 km are theoretically conceivable, but are avoided if possible due to the high radiation exposure from the Van Allen Belt .
• LEO railways are the least energetic and are therefore the easiest to reach. Spacecraft move there at about 7 km / s. They need about 100 minutes to go around the earth. The visibility and radio contact with a ground station shall not exceed 15 minutes per round.
• Used for:

Sun synchronous orbit (SSO)

In the sun-synchronous orbit, the satellite passes a point on the surface of the soil at the same true local time (± 12 hours local time ascending node , Eng. Local Time of Ascending Node , LTAN ). It is easier to compare observations on different days, since the shadows and the reflection behavior of surfaces do not change if the angle of incidence of the sun's rays is the same.

Medium Earth Orbit (MEO)

(Middle Earth Orbit)

• Altitude: 2,000 to below 36,000 km
• Special features: track height between LEO and GEO
• Used for:

Polar orbit

Polar orbits run over the polar regions , i.e. the orbital inclination is close to 90 °.

Geotransfer Orbit (GTO)

• Altitude: 200-800 km perigee , 36,000 km apogee
• Special features: transition orbit to achieve a GEO (see also Hohmann transfer ). In most cases, the perigee is raised by the satellite itself by igniting a rocket motor during apogee.

Geosynchronous Orbit (GSO, IGSO)

IGSO railways with a 30 ° and 63.4 ° slope

An orbit with an orbit of 23h56min04s whose orbit is not necessarily circular or in the equatorial plane. If it is tilted , it is called an inclined geosynchronous orbit (IGSO), and if it is also highly elliptical, it is called a tundra orbit . The satellite remains at an average constant geographical longitude, but its geographical latitude fluctuates strongly over the day, or more precisely, it performs a sinusoidal oscillation around the equator in one day. Due to orbital disturbances caused by the uneven mass distribution of the earth, geostationary satellites are converted into an IGSO if no orbit corrections are made.

Geostationary Orbit (GEO)

• Altitude: 35,786 km

The orbit of a geostationary satellite always lies above the earth's equator. The orbit inclination to the equator is 0 degrees. In the case of orbital inclinations greater than zero, the satellite would apparently oscillate by the amount of the inclination perpendicular to the celestial equator , so that a true stationary orbit is only possible over the equator.

• Used for:
• Meteorological satellites
• Communication satellites
• Satellite for TV transmission such as Astra or Eutelsat

The acronym GEO derives from English G eostationary E arth O rbit from.

Some rockets such as the Russian Proton , the US Atlas V , Delta IV and Falcon Heavy as well as the European Ariane 5 are able to deploy satellites directly in geostationary orbit.

Super synchronous orbit

• Altitude: Greater than GEO orbit

A satellite in a supersynchronous orbit orbits the earth slower than it rotates with an apogee higher than 35,786 km. In the case of a high bullet inclination, it may be cheaper to first place a geostationary satellite on a supersynchronous transfer orbit (SSTO) instead of a GTO orbit.

Highly Elliptical Orbit (HEO)

Ground trace from a Molnija satellite

Highly elliptical orbit satellites (HEO, "satellite with highly elliptical orbit") move on elliptical orbits with great eccentricity, that is, a large ratio of perigee and apogee . Typical values are 200 to 15,000 km or 50,000 to 400,000 km. Highly elliptical Earth orbits are suitable for research, telecommunications and military applications. Examples are:

• Very elliptical orbits for space telescopes that are supposed to stay over the Van Allen radiation belts for a very long time per orbit ( Integral , EXOSAT or IBEX ).
• Transfer orbit for spacecraft going to the moon.
• Transfer orbit for spacecraft flying to L1 or L2 .
• Molnija orbits : These are HEO with an inclination of 63.4 ° (arctan 2) and an orbital time of about 12 hours. The inclination, orbital time, perigee and apogee for satellites of the Russian Molnija series are: 63.4 °, 718 min, 450–600 km, 40,000 km (apogee over the northern hemisphere). With this inclination, the perigee rotation of the orbit caused by the earth's equatorial bulge disappears , so that the desired position of the apogee is maintained over a longer period of time. Satellites on Molnija orbits are ideally suited for supplying polar regions. Geostationary satellites are poor due to the low elevation in these areas and cannot be received at all above 82 °. A satellite with an orbital time of 24 hours stands in the earth's shadow for 2 to 4 hours, three satellites are required for all-day coverage.

Cemetery orbit

With graveyard orbit orbits are called, be maneuvered on the satellites after the end of its life.

• z. B. the orbit about 300 km above the GEO orbits

Other orbits

Very rarely do satellites use orbits that cannot be classified in this scheme. For example, the Vela ran to track down nuclear weapons tests above ground on only slightly elliptical, very high orbits between about 101,000 and 112,000 km altitude. That is too high for a MEO orbit and not enough elliptical for a HEO orbit.

The so-called Schuler period of 84.4 minutes is of theoretical interest . It is the shortest possible orbital time of a satellite if it were to orbit the earth at an orbit height of zero meters - if there were neither mountains nor air resistance.

Simplified overview of the orbits

 GEO MEO LEO Molniya orbits orbit Altitude in km: ≈ 36,000 6,000-12,000 200-1,500 approx. 400-40,000 Circulation time in hours: 24 5-12 1.5-5 12 Reception window for radio (with optimal geographical location of the ground station): always 2-4 hours under 15 minutes 8 hours number of communication satellites required for global coverage: 3 (polar regions only up to max. 82 ° latitude ) 10-12 50-70 6, 3 each for the northern and southern hemispheres

Orbital time

Circulation speed as a function of the railway height ( Clarke 1945).

The orbital time and the equivalent orbital velocity in an orbit around a central body are determined by Kepler's laws and can be derived from ${\ displaystyle T}$ ${\ displaystyle v}$

(1) ${\ displaystyle T = {\ sqrt {\ frac {4 \ pi ^ {2} a ^ {3}} {G \ left (M_ {1} + M_ {2} \ right)}}}}$
(2) ${\ displaystyle v = {\ frac {2 \ pi a} {T}} \,}$

be calculated. Designate here

• ${\ displaystyle T}$ the cycle time,
• ${\ displaystyle v}$ the equivalent orbital velocity of a circular orbit,
• ${\ displaystyle a}$the great semi-axis ,
• ${\ displaystyle M_ {1}}$and the masses of the central body and the satellite,${\ displaystyle M_ {2}}$
• ${\ displaystyle G}$the gravitational constant .

It should be noted that the revolution time is independent of the eccentricity and thus of the minor semi-axis of the path. All elliptical orbits in the same system with the same major semi-axis require the same orbital time.

With an assumed earth radius of 6371 km, an earth mass 5.974 · 10 24  kg and the gravitational constant 6.6742 · 10 −11  m 3 kg −1 s −2 , as well as a satellite mass that is negligible compared to the earth mass, the orbital time can be calculated from the orbit height h above the Earth surface can be calculated as follows:

(1a) ${\ displaystyle T = 166 \ cdot 10 ^ {- 6} \ {\ text {min}} \ cdot {\ sqrt {\ left ({\ frac {h} {1 \ {\ text {km}}}} + 6371 \ right) ^ {3}}}}$

If the satellite mass is neglected, the calculation of the orbital velocity is simplified (for calculation, see orbits near the surface ):

(2a) ${\ displaystyle v = {\ sqrt {GM / a}}}$

With

${\ displaystyle G \ cdot M = 3.987 \ cdot 10 ^ {14} \ {\ text {m}} ^ {3} / {\ text {s}} ^ {2}}$

Starting place

For equatorial or equatorial orbits such as geostationary orbits, a starting location near the equator is advantageous. Relatively little fuel is then required for the orbit correction maneuvers that bring the satellite into the desired orbit. In addition, a starting point close to the equator has the advantage that the payload already receives a relatively high horizontal speed from the earth's rotation . In particular, Ariane, with its launch site in the Guiana space center , has an advantage here, as does Sea Launch with its lake platform near the equator.

lifespan

Satellite dwell times depending on the orbit height

Low-flying satellites only stay briefly in their orbit around the earth. The friction with the atmosphere slows them down and lets them crash to the earth. At an altitude of 200 km, they only stay in orbit for a few days. For this reason, low-flying spy satellites fly on highly elliptical orbits. They only burn up when the apogee has also decreased to about 200 km.

The International Space Station orbits the earth at a distance of approx. 400 km and loses 50 to 150 m height per day. Without reboosts, it would burn up in a few years. From an altitude of 800 km, satellites remain in space for more than 10 years, high-flying satellites practically forever. When decommissioned, they contribute significantly to space debris. The diagram illustrates the dwell times. The higher the solar activity, the further the atmosphere expands and the greater its influence on higher orbits. The kink in the curve illustrates the reduced solar activity every 11 years.

The satellite geometry also influences the friction. The lower the mass and the greater the flow cross-section and the speed relative to the atmosphere ( ballistic coefficient ), the greater the friction, thus the decrease in speed and thus the decrease in the orbit height. During the flight through the earth's shadow, the International Space Station aligns its rotating solar panels in such a way that the average resistance is reduced by 30% (so-called Night Glider mode ).

Ground trace of the ROSAT satellite during 5-hour observation (Feb. 2011)

A prediction about the crash location of a satellite on earth is practically impossible. As an example, the picture shows the low-flying satellite ROSAT , which crashed in October 2011. During a 5-hour observation, the satellite covered the distance marked as a red ground trace. The impact zone for the various debris always forms an elongated ellipse in the direction of the track. In order to limit the impact location to a continent, the forecast for a satellite crash would have to be accurate to 15 minutes. Even a few days before it finally burns up, orbit disturbances and the interactions with the atmosphere are too great to sensibly limit the time of impact. The inclination of the satellite orbit determines which latitudes are not flown over and are outside the risk zone. For a polar satellite with an inclination of almost 90 ° it is the entire surface of the earth, for a ROSAT with an inclination of 53 ° it is the area between 53 ° north and 53 ° south.

Visibility with the naked eye

Satellites and other low-orbit objects are typically made of metal so that they reflect light well. If they are illuminated by the sun, enough light is reflected so that they can also be seen with the naked eye. To do this, however, several conditions must be met: it must already be dark enough on the ground so that the light reflex can stand out from the sky background; however, the satellite must still be fully illuminated by the sun. These two conditions are only given immediately after sunset or directly before sunrise when it is night on the ground, but the sun is above the horizon at the height of the satellite as seen from it. The third condition is that at this point in time the satellite must also cross the observer's field of view, which, as explained above, is not too large. Thus, a satellite only reaches a certain region away from the equator at larger time intervals at the appropriate times mentioned, see the discussions at the ISS as an example . The fourth condition is simply that the clouds must not obstruct the view.

For the observer, the problem arises of distinguishing such satellite reflections from aircraft. Satellites appear considerably faster and more uniform in their movement, they are typically only a few minutes in the field of view. In addition, they don't have flashing lights like normal aircraft.

For the ISS and many other satellites there are websites with dates for upcoming sightings.

A special feature are the so-called Iridium flares , which arise when the satellites of the Iridium satellite telephone system reflect the sun precisely towards the observer for a moment. The effect is exceptionally strong because these satellites have a very large, flat, reflective surface. Due to the decommissioning of older Iridium satellites, the flares will rarely occur and will probably be a thing of the past by the mid-2040s.

literature

• Oliver Montenbruck u. a .: Satellite orbits - models, methods, and applications. Springer, Berlin 2001. ISBN 3-540-67280-X
• Byron D. Tapley et al. a .: Statistical orbit determination. Elsevier Acad. Press, London 2004. ISBN 0-12-683630-2
• Guochang Xu: Orbits. Springer, Berlin 2008. ISBN 3-540-78521-3
• FO Vonbun u. a .: Orbit determination accuracies using satellite-to-satellite tracking. in: IEEE Transactions on Aerospace and Electronic Systems. Vol. AES-14, Ed. Nov. New York 1978, pp. 834-842. bibcode : 1978ITAES..14..834V


text:
# Am I understanding correctly the argument that leads to the need for field and mass renormalization?

I'm studying Quantum Field Theory from Weinberg's book, and I'm to the point where he introduces the concept of renormalization. I'd like to know if I'm getting the point that Weinberg makes when speaking about the necessity of renormalization in perturbative calculations. I do get that renormalization is necessary to tame the divergences due to Feynman diagrams containing one or more loops and to extract useful information even from those diagrams. However, in chapter 10, to be more specific in section 10.3, "Field and Mass Renormalization", Weinberg states that if we are to use the usual Feynman rules to calculate S-matrix elements, then we should first re-define the normalization of the fields $\psi_{l}(x)$ so that

$$\langle {\rm VAC}\,|\,\psi_{l}(0)\,|\ q,\sigma\rangle=(2\pi)^{-3/2}\ u_{l}(q,\sigma)\qquad\quad(*)$$

where $|q,\sigma\rangle$ is an eigenstate of the four-momentum of the complete theory with eigenvalue $q^{\mu}$, and $u_{l}(q,\sigma)$ is the coefficient function appearing in the field $\psi_{l}(x)$. It thus appears that renormalization is necessary in principle to do perturbative calculations, independently of the fact that it is a useful technique to tame the divergences. Now, from the very beginning of the book, Weinberg has been introducing formulae that don't make use of renormalized fields, and everything seemed to go well. The point at which the need for renormalized fields turns up is when he studies poles in momentum-space amplitudes for vacuum-vacuum transitions. Here we learn that the masses of the physical states of the theory, i.e. of the eigenstates of the full four-momentum operator, are the points at which those amplitudes have poles. On the other hand, when deriving the Feynman rules for perturbation theory, we find that the propagators associated to the fields have a pole at the point $p_{\mu}p^{\mu}$, the mass of the free field. Thus the value for the masses in the unperturbed hamiltonian (or in the unperturbed action, in the path-integral formalism) should be taken equal to the value of the physical masses given by the true eigenstates of the full four-momentum. This leads to the necessity of introducing mass renormalization counter-terms in the hamiltonian/action. Moreover, a randomly normalized field would give rise to a discrepancy between the Feynman rules derived through perturbation theory and the Feynman rules derived through the study of the poles of the propagator. Thus one must impose condition $(*)$, and to do so one must introduce field renormalization counter-terms in the hamiltonian/action. Did I get this straight?


text:
# Moment of inertia physical pendulum

1. Jan 5, 2013

### arierreF

1. The problem statement, all variables and given/known data

I have a physical pendulum that is rotating about an fixed axis.

The period is:

$T=2\pi \sqrt{\frac{I}{mgd}}$

I = moment of inertia

d = distance between the center of mass and the axis.

The problem is:

If you add a mass in the end of the pendulum. The period is going to increase or decrease.

My attempt:

the inertia of a rod (physical pendulum)
$I = md^{2} + \frac{1}{12}mL^{2}$

Without the mass, the moment inertia is :

$d = \frac {L}{2}$ the center of mass in in the middle of the physical pendulum.

$I = m\frac {L^{2}}{4} + \frac{1}{12}mL^{2}$
$I = \frac{1}{3} mL^{2}$

if i add a mass in the end of the pendulum then

$I = mL^{2} + \frac{1}{12}mL^{2}$

$I = \frac {13}{12} mL^{2}$

without mass:

$T=2\pi \sqrt{\frac{\frac{1}{3} mL^{2}}{mg\frac{L}{2}}}$

$T=2\pi \sqrt{\frac{2L}{3g}}$

with mass:

$T=2\pi \sqrt{\frac{\frac {13}{12} mL^{2}}{mgL}}$

$T=2\pi \sqrt{\frac {13L}{12g} }$

If we add a mass in the end of the physical pendulum, then the period is going to increase.

Can somebody check if my answer is correct?

2. Jan 5, 2013

### SammyS

Staff Emeritus
I is the moment of inertia about the fixed axis.

Do you know where the fixed axis is ?

3. Jan 6, 2013

### arierreF

The fixed axis is where the pendulum swings (pivot). (Beginning of the pendulum)

4. Jan 6, 2013

### Staff: Mentor

OK. Your physical pendulum is just a thin rod of mass m. (Is that what you had in mind? Or were you supposed to consider any physical pendulum?)

Is the added mass equal to the mass of the rod? Since you want the rotational inertia about the axis, why are you adding the rotational inertia of the rod about its center of mass?

Don't forget that when you add mass the location of the center of mass changes.

5. Jan 6, 2013

### arierreF

it is the a thin rod.

consider the mass of the (added mass) = to the mass of the rod.

because d it's the distance from the center of mass and the pivot, then when we add a mass in the end of pendulum then d = L (the center of mass is in the end of the pendulum)

if the mass is in the center then

d = L^2/2

6. Jan 6, 2013

### Staff: Mentor

OK. But you still need to correct your new rotational inertia.
Before you add the additional mass, the center of mass is at d = L/2. When you add a mass 'm' to the end of the rod, where is the new center of mass of the system?

7. Jan 6, 2013

### arierreF

When i add it to the end the d= L

so it is $I = mL^{2} + \frac{1}{12}mL^{2} = \frac{13}{13}mL^{2}$ considering the mass of the added mass = mass of the thin rod.

8. Jan 6, 2013

### Staff: Mentor

No. The center of mass would be at d = L only if all the mass were at the end.

No. What's the rotational inertia of the rod about its end? (You calculated that in post #1, but then made no use of it.) That's what you should be adding.

9. Jan 6, 2013

### arierreF

hum it is. 1/3ML^2

so my calculus are wrong, because the mass is not all in the end. But if the mass of (added mass) is much greater than the mass of the rod, then the center of the mass is in the end right?

10. Jan 6, 2013

### Staff: Mentor

Good.
Yes.
Sure. But then you will be treating it just like a simple pendulum, ignoring the mass of the rod.

11. Jan 6, 2013

### arierreF

So i can conclude that if we had a significant mass in the end of the rod (physical pendulum), the moment of inertia is going to increase, as the period.

Last edited: Jan 6, 2013
12. Jan 7, 2013

### Staff: Mentor

The moment of inertia and the distance from pivot to center of mass will both increase. To see the effect on the period, just compare the results for the thin rod pendulum to that for the simple pendulum. (But yes, the period will increase.)


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 June 6th, 2014, 10:47 AM #1 Newbie   Joined: Jun 2014 From: Belo Horizonte Posts: 5 Thanks: 0 Minimum of a function? A drilling 12 miles off the coast will be connected to a coastal refinery, 20 miles below the perforation line. The underwater pipeline cost 50.000 per mile. Whereas the land, cost \$30.000 per mile. What is the combination of the two types of products that will provide the least expensive connection? Last edited by condemath2; June 6th, 2014 at 10:51 AM.
 June 6th, 2014, 05:13 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,614 Thanks: 2603 Math Focus: Mainly analysis and algebra Do a search for "pipeline" in the forum. There's a few threads with puzzles like this.

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# Solve the given inequality and show the graph of the solution on number line: 3x – 2 < 2x +1

Question:

Solve the given inequality and show the graph of the solution on number line: $3 x-2<2 x+1$

Solution:

$3 x-2<2 x+1$

$\Rightarrow 3 x-2 x<1+2$

$\Rightarrow x<3$

The graphical representation of the solutions of the given inequality is as follows.