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mersenneforum.org Should I put a plug on Windows 10?
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 2015-06-17, 09:58 #1 jasong     "Jason Goatcher" Mar 2005 5×701 Posts Should I put a plug on Windows 10? As many of you know, Windows 10 is free to Windows 7 and 8 users and available as a download, but the download won't officially install until July 29th of this year. So what do you guys think, should I take the plunge? Or should I wait until 10.1? Btw, here's my humorous take on the whole Windows rule: As many of you know, we have something known in nerdy circles as the "Windows rule" which basically states that only odd-numbered Windows releases are worth buying, and then only after the first major update, so 3.1, 5.1(Win XP), and 7.1(what I have, or had, not sure if it's updated beyond 7.1, don't really track that, though it's been updating every Patch Tuesday). I tentatively suggest that the rule is subtly wrong. It's not simply odd-numbered releases, it's 2mod1 for the good releases, you have to add up the digits before you make the check. So Windows 10 is 1+0=1, meaning Windows 10 will be a reasonably decent release. Otoh, lots of games are being released on Linux, so maybe we're approaching a point where there's no point at all to continuously using Windows. Only time will tell.
 2015-06-17, 11:20 #2 VictordeHolland     "Victor de Hollander" Aug 2011 the Netherlands 22268 Posts Lol, why did Microsoft skip Windows 9? I found the metro design in W8.1 terrible. Every program you install puts half a dozen icons on the already crowded metro tiles. So you end up having to use the search function anyway. Things like the configuration settings are difficult to find. I ended up reverting back to W7 after about a week, but maybe I should have tried it longer to get accustomed with the new design. I heard W10 will see the return of the start button, albeit in a slighly different form. Because the upgrade is free, I feel inclined to give it a try.
 2015-06-17, 12:07 #3 S485122     Sep 2006 Brussels, Belgium 63D16 Posts I have the preview of Windows 10 on a virtual machine at home. The Windows 10 start menu is the "metro" interface spread out over just one column. I did not see a convenient way of creating a hierarchy of folders containing the short cuts to applications. Most of the "Apps" are completely useless for me (Bing News, Bing Stock markets, Bing Food...) As with Windows 10 the "old" start menu folders and shortcuts are stil present on the disk... The interface goes in the direction of Office 2013 but with still more restrictions on themes : instead of having the choice between white, grey and dark grey, the file explorer is white. I found no way to change that and when one has more that one file explorer window open it is very awkward to find its controls (mini-, maximize close...) The funny thing is that the registry entries to customise the colours and appearance of windows still exist, they are just ignored. It seems as if Microsoft had decided that one should not have more than one window open on the desktop. The tablet users (but who uses Windows on a tablet) like the Windows 10 interface and clamour for transparency everywhere. I experience transparency and garish backgrounds to menus only as noise, distracting me from the interface to the programs... I will try to hang on to Windows 7 until I retire and can finally switch to decently interfaced OS. Jacob
 2015-06-17, 12:34 #4 chappy     "Jeff" Feb 2012 St. Louis, Missouri, USA 114810 Posts Windows 9 was skipped because of legacy programming artifacts in the way newer OSs handle some Windows 98 and Windows 95 programming. Some of it is nonsense about numbering. Windows 7 was really just Vista Service Pack 1. Rebranded because of the negative consumer experience with Vista. (my only vista issues were with HP printer drivers and iTunes but Windows 7 was pretty solid.) My only issues with Windows 8 (which I only have on a laptop) is with the metro interface. The rest of the system runs incredible smooth with no issues. (I use an add-on which makes my experience with the OS similar to Windows 7) I can't see any reason not to move from 8 to 10 unless you are a Home version user AND you don't want auto-updates. (Auto-updates will not be optional in the Home version of 10 if I understand the releases) If you are happy with 7 and don't want/need the better widgets of 8/10 you may not need to update. From a cyber-security perspective 7 is good, 8 and presumably 10 are much better. But, most of the cyber security concerns these days for home users are browser related.
 2015-06-17, 12:39 #5 science_man_88     "Forget I exist" Jul 2009 Dumbassville 836910 Posts I also heard that some of the pre-installed games will no longer be pre-installed I think, most of what I do with my computer is go online anyways. but I still don't want a potentially crappy OS either way.
 2015-06-17, 12:58 #6 paulunderwood     Sep 2002 Database er0rr DB016 Posts . Attached Thumbnails
2015-06-18, 04:08   #7
LaurV
Romulan Interpreter

Jun 2011
Thailand

100010111001002 Posts

Quote:
 Originally Posted by VictordeHolland Lol, why did Microsoft skip Windows 9?
Marketing strategy. They have too many old products with 9x or 9.x, imagine the new windows 9 would have some bugs and it would need fixing, then you will have 9.1, 9.2... 9.5, ...9.8, then people will start associating windows 9.5 with windows 95, or 9.8 with 98. And they will do this anyhow, even without stepping up the sub-version.

2015-06-18, 04:31   #8
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

587910 Posts

Quote:
 Originally Posted by LaurV Marketing strategy. They have too many old products with 9x or 9.x, imagine the new windows 9 would have some bugs and it would need fixing, then you will have 9.1, 9.2... 9.5, ...9.8, then people will start associating windows 9.5 with windows 95, or 9.8 with 98. And they will do this anyhow, even without stepping up the sub-version.
Perhaps you are right, but the official reason is that the improvement over 8 was so great that 9 didn't do it justice.

Anyhow, the numbers-for-Windows-versions will have to stop again sometime in the future. There will likely be pressure to be like Chrome and FireFox to increment the version number every few minutes (to show progress) and once they get to the low 90's it's going to get confusing.

2015-06-18, 07:38   #9
tha

Dec 2002

787 Posts

Quote:
 Originally Posted by chappy Windows 9 was skipped because of legacy programming artifacts in the way newer OSs handle some Windows 98 and Windows 95 programming.

This was the code Microsoft advised to use to make applications aware of whether they were hosted on 95/98 or on windows XP:

Code:
if (osName.startsWith("windows 9"))
Now that Microsoft changes from a "pay upfront for a release that we will determine how long it will last" to a "pay as you go, find out about the total later" scheme some people seem to be eager to jump to the new release. Of course, you should go for a real free system.

 2015-06-19, 11:04 #10 jasong     "Jason Goatcher" Mar 2005 5·701 Posts I think the reason they skipped to Windows 10 is that they're moving to Windows as a service soon and saying something's a ten means it's perfect. It's just like always, the public thinks they need it because they've been hypnotized. I use it because I have this strange notion that I'm a serious gamer. What I really want is a computer that can switch reasonably quickly between Linux and Windows while also making use of the same hardware. I have this picture in my head of a 3 OS system, one is Windows, one is Linux and one is a barebones version of Linux for debugging and transferring files between the "bigger" OSes. If you ever needed to restart, you'd still have one or two OSes that stick around while you wait on the reboot. Isn't that one of the things Linux is supposed to offer? Reduced functionality instead of no functionality when you're installing new stuff?
 2015-06-26, 00:36 #11 chappy     "Jeff" Feb 2012 St. Louis, Missouri, USA 22×7×41 Posts Learning is good. Attached Thumbnails

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# Recurrence using Master Theorem and Back-Substitution

Q) For parameters 'a' and 'b' both of which are $$\omega(1)$$, Now, consider the recurrence $$T(n) = T(n^{1/a}) + 1$$ with base condition $$T(b) = 1$$ Then $$T(n)$$ is :

I am not getting the meaning of statement : for parameters 'a' and 'b' both of which are $$\omega(1)$$. Can anyone please explain it. Does it mean 'a' and 'b' both are constants and greater than 1. I have one more doubt if we solve using back- substitution then I am getting answer as $$\Theta(log_alog_b n)$$ but if I use change of variables and Master's theorem then answer is : $$T(n) = T(n^{\frac{1}{a}}) + 1$$ So, why both methods are giving different answers. I know base does not matters in $$\Theta$$ notation here but I want to know the reason of different bases.

Hint.

Another method. Assuming $$a > 0$$, as

$$T\left(a^{\log_a n}\right)=T\left(a^{\log_a (\frac na)}\right)+1$$

calling $$\mathcal{T}(\cdot) = T(a^{(\cdot)})$$ and $$u = \log_a n$$ we have equivalently

$$\mathcal{T}(z)=\mathcal{T}\left(\frac za\right)+1$$

and now calling $$\mathbb{T}(\cdot) = \mathcal{T}(a^{(\cdot)})$$ and $$u = \log_a z$$ we arrive at the recurrence

$$\mathbb{T}(u)=\mathbb{T}(u-1)+1$$

with solution

$$\mathbb{T}(u)=u+C_0$$

following with $$\mathbb{T}\to \mathcal{T}\to T$$

• Thank you for the answer. Can you please explain the meaning of the statement "for parameters 'a' and 'b' both of which are ω(1)." ? and Also can you please explain why 2 methods which I have described in the question are giving different answers ? – ankit Feb 14 at 8:18
• Think that $\log_a(\log_b n) = \frac{\log_2(\log_2 n)-\log_2(\log_2 b)}{\log_2 a}\approx \Theta(\log_2(\log_2 n))$ – Cesareo Feb 14 at 9:19


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Shell Pyramid

Time Limit: 5000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Description

In the 17th century, with thunderous noise, dense smoke and blazing fire, battles on the sea were just the same as those in the modern times. But at that time, the cannon ,were extremely simple. It was just like an iron cylinder, with its rearward end sealed and forward end open. There was a small hole at the rearward end of it, which was used to install the fuse. The cannons on the warships were put on small vehicles which had four wheels and the shells were iron spheres with gunpowder in them.

At that time, it was said that there was an intelligent captain, who was also a mathematician amateur. He liked to connect everything him met to mathematics. Before every battle, he often ordered the soldiers to put the shells on the deck and make those shells to form shell pyramids.

Now let's suppose that a shell pyramid has four layers, and there will be a sequence of ordinal numbers in every layer. They are as the following figure:

In the figure, they are the first layer, the second layer, the third layer and the fourth layer respectively from the left to the right.

In the first layer, there is just 1 shell, and its ordinal number is 1. In the second layer, there are 3 shells, and their ordinal numbers are 1, 2, and 3. In the third layer, there are 6 shells, and their ordinal numbers are 1, 2, 3, 4, 5, and 6. In the fourth layer, there are 10 shells, and their ordinal numbers are shown in the figure above.

There are also serial numbers for the whole shell pyramid. For example, the serial number for the third shell in the second layer is 4, the serial number for the fifth shell in the third layer is 9, and the serial number for the ninth shell in the fourth layer is 19.

There is also a interrelated problem: If given one serial number s, then we can work out the s th shell is in what layer, what row and what column. Assume that the layer number is i, the row number is j and the column number is k, therefore, if s=19, then i=4, j=4 and k=3.

Now let us continue to tell about the story about the captain.
A battle was going to begin. The captain allotted the same amount of shells to every cannon. The shells were piled on the deck which formed the same shell pyramids by the cannon. While the enemy warships were near, the captain ordered to fire simultaneously. Thunderous sound then was heard. The captain listened carefully, then he knew that how many shells were used and how many were left.

At the end of the battle, the captain won. During the break, he asked his subordinate a question: For a shell pyramid, if given the serial number s, how do you calculate the layer number i, the row number j and column number k?

Input

First input a number n,repersent n cases.For each case there a shell pyramid which is big enough, a integer is given, and this integer is the serial number s（s<2^63）. There are several test cases. Input is terminated by the end of file.

Output

For each case, output the corresponding layer number i, row number j and column number k.

Sample Input

2
19
75822050528572544

Sample Output

4 4 3
769099 111570 11179

gaojie

Source

2008 Asia Regional Harbin


text:
# Show finite group is $p$-group given some structure of group

Let $G$ be a finite group. If there exists an $a\in G$ not equal to the identity such that for all $x\in G$,$\phi(x) = axa^{-1}=x^{p+1}$ is an automorphism of $G$ then $G$ is a $p$-group.

This is what I have so for.

The order of $a$ is $p$ since $\phi(a) = a= a^pa\rightarrow a^p=e$ therefore the $order(\phi)|p$

If $order(\phi) = 1$ then for all $x\in G$ $\phi(x) = x=x^{p+1}\rightarrow x^p=e$ . Thus every element has order $p$ therefore $G$ is a $p-group$

For $order(\phi) = p$ I get stuck:

$\phi^p(x) = x = x^{(p+1)^p}$ using the expansion formula and simplifying i reach that the order of each element in $G$ divides $\displaystyle\sum_{k=1}^n \binom{p}{k}p^k=(p+1)^p-1$. But other primes divide this so I can't easily conclude $G$ is a $p$-group.

Suppose for contradiction that $G$ is not a $p$-group. Let $|G| = kp^n$ where $k$ is not a multiple of $p$ ($p\nmid k$). If $k>1$ then take a $q$ in $k$'s prime factorization. So we have $q|k$ and by Cauchy's Theorem $\exists y \in G$ with $y^q = e$ i.e $order(y) = q$.

If $q<p$, applying $\phi$ to $y$ I get that $\langle y\rangle$ has more than $q$ elements since $\phi(y)=y^{p+1}\in \langle y \rangle$ and there are $p$ distint elements achieved by $\phi$

If $p>q$ then....... I cannot reach a contradiction :'(

Question #2: Show each element of $G$ has order $p$

When $order(\phi) = 1$ i get what I want but I'm also stuck when $order(\phi) =p$

I believe $a\in Z(G)$, is there a way I can show this?

Thank you....This is a pretty hard problem. :'(

• An earlier version of the problem did not mention $\theta$ is an inner-automorphism: math.stackexchange.com/questions/823130/… – Jack Schmidt Jun 10 '14 at 2:40
• Sorry about I thought $axa^{-1}$ was just there to show the order of the automorphism is a divisor of $p$ – abe Jun 10 '14 at 2:48
• It is not a good practice to use quantifiers such as $\forall$ and $\exists$ when writing mathematics. It makes the question harder to read. See math.harvard.edu/graduate/advise/kleiman.pdf – Mauricio Tec Jun 10 '14 at 3:04
• :O What! But I took a semester learning this – abe Jun 10 '14 at 3:05
• Should I not be using $\wedge$ $\neg$...either? – abe Jun 10 '14 at 3:06

Here is a plan.

Let $N = \{x \in G \mid x^p = 1\}$.

1. Note that $x \in N$ if and only if $x$ commutes with $a$. It follows that $N$ is a subgroup of $G$. It is easy to see that $N$ is characteristic and therefore normal.

2. Take an arbitrary $y \in G$. We have $a y a^{-1} = y^{p+1}$ in $G$. Project this equality onto $G/N$, it gives us $yN = y^{p+1}N$ ($a$ is gone because $a \in N$). Therefore, $y^p \in N$, therefore $y^{p^2} = 1$. This shows that $G$ is a $p$-group, even that the order of each element divides $p^2$.

Here is a different proof from Dan Shved's:

Lemma: In any such group $G$, $a$ normalizes every subgroup, and $a$ lies in every Sylow $p$-subgroup.

Proof: Let $H$ be a subgroup. Then for $x \in H$, $axa^{-1}=x^{p+1} \in H$, so $aHa^{-1} \leq H$; a similar statement holds for $a^{-1}$ and so $H$ is normal. Let $P$ be a Sylow $p$-subgroup. Since $a$ has order $p$ and $a$ normalizes $P$, $\langle a, P \rangle$ is a $p$-group, and so by maximality of $P$, $\langle a, P \rangle = P$ and $a \in P$. $\square$

Main proof: Reduce to a minimal counterexample:

Let $G$ be a group of smallest order such that $G$ has an element $a$ of order $p$ such that for every $x \in G$, $ax=x^{p+1}a$ and yet $G$ is not a $p$-group. If $H$ is a subgroup of $G$ containing $a$, then $H$ satisfies the same hypothesis. If $H$ is also a proper subgroup, then $H$ is a $p$-group by definition of $G$.

Now consider what this means for an element of prime order $q\neq p$.

Let $g$ be an element of order $q$ for $q \neq p$. By hypothesis, $a$ normalizes $\langle g \rangle$ so $H=\langle a,g\rangle$ has order $pq$ and satisfies the hypothesis of the problem. If $H <G$ then we have a contradiction, so we must $H=G$ and so $G$ is a non-abelian group of order $pq$.

And show that this structure doesn't satisfy the hypothesis:

Let $P=\langle a \rangle$ be a Sylow $p$-subgroup of $G$ and let $Q=\langle g \rangle$ be a Sylow $q$-subgroup. Since $Q$ is normalized by both $P$ and $Q$, $Q$ is normal in $G$. Since $a$ is contained in every Sylow $p$-subgroup by the lemma, but the Sylow $p$-subgroups have order $p$, we have $P$ is also normal. Hence $G = P \times Q$ is abelian. This a contradiction, since then $aga^{-1} aa^{-1}g = g \neq g^{p+1}$ as $g$ has order $q$ not $p$.

$\square$

• (This proof was based on a higher tech proof: $a$ lies in the Baer kernel of $G$, so is contained in the second center, so centralizes Sylow $q$-subgroups, a contradiction. I tried to get the conclusions using earlier ideas.) – Jack Schmidt Jun 10 '14 at 4:53
• Thank you so much :D . I wish I can accept two answers. Sorry about yesterday, i must have driven you nuts. – abe Jun 10 '14 at 16:42


text:
# Arafat’s billion dollars

1. Nov 8, 2004

### GENIERE

Arafat’s personal worth is said to be in excess of billion dollars. Anybody know who advises him on investments? I always wondered why he’s so popular in France. I’m betting he’ll only get 35 virgins though, because he never strapped on a bomb (not to late for him to give it a go).

Incidently, what happens to the virgins after they're er... deflowered, so to speak? Do you get a new supply, are they recycled (virginized), or are you doomed to eternal consort with used material?

Bye-Bye

2. Nov 14, 2004

### kat

Apparently Suhu is (his wife) has attained a financial advisor. One Pierre Rizk, who is better known for having headed the intelligence service of the Phalang during the Lebanese war. Which would also put him in line to have been involved with the Sabra and Shatilla massacre in 1982. Evidently Rizk holds power of attorney for Arafat's wife. To me this gives credence to a long unprovable rumor of involvement of syria and the PLO of the Massacre at Sabra and Shatilla. One can only wonder why there's no resounding outrage by the palestinian people for a member of the Phalang to be involved with Palestinian money.

3. Nov 14, 2004

### tumor

They gonna find Arafat's bilion dollars like they found Sadam's WMD's.

4. Nov 15, 2004

They must have found it, his wife's getting $120,000.00 each month. What does your typical wealthy Palestinian earn each month? 5. Nov 15, 2004 ### revelator That''s a lot of pilfered moneys! Oddly enough, I've heard that the virgins are for acutal.. um. "use" in that sense (if they're meant to perform other functions then why the heck do they need to be virgins though?!). That's just hearsay though, so don't take my word for it. 6. Nov 15, 2004 ### tumor What is your source on Arafat's wife getting 120 thousand a month? Israeli or American(same ****)press? PS. even if Arafat had few milion bucks on his account,it is nothing compared to hundreds of bilions US government is blatantly stealing from ordinary Americans and attacking allready starving nations, and then pay milions to executives and "contractors"(murderers). Last edited: Nov 15, 2004 7. Nov 15, 2004 ### GENIERE According to the Italian news (link below), I was mistaken.. She gets$20,000,000.00 immediately, and then \$35,000.00 each month.

http://www.corriere.it/Primo_Piano/Esteri/2004/11_Novembre/11/tesoro.shtml

...

8. Nov 15, 2004

### tumor

Good for her.
And by the way, I can't read Italian.

Are you sure you are not mistaken this time? you going to say in your next post(according to the Jerusalem Post) that she is getting trilion dollars. :rofl:

Last edited: Nov 15, 2004
9. Nov 15, 2004

### revelator

Geez, everybody makes mikstakes..

10. Nov 15, 2004

### kat

Funny...but I could have sworn I've seen you quote the American press elsewhere on this forum..

We're not talking about the US government here. We're discussing the "Father of the Palestinian People" who apparently has been dancing with their butchers and profiting from their poverty. Without the oppressed Palestinian people Arafat had no wealth. Doesn't it irk you to know that Suha and Arafat had Rizk handling Palestinian money? That's like Bush taking Red Cross donations and U.S. Tax money and asking Bin Laden to control it. :grumpy:

11. Nov 16, 2004

### GENIERE

Apparently he(she?) is not irked.

12. Nov 16, 2004

### revelator

Didn't he have all sorts of money already from having some construction companies, back in his younger days?

13. Nov 16, 2004

### Smurf

That.. didn't... happen?

14. Nov 16, 2004

### kat

Arafat may have owned a small contracting company in Kuwait during the late 50's early 60's but the funds he's taken from the Palestinian people have been fairly well documented, more recently found are the millions he skimmed from the gas funds.
Arafat gave the Palestinian people nationalism and from then on he's done nothing but create death and despair for them, steal their money and have his wife live the high life in Paris. He's not alone in this, much of the PLO leadership skim money from Palestinian funds, send large sums to their families in Paris and send their children to expensive private schools. Palestine becoming a transparent democracy where people can truly speak against the government and have accoutability for public funds... will negatively impact the "Old Guard", so to what benefit is it to them to allow the status quo to actually change?

15. Nov 16, 2004

### tumor

Of course now it is all Arafat's fault! Hello! how about those guys in tanks, F-16 jets, Apache helicopters, constantly killing and terrorizng Palestinians!

When nations are at war(palestinians are actually slaughtered at will) finances are always being misappropriated.
Foreigners from EU who might have gone to Palestine to check on finnances are leaving and are not being let in,no wonder corruption exist.
But guys stop with the Arafat bashing.

Last edited: Nov 16, 2004
16. Nov 16, 2004

### kat

Uhhhh yeah...the "Father of the Palestinians" robbed them blind and the Old Guard he's left behind is revving up to do the same and we should just shut up and take the blame like good little American and Israeli girls and boys. Riiiiiiggghhttt...I don't think that helps the Palestinian people one bit. I also don't think, particularly when I see a response like that, that people who make themselves blind to that care one bit...rather they are more excited to be part of Virulant anti-americanism then caring for what's best for the Palestinian people. Same goes with the Iraqi people...who cares if the hands that are their to help and feed them instead are taking the food and aid right out from under them?

Last edited: Nov 16, 2004
17. Nov 16, 2004

### Tigers2B1

I'll add that I'm glad that Arafat is dead. I'll never have to see that greasy, oily, bug eyed, cheese grinning, face again plying his trade. We'll never see his self-centered 'oneness' - Arafat - the murderer of thousands of innocent people, from athletes at the Olympics to dads, mothers, and children on streets, in buses, or eating lunch. I'll not have to see anymore of Arafat, who was an obstacle to any real peace, refusing many times to negotiate and encouraging violence. I get to say "goodbye" with a big smile to Arafat, the thief of what may amount to multi-millions or billions of dollars now located in foreign accounts, while many of "his people" live in abject poverty. Arafat, who left nothing, built nothing, contributed nothing to this world in 30 years worth of opportunties. And we'll certainly never see Arafat, the "leader" so power hungry he left no succesor or system of succession. The King without a prince. So self-absorbed he probably thought he could take it with him -

Maybe, finally, his screaming, bleach blonde, middle aged wife will be happy now – in the arms of one of her lovers - Otherwise, his death probably changes the lives of very few -

18. Nov 16, 2004

### Staff: Mentor

I won't say I'm glad he's dead, but I will say that his death makes peace more likely than his continued life. Unfortunately, that's often the way of things in such places: progress ocurrs only following the death of a dictator.

19. Nov 16, 2004

### kat

actually...I fear that on the death of dictators mayhem and more dead is usually the status quo. Either the old guard will step up to the plate with strength and somewhat the same status quo will prevail...or the young guard will battle the old...and hope upon hope whoever ends up on top isn't fanatical..but moderate.

20. Nov 16, 2004


text:
# L'Hopital's Rule Problem Help

#### Bzombee

##### New member
Doing a worksheet for class and I was given this problem:
--------
Use l'Hopital's rule to find the exact value of
$\lim_{x \to 0} \left(e^{\frac{5}{x}}-3x\right)^{\frac{x}{2}}$ .
--------
This is what I have done so far:
$\lim_{x \to 0}\ln y=\lim_{x \to 0}\ln\left(e^{\frac{5}{x}}-3x\right)^{\frac{x}{2}}$$\lim_{x \to 0}\ln y=\lim_{x \to 0}\frac{x\ln\left(e^{\frac{5}{x}}-3x\right)}{2}$$\lim_{x \to 0}\frac{x\ln\left(e^{\frac{5}{x}}-3x\right)}{2} = 0$$\lim_{x \to 0}\left(e^{\frac{5}{x}}-3x\right)^{\frac{x}{2}} = \lim_{x \to 0}y = \lim_{x \to 0} e^{\ln y} =e^{0} =1$
The answer key to the worksheet says the answer is actually $e^{2.5}$, not 1 which is what I and the rest of my classmates got. Is the answer key wrong or am I missing something. Thanks!

#### MarkFL

##### Super Moderator
Staff member
I do get what they worksheet key shows. You are letting a still indeterminate form go to zero. You did not apply L'Hopital.

#### Bzombee

##### New member
I do get what they worksheet key shows. You are letting a still indeterminate form go to zero. You did not apply L'Hopital.
Is the $e^{\frac{5}{x}}$ still making it go into an indeterminate form? If so, how do I apply L'Hopital's rule or some other method to solve the problem?

#### MarkFL

##### Super Moderator
Staff member
Is the $e^{\frac{5}{x}}$ still making it go into an indeterminate form? If so, how do I apply L'Hopital's rule or some other method to solve the problem?

Yes, I would write the expression as:

$$\displaystyle \frac{\ln\left(e^{\frac{5}{x}}-3x\right)}{\frac{2}{x}}$$

Now you have a form to which L'Hopital can be applied.

#### Bzombee

##### New member
Yes, I would write the expression as:

$$\displaystyle \frac{\ln\left(e^{\frac{5}{x}}-3x\right)}{\frac{2}{x}}$$

Now you have a form to which L'Hopital can be applied.
As a matter of fact, I did try and write it like that. However, I thought it wouldn't do any good because if you perform l'hopital's rule you get this:
$\frac{\left(\frac{-\frac{5\mathrm{e}^{\frac{5}{x}}}{x^{2}}-3}{\mathrm{e}^{\frac{5}{x}}-3x}\right)}{\left(-\frac{2}{x^{2}}\right)}$ .
This does not get rid of what is leading to indeterminate form which is the e^(5/x).
Putting it into Wolfram Alpha I got the following:

Now this leads me to believe that this question has is flawed because it should have told me to solve from the left or to the right of the limit (which it does not, I double checked).

#### MarkFL

##### Super Moderator
Staff member
I do agree that the problem should have stated from the right. I simplified what you have to:

$$\displaystyle \ln(L)=\lim_{x\to0}\left(\frac{5e^{\frac{5}{x}}+3x^2}{2\left(e^{\frac{5}{x}}-3x\right)}\right)$$

Now, divide numerator and denominator by:

$$\displaystyle e^{\frac{5}{x}}$$...

#### Bzombee

##### New member
So I divided both numerator and denominator by e^(5/x) like you said
$\left(\frac{\left(\frac{5e^{\frac{5}{x}}+3x^{2}}{e^{\frac{5}{x}}}\right)}{\left(\frac{2\left(e^{\frac{5}{x}}-3x\right)}{e^{\frac{5}{x}}}\right)}\right)$Unless I didn't do something right, it brings me back to where we started. A left hand limit at 0 and a right hand limit at 2.5.

#### MarkFL

##### Super Moderator
Staff member
Let's assume we are taking the limit from the right. Doing the division I suggested, the expression becomes:

$$\displaystyle \frac{5+3x^2e^{-\frac{5}{x}}}{2\left(1-3xe^{-\frac{5}{x}}\right)}$$

#### Bzombee

##### New member
Let's assume we are taking the limit from the right. Doing the division I suggested, the expression becomes:

$$\displaystyle \frac{5+3x^2e^{-\frac{5}{x}}}{2\left(1-3xe^{-\frac{5}{x}}\right)}$$
And plugging that in to a graphing calculator gives me the limit as x approaches from the right of 0 gives me a definitive value at 2.5.
Okay, I think I pretty much got what I need now. Thank you for your help

#### Dr.Peterson

##### Elite Member
And plugging that in to a graphing calculator gives me the limit as x approaches from the right of 0 gives me a definitive value at 2.5.
Okay, I think I pretty much got what I need now. Thank you for your help
It's interesting to ponder why it behaves so differently for positive and negative x. Have you considered that? (Hint: which term is most important in each limit, left and right?)

#### Bzombee

##### New member
It's interesting to ponder why it behaves so differently for positive and negative x. Have you considered that? (Hint: which term is most important in each limit, left and right?)

View attachment 35274
When you say positive and negative x, I'm assuming you are referring to the limits? But I did play around with the graphs and I got this which was interesting and I don't know if it was what you are referring to:


text:
## Angular moments of the decay Λ b 0  → Λμ+μ− at low hadronic recoil

Research output: Contribution to journalArticle

Open

### Documents

Original language English 146 Journal of High Energy Physics 1809 26 Sep 2018 https://doi.org/10.1007/JHEP09(2018)146 E-pub ahead of print - 26 Sep 2018

### Abstract

An analysis of the angular distribution of the decay $\Lambda_b^0 \rightarrow \Lambda \mu^{+} \mu^{-}$ is presented, using data collected with the LHCb detector between 2011 and 2016 and corresponding to an integrated luminosity of approximately $5\,fb^{-1}$. Angular observables are determined using a moment analysis of the angular distribution at low hadronic recoil, corresponding to the dimuon invariant mass squared range $15 < q^{2} < 20\, GeV^2/c^4$. The full basis of observables is measured for the first time. The lepton-side, hadron-side and combined forward-backward asymmetries of the decay are determined to be \begin{align} A_{FB}^{l} & = -0.39 \pm 0.04\,\rm{stat} \pm 0.01\, \rm{syst}, \nonumber\\ A_{FB}^{h} & = -0.30 \pm 0.05\,\rm{stat} \pm 0.02\, \rm{syst}, \nonumber\\ A_{FB}^{lh} & = +0.25 \pm 0.04\,\rm{stat} \pm 0.01\, \rm{syst}. \nonumber \end{align} The measurements are consistent with Standard Model predictions.

ID: 90516496


text:
Symbol
Problem
b 1 | If $\begin{cases} a \\ b \\ c \end{cases}$ $1|=2010$ aPnd v $a$ 1 $c-b$ $C-b$ $c^{2}|$ $\begin{cases} c-a \\ a-b \\ b-c \end{cases}$ $a-C$ $111$ $5$ $8$ $0$ $-$ $\bar{8}$ $\infty$ b $a-c$ $a^{2}|=p,$ then the $b-a$ $b-a$ $b^{2}|$ number of positive divisors of p is (a) $36$ (b) $49$ $\left(c\right)$ 64 (d) $81$


text:
# How to access SFTP with SSIS using SSH.NET

## Introduction

For the ones unware of SFTP (SSH File Transfer Protocol), it is a secure channel with SSH2 encryption established between two machines. You can think of it as FTP with encryption and a few more differences. For more information about this you can read this blog post.

There are some alternatives to get this done, ones paid and others free. I will focus on the use of the excellent SSH.NET because I couldn’t find a good guide to get it to work with SSIS. It’s worth mencioning WinSCP as well, and the excellent guide to make it work with Integration Services.

## The GAC

The recommended way of installing SSH.NET is by using Nuget, but the only way to SSIS to be able to access the library is installing the library to the GAC.

3. Install to the GAC with the above PowerShell script as an administrator:
 1 2 3  [System.Reflection.Assembly]::Load("System.EnterpriseServices, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a") $publish = New-Object System.EnterpriseServices.Internal.Publish$publish.GacInstall("Renci.SshNet.dll")

## Create the project

You can see two things in my code that are worthy of note. The first one was the file I had to download and the second the problems with the way security was implemented in the server. For the first one I created the selectedZip variable that contains the name of the correct file to download. The second forced me to create an event to simulate the keyboard typing the password on the server. You can see that on the code below that you can paste to a Script Task. Don’t forget to add the following DLL references:

• System.IO.Compression
• System.IO.Compression.Filesystem
• Renci.Ssh.Net
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64  using Renci.SshNet; using Renci.SshNet.Common; using Renci.SshNet.Sftp; using System; using System.IO; using System.IO.Compression; using System.Linq; using ConnectionInfo = Renci.SshNet.ConnectionInfo; public void Main() { string host = Dts.Variables["$Project::sftpUrl"].Value.ToString(); string username = Dts.Variables["$Project::sftpUser"].Value.ToString(); string password = Dts.Variables["$Project::sftpPassword"].Value.ToString(); string remoteDirectory = Dts.Variables["$Project::sftpRemotePath"].Value.ToString(); string localDirectory = Dts.Variables["$Project::sftpLocalPath"].Value.ToString(); KeyboardInteractiveAuthenticationMethod kauth = new KeyboardInteractiveAuthenticationMethod(username); PasswordAuthenticationMethod pauth = new PasswordAuthenticationMethod(username, password); //Event to simulate keyboard password entry kauth.AuthenticationPrompt += new EventHandler(HandleKeyEvent); ConnectionInfo connectionInfo = new ConnectionInfo(host, 22, username, pauth, kauth); using (var sftp = new SftpClient(connectionInfo)) { sftp.Connect(); var files = sftp.ListDirectory(remoteDirectory); SftpFile selectedZip; //Selecting the most recent file selectedZip = files.Where(x => x.Name.Contains(".zip")).OrderByDescending(x => x.Name).FirstOrDefault(); string localZip = string.Empty; using (var localFile = File.Open(localDirectory + selectedZip.Name, FileMode.Create)) { sftp.DownloadFile(selectedZip.FullName, localFile); localZip = localFile.Name; } using (ZipArchive archive = ZipFile.OpenRead(localZip)) { foreach (ZipArchiveEntry entry in archive.Entries) { entry.ExtractToFile(Path.Combine(localDirectory, entry.FullName), true); } } } Dts.TaskResult = (int)ScriptResults.Success; } void HandleKeyEvent(Object sender, Renci.SshNet.Common.AuthenticationPromptEventArgs e) { foreach (Renci.SshNet.Common.AuthenticationPrompt prompt in e.Prompts) { if (prompt.Request.IndexOf("Password:", StringComparison.InvariantCultureIgnoreCase) != -1) { prompt.Response = Dts.Variables["$Project::sftpPassword"].Value.ToString(); } } }

## Bonus: Test with Docker

If you want to test locally with docker try this command line:

 1   docker run --name "MySFTP" -v C:\Temp\:/home/user/ -p 22:22 -d atmoz/sftp:alpine user:password:1001

Replace with your username and password and directory and you will be able to test locally :)

Have fun!!


text:
# orthogonal gradient descent for continual learning

## Authors

Mehrdad Farajtabar, Navid Azizan, Alex Mott, Ang Li

## Introduction

Neural Networks suffer from catastrophic forgetting: forgetting previously learned tasks when trained to do new ones. Most neural networks can’t learn tasks sequentially despite having the capacity to learn them simultaneously. For example, training a CNN to look at only one label of CIFAR10 at a time results in poor performance for the initially trained labels (catastrophic forgetting). But that same CNN will perform really well if all the labels are trained simultaneously (as is standard). The ability to learn tasks sequentially is called continual learning, and it is crucially important for real world applications of machine learning. For example, a medical imaging classifier might be able to classify a set of base diseases very well, but its utility is limited if it cannot be adapted to learn novel diseases - like local/rare/or new diseases (like Covid-19).

This work introduces a new learning algorithm called Orthogonal Gradient Descent (OGD) that replaces Stochastic Gradient Descent (SGD). In standard SGD, the optimization takes no care to retain performance on any previously learned tasks, which works well when the task is presented all at once and iid. However, in a continual learning setting, when tasks/labels are presented sequentially, SGD fails to retain performance on earlier tasks. This is because when data is presented simultaneously, our goal is to model the underlying joint data distribution $P(X_1,X_2,\ldots, X_n)$, and we can sample batches like $(X_1,X_2,\ldots, X_m)$ iid from this distribution, which is assumed to be "fixed" during training. In continual learning, this distribution typically shifts over time, thus resulting in the failure of SGD. OGD considers previously learned tasks by maintaining a space of previous gradients, such that incoming gradients can be projected onto an orthogonal basis of that space - minimally impacting previously attained performance.

## Previous Work

Continual learning is not a new concept in machine learning, and there are many previous research articles on the subject that can help to get acquainted with the subject ([4], [9], [10] for example). These previous works in continual learning can be summarized into three broad categories. There are expansion based techniques, which add neurons/modules to an existing model to accommodate incoming tasks while leveraging previously learned representations. One of the downsides of this method is the growing size of the model with an increasing number of tasks. There are also regularization based methods, which constraints weight updates according to some important measure for previous tasks. Finally, there are the repetition based methods. These models attempt to artificially interlace data from previous tasks into the training scheme of incoming tasks, mimicking traditional simultaneous learning. This can be done by using memory modules or generative networks.

The key insight to OGD is leveraging the overparameterization of neural networks, meaning they have more parameters than data points. In order to learn new things without forgetting old ones, OGD proposes the intuitive notion of projecting newly found gradients onto an orthogonal basis for the space of previously optimal gradients. Such an orthogonal basis will exist because neural networks are typically overparameterized. Note that moving along the gradient direction results in the biggest change for parameter update, whereas moving orthogonal to the gradient results in the least change, which effectively prevents the predictions of the previous task from changing too much. A small step orthogonal to the gradient of a task should result in little change to the loss for that task, owing again to the overparameterization of the network [5, 6, 7, 8].

More specifically, OGD keeps track of the gradient with respect to each logit (OGD-ALL), since the idea is to project new gradients onto a space which minimally impacts the previous task across all logits. However, they have also done experiments where they only keep track of the gradient with respect to the ground truth logit (ODG-GTL) and with the logits averaged (OGD-AVE). OGD-ALL keeps track of gradients of dimension N*C where N is the size of the previous task and C is the number of classes. OGD-AVE and OGD-GTL only store gradients of dimension N since the class logits are either averaged or ignored respectively. To further manage memory, the authors sample from all the gradients of the old task, and they find that 200 is sufficient - with diminishing returns when using more.

The orthogonal basis for the span of previously attained gradients can be obtained using a simple Gram-Schmidt (or more numerically stable equivalent) iterative method. One such algorithm which can be utilized to improve numerical stability is the modified Gram-Schmidt Orthogonalisation. The issue with the simpler Gram-Schmidt algorithm can be seen in the following:

Suppose we have a matrix $A$ which is to be decomposed into $A=\hat{Q}\hat{R}$ using the Gram-Schmidt algorithm. During the algorithm, columns of $\hat{Q}$ are solved sequentially, where $\hat{\vec{q_j}}$ is the $j^{th}$ column of $\hat{Q}$, and $\hat{r_{ij}}$ which is the $i^{th}$ row and $j^{th}$ column of $\hat{R}$ are solved from left to right and top to bottom for only the elements $\hat{R}$ to result in a upper triangular matrix. Consider when we are calculating the third column of $\hat{Q}$ as follows: $\hat{\vec{q_{3}}}=\vec{a_3} - (\hat{\vec{q_1}}\vec{a_3})\hat{\vec{q_1}} - (\hat{\vec{q_2}}\vec{a_3})\hat{\vec{q_2}}$. $\vec{z_3}=\vec{a_3} - (\hat{\vec{q_1}}\vec{a_3})\hat{\vec{q_1}}$ should not have a component in direction $\hat{\vec{q_1}}$, however, due to numerical stability and catastrophic cancellation [11] this is not always true. The partial result $\vec{z_3}$ ends up having a component in this direction, this leads to a loss in orthogonality in the columns of $\hat{Q}$. To remedy this problem, the modified Gram-Schmidt algorithm replaces $\vec{a_3}$ with $\vec{z_3}$ in $(\hat{\vec{q_2}}\vec{a_3})\hat{\vec{q_2}}$, this helps in ensuring the orthogonality of the columns of $\hat{Q}$ to any loss of numerical significance since we will be orthogonalizing with the vector which already has the loss of significance.

Algorithm 1 shows the precise algorithm for OGD.

And perhaps the easiest way to understand this is pictorially. Here, Task A is the previously learned task and task B is the incoming task. The neural network $f$ has parameters $w$ and is indexed by the $j$th logit.

## Results

Each task was trained for 5 epochs, with tasks derived from the MNIST dataset. The network is a three-layer MLP with 100 hidden units in two layers and 10 logit outputs. The results of OGD-AVE, ODG-GTL, OGD-ALL are compared to SGD, ECW [2], (a regularization method using Fischer information for importance weights), A-GEM [3] (a state-of-the-art replay technique), and MTL (a ground truth "cheat" model which has access to all data throughout training). The experiments were performed for the following three continual learning benchmarks: permuted MNIST, rotated MNIST, and split MNIST.

In permuted MNIST [1], there are five tasks, where each task is a fixed permutation that gets applied to each MNIST digit. The following tables show classification performance for each task after sequentially training on all the tasks. Thus, if solved catastrophic forgetting has been solved, the accuracies should be constant across tasks. If not, then there should be a significant decrease from task 5 through to task 1.

Rotated MNIST is similar except instead of fixed permutation there are fixed rotations. There are five sequential tasks, with MNIST images rotated at 0, 10, 20, 30, and 40 degrees in each task.

Split MNIST defines 5 tasks with mutually disjoint labels [4].

Also, the below table corresponds to the performance of Rotated MNIST and Permuted MNIST as a function of number of gradients stored.

Overall OGD performs much better than ECW, A-GEM, and SGD. The primary metric to look for is decreasing performance in the earlier tasks. As we can see, MTL, which represents the ideal simultaneous learning scenario shows no drop-off across tasks since all the data from previous tasks is available when training incoming tasks. For OGD, we see a decrease, but it is not nearly as severe a decrease as naively doing SGD. OGD performs much better than ECW and slightly better than A-GEM.

## Review

This work presents an interesting and intuitive algorithm for continual learning. It is theoretically well-founded and shows higher performance than competing algorithms. One of the downsides is that the learning rate must be kept very small, in order to respect the assumption that orthogonal gradients do not affect the loss. Furthermore, this algorithm requires maintaining a set of gradients which grows with the number of tasks. The authors mention several directions for future study based on this technique. Finding a way to store more gradients or preauthorize the important directions can result in improved results. Secondly, all the proposed methods including this method fail when the tasks are dissimilar. Finding ways to maintain performance under tasks dissimilarity can be an interesting research direction. Thirdly, solving for learning rate sensitivity will make this method more appealing when large learning rate are desired. Finally, another interesting future work is extending the current method to other types of optimizers such as Adam and Adagrad or even second or even quasi-Newton methods.

One interesting way for increasing learning rate can be considering the gradient magnitude of the parameters for data of the former task. If for some specific parameters, the gradient magnitude for data of task A is low then intuitively it means they have not capture high amount of information from task A. Having this in mind, at least we can increase the learning rate for updating these weights so that we can use them for task B.

A valuable resource for continual learning is the following GitHub page: link continual_learning_papers

## References

[1] Goodfellow, I. J., Mirza, M., Xiao, D., Courville, A., and Bengio, Y. (2013). An empirical investigation of catastrophic forgetting in gradient-based neural networks. arXiv preprint arXiv:1312.6211

[2] Kirkpatrick, J., Pascanu, R., Rabinowitz, N., Veness, J., Desjardins, G., Rusu, A. A., Milan, K., Quan, J., Ramalho, T., Grabska-Barwinska, A., et al. (2017). Overcoming catastrophic forgetting in neural networks. Proceedings of the national academy of sciences, 114(13):3521–3526.

[3] Chaudhry, A., Ranzato, M., Rohrbach, M., and Elhoseiny, M. (2018). Efficient lifelong learning with A-GEM. arXiv preprint arXiv:1812.00420.

[4] Zenke, F., Poole, B., and Ganguli, S. (2017). Continual learning through synaptic intelligence. In Proceedings of the 34th International Conference on Machine Learning-Volume 70, pages 3987–3995. JMLR

[5] Azizan, N. and Hassibi, B. (2018). Stochastic gradient/mirror descent: Minimax optimality and implicit regularization. arXiv preprint arXiv:1806.00952

[6] Li, Y. and Liang, Y. (2018). Learning overparameterized neural networks via stochastic gradient descent on structured data. In Advances in Neural Information Processing Systems, pages 8157–8166.

[7] Allen-Zhu, Z., Li, Y., and Song, Z. (2018). A convergence theory for deep learning via overparameterization. arXiv preprint arXiv:1811.03962.

[8] Azizan, N., Lale, S., and Hassibi, B. (2019). Stochastic mirror descent on overparameterized nonlinear models: Convergence, implicit regularization, and generalization. arXiv preprint arXiv:1906.03830.

[9] Nagy, D. G., & Orban, G. (2017). Episodic memory for continual model learning. ArXiv, Nips.

[10] Nguyen, C. V., Li, Y., Bui, T. D., & Turner, R. E. (2017). Variational continual learning. ArXiv, Vi, 1–18.

[11] Wikipedia: https://en.wikipedia.org/wiki/Loss_of_significance


text:
# Rolling Body mechanics

## Homework Statement

In the picture ,problem 8.29

## Homework Equations

Energy conservation ,velocity relation ,momentum

## The Attempt at a Solution

First i defined the speed of the body relative to the ground as Vb and speed of plane relative to the ground as Vp.
From momentum ,VbcosΘ=-Vp .From the non slipping condition ,ωR-Vb=VpcosΘ.This relation seems to be wrong ,the correct aswer is (Vp+VbcosΘ)/cosΘ=ωR which I am incapable of explaining.I also included my work in the pictures

#### Attachments

• 15019306742351633013540.jpg
58.9 KB · Views: 362
• 1501931174371366449559.jpg
24.2 KB · Views: 330

TSny
Homework Helper
Gold Member
First i defined the speed of the body relative to the ground as Vb and speed of plane relative to the ground as Vp.
From momentum ,VbcosΘ=-Vp .
Shouldn't the masses show up in the momentum equation? Also, VbcosΘ is not the horizontal component of the velocity of the ball relative to the ground since Vb does not make an angle Θ to the horizontal. The velocity of the ball relative to the ground is not in the same direction as the velocity of the ball relative to the incline. You will need to think carefully about relative velocities in this problem.

From the non slipping condition ,ωR-Vb=VpcosΘ.This relation seems to be wrong ,the correct aswer is (Vp+VbcosΘ)/cosΘ=ωR which I am incapable of explaining.I also included my work in the pictures
I don't think either of these two equations is correct.

haruspex
Homework Helper
Gold Member
2020 Award
(Vp+VbcosΘ)/cosΘ=ωR
As TSny wrote, that is incorrect, if Vp is as you have defined it. But it does contain some useful pointers.
In the context of the question, what physical entity does ωR represent?


text:
# How to find the local minimum points in the multi-dimensional numerical data?

I am trying to get some local minima of an n-dimensional function (which only has numerical form). For two-dimensional function data, we can find the local minimum points from the contour plot easily.

Now, I have a three-dimensional function data.

See the data here.

The variables x1,x2,and x3 are from 1 to 50 and the step is 1. Since I have sorted the data based on the f(x1,x2,x3) values, the global minimum point can be easily obtained, that is, -8.13526987(x1=11,x2=50,x3=50).

The problem is, how to find the other local minimum points for this data ?

• What is the domain of the data ? Exhaustive search is the only way to find all the local minima. – image_doctor Aug 20 '15 at 10:13
• With the increase of the x1,x2 and x3, the f(x1,x2,x3) can be more deeper. But the local minimum points are more important for this data. How to exhaustive search ? – Orders Aug 20 '15 at 11:18

You have quite a small data set, so a really inefficient brute force search will still run pretty fast (<1 sec on my computer). I stress that this is STUPID way to do it, and with list manipulation you can surely make it MUCH more efficient. But as I said - it works.

First, transform the data so that you could retrieve the data by calling f[x1,x2,x3], and unset values will be Infinity:

dat = Rest@Import["numdata.txt", "tsv"];
f[i_,j_,k_] := Infinity;
Table[f[d[[1]], d[[2]], d[[3]]] = d[[4]], {d, dat}];

Then simply iterate over all entries, and see which of them is larger than all its neighbors:

Last@Last@Reap@For[i = 1, i <= 50, i++,
For[j = 1, j <= 50, j++,
For[k = 1, k <= 50, k++,
If[
And[
f[i, j, k] < f[i + 1, j, k],
f[i, j, k] < f[i - 1, j, k],
f[i, j, k] < f[i, j + 1, k],
f[i, j, k] < f[i, j - 1, k],
f[i, j, k] < f[i, j, k + 1],
f[i, j, k] < f[i, j, k - 1]
],
Sow[{i, j, k}]
]
]
]
]

Output:

{{4, 3, 3}, {9, 13, 15}, {9, 14, 14}, {10, 17, 2}, {10, 50, 1},
{11, 8, 17}, {11, 15, 2}, {11, 50, 50}, {15, 5, 15}, {15, 10, 3},
{16, 6, 12}, {16, 7, 9}, {16, 8, 7}, {16, 9, 5}, {16, 10, 4},
{50, 50, 50}}

These are the positions of all the local minima.

• Thanks! Yes, it works. Since it is a small data set, is it possible to smooth the multi-dimensional function by fitting based on the present data ? Then, we can find the more exact local minima. – Orders Aug 21 '15 at 0:34


text:
Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# D Y D X + Y Tan X = X N Cos X , N ≠ − 1 - Mathematics

Sum

(dy)/(dx)+ y tan x = x^n cos x, n ne− 1

#### Solution

We have,

(dy)/(dx)+ y tan x = x^n cos x

$\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}$

$P = \tan x$

$Q = x^n \cos x$

Now,

$I . F . = e^{\int\tan x dx}$

$= e^{\log\left( sec x \right)}$

$= \sec x$

So, the solution is given by

$y \times I . F . = \int Q \times I . F . dx + C$

$\Rightarrow y \sec x = \int x^n \cos x \sec x\ dx + C$

$\Rightarrow y \sec x = \int x^n dx + C$

$\Rightarrow y \sec x = \frac{x^{n + 1}}{n + 1} + C$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 61 | Page 146


text:
# History of the Lagrange Inversion Theorem

I'm doing research on the history of the Lagrange inversion theorem. The earliest predecessor I've found is the one referenced by De Morgan; viz. Jo. H. Lambert's construction in Observationes Variae in Mathesin Puram, Acta Helvetica, Vol. 3, 1758, pp. 128-168.

If anyone knows of an earlier construction I'd greatly appreciate hearing about it.

Thanks in advance.

Cheers, Scott

## 1 Answer

If you count any inversion of a power series as a predecessor of Lagrange inversion, then I believe the earliest examples are Newton's inversion of the log series to obtain the exponential series, and inversion of the inverse sine series to obtain the sine series. The exponential series is in his De methodis (1671), p.61, and the sine series is in his De analysi (1669), p. 233, 237.

Also, de Moivre, Philosophical Transactions 20 (1698), pp. 190-193, gave a more general formula in a paper entitled "A method of extracting the root of an infinite equation."

• Newton produced at least the first few terms of the general compostional inversion series for a formal power series that we now relate to the combinatorics of associahedra. See ref in oeis.org/A133437. – Tom Copeland Nov 12 '18 at 8:26


text:
# zbMATH — the first resource for mathematics

Algebraic K-theory of spaces, concordance, and stable homotopy theory. (English) Zbl 0708.19001
Algebraic topology and algebraic K-theory, Proc. Conf., Princeton, NJ (USA), Ann. Math. Stud. 113, 392-417 (1987).
[For the entire collection see Zbl 0694.00022.]
Various splittings, up to homotopy, of the algebraic K-theory space A(X) of Waldhausen include $A(X)\simeq A^ S(X)\times Wh^{DIFF}(X)\text{ and } A^ S(X)\simeq \Omega^{\infty}S^{\infty}(X_+)\times \mu (X).$ The author uses a Kahn-Priddy type theorem to show $$\mu$$ (x) is trivial, and therefore that $A(X)\simeq \Omega^{\infty}S^{\infty}(X_+)\times Wh^{DIFF}(X).$ He also discusses several other proofs of the vanishing of $$\mu$$ (X).
Reviewer: M.R.Stein

##### MSC:
 19D10 Algebraic $$K$$-theory of spaces 55Q45 Stable homotopy of spheres


text:
# Is there a name for this type of CES look-alike utility function?

I do not know whether the following utility functions are used a lot but I was wondering whether there was a common name for this type CES like utility function

$$u(z) = \left[ \left( \sum_{j \in J_1} z_{j}^{\alpha_1}\right)^{\rho/\alpha_1} + \left(\sum_{j \in J_2} z_{j}^{\alpha_2}\right)^{\rho/\alpha_2} \right]^{1/\rho},$$

perhaps it is simply an instance of the CES?

The utility function is defined over two non-overlapping groups $$J_1$$ and $$J_2$$ of goods $$z_j$$. The goods from the two groups are weighted diffrently in the inner CES part by $$\alpha_1$$ and $$\alpha_2$$ respectively.


text:
## Solving Dyson-Schwinger equations

Sunday I posted a paper of mine on arxiv (see here). I was interested on managing a simple interacting theory with the technique of Dyson-Schwinger equations. These are a set of exact equations that permit to compute all the n-point functions of a given theory. The critical point is that a lower order equation depends on higher order n-point functions making the solution of all set quite difficult. The most common approach is to try a truncation at some order relying on some physical insight. Of course, to have a control on such a truncation could be a difficult task and the results of a given computation should be carefully checked. The beauty of these equations relies on their non-perturbative nature to be contrasted with the severe difficulty in solving them.

In my paper I consider a massless $\phi^4$ theory and I solve exactly all the set of Dyson-Schwinger equations. I am able to do this as I know a set of exact solutions of the classical equation of the theory and I am able to solve an apparently difficult equation for the two point function. At the end of the day,  one gets the exact propagator, the spectrum and the beta function. It is seen that this theory has only trivial fixed points. I was able to get these results on another paper of mine. So, it is surely comforting to get identical results with different approaches.

Finally,  I can apply  the mapping theorem with Yang-Mills theories, recently proved thanks also to Terry Tao intervention, to draw conclusions on them in the limit of a very large coupling. In the paper you can find a formulation of this theorem as agreed with Terry, a direct consequence of my latest accepted paper on this matter (see here).

I think this paper adds an important contribution to our understanding of Dyson-Schwinger equations presenting an exact non-trivial solution of them.

### 5 Responses to Solving Dyson-Schwinger equations

1. […] View original here: Solving Dyson-Schwinger equations […]

2. Rafael says:

Dear Marco,

please, give a look at these cosmological implications of infrared QCD propagators:

Are you able to formulate the gauge-to-scalar mapping also in curved backgrounds? What does this imply for ghosts (should they stay coupled as needed in aforementioned papers)?
Best,

Rafael.

• mfrasca says:

Dear Rafael,

Your question is quite interesting as are the papers you put to my attention. Let me clarify what the mapping theorem implies. You can select a set of components of a Yang-Mills field and use them to solve the corresponding equations taking them to be all equal, space-independent and being also solution of a massless quartic scalar field equation. These are Smilga solutions. Otherwise, in the most general case, the mapping is only perturbative giving

$A_\mu^a=\eta_\mu^a\phi(x)+O(1/g)$

holding when the coupling g is taken large enough and being $\phi(x)$ a solution of the equation for the massless quartic scalar field. As you can see, all amounts to a selection of components and so, I cannot expect that being or not in a general relativistic framework should change something. Rather, I would have serious troubles to solve exactly the corresponding equation for the scalar field. Indeed, as you know, cosmologists working with scalar fields in inflationary scenarios apply the so-called slow-roll approximation to avoid this problem. A posteriori they are able to show that this approximation is what one needs in this physical situation.

About the ghost field, I should expect a decoupling also in the most general case because this happens for algebraical reasons. This arises from the antisymmetry of the structure constants of the gauge group and so I cannot see how this field can count here. But, as you may know, the running coupling of the Yang-Mills field has a quite peculiar behavior, as I also discussed in my paper, and this means that the scenario depicted by these authors could be correct provided dark matter is cold enough to be in a kind of intermediate regime where both asymptotic freedom and the low-energy scenario fail.

Finally, let me say that a lattice analysis, using Robertson-Walker metric, could be very interesting here to see how cosmological effects enter into QCD.

Best,

Marco

3. carlbrannen says:

Congratulations on the new paper! (I can’t believe no one has spoken before me.)

Regarding Smilga’s choice, I had this sudden flash that it might be related to a solution to a certain math problem I’ve been pursuing, which has to do with path integrals over finite configuration spaces and the like.

Given a finite group, one defines the “group algebra over the complex numbers” by using the finite group elements as a basis for a vector space with multiplication defined by the finite group multiplication rule. This defines an algebra. Wolfram.com on “Group Algebra”.

Now, given a group algebra, the stable long time propagators for individual particles correspond to the “primitive idempotents” of the algebra. Idempotent means uu = u. Primitive means that it’s not zero, and that it’s not the sum of two non zero idemopotent elements. Example, for spin-1/2, the projection operators for spin in various directions are primitive idempotents. The unit matrix is idempotent but not primitive. In a certain way, complete sets of annihilating group algebra primitive idempotents (i.e. sets where any pair multiply to zero and the sum of all of them is unity (where unity is 1 times the basis element corresponding to the identity of the finite group)) are a generalization of spinors.

Finding the primitive idempotents of a group algebra can be difficult because, with a finite group of size N, the maximum possible number of idempotents is 2^ N.

Anyway, the analogy to Smilga’s choice is that it is always possible to write down one primitive idempotent: The sum over n of g_n/N, where g_n is the basis element corresponding to the nth element of the finite group. It’s easy to show that this is a primitive idempotent.

• mfrasca says:

Hi Carl,

Thank you. Indeed, this is a quite interesting result on a theory generally believed manageable only with perturbation techniques.

Best,

Marco

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text:
## Precalculus (6th Edition) Blitzer

The formula for $_{n}{{C}_{r}}$ has the same numerator as the formula for $_{n}{{P}_{r}}$ but contains an extra factor of $r!$ in the denominator.
We know that permutations of $n$ things taken $r$ at a time can be defined by the formula: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And combinations of $n$ things taken $r$ at a time can be defined by the formula: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ And by comparing both formulas it can be seen that r! is an extra factor. Therefore, there are r! times as many permutations of n things taken r at a time as combinations of n things taken r at a time.


text:
Lens equation

1. Apr 30, 2005

Callisto

Hi Peaple :!!)

How do you determine the minimum distance p+q between the object and image for given focal length given

1/p + 1/q = 1/f

this may seem trivial, but i cant figure it out.

Callisto

2. Apr 30, 2005

whozum

Arent p and q distances....? Just add them.

3. Apr 30, 2005

whozum

If your saying given a fixed focal length, find the relationship between p and q, that case

$$\frac{1}{p} + \frac{1}{q} = k$$

Solve it for either p or q (the result is the same) and graph it. I believe its a logarithmic relation, but I'm not sure.

4. Apr 30, 2005

Integral

Staff Emeritus
What math tools do you have at your disposal?

Actually the thin lens relationship is a hyperbola.

5. Apr 30, 2005

whozum

I should know that, my brother did a presentation on it a week ago.

6. Apr 30, 2005

Callisto

solving 1/p+1/p=1/f for p
i get
-qf/(f-q)
now i substitute this into p+q, then i get -q^2/(f-q).
now if i use calculus to find the minimum value i differentiate -q^2/(f-q) and let it equal zero so
d/dq= -q(2f-q)/(f-q)^2 =0 when q = 0 or 2f

Repeating this process for p then p = 0 or 2f
so the minimum distance is
p+q=4f. why is it not zero

Callisto

7. May 1, 2005

Data

Looks fine to me. You might want to check that 2f is a minimum (not a maximum or point of inflection), to be thorough. Looking at the lens equation itself do you see why the 0 solutions are inadmissible?

Last edited: May 1, 2005
8. May 1, 2005

Callisto

1/0+1/0=0 , 0 doesn't = 1/f
therefore 0's are excluded from the problem, correct?
Callisto

9. May 1, 2005

Data

1/0 + 1/0 is 0? Are you sure?

10. May 1, 2005

Callisto

1/0+1/0=1/0, 1/0 doesn't = 1/f
is this why 0's are excluded?

11. May 1, 2005

Data

1/0 is undefined, as is 1/0+1/0, and no other arithmetic with these quantities is defined either, so if p or q were 0, you could not rearrange the equation the way that you did (without first defining a new arthmetic). In order for the lens equation to make sense with usual definitions, the assumption p, q not zero must be made.

12. May 1, 2005

Callisto

Thanks
2f must be minimum since if the object p is at infinity then 1/p=0 so q = f for an object at infinity.


text:
# Relational algebra operations

Operations in the Relational Data Model are defined by Relational Algebra. Relational Algebra is not based on the attributes, but rather on the order of values (components) in the tuple. There are five basic operations [7]:

1. Union,
2. Difference,
3. Cartesian Product,
4. Projection,
5. Selection.

The Union of relations and is defined as:

 (3.2)

It is the set of tuples that are in or . This operation may be applied to the relations of the same arity only, so all the tuples in the result have the same number of components (values).

The Difference of relations and , defined as:

 (3.3)

It is the set of tuples which are in but not in . It is required that and have the same arity.

The Cartesian Product of relations and is defined as:

 (3.4)

Assume that has arity and has arity . is defined as the set of all possible ()-tuples which first components form a tuple in and last form a tuple in . The result has the arity equal to .

The Projection is denoted as . Assuming that there is a relation of arity , then:

 (3.5)

It denotes the projection of onto components , where are distinct integers in range . That is the set of -tuples such that there is some -tuple in for which for .

For example is computed by taking each tuple from , and forming a 2-tuple set from fourth and second component of . Instead of using locations, attribute names can be used as well. Assuming that is the relation schema for the above relation. Then the above projection can also be denoted as . The resulting relation can be described by the following relation schema: .

The Selection is denoted as:

 (3.6)

Where is a relation and is a formula involving:
• Operands that are constants or component numbers; component is represented by ,
• The arithmetic comparison operators , , , , , ,
• The logical operators (and), (or), (not).
Then is the set of tuples in such that, when for all the -th component of is substituted for any occurrences of in formula , the formula becomes true. As for projection if a relation has named columns, then the formula in the selection can refer to columns by name instead of by number. The arity of is the same as the arity of .

There are also some other algebraic operations defined as: Intersection, Quotient, Join, Natural Join or Semijoin. They are all derived from the basic ones.

The Intersection can be defined as:

 (3.7)

Thus its functionality is provided by difference, that is why it is not considered to be recognized as a basic operation.

The Quotient of relations and is denoted as . Assuming that and are relations of arity and respectively, and , and . Then the quotient is a set of all ()-tuples , such that for all -tuples in , the tuple is in :

 (3.8)

The Join (also known as -Join) of and on columns and is defined as:

 (3.9)

Where is a comparison operator which complies with the definition 3.6, and is the arity of . In other words, these are all the tuples in the Cartesian product of and such that the -th component of stands in relation to the th component of . If is '' then the operation is called the .

The Natural Join is a special case of the Join. It is applicable only when both and have columns that are named by attributes. Assuming that are all the attribute names used for both and , then the Natural Join is defined as:

 (3.10)

such that is the list of all components of , in order, except the components .

The Semijoin of relation by relation , written , is the projection onto the attributes of of the natural join of and :

 (3.11)

where stands for the list of attributes of .

Relational Algebra becomes a natural way to formulate queries. Having selection and projection, a question about a single relation can be asked. Adding Cartesian product, the query can concern more than one relation. Applying union or difference, several queries can be combined, to form a more sophisticated one.

Igor Wojnicki 2005-11-07


text:
(i) The value of tan A is always less than 1.

(ii) sec A = $$\dfrac{12}{5}$$ for some value of angle A.

(iii)cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = $$\dfrac{4}{3}$$ for some angle θ.

Asked by Pragya Singh | 1 year ago |  66

##### Solution :-

Explanation:-

The value of tan A is always less than 1.

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = $$\dfrac{4}{3}$$ which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse

= 5 as it will follow the Pythagoras theorem.

MC2=MN2+NC2

52=32+42

25=9+16

25 = 25

Explanation:-

sec A = $$\dfrac{12}{5}$$ for some value of angle A

Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

MC2=MN2+NC2

(12k)2=(5k)2+NC2

NC2+25k2=144k2

NC2=119k2

Such a triangle is possible as it will follow the Pythagoras theorem.

Explanation:-

cos A is the abbreviation used for the cosecant of angle A.

Abbreviation used for cosecant of angle M is cosec M.

cos M is the abbreviation used for cosine of angle M.

Explanation:-

cot A is the product of cot and A.

cot M is not the product of cot and M. It is the cotangent of ∠M.

Explanation:-

sin θ = $$\dfrac{4}{3}$$ for some angle θ.

sin θ = $$\dfrac{Opposite}{Hypotenuse}$$

We know that in a right angled triangle, Hypotenuse is the longest side.

sin θ will always less than 1 and it can never be $$\dfrac{4}{3}$$ for any value of θ.

Answered by Abhisek | 1 year ago

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