text:
## Traveling

I’ll be doing a lot of traveling over the next two weeks; I’ll pick up with things when I’m back from Barcelona.

## The real theorem

This post is a bit of a detour, but I wanted to point out that Conclusion 3.2A is really just Claim 3.2 grafted onto a nice characterization of ${{\rm pp}(\mu)}$. Shelah implicitly uses this characterization in his proof of Conclusion 3.2A, but I can’t find a formulation of it anywhere in The Book.

Theorem 1 Let ${\mu}$ be a singular cardinal. Then the following three cardinals are equal:

1. ${{\rm pp}(\mu)}$
2. ${\min_{\tau<\mu}\sup\{{\rm cf}(\prod\mathfrak{b}, <):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}\wedge|\mathfrak{b}|\leq{\rm cf}(\mu)\}}$
3. ${\min_{\tau<\mu}\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq(\tau,\mu)\cap{\sf Reg}\wedge |\mathfrak{b}|\leq{\rm cf}(\mu)<\min(\mathfrak{b})\}}$.

Notice that in 2 and 3, the suprema involved can only decrease as ${\tau}$ increases, and this tells us a couple of things. First, the cardinals in 2 and 3 are easily seen to be equal, as without loss of generality ${{\rm cf}(\mu)<\tau}$, hence the collection of relevant sets ${\mathfrak{b}}$ is exactly the same in each case, and we finish using the equality

$\displaystyle {\rm cf}\left(\prod\mathfrak{b}, <\right)=\max{\rm pcf}(\mathfrak{b}) \ \ \ \ \ (1)$

for progressive ${\mathfrak{b}}$.

Second, we know that as ${\tau}$ approaches ${\mu}$, the suprema involve eventually stabilize and hence the minimum value is actually attained by some ${\tau<\mu}$. What the theorem shows is that this stable value is exactly ${{\rm pp}(\mu)}$. Notice as well that although ${{\rm pp}(\mu)}$ is computed using sets cofinal in ${\mu}$, we do not assume this for the ${\mathfrak{b}}$ involved in 2 and 3.

We turn now to the proof:

Proof: We have already noted that 2 and 3 are equal for easy reasons. The fact that 1 is less than or equal to 2 is also trivial from the definition of ${{\rm pp}(\mu)}$:

Suppose ${\eta<{\rm pp}(\mu)}$. Then there is a progressive ${\mathfrak{b}\subseteq \mu\cap{\sf Reg}}$, a cardinal ${\kappa}$, and an ultrafilter ${D}$ on ${\mathfrak{b}}$ such that

• ${|\mathfrak{b}|={\rm cf}(\kappa)}$
• ${D}$ is disjoint to the bounded ideal on ${\mathfrak{b}}$,
• ${\kappa={\rm cf}(\prod\mathfrak{b}/D)}$, and
• ${\eta<\kappa}$.

Given ${\tau<\mu}$, we know ${\kappa\in{\rm pcf}(\mathfrak{b}\cap(\tau,\mu))}$ as ${\mathfrak{b}\cap (\tau,\mu)\in D}$, hence

$\displaystyle \tau<\kappa\leq\max{\rm pcf}(\mathfrak{b})\cap(\tau,\mu). \ \ \ \ \ (2)$

Thus,

$\displaystyle \eta<\kappa\leq\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq(\tau,\mu)\cap{\sf Reg}\wedge |\mathfrak{b}|\leq{\rm cf}(\mu)<\min(\mathfrak{b})\} \ \ \ \ \ (3)$

for each ${\tau<\mu}$ and the desired inequality follows immediately.

What about the other inequality? This is exactly Lemma 1 from our June 17, 2011 post, which was written with this application in mind.

$\Box$

## Conclusion 3.2A (part 4)

So let us now return to the proof of Conclusion 3.2A. We’ll keep notation the same; what remains to be shown is that if ${\mu}$ is singular of cardinality ${\kappa}$ and ${\mu<\aleph_\mu}$, then there is a ${\tau<\mu}$ for which

$\displaystyle {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)\leq{\rm pp}(\mu). \ \ \ \ \ (1)$

We have assumed ${\mu<\aleph_\mu}$, so for all sufficiently large ${\tau<\mu}$ we know ${(\tau,\mu)\cap{\sf Reg}}$ is progressive. Given such a ${\tau}$, we can apply Claim 3.2 (June 15, 2011) to conclude

$\displaystyle {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg}). \ \ \ \ \ (2)$

So we fix can fix ${\tau<\mu}$ large enough so that

• ${\mathfrak{a}:=(\tau,\mu)\cap{\sf Reg}}$ is progressive,
• ${{\rm cf}_\kappa\left(\prod\mathfrak{a}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}$, and
• ${\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq\mathfrak{a}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu)}$,

where for the last statement we use the result of the previous posting.

We will get what we need by comparing the last two bullet points:

Suppose ${\nu\in{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}$. This means that we there is an ultrafilter ${D}$ and a set ${\mathfrak{b}\subseteq \mathfrak{a}}$ of cardinality ${\leq\kappa}$ such that

• ${\mathfrak{b}\in D}$, and
• ${{\rm cf}(\prod\mathfrak{a}/D)=\nu}$.

Clearly we may assume that ${\mathfrak{b}}$ has cardinality exactly ${\kappa}$, and the second bullet implies

$\displaystyle {\rm cf}(\prod\mathfrak{b}/ D)=\nu \ \ \ \ \ (3)$

as well, where we view ${D}$ as an ultrafilter on ${\mathfrak{b}}$. Chaining things together, we have

$\displaystyle \nu={\rm cf}\left(\prod\mathfrak{a}/D\right)={\rm cf}\left(\prod\mathfrak{b}/D\right)\leq\max{\rm pcf}(\mathfrak{b})\leq{\rm pp}(\mu). \ \ \ \ \ (4)$

So for our choice of ${\tau}$, we have

$\displaystyle \sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg})\leq{\rm pp}(\mu), \ \ \ \ \ (5)$

and so (1) follows from (2).

## Conclusion 3.2A (part 3)

This post is an immediate continuation of the previous one, so all notation etc. carries over from there. I noticed last night that Conclusion 3.2A is really two separate results which have been smashed together. The main part actually looks like it holds without the assumption that ${\mu}$ is not a fixed point, so I’m going to work in a slightly more general context today. If things work out, I’ll formulate “the real theorem” underlying things.

Lemma 1 Let ${\mu}$ be a singular cardinal of cofinality ${\kappa}$. Then there is a ${\tau<\mu}$ such that

$\displaystyle \sup\{{\rm cf}\left(\prod\mathfrak{b}, <\right):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (1)$

Proof: First, note that if we restrict ourselves to ${\tau}$ with ${{\rm cf}(\tau)<\tau<\mu}$, then any ${\mathfrak{b}}$ relevant to computing the left-hand side of the above equation is necessarily progressive, and hence the cofinality of ${\prod\mathfrak{b}}$ will just be ${\max{\rm pcf}(\mathfrak{b})}$. Thus, we need only prove that there is a ${\tau}$ in the interval ${(\kappa,\mu)}$ for which

$\displaystyle \sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (2)$

Suppose this is not the case, so for any ${\tau<\mu}$ there is a ${\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}}$ of cardinality ${\kappa}$ with ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$. The following proposition takes this hypothesis and moves us closer to a contradiction:

Proposition 2 Under the above assumptions, for every ${\tau<\mu}$ there is a ${\mathfrak{b}}$ such that

• ${\mathfrak{b}}$ is a progressive subset of ${(\tau,\mu)\cap{\sf Reg}}$,
• ${|\mathfrak{b}|\leq\kappa}$,
• ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$, and
• ${\mathfrak{b}}$ is a bounded subset of ${\mu}$.

Proof: Without loss of generality, we assume ${\kappa<\tau}$ and so we can find (progressive) ${\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}}$ of cardinality ${\kappa}$ such that ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$. If ${\mathfrak{b}}$ is bounded below ${\mu}$ then we are done, so assume ${\mathfrak{b}}$ is unbounded.

Let ${D}$ be an ultrafilter on ${\mathfrak{b}}$ such that

$\displaystyle {\rm pp}(\mu)<{\rm cf}\left(\prod\mathfrak{b}/D\right)=\max{\rm pcf}(\mathfrak{b}). \ \ \ \ \ (3)$

The above equation, taken together with the fact that ${\mathfrak{b}}$ is unbounded in ${\mathfrak{a}}$ of cardinality ${\kappa}$, tells us (by way of the definition of ${{\rm pp}(\mu)}$) that ${D}$ cannot be disjoint to the bounded subsets of ${\mathfrak{b}}$, i.e,. that there is a ${\sigma<\mu}$ such that

$\displaystyle \mathfrak{b}^*:=\mathfrak{b}\cap\sigma \in D. \ \ \ \ \ (4)$

Now this ${\mathfrak{b}^*}$ has all of the required properties. $\Box$

Armed with the above proposition, we can construct sequences ${\langle \mathfrak{b}_i:i<\kappa\rangle}$ and ${\langle \mu_i:i<\kappa\rangle}$ such that

• ${\langle \mu_i:i<\kappa\rangle}$ is a strictly increasing sequence of cardinals cofinal in ${\mu}$
• ${\mu_0=\kappa}$
• ${\mathfrak{b}_i\subseteq (\mu_i, \mu_{i+1})\cap{\sf Reg}}$
• ${|\mathfrak{b}_i|\leq\kappa}$ (hence ${\mathfrak{b}_i}$ is progressive)
• ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b}_i)}$ for each ${i<\kappa}$.

Now define

$\displaystyle \mathfrak{b}=\bigcup_{i<\kappa}\mathfrak{b}_i. \ \ \ \ \ (5)$

By construction, ${\mathfrak{b}}$ is a progressive set of regular cardinals cofinal in ${\mu}$ of cardinality ${{\rm cf}(\mu)}$, and so the definition of ${{\rm pp}(\mu)}$ tells us that if ${D}$ is an ultrafilter on ${\mathfrak{b}}$ disjoint to the bounded ideal, then

$\displaystyle {\rm cf}\left(\prod\mathfrak{b}/D\right)\leq{\rm pp}(\mu). \ \ \ \ \ (6)$

Let ${\sigma={\rm pp}(\mu)^+}$, and consider the ideal ${J}$ on ${\mathfrak{b}}$ generated by ${J_{<\sigma}[\mathfrak{b}]}$ and the bounded subsets of ${\mathfrak{b}}$. It is easy to show that ${J}$ must be a proper ideal on ${\mathfrak{b}}$: if not, then there would be a ${\nu<\mu}$ such that

$\displaystyle \mathfrak{b}\cap (\nu,\mu)\in J_{<\sigma}[\mathfrak{a}], \ \ \ \ \ (7)$

but this is impossible as

$\displaystyle {\rm pp}(\mu)<{\rm pp}(\mu)^+=\sigma\leq\max{\rm pcf}(\mathfrak{b}\cap(\nu,\mu)) \ \ \ \ \ (8)$

by our construction of ${\mathfrak{b}}$. If ${D}$ is an ultrafilter on ${\mathfrak{b}}$ disjoint to ${J}$, then

$\displaystyle {\rm pp}(\mu)<\sigma\leq{\rm cf}\left(\prod\mathfrak{b}/D\right), \ \ \ \ \ (9)$

and we have a contradiction of (6). $\Box$

## Conclusion 3.2A (part 2)

Recall from the last post that I’ve committed myself to providing a proof of the following:

Theorem 1 (Conclusion 3.2A (i) page 378, reformulated) Suppose ${\mu}$ is a singular cardinal of cofinality ${\kappa}$ with ${\mu<\aleph_\mu}$. Then for some ${\tau<\mu}$,

$\displaystyle {\rm pp}(\mu)=^+{\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right). \ \ \ \ \ (1)$

We’ve already discussed what the notation means, so I’m going to plunge right in. I’ll probably split this up into a few posts; the result is true, but the proof given in The Book has the unfortunate property that at a key moment, the reader is given an incorrect reference as explanation, and it might take me a post or two to fill in the necessary details.

Proof: We’ll start with the easy stuff and prove first the inequality

$\displaystyle {\rm pp}(\mu)\leq {\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right) \ \ \ \ \ (2)$

whenever ${\tau<\mu}$ is such that ${(\tau,\mu)\cap{\sf Reg}}$ is progressive.

Let ${\tau}$ be given (such ${\tau}$ exist because ${\mu<\aleph_\mu}$ — see posts of June 15), and define

$\displaystyle \mathfrak{a}:=(\tau,\mu)\cap{\sf Reg}. \ \ \ \ \ (3)$

By Claim 3.2 (June 15, 2011), we know

$\displaystyle {\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right)={\rm cf}^{\aleph_0}_{<\kappa^+}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a}). \ \ \ \ \ (4)$

So let us assume that ${\mathfrak{b}}$ is a progressive set of regular cardinals cofinal in ${\mu}$, ${J}$ is an ideal on ${\mathfrak{b}}$ containing the bounded subsets of ${\mathfrak{b}}$, and

$\displaystyle \nu = {\rm tcf}\left(\prod\mathfrak{b}, <_J\right). \ \ \ \ \ (5)$

To establish our inequality, we must show

$\displaystyle \nu\in{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a}). \ \ \ \ \ (6)$

This is actually quite easy. Let us define an ideal ${I}$ on ${\mathfrak{a}}$ by

$\displaystyle A\in I\Longleftrightarrow A\cap \mathfrak{b}\in J. \ \ \ \ \ (7)$

Set ${\mathfrak{c}:=\mathfrak{b}\cap (\tau,\mu)}$, and note

• ${|\mathfrak{c}|\leq\kappa<\kappa^+}$ (as ${\mathfrak{c}\subseteq\mathfrak{b}}$), and
• ${\mathfrak{a}\setminus\mathfrak{c}\in I}$ (as ${J}$ contains the set ${\mathfrak{b}\cap\tau^+}$).

Thus, to finish this part of the proof we need only check

$\displaystyle \nu = {\rm tcf}\left(\prod\mathfrak{a}, <_I\right), \ \ \ \ \ (8)$

but this follows easily from (5).

I’ll end this post here, and pick up with establishing the other inequality (and the “${+}$” part) tomorrow.

$\Box$

## Conclusion 3.2A (part 1)

In this post, we want to look at the next result in The Book: Conclusion 3.2A on page 378. What I will do is state this exactly as it appears in the book, and then reformulate it in conjunction with providing some commentary. I’ll attempt to prove it in a future post.

Theorem 1 (Conclusion 3.2A page 378) If ${\lambda=\aleph_{\xi(*)+\zeta(*)}}$, ${\zeta(*)<\aleph_{\xi(*)}}$ a limit ordinal, then:

1. for ${i_0<\zeta(*)}$ large enough,

$\displaystyle {\rm pp}(\lambda)=^+{\rm cf}_{{\rm cf}[\zeta(*)]}\left(\prod\{\aleph_{\xi(*)+i+1}: i_0

2. if ${{\rm cf}[\zeta(*)]\leq\kappa\leq |\zeta(*)|^+}$ then for ${i_0<\zeta(*)}$ large enough

$\displaystyle {\rm pp}_\kappa(\lambda)=^+{\rm cf}_\kappa\left(\{\aleph_{\xi(*)+i+1}:i_0

I’m going to concentrate on the first part of the above; here’s my attempt at reformulating it.

Theorem 2 (Conclusion 3.2A (i), reformulated) Suppose ${\mu}$ is a singular cardinal of cofinality ${\kappa}$ with ${\mu<\aleph_\mu}$. Then for some ${\tau<\mu}$,

$\displaystyle {\rm pp}(\mu)=^+{\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right). \ \ \ \ \ (3)$

The right-hand side of the above is not something we’ve explicitly defined yet, but it’s a reasonable variant of the ${{\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})}$ notation we’ve been working with. The notation

$\displaystyle {\rm cf}_\kappa\left(\prod\mathfrak{a}\right) \ \ \ \ \ (4)$

refers to the minimum cardinality of a family ${F\subseteq\prod\mathfrak{a}}$ with the property that for any ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${\kappa}$, the collection

$\displaystyle \{f\upharpoonright\mathfrak{b}:f\in F\} \ \ \ \ \ (5)$

is cofinal in ${(\prod\mathfrak{b}, <)}$.

This definition is robust under small modifications: it is also the minimum cardinality of a family ${F\subseteq\prod\mathfrak{a}}$ with the property that for any ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality at most ${\kappa}$ and any ${g\in\prod\mathfrak{a}}$, there is a finite set ${F_0\subseteq F}$ such that

$\displaystyle g(\eta)<\max\{f(\eta):f\in F_0\}\text{ for all }\eta\in\mathfrak{b}. \ \ \ \ \ (6)$

In other words,

$\displaystyle {\rm cf}_\kappa\left(\prod\mathfrak{a}\right)={\rm cf}_{<\kappa^+}^{\aleph_0}\left(\prod\mathfrak{a}\right), \ \ \ \ \ (7)$

and we are back with the sorts of cardinals we’ve been considering of late.

There is something else implicit in this theorem that I want to bring out: it shows us that if ${\mu}$ is a singular cardinal that isn’t a fixed point, then ${{\rm pp}(\mu)}$ can be calculated just by looking at structure

$\displaystyle \left(\prod(\mu\cap{\sf Reg}), <\right). \ \ \ \ \ (8)$

In this situation ${{\rm pp}(\mu)}$ is the minimum cardinality of a family of functions in ${\prod(\mu\cap{\sf Reg})}$ with the property that for some ${\tau<\mu}$, for any ${\mathfrak{b}\subseteq(\tau,\mu)\cap{\sf Reg}}$ of cardinality ${{\rm cf}(\mu)}$ the family

$\displaystyle \{f\upharpoonright\mathfrak{b}:f\in F\} \ \ \ \ \ (9)$

is cofinal in ${\prod\mathfrak{b}}$. Roughly speaking, if ${\mu}$ is singular and not a fixed point, then ${{\rm pp}(\mu)}$ is the minimum cardinality of a family of functions in ${\prod\mu\cap{\sf Reg}}$ which can cover all products of size ${{\rm cf(\mu)}}$ drawn from some fixed tail of ${\mu\cap{\sf Reg}}$. That seems like it is worth knowing…

## A non fixed point by any other name…

This is a silly post, but looking ahead to Conclusion 3.2A reminded me of one of the frustrations I had when first trying to read The Book. One of the major themes of pcf theory is that it tells us quite a lot about cardinal arithmetic at singular cardinals ${\mu}$ that are not fixed points of the ${\aleph}$ function. The problem is that this is equivalent to lots of other conditions, and very often Shelah will use a different formulation of this property in the statement of a theorem, so it isn’t clear right away that what he’s really talking about is a cardinal that isn’t a fixed point. This little proposition is just an elementary exercise in set theory, but it’s useful to keep in mind when reading The Book. (And let’s hope I haven’t flubbed it since I’m typing from home while babysitting…)

Proposition 1 Let ${\mu}$ be a singular cardinal. Then the following are equivalent:

1. ${\mu}$ is not a fixed point, i.e., ${\mu<\aleph_\mu}$.
2. ${\mu=\aleph_{\xi+\zeta}}$ where ${\zeta}$ is a limit ordinal ${<\aleph_\xi}$.
3. There is a ${\tau<\mu}$ such that ${(\tau,\mu)\cap{\sf Reg}}$ is progressive.

Proof: Suppose ${\mu<\aleph_\mu}$, and fix ${\delta<\mu}$ such that ${\mu=\aleph_\delta}$. Since ${\mu}$ is a limit cardinal, we know ${|\delta|^+<\mu}$, so choose ${\xi<\delta}$ such that

$\displaystyle |\delta|^+=\aleph_\xi. \ \ \ \ \ (4)$

Now fix ${\zeta}$ such that

$\displaystyle \delta=\xi+\zeta. \ \ \ \ \ (5)$

It is clear that ${\zeta}$ must be a limit (as ${\delta}$ is), and the fact that

$\displaystyle \mu=\aleph_\delta=\aleph_{\xi+\zeta} \ \ \ \ \ (6)$

is completely trivial. For the last bit, we know

$\displaystyle |\zeta|\leq |\delta|<|\delta|^+=\aleph_\xi, \ \ \ \ \ (7)$

and we are done with proving that (1) implies (2)

Now assume (2) holds, set ${\tau=\aleph_\xi}$, and define

$\displaystyle \mathfrak{a}=(\tau,\mu)\cap{\sf Reg}. \ \ \ \ \ (8)$

Now note

$\displaystyle |\mathfrak{a}|\leq |(\tau,\mu)|=|\zeta|<\tau<\tau^+=\min(\mathfrak{a}), \ \ \ \ \ (9)$

and we have (3).

Moving on to the last implication, let us suppose ${\tau<\mu}$ and ${\mathfrak{a}=(\tau,\mu)\cap{\sf Reg}}$ is progressive, i.e., ${|\mathfrak{a}|\leq\tau}$. But then ${|(\tau,\mu)\cap{\sf Card}|\leq\tau}$ as well, and hence

$\displaystyle |\mu\cap{\sf Card}|\leq\tau. \ \ \ \ \ (10)$

Thus, we have

$\displaystyle \mu\leq\aleph_{\tau^+}<\aleph_\mu, \ \ \ \ \ (11)$

as required.

$\Box$

## Claim 3.2 page 377

Last time, we left off with the statement of the following theorem:

Theorem 1 Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals, and

$\displaystyle \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (1)$

Then

$\displaystyle {\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}). \ \ \ \ \ (2)$

The above is just Claim 3.2 on page 377 of The Book with the notation changed a little. Today I want to give a proof of this theorem. If you’ve already read through our June 10th post, then today will be simple because it’s essentially the same proof.

Proof: Start by letting ${F\subseteq\prod\mathfrak{a}}$ be ${<\sigma}$-cofinal in ${\prod\mathfrak{b}}$ for any ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$, and suppose ${\kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$. We will show that ${\kappa\leq|F|}$, thereby establishing the “${\geq}$” part of the theorem.

Assume by way of contradiction that ${|F|<\kappa}$. Let ${J}$ and ${\mathfrak{b}}$ witness that ${\kappa}$ is in ${{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$, i.e.,

• ${J}$ is a ${\sigma}$-complete ideal on ${\mathfrak{a}}$,
• ${(\prod\mathfrak{a}, <_J)}$ has true cofinality ${\kappa}$,
• ${|\mathfrak{b}|<\theta}$, and
• ${\mathfrak{a}\setminus\mathfrak{b}\in J}$.

Since ${|F|<\kappa}$, there is a function ${g\in\prod\mathfrak{a}}$ such that

$\displaystyle f<_J g\text{ for all}f\in F. \ \ \ \ \ (3)$

Since ${|\mathfrak{b}|<\theta}$, we can find ${F_0\subseteq F}$ of cardinality ${<\sigma}$ such that

$\displaystyle g(\eta)<\sup\{f(\eta):f\in F_0\}\text{ for all }\eta\in\mathfrak{b}. \ \ \ \ \ (4)$

Given ${f\in F_0}$, we define

$\displaystyle A_f:=\{\eta\in\mathfrak{a}: g(\eta)

Then (4) tells us

$\displaystyle \mathfrak{b}\subseteq\bigcup\{A_f:f\in F_0\}. \ \ \ \ \ (6)$

A couple of things now come to our rescue: since ${\mathfrak{a}\setminus\mathfrak{b}\in J}$ and ${J}$ is ${\sigma}$-complete, there must be an ${f^*\in F_0}$ such that ${A_f}$ is not in ${J}$, and this tells us

$\displaystyle \{\eta\in\mathfrak{a}:g(\eta)

This is absurd, of course, for ${g}$ was an upper bound of ${F}$ in ${(\prod\mathfrak{a}, <_J)}$. Thus, ${\kappa\leq |F|}$ and the “${\geq}$” part of the theorem has been established.

For the other direction, we once again mirror our June 10th post. Given ${\kappa\in{\rm pcf}(\mathfrak{a})}$, let ${J_{<\kappa}}$ and ${B_{\kappa}}$ be the associated ideal and generating set. Basic pcf theory tells us

$\displaystyle {\rm tcf}\left(\prod\mathfrak{a}, <_{J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa)}\right)=\kappa, \ \ \ \ \ (8)$

and we let ${F_\kappa=\{ f^\kappa_\alpha:\alpha<\kappa\}}$ serve to witness this.

Now define

$\displaystyle F:=\bigcup\{F_\kappa:\kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})\}. \ \ \ \ \ (9)$

Clearly ${|F|=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$, so we will be done if we can show ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{b}}$ for every ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$.

Suppose this is not the case. Then we can find a set ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$ and a function ${g\in\prod\mathfrak{a}}$ such that the restriction of ${g}$ to ${\mathfrak{b}}$ cannot be covered by fewer than ${\sigma}$ functions in ${F}$.

Let us now define a collection ${J}$ of subsets of ${\mathfrak{a}}$ by putting ${A}$ into ${J}$ if and only if there is a family ${F_0\subseteq F}$ of cardinality ${<\sigma}$ such that

$\displaystyle \eta\in A\cap\mathfrak{b}\Longrightarrow g(\eta)<\sup\{f(\eta):f\in F_0\}. \ \ \ \ \ (10)$

It’s easy to check that ${J}$ is an ideal, and ${\mathfrak{a}\setminus\mathfrak{b}\in J}$ for trivial reasons. Our choice of ${g}$ guarantees that ${J}$ is a ${\sigma}$-complete proper ideal as well.

Now let ${\kappa\in{\rm pcf}(\mathfrak{a})}$ be minimal with ${B_\kappa\notin J}$, and define

$\displaystyle I=J+(\mathfrak{a}\setminus B_\kappa). \ \ \ \ \ (11)$

The ideal ${I}$ is essentially ${J}$ restricted to ${B_\kappa}$; it is easy to see that ${I}$ is a proper (by choice of ${\kappa}$) ${\sigma}$-complete (as ${J}$ is) ideal on ${\mathfrak{a}}$. Clearly ${\mathfrak{a}\setminus\mathfrak{b}\in I}$ as well, since this set is in ${J}$.

The minimality of ${\kappa}$ ensures

$\displaystyle J_{<\kappa}\subseteq J, \ \ \ \ \ (12)$

which immediately implies

$\displaystyle J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa)\subseteq I. \ \ \ \ \ (13)$

Thus, the functions ${\langle f^\kappa_\alpha:\alpha<\kappa\rangle}$ witness

$\displaystyle \kappa={\rm tcf}\left(\prod\mathfrak{a}, <_I\right). \ \ \ \ \ (14)$

Putting all of the above together, we see

$\displaystyle \kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}) \ \ \ \ \ (15)$

by way of our ideal ${I}$.

Our goal now is to reach a contradiction exactly as in the June 10th post by demonstrating that ${B_\kappa\in J}$, and ${B_\kappa\notin J}$.

Part of this is trivial, as ${\kappa}$ was selected so that ${B_\kappa\notin J}$. For the other part, we know that given our choice of ${F_\kappa}$, there is an ${f\in F_\kappa}$ such that

$\displaystyle g

which means,

$\displaystyle \{\eta\in B_\kappa: f(\eta)\leq g(\eta)\}\in J_{<\kappa}\subseteq J. \ \ \ \ \ (17)$

But since ${F_\kappa\subseteq F}$, the definition of ${J}$ tells us

$\displaystyle \{\eta\in B_\kappa\cap \mathfrak{b}: g(\eta)

Since ${\mathfrak{a}\setminus\mathfrak{b}\in J}$, we have

$\displaystyle B_\kappa\setminus\mathfrak{b}\subseteq \mathfrak{a}\setminus\mathfrak{b}\in J, \ \ \ \ \ (19)$

and now the conjunction of (17), (18), and (19) together with the fact that ${J}$ is an ideal shows us that ${B_\kappa\in J}$. Contradiction. $\Box$

## Some terminology

Having set the mood with our last post, we now return to “The Book” and examine “covering all small products”. I want to use this post to try out some terminology to see if it’s possible to make the material a little more palatable.

Definition 1 Let ${\mathfrak{a}}$ be a set of regular cardinals, let ${\sigma}$ be a regular cardinal, and let ${F\subseteq\prod\mathfrak{a}}$. Given a set ${\mathfrak{b}\subseteq\mathfrak{a}}$, we say that ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{b}}$ if for any function ${f\in\prod\mathfrak{a}}$, there is a family ${F_0}$ of fewer than ${\sigma}$ elements of ${F}$ such that

$\displaystyle f(\eta)<\sup\{g(\eta):g\in F_0\}\text{ for all }\eta\in \mathfrak{b}. \ \ \ \ \ (1)$

In the last post, we characterized the minimum cardinality of a ${<\sigma}$-cofinal subset of ${\prod\mathfrak{a}}$ in the case where ${\mathfrak{a}}$ is progressive and ${\sigma<|\mathfrak{a}|}$. Today, I want to consider a related question:

What is the minimum cardinality of a family ${F\subseteq\prod\mathfrak{a}}$ such that ${F}$ is ${<\sigma}$-cofinal in ${\mathfrak{b}}$ for every ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$?

Let us agree to denote the above cardinal by ${{\rm cf}^{\sigma}_{<\theta}(\prod\mathfrak{a})}$ (again, don’t blame me for the notation!).

Ok, so why on earth would I want to consider such a cardinal? Well, we’re going to see that this is actually quite relevant for cardinal arithmetic — we will show that such cardinals are intimately tied up with Shelah’s covering numbers. Our first task, though, will be to characterize such cardinals in terms of pcf theory. The characterization will require a little terminology.

Definition 2 Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals, and let ${\sigma}$ and ${\theta}$ be cardinals such that

$\displaystyle \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (2)$

We define the set ${{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$ to consist of all (regular) cardinals ${\kappa}$ for which there is an ideal ${J}$ on ${\mathfrak{a}}$ such that

• ${J}$ is ${\sigma}$-complete,
• ${(\prod\mathfrak{a}, <_J)}$ has true cofinality ${\kappa}$, and
• there is a ${\mathfrak{b}\subseteq\mathfrak{a}}$ such that ${|\mathfrak{b}|<\theta}$ and ${\mathfrak{a}\setminus\mathfrak{b}\in J}$.

Notice that the above is very close to the definition of ${{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$; the difference is that the last clause further restricts the acceptable ideals to those which also “reduce” ${\prod\mathfrak{a}}$ to something of the form ${\prod\mathfrak{b}}$ where ${|\mathfrak{b}|<\theta}$.

We’ll end this post with the statement of a theorem; we’ll prove the theorem in the next post.

Theorem 3 Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals, and

$\displaystyle \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (3)$

Then

$\displaystyle {\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}). \ \ \ \ \ (4)$

## Characterize the sup of pcf sigma-complete

This post should be considered a warmup. To start, I want to recall one of the standard results of pcf theory, namely that if ${\mathfrak{a}}$ is a progressive set of regular cardinal, then

$\displaystyle \max{\rm pcf}(\mathfrak{a})={\rm cf}(\prod\mathfrak{a}, <), \ \ \ \ \ (1)$

where the ordering on ${\prod\mathfrak{a}}$ is given by

$\displaystyle f

My aim here is to prove an analogous result characterizing ${\sup{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$. For those new to this, a cardinal ${\kappa}$ is in ${{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$ if there is a ${\sigma}$-complete ideal ${J}$ on ${\mathfrak{a}}$ with

$\displaystyle \kappa={\rm tcf}(\prod\mathfrak{a}, <_J), \ \ \ \ \ (3)$

which means that there is a ${<_J}$-increasing ${\kappa}$-sequence of functions in ${\prod\mathfrak{a}}$ cofinal in ${\prod\mathfrak{a}}$. In other words, (3) means that ${(\prod\mathfrak{a}, <_J)}$ has cofinality ${\kappa}$, AND every subset of ${\prod\mathfrak{a}}$ of cardinality less than ${\kappa}$ has an upper bound in ${(\prod\mathfrak{a}, <_J)}$.

The characterization we give involves the following notion:

Definition 1 Let ${\sigma}$ be a regular cardinal, and suppose ${\mathfrak{a}}$ is a set of regular cardinals. We say that a family ${F\subseteq\prod\mathfrak{a}}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{a}}$ if for every ${f\in\prod\mathfrak{a}}$, there is a family ${F_0\subseteq F}$ of size less than ${\sigma}$ such that ${f<\sup F_0}$, i.e., such that

$\displaystyle f(\eta)<\sup\{g(\eta):g\in F_0\} \ \ \ \ \ (4)$

for all ${\eta\in \mathfrak{a}}$. The ${<\sigma}$-cofinality of ${\mathfrak{a}}$ is defined to be the minimum cardinality of a ${<\sigma}$-cofinal subset of ${\mathfrak{a}}$. We will denote this cardinal by ${{\rm cf}^{\sigma}\prod(\mathfrak{a})}$.

So now we come to the promised characterization:

Theorem 2 If ${\mathfrak{a}}$ is a progressive set of regular cardinals, and ${\sigma}$ is a regular cardinal with ${\sigma\leq|\mathfrak{a}|}$, then

$\displaystyle {\rm cf}^{\sigma}(\prod\mathfrak{a}) = \sup{\rm pcf}_{\sigma{\rm -complete}}(\mathfrak{a}). \ \ \ \ \ (5)$

Proof: We prove the ${\geq}$ direction first, so assume ${F}$ is a ${<\sigma}$-cofinal family in ${\prod\mathfrak{a}}$, and ${\kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$. Let ${J}$ be a ${\sigma}$-complete ideal on ${\mathfrak{a}}$ such that

$\displaystyle {\rm tcf}(\prod\mathfrak{a}, <_J)=\kappa, \ \ \ \ \ (6)$

and assume by way of contradiction that ${|F|<\kappa}$. Given this, (15) implies that ${F}$ is bounded modulo ${J}$ in ${\prod\mathfrak{a}}$, so there is a ${g\in\prod \mathfrak{a}}$ with

$\displaystyle f<_J g\text{ for all }f\in F. \ \ \ \ \ (7)$

We have assumed, however, that ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{a}}$, so we can fix ${\alpha<\sigma}$ and ${F_0=\{f_\beta:\beta<\alpha\}\subseteq F}$ with

$\displaystyle g<\sup F_0. \ \ \ \ \ (8)$

Given ${\beta<\alpha}$, define

$\displaystyle A_\beta=\{\eta\in\mathfrak{a}: g(\eta)

By our choice of ${F_0}$, we have

$\displaystyle \mathfrak{a}=\bigcup_{\beta<\alpha}A_\beta. \ \ \ \ \ (10)$

Since ${J}$ is ${\sigma}$-complete and ${\alpha<\sigma}$, there is a ${\beta<\alpha}$ with ${A_\beta\in J^+}$. But this is a contradiction as ${g}$ was an upper bound for ${F}$ in ${(\prod\mathfrak{a}, <_J)}$. Thus, ${\kappa\leq |F|}$ and the inequality “${\geq}$” has been established.

For the other direction, we are going to need a little pcf theory. In general, for each ${\kappa\in{\rm pcf}(\mathfrak{a})}$ we have an associated ideal ${J_{<\kappa}[\mathfrak{a}]}$ and generator ${B_{\kappa}[\mathfrak{a}]}$, and

$\displaystyle {\rm tcf}(\prod\mathfrak{a}, <_{J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa[\mathfrak{a}])})=\kappa. \ \ \ \ \ (11)$

This entails the existence of a sequence of functions ${\langle f^\kappa_\alpha:\alpha<\kappa\rangle}$ witnessing the true cofinality of the partial order, and we let ${F_\kappa}$ denote the collection ${\{f^\kappa_\alpha:\alpha<\kappa\}}$.

Let us define

$\displaystyle F:=\bigcup\{F_\kappa: \kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})\}. \ \ \ \ \ (12)$

Clearly ${|F|=\sup{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$. We claim that ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{a}}$.

Suppose this is not the case. Then there is a function ${g\in\prod\mathfrak{a}}$ that cannot be covered by fewer than ${\sigma}$ functions from ${F}$. We define an collection ${J}$ of subsets of ${\mathfrak{a}}$ by putting ${A}$ into ${J}$ if and only if there is a family ${F_0\subseteq F}$ of cardinality ${<\sigma}$ such that

$\displaystyle \eta\in A\Longrightarrow g(\eta)<\sup\{f(\eta):f\in F_0\}. \ \ \ \ \ (13)$

It is easy to check that ${J}$ is a ${\sigma}$-complete ideal on ${\mathfrak{a}}$, and this ideal is proper because of our choice of ${g}$.

Let ${\kappa\in{\rm pcf}(\mathfrak{a})}$ be minimal with ${B_\kappa[\mathfrak{a}]\notin J}$, and define

$\displaystyle I= J + (\mathfrak{a}\setminus B_\kappa[\mathfrak{a}]). \ \ \ \ \ (14)$

Since ${J}$ is ${\sigma}$-complete, it follows easily that ${I}$ is as well, and our choice of ${\kappa}$ guarantees that ${I}$ is a proper ideal on ${\mathfrak{a}}$. Our choice of ${\kappa}$ also guarantees

$\displaystyle J_{<\kappa}[\mathfrak{a}]\subseteq J, \ \ \ \ \ (15)$

and hence

$\displaystyle J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa[\mathfrak{a}])\subseteq I. \ \ \ \ \ (16)$

Thus, the functions ${\langle f^\kappa_\alpha:\alpha<\kappa\rangle}$ show us

$\displaystyle \kappa={\rm tcf}(\prod\mathfrak{a}, <_I), \ \ \ \ \ (17)$

and therefore

$\displaystyle \kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a}). \ \ \ \ \ (18)$

We begin to close in on a contradiction. We know that ${J}$ is a proper ideal on ${\mathfrak{a}}$ and therefore ${B_\kappa}$ (we’ll drop the “${[\mathfrak{a}]}$”) from now on) is not in ${J}$; we will finish the proof by demonstrating that in fact ${B_\kappa}$ MUST be in ${J}$.

Given the definition of ${F_\kappa}$, we know there is an ${f\in F_\kappa}$ for which

$\displaystyle g< f\mod J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa). \ \ \ \ \ (19)$

Said another way,

$\displaystyle \{\eta\in B_\kappa: f(\eta)\leq g(\eta)\}\in J_{<\kappa}[\mathfrak{a}]\subseteq J, \ \ \ \ \ (20)$

where the last inclusion is from (15).

But since ${F_\kappa\subseteq F}$, we know

$\displaystyle \{\eta\in B_{\kappa}: g(\eta)

as well, by the very definition of ${J}$.

Thus, ${B_\kappa}$ is the union of two sets in ${J}$, hence an element of ${J}$. Contradiction.

$\Box$

Next Page »


text:
0
TECHNICAL PAPERS

# Riser-Soil Interaction: Local Dynamics at TDP and a Discussion on the Eigenvalue and the VIV Problems

[+] Author and Article Information
Celso P. Pesce

Fluid-Structure Interaction and Offshore Mechanics Laboratory, Department of Mechanical Engineering, Escola Politécnica, University of São Paulo, São Paulo 05508-900, Brazilceppesce@usp.br

Clóvis de A. Martins

Fluid-Structure Interaction and Offshore Mechanics Laboratory, Department of Mechanical Engineering, Escola Politécnica, University of São Paulo, São Paulo 05508-900, Brazil and NDF—Nucleus for Dynamics and Fluids, Av. Prof. Mello Moraes 2231, São Paulo, SP 05508-900, Brazil

Lauro M. da Silveira

NDF—Nucleus for Dynamics and Fluids, Av. Prof. Mello Moraes 2231, São Paulo, SP 05508-900, Brazil

If compared to the dynamics of the supported part.

For dynamic compression, see (15-16).

$ηI$ indicates derivative with respect to the local coordinate $ξ$.

ORCAFLEX®, version 8.6.

As shown in Aranha et al. (8), the nonlinear hydrodynamic damping along the riser length causes clockwise and anti-clockwise motions to provoke quite different responses in the dynamic curvature at TDP.

See, e.g., Burridge et al. (17) for the treatment of a similar problem.

Triantafyllou uses the terms “slowly” and “fast” varying in space.

See, also, (19).

Though relatively poor, this estimate is generally in the safe side, as dynamic tension usually reduces dynamic curvature.

Actually, the transverse current makes the elastica to depart from the vertical plane.

In fact, the reduced velocity takes the value $Vr25=2πU∕Ω25D≅5.3$.

J. Offshore Mech. Arct. Eng 128(1), 39-55 (Aug 31, 2005) (17 pages) doi:10.1115/1.2151205 History: Received October 18, 2004; Revised August 31, 2005

## Abstract

The eigenvalue problem of risers is of utmost importance, particularly if vortex-induced vibration (VIV) is concerned. Design procedures rely on the determination of eigenvalues and eigenmodes. Natural frequencies are not too sensitive to the proper consideration of boundary condition, within a certain extent where dynamics at the touchdown area (TDA) may be modeled as dominated by the dynamics of the suspended part. However, eigenmodes may be strongly affected in this region because, strictly speaking, this is a nonlinear one-side (contact-type) boundary condition. Actually, we shall consider a nonlinear eigenvalue problem. Locally, at TDA, riser flexural rigidity and soil interaction play important roles and may affect the dynamic curvature. Extending and merging former analytical solutions on touchdown point (TDP) dynamics and on the eigenvalue problem, obtained through asymptotic and perturbation methods, the present work critically address soil and bending stiffness effects a little further. As far as linear soil stiffness and planar dynamics hypotheses may be considered valid, it is shown that penetration in the soil is small and that its effect does not change significantly the bending loading that is mainly caused by the cyclic excursion of the TDP and corresponding dynamic tension. A comparison of the analytical results with a full nonlinear time-domain simulation shows a remarkable agreement for a typical steel catenary riser. The WKB approximation for the eigenvalue problem gives good estimates for TDP excursion. As the dynamic tension caused solely by VIV is very small, the merged analytical solution may be used as a first estimate of the curvature variation at TDP in the cases of current perpendicular to the “riser plane.”

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## Figures

Figure 1

Two-dimensional catenary riser problem. Displacements exaggerated

Figure 2

Normalized elastic curve, K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1, β0=0.2.

Figure 3

Normalized angle, K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1, β0=0.2.

Figure 4

Normalized curvature, K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1, β0=0.2.

Figure 5

Normalized shear force, K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1, β0=0.2.

Figure 6

Normalized TDP excursion and dynamic tension in time: a0=1, β0=0.2, K=10

Figure 7

Normalized curvature in time at various sections: x∕λ=−6,−5,…,5,6; a0=1; β0=0.2; K=10

Figure 8

Normalized TDP excursion and dynamic tension in time: a0=1, β0=0.7, K=10

Figure 9

Normalized curvature in time at various sections: x∕λ=−6,−5,…,5,6; a0=1; β0=0.7; K=10

Figure 10

Normalized elastic curve. K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1; β0=0.7.

Figure 11

Normalized curvature. K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1; β0=0.7.

Figure 12

Normalized curvature in time at various sections: x∕λ=−6,−5,…,5,6; a0=1; β0=0.2; K=10,000

Figure 13

Normalized elastic curve, K=10,000. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1; β0=0.2.

Figure 14

Normalized curvature, K=10,000. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=1; β0=0.2.

Figure 15

Nonlinear numerical simulation. Normalized TDP excursion, x0(t)∕λ, of a typical SCR.

Figure 16

Nonlinear numerical simulation. Normalized dynamic tension, τ0(t)∕T0, at TDP of a typical SCR.

Figure 17

Curvature time histories at x∕λ=4.0, at the critical section. Analytical and numerical solutions, K=10.

Figure 18

Curvature time histories at x∕λ=−3, a section that rests on the soil cyclically. Analytical and numerical solutions, K=10.

Figure 19

Curvature time histories at x∕λ=−3. Analytical and numerical solutions, K=1000.

Figure 20

Normalized curvature. Minimum value: (a) along the length and (b) as a function of the soil rigidity parameter K.

Figure 21

WKB approximate solution compared to numerical results obtained by a standard finite element method. Free-hanging SCR. No current. θL=70deg. Assessing the extensibility effect.

Figure 22

Natural frequencies of a SCR. Assessing the extensibility effect. Numerical solution, with three different values of axial rigidity, compared to the WKB analytical approximation. No current; θL=70deg.

Figure 23

Normalized elastic curve, K=10. Snapshots for t∕T=0,0.1,0.2,0.3,0.4,0.5; T=2πω−1. a0=2.08; β0≈O(10−6).

Figure 24

Snapshots of the normalized curvature, K=10. a0=2.08; β0≈O(10−6).

Figure 25

Snapshots of the normalized shear force, K=10. a0=2.08; β0≈O(10−6).

Figure 26

Normalized curvature in time at various sections, x∕λ=−6,−5,…,5,6; a0=2.08; β0≈O(10−6); K=10

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text:
# docker-compose collisions

Here’s a description of collisions between groups of services described in similar docker-compose files, in different directories, when they are brought up on the same host with docker-compose up

I think of them as “collisions” because when using different docker-compose.yml files in different directories, I tend to think of them as different instances, even if they are of the same groups of services.

docker-compose actually identifies services using a combination of service name (declared in the file) and the project name, which defaults to the directory containing docker-compose.yml.

Consider a simple service described in a docker-compose.yml file, in service-instance-a/: (let this be the first instance that later scenarios refer to)

$cat service-instance-a/docker-compose.yml version: "3" services: nginx: image: nginx:stable-alpine$

Start that, and it should start as expected.

Now if you copy that entire directory cp -r service-instance-a/ service-instance-b/, and docker-compose up from service-instance-b/, the same service starts as expected, and both instances can be stopped independently. (Keep this instance running, it should not be affected by the scenarios below)

Now once the containing directory is the same, docker-compose starts to treat them as the same project.

Suppose you have

$cat same-instance-a/service-instance-a/docker-compose.yml version: "3" services: nginx: image: nginx:stable-alpine$

And now you cd same-instance-a/service-instance-a/ and do docker-compose up, you’ll see that it starts fine, but it’s actually a related instance to the first one (service-instance-a/docker-compose.yml). If you start both, send Ctrl-C on either one, and the other will go down as well.

If you try and fix that problem by changing the container name, that’s actually worse - docker-compose thinks you have changed a configuration of the previous instance, and promptly shuts down the first instance.

$cat different-container-name/service-instance-a/docker-compose.yml version: "3" services: nginx: image: nginx:stable-alpine container_name: different-nginx$

This also applies if you change any other configuration, like if you decided to try and fix the problem by trying to connect the second instance to another network - docker-compose also shuts down the first instance.

$cat different-network/service-instance-a/docker-compose.yml version: "3" services: nginx: image: nginx:stable-alpine networks: - foobar networks: foobar:$

Finally, if you try and rename the service, docker-compose still thinks it’s part of the same project!

$cat different-service-name/service-instance-a/docker-compose.yml version: "3" services: nginx-renamed: image: nginx:stable-alpine$

Now if you start this one, you get

\$ docker-compose up
WARNING: Found orphan containers (service-instance-a_nginx_1) for this project. If you removed or renamed this service in your compose file, you can run this command with the --remove-orphans flag to clean it up.
Creating service-instance-a_nginx-renamed_1 ... done
Attaching to service-instance-a_nginx-renamed_1

docker-compose thinks you renamed a container in the same project as the other running one, and recommends that you clean up the orphans. On the bright side, the other container is still up and running.

Here’s a repo with all the scenarios in case you want to try it out: https://github.com/ackerleytng/docker-compose-collisions

I ran into this issue while assuming that making a copy of docker-compose.yml in another directory with the same name would allow me to run two instances of the same service, allowing me to keep one service running while trying out another configuration - except I ended up bringing both services down :(


text:
Look up a non-US stock price

I need to get historical prices of certain non-US stocks (specifically, they are traded at Prague Stock Exchange, Czech Republic) but I am not able to find the proper "name" that Mathematica would accept. For example, is there a way to look up the Mathematica name for a stock if I know the ISIN code or the exchange (but probably not the exchange's Mathematica symbol) where the stock is traded?

-
FinancialData["^PSE", "Members"] indicates that stocks on the Prague Stock Exchange are not supported. You can try using Yahoo Finance instead. – Pickett Jun 10 '14 at 9:10


text:
Chapter 37

### INTRODUCTION

The finite life span of most mature blood cells requires their continuous replacement, a process termed hematopoiesis. New cell production must respond to basal needs and states of increased demand. Red blood cell production can increase >20-fold in response to anemia or hypoxemia, white blood cell production increases dramatically in response to a systemic infection, and platelet production can increase 10- to 20-fold when platelet consumption results in thrombocytopenia.

The regulation of blood cell production is complex. Hematopoietic stem cells are rare bone marrow cells that manifest self-renewal and lineage commitment, resulting in cells destined to differentiate into the nine distinct blood-cell lineages. For the most part, this process occurs in the marrow cavities of the skull, vertebral bodies, pelvis, and proximal long bones; it involves interactions among hematopoietic stem and progenitor cells and the cells and complex macromolecules of the marrow stroma, and is influenced by a number of soluble and membrane-bound hematopoietic growth factors. A number of these hormones and cytokines have been identified and cloned, permitting their production in quantities sufficient for therapeutic use. Clinical applications range from the treatment of primary hematologic diseases to use as adjuncts in the treatment of severe infections and in the management of patients who are undergoing cancer chemotherapy or marrow transplantation.

Hematopoiesis also requires an adequate supply of minerals (e.g., iron, cobalt, and copper) and vitamins (e.g., folic acid, vitamin B12, pyridoxine, ascorbic acid, and riboflavin); deficiencies generally result in characteristic anemias, or, less frequently, a general failure of hematopoiesis (Hoffbrand and Herbert, 1999; Wrighting and Andrews, 2008). Therapeutic correction of a specific deficiency state depends on the accurate diagnosis of the anemic state, knowledge about the correct dose, the use of these agents in various combinations, and the expected response. This chapter deals with the growth factors, vitamins, minerals, and drugs that affect the blood and blood-forming organs.

### HEMATOPOIETIC GROWTH FACTORS

History. Modern concepts of hematopoietic cell growth and differentiation arose in the 1950s when cells from the spleen and marrow were shown to play an important role in the restoration of hematopoietic tissue in irradiated animals. In 1961, Till and McCulloch demonstrated that individual hematopoietic cells could form macroscopic hematopoietic colonies in the spleens of irradiated mice. Their work established the concept of discrete hematopoietic stem cells, which can be experimentally identified, albeit in retrospect (i.e., the presence of a multilineage clonal splenic colony appearing 11 days after transplantation implied that a single cell lodged and expanded into several cell lineages). This concept now has been expanded to include normal human marrow cells. Moreover, such cells now can be prospectively identified.

The basis for identifying soluble growth factors was provided by Sachs and independently by Metcalf, who developed clonal, in vitro assays for hematopoietic progenitor cells. Initially, such hematopoietic colonies developed only in the presence of conditioned culture medium from leukocytes or tumor ...

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text:
Mangoes and Blueberries

• Published in 1999
In the collections
We prove the following conjecture of Erdős and Hajnal: For every integer $k$ there is an $f(k)$ such that if for a graph $G$, every subgraph $H$ of $G$ has a stable set containing vertices, then $G$ contains a set $X$ of at most $f(k)$ vertices such that $G−X$ is bipartite. This conjecture was related to me by Paul Erdős at a conference held in Annecy during July of 1996. I regret not being able to share the answer with him.

Other information

key
MangoesandBlueberries
type
article
2020-09-21
date_published
1999-11-14
publisher
Bolyai Society – Springer-Verlag
fulltext_html_url
journal
Combinatorica
issn
1439-6912
volume
19
issue
2
identifier
doi:10.1007/s004930050056
doi
10.1007/s004930050056
pages
267-296

BibTeX entry

@article{MangoesandBlueberries,
key = {MangoesandBlueberries},
type = {article},
title = {Mangoes and Blueberries},
author = {Bruce Reed},
abstract = {We prove the following conjecture of Erd{\H{o}}s and Hajnal:

For every integer $k$ there is an $f(k)$ such that if for a graph $G$, every subgraph $H$ of $G$ has a stable set containing vertices, then $G$ contains a set $X$ of at most $f(k)$ vertices such that $G−X$ is bipartite.

This conjecture was related to me by Paul Erd{\H{o}}s at a conference held in Annecy during July of 1996. I regret not being able to share the answer with him.},
comment = {},
date_published = {1999-11-14},
collections = {attention-grabbing-titles,food},
}


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Solving Systems by Multiplying Both Equations

## Multiply both equations and add to cancel one variable

Estimated10 minsto complete
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Practice Solving Systems by Multiplying Both Equations
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Solving Systems by Multiplying Both Equations to Cancel a Variable

A one pound mix consisting of 30% cashews and 70% pistachios sells for $6.25. A one pound mix consisting of 80% cashews and 20% pistachios sells for$7.50. How would a mix consisting of 50% of each type of nut sell for?

### Solving Systems by Multiplying Both Equations

In the linear systems in this lesson, we will need to multiply both equations by a constant in order to have opposite coefficients of one of the variables. In order to determine what numbers to multiply by, we will be finding the least common multiple of the given coefficients. Recall that the least common multiple of two numbers is the smallest number which is divisible by both of the given numbers. For example, 12 is the least common multiple of 4 and 6 because it is the smallest number divisible by both 4 and 6.

Let's solve the following systems using linear combinations.

1.

\begin{align*}2x-5y &= 15\\ 3x+7y &= 8\end{align*}

In this problem we cannot simply multiply one equation by a constant to get opposite coefficients for one of the variables. Here we will need to identify the least common multiple of the coefficients of one of the variables and use this value to determine what to multiple each equation by. If we look at the coefficients of \begin{align*}x\end{align*}, 2 and 3, the least common multiple of these numbers is 6. So, we want to have the coefficients of \begin{align*}x\end{align*} be 6 and -6 so that they are opposites and will cancel when we add the two equations together. In order to get coefficients of 6 and -6 we can multiply the first equation by 3 and the second equation by -2 (it doesn’t matter which one we make negative.)

\begin{align*}&\quad \ 3(2x-5y=15) \quad \Rightarrow \qquad \ \cancel{6x}-15y = 45\\ & -2(3x+7y=8) \qquad \quad \ + \ \underline{- \cancel{6x}-14y = -16 \; \; \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad -29y=29\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad y=-1\end{align*}

Now find \begin{align*}x\end{align*}:

\begin{align*}2(x)-5(-1) &= 15\\ 2x+5 &= 15\\ 2x &= 10 \\ x &= 5 \end{align*}

Solution: (5, -1)

\begin{align*}2(5)-5(-1) &= 10 + 5 =15\\ 3(5)+7(-1) &= 15-7=8 \end{align*}

\begin{align*}^*\end{align*} This problem could also be solved by eliminating the \begin{align*}y\end{align*} variables first. To do this, find the least common multiple of the coefficients of \begin{align*}y\end{align*}, 5 and 7. The least common multiple is 35, so we would multiply the first equation by 7 and the second equation by 5. Since one of them is already negative, we don’t have to multiply by a negative.

1.

\begin{align*}7x+20y &= -9\\ -2x-3y &= 8\end{align*}

The first step is to decide which variable to eliminate. Either one can be eliminated but sometimes it is helpful to look at what we need to multiply by to eliminate each one and determine which is easier to eliminate. In general, it is easier to work with smaller numbers so in this case it makes sense to eliminate \begin{align*}x\end{align*} first. The Least Common Multiple (LCM) of 7 and 2 is 14. To get 14 as the coefficient of each term, we need to multiple the first equation by 2 and the second equation by 7:

\begin{align*}& 2(7x+20y=-9) \quad \Rightarrow \quad \ \ \cancel{14x}+40y= -18 \\ & 7(-2x-3y=8) \qquad \qquad \ \underline{- \cancel{14x}-21y=56 \; \; \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 19y = 38\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ y=2\end{align*}

Now find \begin{align*}x\end{align*}:

\begin{align*}-2x-3(2) & = 8\\ -2x-6 &= 8\\ -2x &= 14\\ x &= -7\end{align*}

Solution: (-7, 2)

\begin{align*}7(-7)+20(2) &= -49 + 40 = -9\\ -2(-7)-3(2) & =14-6=8 \end{align*}

1.

\begin{align*}14x-6y &= -3\\ 16x-9y &= -7\end{align*}

This time, we will eliminate \begin{align*}y\end{align*}. We need to find the LCM of 6 and 9. The LCM is 18, so we will multiply the first equation by 3 and the second equation by -2. Again, it doesn’t matter which equation we multiply by a negative value.

\begin{align*}& \quad 3(14x-6y=-3) \quad \Rightarrow \quad \ + 42x - \cancel{18y} = -9\\ & -2(16x-9y=-7) \qquad \quad \ \ \underline{-32x+ \cancel{18y} = 14 \; \; \; \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 10x = 5\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x=2\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*} 14 \left(\frac{1}{2} \right)-6y &= -3\\ 7-6y &= -3\\ -6y &= -10\\ y & = \frac{10}{6} = \frac{5}{3}\end{align*}

Solution: \begin{align*}\left(\frac{1}{2}, \frac{5}{3}\right)\end{align*}

\begin{align*}14 \left(\frac{1}{2}\right) - 6 \left(\frac{5}{3}\right) &= 7-10=-3\\ 16 \left(\frac{1}{2}\right) - 9 \left(\frac{5}{3}\right) &= 8-15=-7 \end{align*}

### Examples

#### Example 1

Earlier, you were asked to find how much a mix consisting of 50% of each type of nut would sell for.

Write a system of linear equations to represent the given information. Let \begin{align*}x=\end{align*} the cost of the cashews per pound and let \begin{align*}y=\end{align*} the cost of the pistachios per pound. Now we can write two equations to represent the two different mixes of nuts:

\begin{align*}0.3x+0.7y &= 6.25\\ 0.8x+0.2y &= 7.50\end{align*}

Solve this system to determine the cost of each type of nut per pound. If we eliminate \begin{align*}y\end{align*}, we will need to multiple the first equation by 2 and the second equation by -7:

\begin{align*}& \quad 2(0.3x-0.7y=6.25) \quad \Rightarrow \quad +0.6x - \cancel{1.4y} = 12.5\\ & -7(0.8x+0.2y=7.50) \qquad \quad \ \underline{-5.6x+ \cancel{1.4y} = -52.5 \; \; \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ -5x = -40\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x=8\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*}0.3(8)+0.7y &= 6.25\\ 2.4+0.7y &= 6.25\\ 0.7y &= 3.85\\ y &= 5.5\end{align*}

So, we have determined that the cost of the cashews is $8 per pound and the cost of the pistachios is$5.50 per pound. Now we can determine the cost of the 50% mix as follows:

\begin{align*}0.5(8.00)+0.5(5.50)=4.00+2.75=6.75\end{align*} So, the new mix is 6.75 per pound. Solve the following systems using linear combinations. #### Example 2\begin{align*}6x+5y &= 3\\ -4x-2y &= -14\end{align*} We can eliminate either variable here. To eliminate \begin{align*}x\end{align*}, we can multiple the first equations by 2 and the second equation by 3 to get 12 - the LCM of 6 and 4. \begin{align*}& \ \ 2(6x+5y=3) \quad \ \Rightarrow \quad \ +\cancel{12x} +10y = 6\\ & 3(-4x-2y=-14) \qquad \quad \quad \underline{ - \cancel{12x}-6y = -42 \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 4y = -36\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ y=-9\end{align*} Now find \begin{align*}x\end{align*}: \begin{align*}6x+5(-9) &= 3\\ 6x-45 &= 3\\ 6x &= 48\\ x &= 8\end{align*} Solution: (8, -9) #### Example 3 \begin{align*}9x-7y &= -19\\ 5x-3y &= -15\end{align*} Again we can eliminate either variable. To eliminate \begin{align*}y\end{align*}, we can multiply the first equation by 3 and the second equation by -7: \begin{align*} & \quad \ 3(9x-7y=-19) \quad \Rightarrow \ \ \quad +27x - \cancel{21y} = -57\\ & -7(5x-3y=-15) \qquad \quad + \ \underline{-35x+\cancel{21y} = 105 \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ -8x = 48\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x=-6\end{align*} Now find \begin{align*}y\end{align*}: \begin{align*}5(-6)-3y &= -15\\ -30-3y &= -15\\ -3y &= 15\\ y &= -5\end{align*} Solution: (-6, -5) #### Example 4\begin{align*}15x-21y &= -63\\ 7y &= 5x+21\end{align*} To start this one we need to get the second equation in standard form. The resulting system will be: \begin{align*}15x-21y &= -63\\ -5x+7y &= 21\end{align*} This time we just need to multiply the second equation by 3 to eliminate \begin{align*}x\end{align*}: \begin{align*}& \ \ 15x-21y=-63 \quad \Rightarrow \quad + \cancel{15x} - 21y = -63\\ & 3(-5x+7y=21) \qquad \ + \ \underline{- \cancel{15x}+21y = -63 \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 0y = 0\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0=0 \end{align*} Solution: There are infinite solutions. ### Review Solve the systems using linear combinations. 1. \begin{align*}17x-5y &= 4\\ 2x+7y &= 46\end{align*} 2. \begin{align*}9x+2y &= -13\\ 11x+5y &= 2\end{align*} 3. \begin{align*}3x+4y &= -16\\ 5x+5y &= -5\end{align*} 4. \begin{align*}2x-8y &= -8\\ 3x+7y &= 45\end{align*} 5. \begin{align*}5x-10y &= 60\\ 6x+3y &= -33\end{align*} 6. \begin{align*}3x+10y &= -50\\ -5x-7y &= 6\end{align*} 7. \begin{align*}11x+6y &= 30\\ 13x-5y &= -25\end{align*} 8. \begin{align*}15x+2y &= 23\\ 18x-9y &= -18\end{align*} 9. \begin{align*}12x+8y &= 64\\ 17x-12y &= 9\end{align*} 10. \begin{align*}11x-3y &= 12\\ 33x-36 &= 9y\end{align*} 11. \begin{align*}4x+3y &= 0\\ 6x-13y &= 35\end{align*} 12. \begin{align*}18x+2y &= -2\\ -12x-3y &= -7\end{align*} 13. \begin{align*}-6x+11y &= -109\\ 8x-15y &= 149\end{align*} 14. \begin{align*}8x &= -5y-1\\ -32x+20y &= 8\end{align*} 15. \begin{align*}10x-16y &= -12\\ -15x+14y &= -27\end{align*} Set up and solve a system of equations to answer the following questions. 1. A mix of 35% almonds and 65% peanuts sells for5.70. A mix of 75% almonds and 25% peanuts sells for $6.50. How much should a mix of 60% almonds and 40% peanuts sell for? 2. The Robinson family pays$19.75 at the movie theater for 3 medium popcorns and 4 medium drinks. The Jamison family pays $33.50 at the same theater for 5 medium popcorns and 7 medium drinks. How much would it cost for a couple to get 2 medium drinks and 2 medium popcorns? 3. A cell phone company charges extra when users exceed their included call time and text message limits. One user paid$3.24 extra having talked for 240 extra minutes and sending 12 additional texts. A second user talked for 120 extra minutes and sent 150 additional texts and was charged \$4.50 above the regular fee. How much extra would a user be charged for talking 140 extra minutes and sending 200 additional texts?

To see the Review answers, open this PDF file and look for section 3.8.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes


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## Probability and a Maximum

A large sock drawer contains red and blue socks. The total number of socks is at most 2012. It just so happens that if two socks are chosen at random without replacement from the drawer, the probability is 0.5 that both are blue. What is the maximum number of blue socks in the drawer? Be clear in your explanation and solution.


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Difference between revisions of "Ptolemy's Theorem"

Ptolemy's theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of the Ptolemy inequality. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures.

Definition

Given a cyclic quadrilateral $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$:

$ac+bd=ef$.

Proof

Given cyclic quadrilateral $\displaystyle ABCD,$ extend $\displaystyle CD$ to $\displaystyle P$ such that $\angle BAC=\angle DAP.$

Since quadrilateral $\displaystyle ABCD,$ is cyclic, $\displaystyle m\angle ABC+m\angle ADC=180^\circ .$ However, $\displaystyle \angle ADP$ is also supplementary to $\angle ADC,$ so $\displaystyle \angle ADP=\angle ABC$ Hence, $\displaystyle \triangle ABC \sim \triangle ADP$ by AA similarity and $\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.$

Now, note that $\angle ABD=\angle ACD$ (subtend the same arc) and $\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\angle CAP,$ so $\triangle BAD\sim \triangle CAP.$ This yields $\frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.$

However, $\displaystyle CP= CD+DP.$ Substituting in our expressions for $\displaystyle CP$ and $\displaystyle DP,$ $\frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.$ Multiplying by $\displaystyle AB$ yields $\displaystyle (AC)(BD)=(AB)(CD)+(AD)(BC)$

--4everwise 14:09, 22 June 2006 (EDT)

Example

In a regular heptagon ABCDEFG, prove that: 1/AB = 1/AC + 1/AD.

Solution: Let ABCDEFG be the regular heptagon. Consider the quadrilateral ABCE. If a, b, and c represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of ABCE are a, a, b and c; and the diagonals of ABCE are b and c, respectively.

Now Ptolemy's theorem states that ab + ac = bc, which is equivalent to 1/a=1/b+1/c.


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# Karnataka 1st PUC Maths Important Questions

For a student of Karnataka Board 1st Year PUC class, mathematics can be a challenging subject as it deals with a wide range of topics from sets, relations and functions to complex numbers and quadratic equations, sequences and series, mathematical reasoning and so on. If a student needs to get a good foundation in class 11 mathematics they would need to do well in the classes and for the exams. To prepare for these exams, students can access the Karnataka 1st PUC Maths Important Questions get practice. This will be a good supplementary study material for the students apart from their Karnataka Board 1st PUC Question Papers, 1st PUC textbooks and so on.

Students can also get an overview of the list of important questions here:

Karnataka Board 1st PUC Maths Important Questions List

1. If $\frac{a}{b}, \frac{c}{d}, \frac{e}{f}$ are the proper fractions, then LCM is

given by ___________

1. Three vessels can hold 9, 15, 24 litres of water. Find the least quantity of water which can be filled by these vessels an exact number of times.
2. Find the H.C.F. of 165, 225 and 435

4. The cost of a chair is 600 and the cost of a table is 900. Find the least sum of money that a person must possess in order to purchase a whole number of chairs or tables.

5. Express each of the following rational numbers as decimals:

1. $\frac{3}{4}$ (b) $\frac{15}{8}$ (c) $\frac{8}{125}$

6. Without actually performing the long division state whether the following rational numbers will have a terminating decimal expansion or non- terminating repeating decimal expansion.

1. $\frac{17}{3125}$ (b) (b) $\frac{13}{8}$

7. Prove that 2 + 3 $\sqrt{5}$ is an irrational number

8. Represent the following sets in both Roster form and Rule method:

(a) Set of factors of 20

(b) Set of all prime Numbers less than 10

9. Out of 250 people, 160 drink coffee, 90 drink tea, 85 drink milk, 45 drink

coffee and tea, 35 drink tea and milk, 20 drink all the three. How many will

drink Coffee and Milk?.

10. If A = {5, 6, 7}, Find P (A)

11. If (2x, x + y) = (8, 4). Find x and y

12. R is a relation on the set of natural numbers N standing for x is

related to y. if x = 3y, Find R

13. Find the domain and range of the function $F(x)\frac{x^{2}-2x+1}{x^{2}-9x+13}$, where $x\varepsilon N$

14. Simplify $2(3^{-2})+\frac{1}{3}^{-3}+3^{2}$

15. Prove that 2log$\frac{3}{7}$ + log$\frac{49}{9}$=0

16. If log 5 = 0.6990, find the number of digits in the integral part of $5^{23}$

17. Find the 11th term of the A.P. 3, 5, 7, 9 – – – –

18. Imrez buys a used car for 1,50,000 he pays 1,00,000 Cash and agrees to pay the balance in annual installments of `5,000 plus 8% interest on the unpaid amount. How much will the Car Cost for him?

19. If $n(\cup)$=700, $n(A)$=200, $n(B)$ =300 and $n(A\cap B)$=100, then find $n(A’\cap B’)$

20. Prove that $\frac{1}{1+cos A} +\frac{1}{1-cos A}= 2cosec^{2}A$

### Why Solve Karnataka 1st PUC Maths Important Questions?

These important questions covers almost all the important topics that come under the class 11 Maths syllabus. Hence, it also help students to learn a subject properly. Other benefits are given here:

• Help students to get proper information and expectation about the exam pattern
• Help them to self analyze their failings and knowledge gap of the subject
• If the find any question more tough, they can spend more time on that topic
• It is a good tool for revision
• Get practice doing mathematical problems

Students can also come back to BYJU’S for more such interesting resources and study material like question papers or textbooks of Karnataka 1st PUC.

#### Practise This Question

Identify the terms of expression 4pq + p - q - r.


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# Beaufort scale

Last updated

The Beaufort scale is an empirical measure that relates wind speed to observed conditions at sea or on land. Its full name is the Beaufort wind force scale.

## History

The scale was devised in 1805 by the Irish hydrographer Francis Beaufort (later Rear Admiral), a Royal Navy officer, while serving on HMS Woolwich. The scale that carries Beaufort's name had a long and complex evolution from the previous work of others (including Daniel Defoe the century before) to when Beaufort was Hydrographer of the Navy in the 1830s, when it was adopted officially and first used during the voyage of HMS Beagle under Captain Robert FitzRoy, who was later to set up the first Meteorological Office (Met Office) in Britain giving regular weather forecasts. [1] In the 18th century, naval officers made regular weather observations, but there was no standard scale and so they could be very subjective – one man's "stiff breeze" might be another's "soft breeze". Beaufort succeeded in standardising the scale.

The initial scale of 13 classes (zero to 12) did not reference wind speed numbers but related qualitative wind conditions to effects on the sails of a frigate, then the main ship of the Royal Navy, from "just sufficient to give steerage" to "that which no canvas sails could withstand". [2]

The scale was made a standard for ship's log entries on Royal Navy vessels in the late 1830s and was adapted to non-naval use from the 1850s, with scale numbers corresponding to cup anemometer rotations. In 1853, the Beaufort scale was accepted as generally applicable at the First International Meteorological Conference in Brussels. [3]

In 1916, to accommodate the growth of steam power, the descriptions were changed to how the sea, not the sails, behaved and extended to land observations. Rotations to scale numbers were standardised only in 1923. George Simpson, CBE (later Sir George Simpson), director of the UK Meteorological Office, was responsible for this and for the addition of the land-based descriptors. [1] The measures were slightly altered some decades later to improve its utility for meteorologists. Nowadays, meteorologists typically express wind speed in kilometres or miles per hour or, for maritime and aviation purposes, knots; but Beaufort scale terminology is still sometimes used in weather forecasts for shipping [4] and the severe weather warnings given to the public. [5]

### Extended scale

The Beaufort scale was extended in 1946 when forces 13 to 17 were added. [3] However, forces 13 to 17 were intended to apply only to special cases, such as tropical cyclones. Nowadays, the extended scale is only used in Taiwan and mainland China, which are often affected by typhoons. Internationally, WMO Manual on Marine Meteorological Services (2012 edition) defined the Beaufort Scale only up to force 12 and there was no recommendation on the use of the extended scale. [6]

Wind speed on the 1946 Beaufort scale is based on the empirical relationship: [7]

• v = 0.836 B3/2 m/s
• v = 1.625 B3/2 knots (${\displaystyle ={\frac {13}{8}}{\sqrt {B^{3}}}}$)

where v is the equivalent wind speed at 10 metres above the sea surface and B is Beaufort scale number. For example, B = 9.5 is related to 24.5 m/s which is equal to the lower limit of "10 Beaufort". Using this formula the highest winds in hurricanes would be 23 in the scale. F1 tornadoes on the Fujita scale and T2 TORRO scale also begin roughly at the end of level 12 of the Beaufort scale, but are independent scales – although the TORRO scale wind values are based on the 3/2 power law relating wind velocity to Beaufort force. [8]

Wave heights in the scale are for conditions in the open ocean, not along the shore.

## Modern scale

Beaufort scale [9] [10] [11] [12]
Beaufort
number
DescriptionWind speedWave
height
Sea conditionsLand conditionsSea conditions
(photo)
Associated
warning flag
0Calm< 1  knot
< 1  mph
< 2  km/h
< 0.5  m/s
0 ft
0 m
Sea like a mirrorSmoke rises vertically
1Light air1–3 knots
1–3 mph
2–5 km/h
0.5–1.5 m/s
0–1 ft
0–0.3 m
Ripples with appearance of scales are formed, without foam crestsDirection shown by smoke drift but not by wind vanes
2Light breeze4–6 knots
4–7 mph
6–11 km/h
1.6–3.3 m/s
1–2 ft
0.3–0.6 m
Small wavelets still short but more pronounced; crests have a glassy appearance but do not breakWind felt on face; leaves rustle; wind vane moved by wind
3Gentle breeze7–10 knots
8–12 mph
12–19 km/h
3.4–5.5 m/s
2–4 ft
0.6–1.2 m
Large wavelets; crests begin to break; foam of glassy appearance; perhaps scattered white horsesLeaves and small twigs in constant motion; light flags extended
4Moderate breeze11–16 knots
13–18 mph
20–28 km/h
5.5–7.9 m/s
3.5–6 ft
1–2 m
Small waves becoming longer; fairly frequent white horsesRaises dust and loose paper; small branches moved
5Fresh breeze17–21 knots
19–24 mph
29–38 km/h
8–10.7 m/s
6–10 ft
2–3 m
Moderate waves taking a more pronounced long form; many white horses are formed; chance of some spraySmall trees in leaf begin to sway; crested wavelets form on inland waters
6Strong breeze22–27 knots
25–31 mph
39–49 km/h
10.8–13.8 m/s
9–13 ft
3–4 m
Large waves begin to form; the white foam crests are more extensive everywhere; probably some sprayLarge branches in motion; whistling heard in telegraph wires; umbrellas used with difficulty
7High wind,
moderate gale,
near gale
28–33 knots
32–38 mph
50–61 km/h
13.9–17.1 m/s
13–19 ft
4–5.5 m
Sea heaps up and white foam from breaking waves begins to be blown in streaks along the direction of the wind; spindrift begins to be seenWhole trees in motion; inconvenience felt when walking against the wind
8 Gale,
fresh gale
34–40 knots
39–46 mph
62–74 km/h
17.2–20.7 m/s
18–25 ft
5.5–7.5 m
Moderately high waves of greater length; edges of crests break into spindrift; foam is blown in well-marked streaks along the direction of the windTwigs break off trees; generally impedes progress
9Strong/severe gale41–47 knots
47–54 mph
75–88 km/h
20.8–24.4 m/s
23–32 ft
7–10 m
High waves; dense streaks of foam along the direction of the wind; sea begins to roll; spray affects visibilitySlight structural damage (chimney pots and slates removed)
10 Storm, [13]
whole gale
48–55 knots
55–63 mph
89–102 km/h
24.5–28.4 m/s
29–41 ft
9–12.5 m
Very high waves with long overhanging crests; resulting foam in great patches is blown in dense white streaks along the direction of the wind; on the whole the surface of the sea takes on a white appearance; rolling of the sea becomes heavy; visibility affectedSeldom experienced inland; trees uprooted; considerable structural damage
11Violent storm56–63 knots
64–72 mph
103–117 km/h
28.5–32.6 m/s
37–52 ft
11.5–16 m
Exceptionally high waves; small- and medium-sized ships might be for a long time lost to view behind the waves; sea is covered with long white patches of foam; everywhere the edges of the wave crests are blown into foam; visibility affectedVery rarely experienced; accompanied by widespread damage
12 Hurricane-force [13] ≥ 64 knots
≥ 73 mph
≥ 118 km/h
≥ 32.7 m/s
≥ 46 ft
≥ 14 m
The air is filled with foam and spray; sea is completely white with driving spray; visibility very seriously affectedDevastation

The Beaufort scale is not an exact nor an objective scale; it was based on visual and subjective observation of a ship and of the sea. The corresponding integral wind speeds were determined later, but the values in different units were never made equivalent.

The scale is used in the Shipping Forecasts broadcast on BBC Radio 4 in the United Kingdom, and in the Sea Area Forecast from Met Éireann, the Irish Meteorological Service. Met Éireann issues a "Small Craft Warning" if winds of Beaufort force 6 (mean wind speed exceeding 22 knots) are expected up to 10 nautical miles offshore. Other warnings are issued by Met Éireann for Irish coastal waters, which are regarded as extending 30 miles out from the coastline, and the Irish Sea or part thereof: "Gale Warnings" are issued if winds of Beaufort force 8 are expected; "Strong Gale Warnings" are issued if winds of Beaufort force 9 or frequent gusts of at least 52 knots are expected.; "Storm Force Warnings" are issued if Beaufort force 10 or frequent gusts of at least 61 knots are expected; "Violent Storm Force Warnings" are issued if Beaufort force 11 or frequent gusts of at least 69 knots are expected; "Hurricane Force Warnings" are issued if winds of greater than 64 knots are expected.

This scale is also widely used in the Netherlands, Germany, [14] Greece, China, Taiwan, Hong Kong, Malta, and Macau, although with some differences between them. Taiwan uses the Beaufort scale with the extension to 17 noted above. China also switched to this extended version without prior notice on the morning of 15 May 2006, [15] and the extended scale was immediately put to use for Typhoon Chanchu. Hong Kong and Macau retain force 12 as the maximum.

In the United States of America, winds of force 6 or 7 result in the issuance of a small craft advisory, with force 8 or 9 winds bringing about a gale warning, force 10 or 11 a storm warning ("a tropical storm warning" being issued instead of the latter two if the winds relate to a tropical cyclone), and force 12 a hurricane-force wind warning (or hurricane warning if related to a tropical cyclone). A set of red warning flags (daylight) and red warning lights (night time) is displayed at shore establishments which coincide with the various levels of warning.[ citation needed ]

In Canada, maritime winds forecast to be in the range of 6 to 7 are designated as "strong"; 8 to 9 "gale force"; 10 to 11 "storm force"; 12 "hurricane force". Appropriate wind warnings are issued by Environment Canada's Meteorological Service of Canada: strong wind warning, gale (force wind) warning, storm (force wind) warning and hurricane-force wind warning. These designations were standardised nationally in 2008, whereas "light wind" can refer to 0 to 12 or 0 to 15 knots and "moderate wind" 12 to 19 or 16 to 19 knots, depending on regional custom, definition or practice. Prior to 2008, a "strong wind warning" would have been referred to as a "small craft warning" by Environment Canada, similar to US terminology. (Canada and the USA have the Great Lakes in common.)[ citation needed ]

## Weather scale

Beaufort's name was also attached to the Beaufort scale for weather reporting:

SymbolInterpretation
bblue sky
cdetached clouds
ddrizzling rain
ffog
gdark, gloomy
hhail
llightning
mmisty (hazy)
oovercast
ppassing showers
qsqually
rrain
ssnow
tthunder
uugly (threatening)
vvisibility (unusual transparency)
wwet, dew

In this scale the weather could be reported as "s.c." for snow and detached cloud or "g.r.q." for dark, rain and squally. [16]

## Related Research Articles

In meteorology, wind speed, or wind flow speed, is a fundamental atmospheric quantity caused by air moving from high to low pressure, usually due to changes in temperature. Wind speed is now commonly measured with an anemometer.

A gale is a strong wind; the word is typically used as a descriptor in nautical contexts. The U.S. National Weather Service defines a gale as sustained surface winds moving at a speed of between 34 and 47 knots. Forecasters typically issue gale warnings when winds of this strength are expected. In the United States, a gale warning is specifically a maritime warning; the land-based equivalent in National Weather Service warning products is a wind advisory.

A small craft advisory is a type of wind warning issued by the National Weather Service in the United States. In Canada a similar warning is issued by Environment Canada. It is issued when winds have reached, or are expected to reach within 12 hours, a speed marginally less than gale force. A Small Craft Advisory may also be issued when sea or lake ice exists that could be hazardous to small boats.

The great storm of 1987 was a violent extratropical cyclone that occurred on the night of 15–16 October, with hurricane-force winds causing casualties in the United Kingdom, France, and the Channel Islands as a severe depression in the Bay of Biscay moved northeast. Among the most damaged areas were Greater London, the East Anglian coast, the Home Counties, the west of Brittany, and the Cotentin Peninsula of Normandy, all of which weathered gusts typically with a return period of 1 in 200 years.

Tropical cyclone warnings and watches are alerts issued by national weather forecasting bodies to coastal areas threatened by the imminent approach of a tropical cyclone of tropical storm or hurricane intensity. They are notices to the local population and civil authorities to make appropriate preparation for the cyclone, including evacuation of vulnerable areas where necessary. It is important that interests throughout the area of an alert make preparations to protect life and property, and do not disregard it on the strength of the detailed forecast track.

A gale warning is an alert issued by national weather forecasting agencies around the world in an event that maritime locations currently or imminently experiencing winds of gale force on the Beaufort scale. Gale warnings allow mariners to take precautionary actions to ensure their safety at sea or to seek safe anchorage and ride out the storm on land. Though usually associated with deep low-pressure areas, winds strong enough to catalyze a gale warning can occur in other conditions too, including from anticyclones, or high-pressure systems, in the continental interior. The winds are not directly associated with a tropical cyclone.

This article describes severe weather terminology used by the National Weather Service (NWS) in the United States. The NWS, a government agency operating as an arm of the National Oceanic and Atmospheric Administration (NOAA) branch of the United States Department of Commerce (DoC), defines precise meanings for nearly all of its weather terms.

The Hong Kong tropical cyclone warning signals, or informally typhoon signals, are a set of signals used to indicate the threat or effects of a tropical cyclone. The Hong Kong Observatory issues the warning signal if a tropical cyclone approaches within 800 kilometres of Hong Kong and poses a threat of deteriorating conditions in Hong Kong.

Tropical cyclones are ranked on one of five tropical cyclone intensity scales, according to their maximum sustained winds and which tropical cyclone basins they are located in. Only a few scales of classifications are used officially by the meteorological agencies monitoring the tropical cyclones, but other scales also exist, such as accumulated cyclone energy, the Power Dissipation Index, the Integrated Kinetic Energy Index, and the Hurricane Severity Index.

Cyclone Percy was the seventh named storm of the 2004–05 South Pacific cyclone season and the fourth and final severe tropical cyclone to form during the 2004–05 South Pacific cyclone season. Cyclone Percy originated as a tropical disturbance on February 23. Over the course of the next few days, the system organized while moving east-southeastward, before intensifying into a Category 1 tropical cyclone on the Australian region scale on February 26. The system quickly intensified, reaching Category 4 status later that day. On the next day, Percy was steered southward by a blocking ridge of high pressure, while stretched out the structure of the storm into an elliptical shape, weakening it back to Category 3 status. Afterward, the storm rapidly reintensified, reaching its peak intensity as a Category 5 tropical cyclone on March 2. Afterward, Percy encountered increasing wind shear and weakened once again, turning southeastward on the next day. On March 5, Percy transitioned into an extratropical storm, before dissipating soon afterward.

This article describes severe weather terminology used by the Meteorological Service of Canada, a branch within Environment and Climate Change Canada. The article primarily describes various weather warnings, and their criteria. Related weather scales and general weather terms are also addressed in this article. Some terms are specific to certain regions.

The 2005–06 Australian region cyclone season was an above average tropical cyclone season. It began on 1 November 2005 and ended on 30 April 2006. The regional tropical cyclone operational plan also defines a tropical cyclone year separately from a tropical cyclone season, which runs from 1 July 2005 to 30 June 2006.

The 1924 Atlantic hurricane season featured the earliest known Category 5 hurricane – a tropical cyclone with maximum sustained winds exceeding 155 mph (250 km/h). The first system, Tropical Storm One, was first detected in the northwestern Caribbean Sea on June 18. The final system, an unnumbered tropical depression, dissipated on November 24. These dates fall within the period with the most tropical cyclone activity in the Atlantic. Of the 13 tropical cyclones of the season, six existed simultaneously. The season was average with 11 tropical storms, five of which strengthened into hurricanes. Further, two of those five intensified into major hurricanes, which are Category 3 or higher on the Saffir–Simpson hurricane wind scale.

The following is a glossary of tropical cyclone terms.

Hurricane Katia was a fairly intense Cape Verde hurricane that had substantial impact across Europe as a post-tropical cyclone. The eleventh named storm, second hurricane, and second major hurricane of the active 2011 Atlantic hurricane season, Katia originated as a tropical depression from a tropical wave over the eastern Atlantic on August 29. It intensified into a tropical storm the following day and further developed into a hurricane by September 1, although unfavorable atmospheric conditions hindered strengthening thereafter. As the storm began to recurve over the western Atlantic, a more hospitable regime allowed Katia to become a major hurricane by September 5 and peak as a Category 4 hurricane with winds of 140 mph (220 km/h) that afternoon. Internal core processes, increased wind shear, an impinging cold front, and increasingly cool ocean temperatures all prompted the cyclone to weaken almost immediately after peak, and Katia ultimately transitioned into an extratropical cyclone on September 10.

Severe Tropical Cyclone Ofa was a powerful tropical cyclone that caused severe damage in Polynesia in February of 1990. The system was first noted on January 27, 1990, near Tuvalu, as a shallow tropical depression that had developed within the South Pacific Convergence Zone. The cloud pattern slowly organized, and on January 31, while located east of Tuvalu, Ofa attained cyclone intensity. Moving slowly southeast, Ofa developed storm-force winds. It attained hurricane-force winds on February 2. Cyclone Ofa reached peak intensity on February 4. Shortly after, its peak Ofa began to weaken over a less favourable environment. Ofa was declared an extratropical cyclone on February 8, though the system was still tracked by meteorologists until February 10.

The following is a glossary of tornado terms. It includes scientific as well as selected informal terminology.

The following is a list of tropical cyclones by year. Since the year 957, there have been at least 12,791 recorded tropical or subtropical cyclones in the Atlantic, Pacific, and Indian Oceans, which are known as basins. Collectively, tropical cyclones caused more than US\$1.2 trillion in damage, unadjusted for inflation, and have killed more than 2.6 million people. Most of these deaths were caused by a few deadly cyclones, including the 1737 Calcutta cyclone, the 1839 Coringa cyclone, the 1931 Shanghai typhoon, the 1970 Bhola cyclone, Typhoon Nina in 1975, the 1991 Bangladesh cyclone, and Cyclone Nargis in 2008.

The Tropical Cyclone Wind Signals are tropical cyclone alert levels issued by the Philippine Atmospheric, Geophysical, and Astronomical Services Administration (PAGASA) to areas within the Philippines that may be affected by tropical cyclone winds and their associated hazards.

## References

• Huler, Scott (2004). . Crown. ISBN   1-4000-4884-2.
1. "National Meteorological Library and Archive Fact sheet 6 – The Beaufort Scale" (PDF). Met Office. Archived from the original (PDF) on 2 October 2012. Retrieved 13 May 2011.
2. Oliver, John E. (2005). Encyclopedia of world climatology. Springer.
3. Saucier, Walter Joseph (1955). Principles of Meteorological Analysis. Chicago: The University of Chicago Press. OCLC   1082907714.
4. McIlveen, Robin (1991). Fundamentals of Weather and Climate. Cheltenham, England: Stanley Thornes. p.  40. ISBN   978-0-7487-4079-6.
5. Hay, William W. (2016). Experimenting on a Small Planet: A History of Scientific Discoveries, a Future of Climate Change and Global Warming (second ed.). Cham, Switzerland: Springer Verlag. p.  26. ISBN   978-3-319-27402-7.
6. Manual on Marine Meteorological Services: Volume I – Global Aspect (PDF). World Meteorological Organization. 2012. Archived from the original (PDF) on 5 May 2017.
7. Beer, Tom (1997). Environmental Oceanography. CRC Press. ISBN   0-8493-8425-7.
8. Maiden, Terence. "T-Scale: Origins and Scientific Basis". TORRO. Archived from the original on 5 February 2012. Retrieved 4 January 2012.
9. "The Beaufort Scale". RMetS. Retrieved 6 July 2021.
10. "Beaufort wind force scale". Met Office. Retrieved 27 November 2015.
11. "Beaufort Scale". Royal Meteorological Society. Retrieved 27 November 2015.
12. "Beaufort Scale". Encyclopædia Britannica. Retrieved 27 November 2015.
13. The names "storm" and "hurricane" on the Beaufort scale refer only to wind strength, and do not necessarily mean that other severe weather (for instance, a thunderstorm or tropical cyclone) is present. To avoid confusion, strong wind warnings will often speak of e.g. "hurricane-force winds".
14. "Wetterlexikon - Beaufort-Skala" (in German). Deutscher Wetterdienst. Archived from the original on 12 December 2013. Retrieved 14 February 2014.
15. "昨日实行新标准"珍珠"属强台风_新闻中心_新浪网". news.sina.com.cn.
16. . The Times . 29 April 1873. p. 10. ISSN   0140-0460 . Retrieved 3 July 2020.


text:
Sharp ill-posedness and well-posedness results for the KdV-Burgers equation: the periodic case

Abstract : We prove that the KdV-Burgers is globally well-posed in $H^{-1}(\T)$ with a solution-map that is analytic from $H^{-1}(\T)$ to $C([0,T];H^{-1}(\T))$ whereas it is ill-posed in $H^s(\T)$, as soon as $s<-1$, in the sense that the flow-map $u_0\mapsto u(t)$ cannot be continuous from $H^s(\T)$ to even ${\cal D}'(\T)$ at any fixed $t>0$ small enough. In view of the result of Kappeler and Topalov for KdV it thus appears that even if the dissipation part of the KdV-Burgers equation allows to lower the $C^\infty$ critical index with respect to the KdV equation, it does not permit to improve the $C^0$ critical index .
Keywords :

Cited literature [19 references]

https://hal.archives-ouvertes.fr/hal-00467657
Contributor : Luc Molinet <>
Submitted on : Tuesday, March 30, 2010 - 9:17:58 AM
Last modification on : Tuesday, May 5, 2020 - 1:03:21 PM
Document(s) archivé(s) le : Thursday, September 23, 2010 - 12:32:20 PM

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• HAL Id : hal-00467657, version 2
• ARXIV : 1005.4805

Citation

Luc Molinet, Stéphane Vento. Sharp ill-posedness and well-posedness results for the KdV-Burgers equation: the periodic case. Transactions of the American Mathematical Society, American Mathematical Society, 2013, 365 (1), pp.123-141. ⟨hal-00467657v2⟩

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text:
Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6

1. Jun 27, 2008

staples82

1. The problem statement, all variables and given/known data
Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6

2. Relevant equations
n/a

3. The attempt at a solution
I'm using y'=(dy/dx)

I found the derivative of the above problem.
12x+8xy'+10y=0 (I used the product rule to find the derivative of 8xy')

12x+8xy'+10y=0

8xy'=-12x-10y

xy'=(-12x-10y)/(8x)

I checked the answer in the back of the book and it was y'=(-6x-4y)/(4x+y)

I'm not sure where I missed a step or a calculation error

Thanks!!

2. Jun 27, 2008

rocomath

You forgot the chain rule on your y^2 term.

3. Jun 27, 2008

rootX

6x^2+8xy+y^2=6

12x+8xy' +8y + 2y**=0

This is wrong ...

4. Jun 27, 2008

jaredmt

i see what you tried to do. u tried to combine the 8y and 2y(dy/dx). they cannot be added together.

p.s.
next time u get stuck on something like this u can always work backwards. you could take the book's answer and multiply the top and bottom by 2. then u can figure it out from there

5. Jun 27, 2008

staples82

First of, thanks for all the help on my previous question, I have another one!

1. The problem statement, all variables and given/known data
Find dy/dx by implicit differentiation: 3x^2=(2-y)/(2+y)

2. Relevant equations
n/a

3. The attempt at a solution
I'm using y'=(dy/dx)

I found the derivative of the above problem.
First I used the quotient rule...
6x=[(2+y)(y')-(2-y)(y')]/(2+y)^2

6x=(2y'+yy'-2y'+yy')

6x=2yy'/(2+yy')^2

After this, I got into a jumbled mess, I wasn't sure if I had to add y' to the bottom of the quotient rule.

I checked the answer in the back of the book and it was y'=-3x(2+y)^2/2

Thanks!!

6. Jun 27, 2008

dynamicsolo

Don't forget that the first term in the numerator involves the derivative of (2 - y), so the quotient should be

[(2+y)(-y')-(2-y)(y')]/(2+y)^2 .

You could multiply by (2+y)^2 to get

6x · [(2+y)^2] = -2y' - yy' - 2y' + yy' = -4y'

and solve for y' .

7. Jun 27, 2008

Redbelly98

Staff Emeritus
Sign error: what's the derivitive of (2-y) ?

What happened to the denominator on the RHS?

How did (2+y)^2 from before become (2+yy')^2 ?