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# Solving $y'' - y = 0$

I am attemtping to solve $y'' - y = 0$

I come to this solution, by using something like $\frac{dy}{dx} = p$

So it does $\frac{dp}{dy} \cdot \frac{dy}{dx} - y = 0$

Which gives $\frac{dp}{dy} \cdot p = y$

After all the transformations, integrating and all, I end up with this expression!

$c_{1}e^{x} = y + \sqrt{y^{2} + c_{2}}$

wow... How am I supposed, from there, to obtain the expected solution, that is, $y(x) = c_{1}e^{x}+c_{2}e^{-x}$

(Note c1 and c2 are unrelated to the other equation)

Thank you

• Hint: Does the auxiliary polynomial, or characteristic polynomial ring a bell? y''-y=0 gives us the polynomial r^2-1=0 – EpicMochi Nov 6 '13 at 9:43
• do you just want the solution (in some other manner) or you want to use just this method... :O – user87543 Nov 6 '13 at 9:44
• Thank you for your comments but... I am supposed to get it by this solution. I mean... the solutions book arrived to the same thing than I. They just... come from the solution I have (well, some constants differ but its essentially the same), but they do not show how they went from the euqation I gave to the final solution :( So basically, yes, I'm supposed to solve it this way – Yannick Nov 6 '13 at 9:46

Starting from $c_1e^x= y+\sqrt{y^2+c_2}$:

$$c_1e^x -y= \sqrt{y^2+c_2}$$ Taking square of both sides

$$c_1e^{2x} + y^2 - 2c_1e^xy = y^2 + c_2$$ Finally it comes to $$y = \frac{c_1^2e^{2x} + c_2}{2c_1e^x}$$

• Sigh. I had it in my face all this time. I dunno why, but from first glace I told myself "you cannot get rid of y^2". Guess I am tired. Thank you – Yannick Nov 6 '13 at 10:27

General hint:

Assume your solution $y(x)$ has a general form like $e^{mx}$ for some $m$. So, set $y=e^{mx}$ and then satisfy it into the ODE to find the probable value of $m$. Note that the superposition principle will help us to find the general solution for this ODE.

• Yes, I have seen this mentionned multiples times before. Well, on different sites and all. It brings an interesting subject actually and a question I had about it : on which cases can we apply this assumption? On equations of all forms or... only this particular form?! – Yannick Nov 6 '13 at 10:01
• @Yannick: You can use this way for a linear $n$th-order OE with constant coefficients. – Mikasa Nov 6 '13 at 10:08
• Ohhhh I actually get it now. I shud have read a beeet moar before trying to solve a problem :) Thanks mate – Yannick Nov 6 '13 at 14:01
• @Yannick: Thanks. However, other posts completed the answer. :-) – Mikasa Nov 6 '13 at 14:04
• $\ddot\smile \rightarrow +1$ – amWhy Nov 6 '13 at 14:22

One possible way would be , let $$\displaystyle \frac{d}{dx}y(x) + y(x) = g(x) \hspace{1cm} (1) \\ \displaystyle \frac{d}{dx}g(x) - g(x) = 0 \hspace{1cm} (2)$$ first solve $(2)$ for $g(x)$ and then put it's value on $(1)$ and solve $y(x)$ in first order equation.

Equation $(2)$ gives you $\displaystyle g(x) = c_1 e^{x}$ and Equation $(2)$ gives you $\displaystyle y(x) = c_1 e^x + c_2 e^{-x}$. It also gives you an idea that solution will be of form $y(x)=e^{\lambda x}$.

The idea is to represent the equation in terms of operator. i.e. $(D^2 -1) y(x) = 0$. Then $(D^2 - 1)y = (D-1)(D+1)y$, let $(D+1)y = g(x)$ then you have $(D-1)g(x) = 0$ and solve like above.

• I did not think about this one. Very nice. I was actually sticking with solutions similar with what I've read so far, but I guess we can use any substitution. Thanks for this one. – Yannick Nov 6 '13 at 10:24
• @Yannick you are welcome :) – Santosh Linkha Nov 6 '13 at 10:27

• That's pretty fantastic, I'd never thought of exponentiating matrices as a method of solving differential equations! – user18862 Jan 9 '14 at 4:47
• @NeuroFuzzy This is a good way to have first order equations. The price to pay is "matrices". Thanks. – Felix Marin Jan 9 '14 at 4:49


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# What to do when a player dislikes his character, that is vital for the story?

One of the PCs (a greedy, quirky dwarf fighter) in the game I am hosting recently died, and his player chose to join the group as a new character, a human paladin. After two sessions the player is not happy with his choice of character, and often mentions that he loved his dwarf and loathes the paladin character due to his good/lawful alignment.

Unfortunately the PC has to be able to stick around as part of the plot, for at least the next four or five sessions, there is no way around that. The PC has had an encounter with certain NPCs who will play an important role later on - and the NPCs need this PC. Before the previous dwarf's death, I had a similar encounter which I had to reconstruct for the paladin - if I do it a third time my players will think I'm mad!

How can I handle this player being unhappy with his present character? How can I make the player happy again, and avoid having to kill the paladin?

• "due to his good/lawful alignment"? You are, imposing roleplay requirements on this character? Nobody (well, nearly nobody) likes playing Lawful Stupid. – AJMansfield Apr 21 '14 at 20:50
• @AJMansfield - I think it's the player's belief to play his character "lawful stupid". I'll show him the tvtropes page you mentioned, maybe he'll reconsider. Thanks! – mawimawi Apr 22 '14 at 8:21

Address the problem at the source: Retcon1 the story. If your players demand an in-story explanation, remember the origins of the owlbear: "A wizard did it."

At the end of the day, all the participants involved are aware that the game that they are playing is a story. The cleanest solution, therefore, to an external (non-narrative) story influence that is just messing things up is to fix it outside of the story.

Simply say "Hey, this encounter? Here's how it really happened. This way, $Player can play$character and everyone can have more fun." It's the honest way, it hurts the least, and it allows everyone to just get on with things.

As a matter of protocol, I give all new characters a "retcon cookie" (but only one) that can be used at any time to adjust their character sheet in any rules-valid way. There are elements of a character that only come out during play that can just be seriously annoying for everyone involved. There is a tacit understanding that this cookie should be used to resolve a player's issue with her character, not as a "oh, we need to get around this in-story problem... I was a rogue all around." But it's quite acceptable if that restriction is spelled out.

1 A definition of Retcon:

/ret'kon/ [short for retroactive continuity, from the Usenet newsgroup rec.arts.comics] 1. n. The common situation in pulp fiction (esp. comics or soap operas) where a new story 'reveals' things about events in previous stories, usually leaving the 'facts' the same (thus preserving continuity) while completely changing their interpretation. For example, revealing that a whole season of "Dallas" was a dream was a retcon.

For your purposes, the retcon would leave the events of the story the same, but simply indicate that $Player's new character did them instead of the old character, thereby preserving continuity. • This is a really important technique that many people don't realize they have. RPGs are much more personal than other kinds of media. It's totally fine to say "Hey, I screwed something in the plot, mind if I retcon some things?" – Hovercouch Apr 29 at 19:38 How can I make the player happy again, and avoid having to kill the paladin? It seems to be a drastic change, so the best way seems to be to create a new character and play that. I'm not in favor of killing the old one though. There are a million ways to part ways and why would it always be death? As a paladin, he could be ordered to help fighting evil in a remote part of the world by his Order. Maybe he is honored for his deeds and becomes a member of the royal guard. Standing watch over the king all day does not make a good adventurer either. Unfortunately the PC has to be able to stick around as part of the plot This seems to be your core problem. And it's not because the player wants to play another character. The character is basically unkillable, too. If you cannot commence play without a certain character, you painted yourself into a corner. There are many ways out of that corner, you just have to pick one: maybe the paladin is still along as an NPC to the party. Maybe the party as a whole encounters the NPCs again. Maybe the paladin does die and as a last action gives the characters some token of appreciation that the NPCs will recognize. He could do the same on parting ways when he is joining the royal guards. I'd say instead of spending energy somehow making the character work, let him create a new character and instead focus on making your story work. "there is no way around that". There probably is a way around it, but it depends exactly what you've set up. The PC could be eaten by a grue next session due to a series of bad decisions and/or dice rolls. The world would continue, even if those NPCs all die due to lack of that particular paladin. Options that might work depending on your plot and your players: • Play that paladin as an NPC for 4-5 sessions. The party (including the player's new character) hang around with him for some reason, he hooks them up with the plot and then leaves • Let the NPCs find their own paladin, use a different way to get the party into the plot • Let the NPCs do what they like, just scratch those 4-5 sessions. Obviously that's a shame if you've done a lot of work but it becomes a question of railroading. Is the party obliged to run through every scene you've prepared? • If "yes", your plot is on rails, then think of any old absurd reason to shoehorn the party into the plot, and be done with it. Maybe the NPCs approach the group assuming that the paladin they need will be with them, when in fact he's left. Maybe the PCs play along. Sucks to be those NPCs. The next plot hook will be neater ;-) • Ask the player, as a favour to you, to play the paladin for 4-5 more sessions. Find some way to compensate or make it fun. For example you might agree that the player will play the most stuck-up, nit-picking, irritating paladin imaginable, as a parody of paladins. • Pick a different player and say, "hey, do you mind if your character goes on vacation for 5 sessions and you play this paladin instead? All XP counts to your real character". Normally I wouldn't advise passing a character from one player to another, and I'd certainly check the original player is OK with it. But who cares if someone else takes a character you played for 2 sessions and hate anyway? Basically, first think "there must be a way around this", and then figure out what it is. # The cure I think the easiest remedy is to let him pick a new character, and make the paladin go with the group while you close all his plots and prepare a transition. You can also make the player control the two characters, or give the paladin to someone else. Talk to your players about the approach and tell them the situation is temporary. # The prevention Never make the plot depend on a single factor (for instance, a character). What if the paladin had died? What if he is lost? What if the player has a problem and can't come to play for a few months. Design your plots to rely on different factors, but don't make any of them indispensable. • I think the second point about prevention is really important. If you are truly giving your players the opportunity to affect the direction that a story goes in and rolling with the punches that random dice rolls can give you, then relying on a specific anything happening or not happening is pretty much futile. – Wibbs Apr 20 '14 at 12:41 • There are very few players that genuinely only can play a few character concepts. Usually, it's used as a crutch to avoid having to get into character psychology and playing interesting roles. I disagree with your Cure heading, but agree with the Prevention, so +0. – user2754 Apr 20 '14 at 14:52 The above advice is all very good, but I would also like to add another option. Over the years (been GMing since the early 80's) I've occasionally had players make characters that seemed like fun, but once they got into the game discover that they, for whatever reason, just didn't enjoy the character, they would let me know and I would usually go with one of two options: 1) Normally I would just allow them to roll up a new character , but If for some reason the character was already plot integral I would ask them if they wanted to continue until this was no longer the case and then make a new character, some did as a role playing challenge, but most were uninterested. In these cases I just took option 2 2) I simply take over the character as a NPC and allowed the player to introduce a new character, it's not difficult and gives you very fleshed out npc for future use. Often a good way to avoid these situation is to always have good contingency plots, this is particularly important when making a player integral to a plot line, I've learned over the years that players have a terrible knack for getting killed at inopportune moments or just not doing what you thought they would, which makes (to me at least) the game fun from a GM perspective, having these contingency plots prepared makes your live as a GM easier, allows you players to feel like they're in a living breathing world free to do whatever they want without feeling railroaded and makes you seem like a badass GM with a never ending supply of stories and adventures. This gets even easier as you develop a world of your own and it becomes easier to have multiple plots in action at once, switching to new plots because your PCs did something unexpected becomes easy and organic after enough time because like the real world, there's always something interesting going on. ## Something Completely Different 1. "Listen.. cough... the Yosritsi Prophecy... must not.. be fulfilled! * dies *" -Use it to add to the challenge - now they have to race to find out what the NPCs told the paladin that he struggled to give them some clues before he died! 2. "Heh. Like you fools could stop the Yosritsi Prophecy from being fulfilled.. clearly I was wasting my time here! * evil vanish *" - Gasp!!!! The NPCs come running to the party - they were wrong! The chosen one is one of the PCs! 3. "YOU WERE THE CHOSEN ONE!" - The NPCs are wrong. It turns out something else was going on the whole time! 4. "Yes... excellent.. just according to plan... * /Mr. Burns *" The NPCs were lying, and their lies sent the paladin to his death. Vengeance! 5. "Real paladin? There was no real paladin. It was smoke and lies and illusions, and I was the ringmaster to it all! You suffered at my hands, adventurers, and now you are far too late to stop the Yosritsi Prophecy!" The NPCs never existed, and the Paladin is actually the BBEG. 6. "Never fear friends! We're nearly there! The Yosritsi Prophecy CAN be stopped!" Man, ever since Jim the Halfling Thief showed up, this paladin has been a pain in the ass. (nothing stops players from grumbling about NPCs getting the spotlight like making them hella annoying) 7. "Wait, so the Yosritsi Prophecy says that a group of people - that look exactly like us - will one day end the world, and must be stopped? And you've taken on the holy vow of fulfilling the Yosritsi Prophecy, no matter what?""Yes.""$@#%."

Killing the character is so... straightforward. It lacks imagination, and imagination is what D&D is all about. Also, just killing them would rob you of a chance to do every DM's favourite thing: put your characters through hell.

Notice, that's the character going through hell, not the player. If done right, the player will love it.

Talk to the player and figure out exactly what the gap is between what the character is now, and what the player would rather be playing. It sounds like you have a pretty good handle on this already.

Then, see if you can find a way to close that gap.

Tell the player that you're going to try to do this. Don't tell them how you plan to do it, but you could drop hints if you want. So, the player dislikes the Lawful Good element? Which do they dislike most? The Lawful, or the Good?

If it's the Lawful, then put them in tough spots where their laws are forcing them to do something that they have in-character reasons to not want to do, or where their laws conflict. Random example: an NPC is trying to find out who killed their father, and have vowed to kill whoever it is. The paladin knows who did it, but it happens to be someone the paladin respects, and they did it because the father was secretly evil. The paladin can't lie, but he also has to uphold justice, so now he has a problem. If he tells the truth, someone innocent may be killed, but if they lie, then they could prevent an unjust death. (Yes, this is basically the tale of Spiderman/Green Goblin/Harry Osborne. Any good plot deserves to be stolen and turned into a D&D campaign.)

If it's the Good, then put them in tough spots where their good alignment forces them to do something they don't want to. Random example: an ancient and powerful curse is awakened, and threatens to unleash hellish destruction on a whole town... unless an innocent life is sacrificed somehow. And lo and behold, the party - particularly the paladin - ends up in the right place, at the right time, with a suitable NPC available. If they kill the innocent, it'll haunt them, but they'll save the town. If they spare the innocent, then they have a lot of work and a lot of deaths to contend with.

If it's both, then do both!

Now comes the important bit: with enough of these big tough situations, their alignment can shift. This takes a while, which is why it's important to tell the player that it'll be happening and explain that it may take a while, but it'll be worth it and you'll make it as cool as you can. That way, they hang in there, waiting for the next step on the road, reshaping the character as they want them to be and generating a really cool story about the terrible hardships that broke the paladin's spirit, causing him to forsake his vows and his code of conduct, and... well, your player gets to choose what happens next.

We have this happening with multiple characters in our current campaign. We have a wizard whose lust for power is slowly driving him to forsake his friends and commitments and slide towards the Dark Side... I mean, Chaotic Evil. We have a rogue who's been through such hardships with the party that it forged incredible friendships, eventually inspiring him to disregard his own interests to selflessly risk himself to save his friends, shifting him towards a Good alignment. And we have a bard who has seen one too many comrades die despite her best efforts to save them, and is slipping into madness and becoming increasingly Chaotic, doing whatever it takes to keep the party alive.

When done well, character evolution is an incredible driver of good plot and player enjoyment. You've got some real potential here; work with your player to have some serious fun with it.

Brian stated it far better that I could ever could. If a certain mechanical aspect blocks your story and hurts the fun of your players, throw it away. But I do believe that there are other things to solve this idea which may be a little less drastic.

## Talk with the player

It is my number one solution to most of the problems I've encountered: have a conversation with her and see if she can manage to play the character for a few more sessions. At the end of the day, this is the cleanest way to solve it. If the player says no we don't have any other choice but to retcon the story. If, on the other hand, she says yes we can move forward to solving the problem in the short (or not so short) term.

Through the conversation you should get another thing, which is what drew the player to this particular class. Maybe it was the powers and abilities of the paladin or maybe it was a certain part of the fluff. By knowing what drew the player to class you will know how to bring back some of the original sense of enjoyment, enough for at least a few sessions.

## Be willing to tweak some of the rules

As it looks from here, one of the leading problems was the alignment problem: your player doesn't enjoy playing an LG character. I see 2 ways to solve it: Changing the alignment restriction or changing the sense of the alignment itself. Changing the alignment restriction is probably easier and it has been used for quite a time in many many places. The Paladin is a hero for good, a fighter for the holiest of causes, a person who fights for the greater cause, the one that will bring the highest amount of good for the world. Not all good is LG, and as such we can have a paladin who fights for good from the perspective of CG, for example. CGs are still good, and the example of Robin Hood is used to illustrate this alignment mainly because he used a less honest way to create good, so there is no badness or harm made to the concept.

A far preferable way to solve this, at least for me, is to change the meaning of LG. Alignments in D&D were always spectrums which usually touched one another in one way or another, in one place or another. OOTS actually gives a great example of city of LG Paladins who are still LG but are still quite different from one another. Some of them actually commit actions that would be perceived as villainous even though made for the sake of the greater good and many a time even succeed in that.

## Focus on what interests the player

There are still other ways to engage a player and helping her enjoy the game even without helping her mechanically. There are drives for each and every one of us, drives that makes us wanna play, reasons that we come every week. Those reasons are the things we're looking for in order to have fun. Have in your session more of those drives of the particular player. She makes a huge sacrifice here, so make it pay for her in other terms.

If the player really likes fights, have more fights in the game 'till she changes character. If she prefers interrogation scenes have more of those. It doesn't mean forgetting the other players, or always letting her win, but just to raise the number of times that you do the things she likes to see in the game the most.

## Change the next parts of the adventure so she'll be able to change class

A little bit harder to pull right but a far more useful way to retcon the story is to actually retcon the next parts of the story, those that they haven't reached for yet. Postpone the plot that you've written for a session or 2 and let the other characters meet those NPCs. Then you can quite easily give her the time and space to switch characters between the sessions. They can even serve as a plot-reason for the character change, deciding to switch some of their employees or whatever from one group to the other.

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# Boundary regularity of weighted p-Laplace equation

It seems that I need to find some regularity results for the weighted p-Laplace equation, namely $\nabla \cdot (\gamma |\nabla u|^{p-2} \nabla u) = 0$ with Dirichlet boundary values.

Suppose that we have a smooth bounded domain $\Omega$ and smooth Dirichlet boundary values. Take the weight $\gamma$ to be smooth and positive in the closure of $\Omega$. Take p in the interval $]1,\infty[.$

What can we say about the regularity of the solution u on the boundary $\partial \Omega$?

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Some references I have found:

An article by Xiangling Fan titled "Global $C^{1,\alpha}$ regularity for variable exponent elliptic equations in divergence form" states that with certain assumptions on the domain, weight and boundary conditions we have $u \in C^{1,\alpha}(\overline{\Omega})$. The article focuses on the variable exponent case, though, so better results might exist for static $p$.

The work of Kaj Nyström et al concerns the boundary behaviour of the p-laplace equation, but they assume the solution is everywhere positive and zero near the part of boundary they are investigating. These assumptions are too strict for my purposes.

The lecture notes of Peter Lindqvist seem to focus more on interior regularity and non-weighted case.

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I also asked this on math.SE: http://math.stackexchange.com/questions/364404/regularity-of-weighted-p-laplace-equation-up-to-boundary

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does the "convolution theorem" apply to weaker algebraic structures? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:35:18Z http://mathoverflow.net/feeds/question/10237 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10237/does-the-convolution-theorem-apply-to-weaker-algebraic-structures does the "convolution theorem" apply to weaker algebraic structures? Adam 2009-12-31T06:28:51Z 2010-02-04T23:19:48Z
The Convolution Theorem is often exploited to compute the convolution of two sequences efficiently: take the (discrete) Fourier transform of each sequence, multiply them, and then perform the inverse transform on the result.

The same thing can be done for convolutions in the quotient ring Z/pZ via the analogous Number Theoretic Transform.

Does this procedure generalize to other algebraic structures? Arbitrary rings? Semirings? Fast convolutions over the semiring (min,+) would be particularly useful.

I'm led to suspect this because both sorting and FFT can be computed using the same butterfly-like network with simple operations like "min", "max", "+", and "*" at the nodes of the network, and sorting can be thought of as a kind of convolution.
 http://mathoverflow.net/questions/10237/does-the-convolution-theorem-apply-to-weaker-algebraic-structures/11606#11606 Answer by Ryan Williams for does the "convolution theorem" apply to weaker algebraic structures? Ryan Williams 2010-01-13T02:12:15Z 2010-01-13T02:12:15Z
In general, it is a major open question in discrete algorithms as to which algebraic structures admit fast convolution algorithms and which do not. (To be concrete, I define the $(\oplus,\otimes)$ convolution of two $n$-vectors $[x_0,\ldots,x_{n-1}]$ and $[y_0,\ldots,y_{n-1}]$, to be the vector $[z_0,\ldots,z_{n-1}]$ with $$z_k = (x_0 \otimes y_k) \oplus (x_1 \otimes y_{k-1}) \oplus \cdots \oplus (x_k \otimes y_0).$$ Here, $\otimes$ and $\oplus$ are the multiplication and addition operations of some underlying semiring.)

For any $\otimes$ and $\oplus$, the convolution can be computed trivially in $O(n^2)$ operations. As you note, when $\otimes = \times$, $\oplus = +$, and we work over the integers, this convolution can be done efficiently, in $O(n \log n)$ operations.

But for more complex operations, we do not know efficient algorithms, and we do not know good lower bounds. The best algorithm for $(\min,+)$ convolution is $O(n^2/\log n)$ operations, due to Chan. See

David Bremner, Timothy M. Chan, Erik D. Demaine, Jeff Erickson, Ferran Hurtado, John Iacono, Stefan Langerman, Perouz Taslakian: Necklaces, Convolutions, and X + Y. ESA 2006: 160-171

A substantially subquadratic algorithm for $(\min,+)$ convolution would (to my knowledge) imply a subcubic algorithm all-pairs shortest paths in general graphs, a longstanding open problem. The above reference also gives a $O(n^2 (\log \log n)^2/\log n)$ algorithm for a "(median,+) convolution".

The situation is subtle. It's not clear when convolution over a semiring is easy and when it's hard. For instance, the $(\min,\max)$ convolution can be computed in subquadratic time: I believe that $O(n^{3/2} \log n)$ operations suffice. This can be obtained from adapting the $(\min,\max)$ matrix multiplication algorithm in my work with Vassilevska and Yuster on all-pairs bottleneck paths. Basically you reduce the problem to computing $\sqrt{n}$ instances of a $(+,\times)$ ring convolution.
 http://mathoverflow.net/questions/10237/does-the-convolution-theorem-apply-to-weaker-algebraic-structures/14199#14199 Answer by A. Johnston for does the "convolution theorem" apply to weaker algebraic structures? A. Johnston 2010-02-04T22:43:41Z 2010-02-04T22:43:41Z
Look to the Chinese remainder theorem over rings. The discrete Fourier transform is just one example.

The discrete Fourier transform over a field $\mathbb{F}$ (for example $\mathbb{C}$ or $\mathbb{F}_{q}$) using an $n^{th}$ primitive root of unity $\alpha$ (in $\mathbb{C}$, $\alpha=e^{2\pi i/n}$) of a vector $\left[a_{n-1}\ a_{n-2}\ \ldots\ a_0\right]$ is equivalent to evaluating the polynomial $\sum^{(n-1)}_{j=0}a_jx^k$ at $\alpha^k$ for $k=0,1,\ldots, (k-1)$: $\mathcal{F}\left( \sum^{(n-1)}_{j=0}a_jx^k\right)= \left[\sum^{(n-1)}_{j=0}a_j(\alpha^k)^j\right]_{j=0}^{n-1}$ and the inverse DFT can be equated to (or easily derived from) Lagrangian interpolation.

Interpolation is a special case of the Chinese remainder theorem over $\mathbb{F}[x]$. Reducing a polynomial $f(x)$ modulo $(x-\alpha)$ is identical to evaluating $f(x)$ at $\alpha$ ($x\equiv \alpha \bmod{(x-\alpha)}$). Interpolation of the set of distinct points $\left\{(\alpha_k,\beta_k),\ k=0,1,\ldots,(n-1)\right\}$ is equivalent to finding the polynomial $f(x)\in \mathbb{F}[x]/\left(\prod_{k=0}^{n-1}(x-\alpha_k)\right)\mathbb{F}[x]$ such that $f(x)\equiv \beta_k\bmod{(x-\alpha_k)}$: $\begin{array}{rl}f(x)&\equiv\sum_{k=0}^{n-1}\left(\prod_{j\neq k}(x-\alpha_j)\right)\left(\left(\prod_{j\neq k}(x-\alpha_j)\right)^{-1}\beta_k\bmod{(x-\alpha_k)}\right)\\ &\equiv \sum_{k=0}^{n-1}\left(\frac{\prod_{j\neq k}(x-\alpha_j)}{\prod_{j\neq k}(\alpha_k-\alpha_j)}\right)\beta_k \bmod{\prod_{k=0}^{n-1}(x-\alpha_k)}\end{array}$ The second line of that equation, minus the mod, is the Lagrangian interpolation formula. The mod just means that any polynomial of the form $f(x)+h(x)\prod_{k=0}^{n-1}(x-\alpha_k)$, where $h(x)$ is any polynomial over $\mathbb{F}$, will pass through these same points.

For the DFT the final modulus product is $\prod_{k=0}^{n-1}(x-\alpha^n)=(x^n-1)$ and polynomial multiplication modulo $(x^n-1)$ is equivalent to the convolution of two $n$ long vectors.

The butterfly effect of the FFT comes from using a $2^n$-th primitive root of unity, $\alpha$. A similar effect (though not as elegant) can be achieved using $3^n$-th primitive roots of unity (or any other $p^n$) and combining specified sets of three ($p$).

What the Chinese remainder theorem is really giving you is an isomorphism between the factor rings $R/\bigcap_{k=0}^{n-1}I_k\cong \bigotimes_{k=0}^{n-1}R/I_k$ where $I_k<R$ are ideals in $R$ with special properties. If $R$ is a Euclidian domain, such as $\mathbb{Z}$ or $\mathbb{F}[x]$, these special properties amount to the ideals being generated by relatively prime elements. For example over $\mathbb{Z}$ the ideals would look like $I_k=p_k\mathbb{Z}$ with $GCD(p_k,p_j)=1$ for all distinct pairs of $(k,j)$, and the intersection of the ideals would be $\bigcap I_k=\left(\prod p_k\right)\mathbb{Z}$. Any operations which done modulo the product can be performed in parallel modulo each factor.
 http://mathoverflow.net/questions/10237/does-the-convolution-theorem-apply-to-weaker-algebraic-structures/14204#14204 Answer by Dan Piponi for does the "convolution theorem" apply to weaker algebraic structures? Dan Piponi 2010-02-04T23:19:48Z 2010-02-04T23:19:48Z
You can do infimal convolution using the Legendre transform. You can certainly implement a reasonably fast algorithm for infimal convolution of convex piecewise linear functions though I don't know about the fastest known algorithms that don't use the Legendre transform. The Legendre transform is also called the Fenchel conjugate in optimisation papers so that's another good term to google on.


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# A proof of the Arithmetic-Geometric mean inequality.

· Uncategorized

The following result is essential to the proof.

If xεR, x≥0, then (1+x/(n-1))n-1≤(1+x/n)n, nεω, n≥2. Equality holds iff x=0. (*)

Proof

If xεR, x≥0, and nεω, n≥2, then the following expression may be expanded using the binomial Theorem …

(1+x/n)n-(1+x/(n-1))n-1

= 1 +∑1≤r≤n n(n-1) … (n-r+1)/r! (x/n)r

-(1 +∑1≤r≤n-1 (n-1)(n-2) … (n-r)/r! (x/(n-1))r)

=∑1≤r≤n-1 xr/r![(n-1)/n … (n-r+1)/n – (n-2)/(n-1) … (n-r)/(n-1)]

+ (x/n)n.

It is straightforward to show that …

[(n-1)/n … (n-r+1)/n – (n-2)/(n-1) … (n-r)/(n-1)]>0 since, (n-1)/n > (n-2)/(n-1) … and (n-r+1)/n > (n-r)/(n-1).

Hence it follows that (1+x/n)n-(1+x/(n-1))n-1≥(x/n)n if x≥0.

The result, i.e. (1+x/(n-1))n-1≤(1+x/n)n, follows easily from this, and if equality holds, then (x/n)n = 0, and so x=0.

Statement and proof of the Arithmetic-Geometric Mean inequality.

For each nεω, let P(n) be the proposition that for any set of non-negative numbers a1, a2, … an,

(a1 + a2 + …. + an)/n ≥(a1a2 … an)1/n

with equality if and only if a1 = a2 = a3 … = an.

Proof

P(1) is trivially true, i.e. a1/1 = a11/1.

P(2) If a1 and a2 are ≥0 then it is easily shown that …

(a1 + a2)2≥ 4a1a2 and so (a1 + a2)/2 ≥ √(a1a2).

If (a1+ a2)/2 = √(a1a2) then (a1 – a2)2 = 0 and so a1 = a2.

Now suppose that P(n-1) is true for n-1 ε ω and n – 1≥ 2.

Suppose that a1, a2, … an are n non-negative numbers. We can assume that a1≤a2≤a3 …≤ an. If not then we can choose a re-arrangement of a1, a2, … an that does have this property. It now follows that …

(a2 + a3 + … an)/(n-1)≥(a2a3 … an)1/(n-1)or

a2a3… an≤((a2+ a3+ … an)/(n-1))(n-1) with equality iff a2 = a3 = a4 … = an.

Now let D = (a2 – a1) + (a3 – a1) + … + (an – a1). Clearly D≥0 with equality iff a1=a2=a3 … = an.

D = a2 + a3 + … +an – (n-1)a1 and so, a1 + D/(n-1) = (a2 + a3 + … + an)/(n-1).

So, a2a3… an≤(a1 + D/(n-1))(n-1) with equality iff a2= a3= a4… = an.

Hence, multiplying by a1 we get a1a2a3… an≤a1(a1+ D/(n-1))(n-1).

Applying (*) above we get a1a2a3… an≤a1(a1+ D/(n-1))(n-1)≤(a1+ D/n)n, or

a1a2a3… an≤(a1+ D/n)n=((a1 + a2 + … an)/n)n, with equality iff D=0 or iff a1=a2=a3… = an.

Hence it follows that (a1 + a2 + … +an)/n≥ n√(a1a2 … an) with equality iff a1=a2=a3… = an. Thus P(n) is true and the result follows by the principle of Mathematical Induction.

Applications of the AGM to the sequence n1/n, where nεω, n≥1.

Numerical computation of the first few terms of this sequence shows that the first term 11/1 = 1, the second is 21/2 = √2 and the third term is 3√3 ( the maximum value of the sequence). For n≥3 the sequence appears to be monotone decreasing, tending to a limit of 1.

Firstly, 2 subsequent values cannot be equal. Otherwise n1/n = (n+1)1/(n+1) for some n. From this equation (if true) it follows that n(n+1) = (n+1)n. If n=1 then it follows that 1 = 2 which is obviously false. If n>1 then n and n+1 must have a common prime factor, again false since n and n+1 are coprime.

So, for all nεω, n≥1, either n1/n> (n+1)1/(n+1) or n1/n< (n+1)1/(n+1).

Suppose that n1/n> (n+1)1/(n+1). Then, n(n+1)> (n+1)n.

So, n(n+1)– nn> (n+1)n-nn and from this it follows that …

nn(n-1)> (n+1)n-1 + (n+1)n-2n + … + nn-1 >n(n(n+1))(n-1)/2 (applying the AGM).

Simplifying nn(n-1)>n(n(n+1))(n-1)/2gives n-1>((n+1)/n)(n-1)/2>1. Hence it follows that n≥3.

So, for all nεω, n≥3, n1/n> (n+1)1/(n+1).

To prove that lim n→∞ n1/n = 1, first observe that n1/n> 1 for all n> 1, otherwise n1/n≤1 for some n>1. n1/n≤ 1 implies that n≤1, which clearly gives a contradiction, so n1/n> 1 for all n> 1.

Next, if n>2, we apply the AGM inequality to the set of n positive numbers 1, 1, … ,1 (n-2 times),√n,√n.

This gives (n-2+2√n)/n>n1/n, or 1-2/n+2/√n>n1/n. This last inequality implies that as n→∞, n1/n approaches arbitrarily close to 1. Hence the result limn→∞ n1/n= 1.


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# Why have both: BJT and FET transistors on IC output?

This is the structure of FAN3100 gate driver IC:

(taken from its datasheet)

As you can see - there are two ouput switches: CMOS and BJT.

Why they put them both?

• Another question that arises is why the bottom NPN is NPN and not PNP – Harry Svensson Aug 2 '19 at 13:36
• Notice the differential inputs. That makes this driver more immune to Ground upsets. – analogsystemsrf Aug 2 '19 at 15:12

## 3 Answers

Paragraph 2 of the description says:

FAN3100 drivers incorporate MillerDriveTM architecture for the final output stage. This bipolar-MOSFET combination provides high peak current during the Miller plateau stage of the MOSFET turn-on / turn-off process to minimize switching loss, while providing rail-to-rail voltage swing and reverse current capability.

At the bottom of page 14 in the section *MillerDrive Gate Drive Technology" it goes on to explain:

The purpose of the MillerDrive architecture is to speed up switching by providing the highest current during the Miller plateau region when the gate-drain capacitance of the MOSFET is being charged or dischared as part of the turn-on / turn-off precess. For applications that have zero voltage switching during the MOSFET turn-on or turn-off interval the driver supplies high peak current for fast switching even though the Miller plateau is not present. This situation often occurs in synchronous rectifier applications ecause the body diode is generally conducting before the MOSFET is switched on.

The answer to "Who can tell me about Miller Plateau?" explains it thus:

When you look at the datasheet for a MOSFET, in the gate charge characteristic you will see a flat, horizontal portion. That is the so-called Miller plateau. When the device switches, the gate voltage is actually clamped to the plateau voltage and stays there until sufficient charge has been added/ removed for the device to switch. It is useful in estimating the driving requirements, because it tells you the voltage of the plateau and the required charge to switch the device. Thus, you can calculate the actual gate drive resistor, for a given switching time.

The BJTs are able to get the output moving while the MOSFETs are ramping up. The MOSFETS can then provide the rail to rail voltage swing.

• Interesting topology, but I don't understand a thing: how can the lower NMOS turn on, since its Vgs is clamped to ~0,7V by the lower NPN BJT? It will work if the lower mosfet has a very low Vgs(th), but can they make a, say, ~100mV threshold NMOS? I understand that is a simplified schematic, so something could have been left out in that respect, however why not put a buffer symbol before the NPN base if it's there, after all in the positive rail driver there is an inverting buffer before the upper NPN. Not drawing one when there is one seems a silly simplification. – Lorenzo Donati -- Codidact.com Aug 4 '19 at 15:57
• I haven't a clue. I found the question interesting, didn't have a definite answer, did a little research and, to my surprise, had my answer accepted and upvoted. As you say, the block diagram is probably a simplification, the NPN may not be a very good one and there could be some resistance or current limit in its base. – Transistor Aug 4 '19 at 16:01
• Problem solved, thanks! I delved into the datasheet and indeed further down there is a figure (figure 42) that shows the details of the MillerDrive architecture. It shows that both the upper and the lower BJTs have their own driving circuitry, comprised of a couple of MOSFETs. – Lorenzo Donati -- Codidact.com Aug 4 '19 at 21:39
• @Lorenzo, thanks for the feedback. I had scanned the datasheet while researching the answer but missed the significance of that diagram. – Transistor Aug 4 '19 at 21:41
• You're welcome! I still find utterly silly the "simplified diagram". It is not "simplified", it's wrong! If they didn't want to show just that 4 additional MOSFETs for fear of overcomplicate things, it would have sufficed to place a box before the bases of the BJTs with "driver" written on it. Meh! – Lorenzo Donati -- Codidact.com Aug 4 '19 at 21:46

The CMOS and BJT output stages are combined to from one stage, the manufacturer calls this a "MillerDrive(tm)".

Why they do this is explained in the datasheet:

My guess is that they want to achieve a certain (output drive) performance that cannot be achieved by only using CMOS transistors or only using the NPNs with the manufacturing process that they're using for this chip.

The CMOS part helps pulling the output to GND and VDD, the NPNs cannot do that so well as there will always be a $$\V_{CE,sat}\$$ at the GND side and a $$\V_{BE}\$$ at the VDD side.

The NPNs are very likely able to deliver more current and will switch faster. This might be a consequence of the manufacturing process they're using as it is possible that in a different process the MOSFETs are so much better that similar performance could be achieved using CMOS only. Such a process might be more expensive though.

Notice how the top NPN can only make the output reach VDD-0.7 V, I assume it is the job of the mosfet to take care of the last 0.7 V.

It looks as if the BJT's are doing most of the grunt work and the mosfets are taking care of making the output reach VDD and a strong GND.

I could be wrong though.


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# Plus and minus 18 volt DC power supply

#### crankcase

Joined Sep 6, 2011
2
Hi All

I’m looking to rig up a power supply for a set of computer speakers that is missing its original. The speakers use + 18 volt DC, -18 volt DC, and ground (I think) power supply. I plan on getting four 9 volt batteries to test them with but am thinking of how to achieve a more permanent power supply. I guess the amp in these speakers need well matched voltages?

Any suggestions?

I’ve been thinking of a 36 volt power supply hooked up to a 2 resistor voltage divider.
Maybe two 18 Volt supplies (my favorite cheap approach).
Possibly a computer supply to see if the +-12 Volts and +-5 Volts could be hooked together without destroying the supply.
As a last resort, I suppose I could build something if I could get it done for less than $20.00, especially If someone could suggest a complete project design (I would really appreciated it). Also, the supply is no longer available new, none is on ebay, and I have not found a work around on the web. Thanks #### R!f@@ Joined Apr 2, 2009 9,734 Welcome to the forum There are two things to keep in mind when finding a suitable transformer for an amp. 1. Find it's rated voltage 2. Figure out it's needed amperage. 1. Finding voltage. If it is a transistor based amp (output stage) it is a little difficult and I need to know the circuitry. But there is another approach which to find the working voltage of the filter caps. The voltage needed for the Amp to work will be around 10% less than the rated voltage on the filter caps. If the filter caps are not there, then u seriously needs to know SAO of the output devices. If the amp is an IC driving the speaker, then all y need is the data sheet of that chip. 2. Amperage. If like above, if the amp is an IC, then the amps needed can be found using the data specs. By using ohms' law will give u a rough idea on how much amp it will draw at peaks. Same goes for the transistor output stage types. The Ic of the out put device can give a rough idea on how much it can handle depending on the load used. #### SgtWookie Joined Jul 17, 2007 22,220 MPJA carries these regulated 18vdc, 1A plug supplies for$6.95/ea:
http://www.mpja.com/prodinfo.asp?number=18253+PS
That would be a fairly cheap way of getting your ±18v supply. Lots easier and more efficient than trying to build your own, too.

#### crankcase

Joined Sep 6, 2011
2

Thanks for the ideas, but alas, I blew up the amp using two 18 volt cordless drill batteries. I'm ok with how I had it wired, I think I just guessed wrong at the polarities. I had the + and - tied together as the "ground" which left me with a - 18 and + 18 remaining. It measured 36 volts from plus to minus. If I had fused the lines perhaps no smoke would have escaped.

I blew an electrolytic cap. and a larger NPN transistor. The NPN seemed to be matched with a PNP that did not physically look blown. I'm not sure what kind of amp it was, but there was one chip with about 40 pins sitting under a large heat sink. I assume it must be one of the "new" class D jobs. I'm not real familiar with these, but I guess they work similar to a switching power supply. I wish I could have known an easy way to determine polarity.

I wish I knew more of what was going on, but the amp was highly proprietary, and judging from the info. on the net, the manufacturer was perfectly happy with people blowing up their amps since they they no longer sold replacement power supplies anyway.

I'm going to throw a couple of more dollars at it, see if I can learn something, and call it a day. I don't understand the exact reasons for the + and - power supply nor how to determine polarity when no technical documentation is available.

Thanks

#### SgtWookie

Joined Jul 17, 2007
22,220
The markings on the electrolytic capacitors is a big hint as to what the polarity should be. Electrolytic caps won't tolerate reverse voltages; they will have a very high leakage rate (sort of like a forward-biased diode), and that will cause heating of the electrolyte, which will boil and cause the enclosure to rupture, spewing conductive confetti over your circuit board.

Class "D" amplifiers use rather large inductors in their output section, usually toroids.

You have not provided a manufacturer, model #, photo, or any other such information for the amplifier, so there is not much more we can do to help you.

#### RRITESH KAKKAR

Joined Jun 29, 2010
2,829
The voltage needed for the Amp to work will be around 10% less than the rated voltage on the filter caps......
In case the rating of capacitor is higher or lower then, what to do..??

#### R!f@@

Joined Apr 2, 2009
9,734
Only a moron would use caps with lower rating Pritesh.
It is advisable to use higher voltage rating capacitors for long service life.


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## Tuesday, 19 March 2013

### DESIGN PRINCIPLES RESEARCH

1. What are Points, Pixels and Picas?

A pixel is a picture element, and is the same thing as a dot, a printer's smallest measure, i.e. ppi or dpi.

(A pixel, point or dot, are all the same thing)

Pixels were created and are ideal for website designs, and digital imagery. It has no relevancy to print. It is a unit of resolution, rather than a uni of measurement.

These are the standard resolutions for digital imagery:

71 dpi (Mac)

96 ppi (Windows)

An image that is 144x144 pixels displays as 2 inches square, though inches are irrelevant on screen or on the web.

In printers’ measures, there are 12 points to a pica and 6 picas in an inch. So, 12 x 6 = 72; there are 72 points in an inch. And since points and pixels are the same thing, there are 72 pixels in an inch, too.

So, if you want something on screen to be 4 inches wide: 4 x 72 = 288. An element you want to have take up 4 inches horizontally on the screen should be made 288 pixels wide.

Why 72 dpi? Because again, to maintain the basic units of scale between print and monitor, a point is a pixel is a dot, and there are 72 points to an inch — which means there are 72 dots to an inch. If we want something that measures one inch wide to be one inch wide on the screen, we need to set its resolution to 72 dpi.

Here are some examples of the effect of different resolutions on the same image. First, a 250-pixel square at 72 dpi.

Let’s look at the same image at 150 dpi. It should appear on screen slightly more than twice the size of the square above.

And here’s the same square at 35 dpi. This time, it’s going to appear to be about half the size of the first square.

The more dots per inch, or pixels per inch, the high the resolution, due to the higher quality image. Because the number of dots / points in an inch is the true measure of resolution, the more dots I can fit in an inch, the sharper the picture will be in print.

Pixels and Computer Graphics:

16 Pixels = 1 Pica
1 Pixel = 0.75 Points

So a 12 point font is equvilent to 16 pixels.

1 Pixel = 0.010416667 inches = 96 Pixels = 1 inch

Picas:

The standard unit of measurement used in Graphic Design, desktop publishing and in other design and print industries.

1/6th of an inch = 0.166

Dimensions of less than full pica increments are measured in points, which should be familiar to anyone types on a computer - linking to the sizes of type.

12 Points = 1 Pica
1 Point = 0.013837 inches
72 Points (6 Picas) = 0.996264 inches (1 Inch)

Hence 72dpi

The actual difference between 72 points or 6 picas to 1 inch is actually 1/268.

Each interior layout consists of sharp margins between photos and blocks of text all at 1 Pica.

"pica is a typographic unit of measure corresponding to 1/72 of its respective foot, and therefore to 1/6 of an inch. The pica contains 12 point units of measure.

The pica originated around 1785, when François-Ambrose "L'éclat" Didot (1730–1804) refined the typographic measures system created by Pierre Simon Fournier le Jeune (1712–1768). He replaced the traditional measures of cicéro, Petit-Roman, and Gros-Text with “ten-point”, “twelve-point”, etc.
To date, in printing these three pica measures are used:
• The French pica of 12 Didot points (also called cicéro) generally is: 12 × 0.376 = 4.512mm (0.177in).
• The American pica measure of 0.013837 ft. (1/72.27 ft.). Thus, a pica is 0.166044in. (4.2175mm)
• The contemporary computer pica is 1/72 of the Anglo-Saxon compromise foot of 1959, i.e. 4.233mm or 0.166in. Notably, Adobe PostScript promoted the pica unit of measure that is the standard in contemporary printing, as in home computers and printers.
Note that these definitions are different from a typewriter's pica setting, which denotes a type size of ten characters per horizontal inch.
Usually, pica measurements are represented with an upper-case "P" with an upper-right-to-lower-left virgule (slash) starting in the upper right portion of the "P" and ending at the lower left of the upright portion of the "P"; essentially drawing a virgule ( / ) through a "P". (P̸)Likewise, points are represented with number of points before a lower-case "p", for example, 5p represents “5 points”, and 6P̸2p represents “6 picas and 2 points”, and 1P̸1 represents “13 points”, which is converted to a mixed fraction of 1 pica and 1 point.
Publishing applications such as Adobe InDesign and QuarkXPress represent pica measurements with whole-number picas left of a lower-case "p", followed by the points-number, for example: 5p6, represents 5 picas and 6 points, or 5½ picas.
Cascading Style Sheets defined by the World Wide Web Consortium use "pc" as the abbreviation for pica (1/6 of an inch), and "pt" for point (1/72 of an inch)."

2. What are Grids, Columns and Gutters?
1. Columns: Vertical divisions of space on a page.
2. Rows: A series of flowlines that create horizontal divisions of space on a page.
3. Gutters: Space that separates rows and columns or two facing pages.
Grids are usually the 1st step a designer takes when designing a layout, they are invisible when finished/printed, and can be moved around and deleted as pleased. They are made up of vertical lines (columns) and horizontal lines (rows), and constructed through the use of rulers, giving order to Graphic Design, and simplifying the designers job.

There are various types of grids which can be used.

Perhaps the simplest method is to use the rule of thirds. This involves dividing a page into thirds with two equally spaced vertical and/or horizontal lines so that important compositional elements can be placed along these lines or intersections. This method is also commonly used in photography. Another popular method uses the golden ratio (approximately 1.618)  for proportioning. This ratio has been studied by mathematicians for its presence throughout the universe and has been used for centuries by architects, artists and book designers for it’s aesthetically pleasing qualities.

Most newspapers and magazines use equally sized columns with a baseline grid underneath which all typography and body copy sits on, so it is aligned accurately.

There is no right or wrong grid system I don't think, as long as it maintains function and purpose, as well as being an aesthetically pleasing, legible and readable piece of design.

Depending on the content that they will contain, grids vary in size, shape and complexity. However, every grid is made up of the following parts:
1. Margins: The space that separates the content from the edge of the page.
2. Flowlines: Alignments that break the space into horizontal bands.
3. Columns: Vertical divisions of space on a page.
4. Rows: A series of flowlines that create horizontal divisions of space on a page.
5. Gutters: Space that separates rows and columns or two facing pages.
6. Modules: Individual units of space created from intersecting rows and columns.
7. Spatial Zones: Groups of modules that cross multiple rows and columns.
Many designers have come up with grid systems to simply their workload and make their job easier. Below is an example of one of these grid systems: The Golden Grid System for Responsive Design.

The conventional use of grids can be broken, which I personally think can create an edgier, more creative layout design depending on the content and context of the work being created. However when breaking these 'rules' important consideration is needed.

3. What is a DPS and what does it do?

A DPS, is also known as a Double Page Spread. It revolves around aesthetics of a layout or publication, as well as functionality, and costs.

However, it allows for larger images, and more body copy, whilst also encouraging the viewer to maintain interest.

I personally really like double page spreads, I feel that they work with editorial publications especially really well, and with newspapers. They are so commonly used, it is unlikely to see a publication with single leaves. However, like with the grid system, if broken creative ambitions can be shown, leading to unique designs and style, as long as it is relative to content, context and format.

4. What is a Golden Section?

Phi, 1.618 (also known as the Golden Ratio, Golden Mean, Golden Section or Divine Proportion, in mathematicsgeometrylife and the universe and shows you how to apply it, and its applications are limitless:

## The Golden Section is a ratio based on a the number Phi, 1.618…

The Golden Section or Ratio is is a ratio or proportion defined by the number Phi (= 1.618033988749895… )
It can be derived with a number of geometric constructions, each of which divides a line segment at the unique point where:
the ratio of the whole line (A) to the large segment (B)
is the same as
the ratio of the large segment (B) to the small segment (C).
In other words, A is to B as B is to C.
This occurs only where A is 1.618 … times B and B is 1.618 … times C.
It was used in the design of Notre Dame in Paris, and continues today in many examples of art, architecture and design.
golden rectangle with longer side a and shorter side b, when placed adjacent to a square with sides of length a, will produce asimilar golden rectangle with longer side a + b and shorter side a. This illustrates the relationship $\frac{a+b}{a} = \frac{a}{b} \equiv \varphi$.

The drawing of a man's body in a pentagram suggests relationships to the golden ratio.

The 16th-century philosopher Heinrich Agrippa drew a man over a pentagram inside a circle, implying a relationship to the golden ratio.
Leonardo da Vinci's illustrations of polyhedra in De divina proportione (On the Divine Proportion) and his views that some bodily proportions exhibit the golden ratio have led some scholars to speculate that he incorporated the golden ratio in his paintings.

There was a time when deviations from the truly beautiful page proportions 2:3, 1:√3, and the Golden Section were rare. Many books produced between 1550 and 1770 show these proportions exactly, to within half a millimeter.

If φ were rational, then it would be the ratio of sides of a rectangle with integer sides. But it is also a ratio of sides, which are also integers, of the smaller rectangle obtained by deleting a square. The sequence of decreasing integer side lengths formed by deleting squares cannot be continued indefinitely, so φ cannot be rational.

I think the evidence and examples shown above, prove that the golden section is a fool proof way of constructing page proportions. Using divine proportions have been proven to work and will always consist doing so.

5. What are rulers, boxes, folio numbers, and drop caps?

Rulers: Either a physical ruler, or digitally on design software to set guides for grids and layout of design elements.
Boxes: Square box format, which are used for text, image or vectors.
Folio Numbers: Usually on the front side of the leaf/page number.
Drop Caps: The first letter of a paragraph that is enlarged to "drop" down two or more lines, as in the next paragraph. Drop caps are often seen at the beginning of novels, where the top of the first letter of the first word lines up with the top of the first sentence and drops down to the four or fifth sentence to the beginning of a section.

6. What are sub-heads, paragraphs, captions and ligatures?


text:
# Zebrafish reveal drugs that may improve bone marrow transplant

July 22, 2015

Compounds boost stem cell engraftment; could allow more matches for patients with cancer and blood diseases

By Nancy Fliesler
Boston Children’s Hospital

Using large-scale zebrafish drug-screening models, Harvard Stem Cell Institute (HSCI) researchers at Boston Children’s Hospital have identified a potent group of chemicals that helps bone marrow transplants engraft or “take.” The findings, featured on the cover of the today’s issue of Nature, could lead to human trials in patients with cancer and blood disorders within a year or two, says senior investigator Leonard Zon, MD, a member of the HSCI Executive Committee and a Professor in Harvard’s Department of Stem Cell and Regenerative Biology.

The compounds, known as epoxyeicosatrienoic acids, or EETs, boosted stem cell engraftment in both zebrafish and mice and could make human bone marrow transplant more efficient. Better engraftment could also allow umbilical cord blood to be used as an alternative to marrow as a source of blood stem cells, greatly increasing a patient’s chances of finding a matched donor and enhancing safety.

“Ninety percent of cord blood units can’t be used because they’re too small,” explains Zon, who directs the Stem Cell Research Program at Boston Children’s. “If you add these chemicals, you might be able to use more units. Being able to get engraftment allows you to pick a smaller cord blood sample that might be a better match.”

EETs are fats that appear to work by stimulating cell migration. They were among the top hits in a screen of 500 known compounds conducted in Boston Children’s newly upgraded Karp Aquatics Facility. While zebrafish have previously led Zon’s team to compounds that boost blood stem cell number, such as prostaglandin (currently in several clinical trials under the name ProHema), the new drug screen specifically tested the stem cells’ transplantability and engraftment.

Red fish, green fish

The screen was done in a lab-created strain of zebrafish called Casper. Because Casper is see-through, Zon and colleagues could visually compare engraftment of transplanted blood stem cells chemically tagged to glow green or red—in what they’ve dubbed the “Dr. Seuss experiment.”

Led by co-first authors Pulin Li, PhD, Jamie Lahvic, and Vera Binder, MD, the researchers first used tagging to color the fishes’ marrow either red or green, then removed blood stem cells for transplantation. The green cells were incubated with various chemicals, while the red cells were left untreated. Next, the researchers injected a mixture of green and red marrow cells into other groups of zebrafish (10 fish per test chemical). The team then visually tracked the cells’ activity in the transplant recipients and measured the green:red ratio.

“We call this a competitive transplant model because we can literally compete a green stem cell against a red stem cell and see what wins,” says Zon. “The expectation was that if chemical didn’t increase engraftment, all the fish would be equal parts red and green. But if it was effective, green marrow would predominate.”

That was the case for green marrow incubated with EETs, a finding that held up over thousands of marrow transplants. “In a mouse system, this experiment would cost $3 million,” notes Zon. “In fish, it cost about$150,000.”

In a smaller-scale set of mouse experiments, the team confirmed EETs’ efficacy in promoting homing and engraftment of transplanted blood stem cells in mammals.

Although EETs are chemical cousins of prostaglandin (both are made from arachidonic acid, and both are made during inflammation), EETs work in a different way, by activating a pathway known as PI3K.  EETs also enhanced PI3K activity in human blood vessel cells in a dish.

After more studies in human cells to tease out how EETs work, Zon hopes to begin clinical trials of EETs within the next two years, likely in the setting of cord blood transplant. The lab is also investigating its other top hits from the zebrafish screen.

“Every new pathway that we find has the chance of making stem cell engraftment and migration even better,” says Zon. “I think we’ll end up being able to manipulate this process.”

Supporters of the study include the Howard Hughes Medical Institute, the National Institutes of Health (NIH 1R01HL04880, Z01 ES025034, NIH-P50-NS40828, NIH-P30-HD18655), the National Cancer Institute (ROCA148633-01A5), DFG and the Care-for-Rare Foundation. Zon is a founder and stockholder of Fate, Inc. and a scientific advisor for Stemgent.


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# Simplifying the approach, to finding roots of polynomials.

In some cases, the aim of my postings is to say, ‘I am able to solve a certain problem – more or less – and therefore, the problem is solvable.’ It follows from this position that my solutions are not assumed to be better by any means, than mainstream solutions. So recently, I suggested an approach to finding the roots of polynomials numerically, again just to prove that it can be done. And then one observation which my readers might have made would be, that my approach is only accurate to within (10-12), while mainstream solutions are accurate to within (10-16). And one possible explanation for this would be, that the mainstream solutions polish their roots, which I did not get into. (:1)

(Edit 2/8/2019, 6h40 : )

A detail which some of my readers might have missed is, that when I refer to a ‘numerical solution’, I’m generally referring to an approximation.

(End of Edit, 2/8/2019, 6h40 . )

But another observation which I made, was that Mainstream Code Examples are much tighter, than what I suggested, which poses the obvious question: ‘Why can mainstream programmers do so much, with much less code complexity?’ And I think I know one reason.

The mainstream example I just linked to, bypasses a concept which I had suggested, which was to combine conjugate complex roots into quadratic terms, which could be factorized out of the original polynomial as such. What the mainstream example does is to assume that the coefficients of the derived polynomials could be complex, even though the original one only has real coefficients. And then, if a complex root has been found, factorizing it out results in such a polynomial with complex coefficients, after which to factorize out the conjugate, causes the coefficients of the quotient to become real again.

(Edited 1/30/2019, 8h50… )

(Updated 2/9/2019, 23h50… )

I’ve just written some source-code of my own, to test my premises…

(As of 1/26/2019, 20h15: )

At first glance this might cause the object-code to seem more complicated. But it causes simpler source code because a variable can just be declared to be of complex type, both in the FORTRAN example linked to, and in other high-level programming languages, such as in C++.

1:) “Polishing A Root” is usually a practical trick used when solving them, which uses the remainder after the synthetic division (see “Deflation” according to the reference document linked to above), which would be zero if the divisor was exact, in order to make the divisor more exact, rather than to rely entirely on the search algorithm, to generate a root as accurately as possible.

(Edit 1/30/2019, 8h50 : )

I realize that this way of polishing the root differs, from what the reference document suggests, which I linked to above. (:3)

I have just gotten the full concept to work. The trick that I needed was, to use the root implied by the divisor, and to plug that in to the quotient, as my basis for estimating what the derivative of the remainder could be, with respect to the divisor. I make two passes.

This results in yet another situation, where a doubled root would prevent my underlying strategies from working – since they’d result in a derivative of zero as well – except for the fact that I’ve dealt with them explicitly, according to my own ‘Hypothetical Algorithm’, linked to above, which is no longer hypothetical.

The reason for which this works, could be as follows:

At the root of the divisor, the divisor should be equal to zero. But, assuming that the root is not doubled, the quotient should be non-zero. In theory, the two would be multiplied, to result in the dividend. Hence, the effect of small changes in the divisor, on the dividend, should be equal to the quotient. The only fact which makes this harder to understand, is that the divisor does not exactly correspond to a root of the dividend, for which reason there is a non-zero remainder

What can be observed in the case of the first-order dividend, which is that the derivative is always (-1) when the quotient is the constant term, is also to be applied in non-trivial cases, by using the negative of this quotient-value, at the root implied by the divisor. The remainder needs to be divided by this derivative, and then either subtracted from the divisor, or added to the estimated root. (:2)

What results is output accurate to within (10-14) instead of only to within (10-12).

http://dirkmittler.homeip.net/text/roots_1.cpp

(Code Updated 2/1/2019, 13h45. )

The source-file ‘roots_1.cpp’ is hard-coded, only to produce the test output below.

dirk@Plato:~/Programs/Dirk_Roots$./roots_1 x^3 -3x^2 -2x + 6 = 0 : 1.4142135623731 + 0I -1.4142135623731 + 0I 3 + 0I x^3 -3x^2 -2x + 15 = 0 : 2.4768359735688 + -1.2422292374924I 2.4768359735688 + 1.2422292374924I -1.9536719471376 + 0I x^4 +2x^2 + 1 = 0 : 0 + 1I 0 + 1I 0 + -1I 0 + -1I x^5 = 0 : 0 + 0I 0 + 0I 0 + 0I 0 + 0I 0 + 0I x^8 + 3x^6 -3x^5 +x^4 +x^3 +2x^2 + 1 = 0 : -0.44236729147741 + 1.8678073968046I -0.44236729147741 + -1.8678073968046I 0.12390731304389 + 0.59992017540847I 0.12390731304389 + -0.59992017540847I -0.60417361259701 + 0.49571073126293I -0.60417361259701 + -0.49571073126293I 0.92263359103053 + -0.57704637832846I 0.92263359103053 + 0.57704637832846I dirk@Plato:~/Programs/Dirk_Roots$

Now, in case the reader prefers a C++ program which, when compiled, gives a command-line from which to type in arbitrary polynomials, I have created a rudimentary command-line version of the same program:

http://dirkmittler.homeip.net/text/roots_2.cpp

(Comment 2/5/2019, 20h35 : )

Some readers might find it odd, that I used the header file <complex.h> instead of <complex>, in roots_1.cpp, along with the old function ‘cexp()’. The reason I did that is the fact that As of December 2016, the standard which Linux builds under was changed to GNU++14, replacing C++14, and creating issues.

I’m fully able to compile the code with the more-correct function overloads, if I set the flag ‘-std=c++14′. But what I really wanted, was some quick-and-dirty source-code, which would compile without my readers having to set compiler flags. If using the header file <complex>, in the function ‘search_root()’, near Line 148, the relevant line of code actually needs to be changed to:

CC xb = exp(M_PI * 1i / CC(N + 1.0, 0.0));

Along the same lines, the operation is not always defined, to divide a complex number by an integer, or to multiply one so. Whether that’s defined or not may depend on one specific header file… Therefore, the more-correct thing to do, is to convert the integer to a complex number first, and then to divide by it…

(End of Comment, 2/5/2019, 20h35 . )

(Comment, 3/2/2019, 19h10 : )

In case the reader did not understand, the way to compile the second of the two source-files, which is the useful one, on an up-to-date Linux computer, is:

$g++ -std=c++14 -Wall -o roots_2 roots_2.cpp   (End of Comment, 3/2/2019, 19h10. ) (Comment, 3/2/2019, 23h00 : ) The source code in ‘roots_1.cpp’ previously only compiled properly on an up-to-date, Debian / Stretch computer with ‘-std=gnu++14′. Because it’s generally important to be able to reproduce even test results, I have now edited this code slightly, so that it will also compile and run properly on a Debian / Jessie machine with: $ g++ -std=gnu++11 -Wall -o roots_1 roots_1.cpp

(End of Comment, 3/2/2019, 23h00. )

(Warning, 2/9/2019, 23h50 : )

There exists a bizarre bug in the header files and/or libraries of some Linux computers, when trying to use the (correct) include <complex>. The overloaded absolute function ‘abs()’ now works for complex parameters, but no longer for real parameters! This affected how the secondary polishing works on Debian / Jessie computers. The effect is supposed to be, that imaginary or real components less than ‘epsilon’ be replaced with zero. The effect was that components less than 1.0 were replaced with zero.

Just in case the same bug exists on other architectures, I’ve written a workaround.

(End of Warning, 2/9/2019, 23h50 . )

(Update 2/2/2019, 10h55 : )

2:)

I suppose that a valid question which my reader could have, since the art of guessing the derivative of the remainder, with respect to the divisor, for the purpose of polishing the root, is admittedly guesswork on my part, is how I can be reasonably confident that it ‘works’.

And the answer is relatively straightforward. My C++ program actually tests each root it estimates, and does not just estimate a root, by plugging that root in to the polynomial, and verifying that the value of (y) which results is extremely small – small to whatever extent matches the level of accuracy, with which (x) was to be found. If the value of (y) is not as small as the situation requires, the algorithm stops computing any roots, and also stops outputting any results, for any given polynomial. The result is then no output.

Well, when I had tried other methods of estimating the derivative of the remainder, and had set the Boolean ‘polish’ to ‘true’, I was getting no output, for certain polynomials. But if I set ‘polish’ to ‘false’, I got output. What this means is, that earlier attempts to polish were throwing the value of (x) further away from the exact root, than (x) already was, so that a  value of (y) followed, which was no longer as close to zero as it was supposed to be. Then, those roots were omitted from the output because polishing them had actually made them inaccurate.

So what I know now is, that my present method of polishing the roots, does not make them less accurate than they were before, even though I’m doing so twice.

As for knowing that my accuracy has improved:

I can set the level of precision, with which the roots are printed, to 14 decimal digits. Without polishing, even in the case where I knew digits would be repeated, the last 2 digits of output, of real roots, were not repeated exactly. This would be consistent with having set ‘epsilon’ to (10-12). But if I set ‘polish’ to true, then all 14 printed decimal digits are consistent.

(Update 2/3/2019, 5h50 : )

3:)

I suppose that a valid question to ask might be, ‘Why doesn’t Dirk use the method described in the document linked to, to polish the roots?’

The answer is the fact that the document recommends, that the last root found by the search algorithm, should be plugged in to a second invocation of the search algorithm, as the neighbourhood of the root to find. The problem with that is the fact, that not all search algorithms become more-accurate when this is done. It just so happens that the LaGuerre method outputs a slightly inaccurate result, if the initial guess as to where the root might be, was greatly inaccurate.

But my custom search algorithm is not more or less accurate, based on whether the initial guess was close to the exact root or not. So in order to polish the root, I’d either need to implement two search algorithms, or devise a different way to polish the root – my chosen way being, to try to use the remainder of the deflation / factorization, which is not a foolproof method by nature.

For this reason, the command-line version of my C++ program allows the user to turn polishing off.

Similarly, the question could be asked of why, I use the complicated search algorithm to solve a linear equation, i.e., a polynomial of degree (1), as this was not what that algorithm was initially meant to do.

And the answer would be the fact that although a separate solver could easily be written, just for the case of degree-1 equations, I’d also need to put code that tests for this case, so that the additional solver will get used. This would tend to make my source-code longer overall, even though it would ultimately cause fewer CPU cycles to get spent.

Because my complicated search-algorithm can solve such linear equations in one iteration, its output is more-exact than the output would be for equations of a higher degree, and that solver can remain in-use, while also keeping my source-code simpler.

Dirk

## 2 thoughts on “Simplifying the approach, to finding roots of polynomials.”

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# Mistuned Toeplitz operators and applications

 Orateur: BOULTON Lyonell Localisation: Université Heriot-Watt, Royaume-Uni Type: Séminaire problèmes spectraux en physique mathématique Site: IHP Salle: 314 Date de début: 10/06/2013 - 15:30 Date de fin: 10/06/2013 - 15:30

The study of a new family of highly non-self-adjoint operators which is reminiscent of the classical Toeplitz operators arises naturally in the context of evolution problems connected to sandpiles/slow-fast diﬀusion. The aim of this talk is to describe the diﬃculties involved in determining general spectral properties of operators in this family and show how to overcome these diﬃculties. In turns we will establish basisness/non-basisness theorems for general families of 1D periodic functions. We will then apply the latter, in order to examine non-orthogonal projection methods for the $p$-poisson parabolic time-evolution initial value problem with stochastic forcing.


text:
# Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 2

$\frac{9}{2}$

#### Work Step by Step

Let $I= \int^{\pi}_{0}3sin\frac{x}{3}dx$ $\int 3sin\frac{x}{3}dx= 3\int sin\frac{x}{3}dx$ $= 3\times(\frac{-cos\frac{x}{3}}{\frac{1}{3}})+C$ $= -9(cos\frac{x}{3})+C= F(x)$ By the second fundamental theorem of calculus, we have $I= F(\pi)-F(0)$ $= -9(cos\frac{\pi}{3})-[-9(cos\frac{0}{3})]$ $= (-9\times\frac{1}{2})-(-9\times1)$ $= -\frac{9}{2}+9 = \frac{9}{2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Working Principle of Autotransformer Starter

Digital ElectronicsElectronElectronics & Electrical

## Circuit and Working Principle of Autotransformer Starter

The circuit diagram of an autotransformer starter for starting a 3-phase induction motor is shown in the figure. The autotransformer starter can be used for starting both star and delta connected 3-phase induction motors. In this method, the starting current of the motor is limited by using a 3-phase autotransformer to decrease the initial applied voltage to the stator. The autotransformer is provided with a number of tappings to obtain the variable voltage.

In the autotransformer starting method, the starter is connected to a particular tapping of the autotransformer to obtain the most suitable starting voltage. A changeover switch S is used to connect the autotransformer in the circuit for starting the motor.

When the handle (H) of the switch S is at the START position, the primary winding of the autotransformer is connected to supply line and the induction motor is connected to the secondary winding of the autotransformer. When the motor picks up the speed (about 80 % of the rated speed), then the handle H is thrown to the RUN position. Consequently, the autotransformer is isolated from the circuit and the motor is now directly connected to the supply line and gets its rated voltage. The handle is held in the RUN position by the under-voltage relay. When the supply voltage fails or falls below a certain value, then the handle is released and comes back to the OFF position. To provide the overload protection to the motor, a thermal overload relay is used in the motor circuit.

## Theory of Autotransformer Starter

The figure below shows the circuit connections for the direct switching of the motor and the autotransformer starting of the motor.

When direct switching of the motor to the supply line −

Let,

𝑍𝑒10 = Per phase equivalent impedance of motor referred to stator at standstill

𝑉1 = Supply voltage per phase

Thus, when the full supply voltage per phase (V1) is applied to the motor with direct switching, then the starting line current drawn by the motor from the supply line is given by,

$$\mathrm{𝐼_{𝑠𝑡𝑙} =\frac{𝑉_{1}}{𝑍_{𝑒10}}… (1)}$$

Now, with autotransformer starting, if a tapping of autotransformer with transformation ratio kis used, then the voltage available across the secondary winding terminals of the autotransformer (i.e., the voltage per phase across the motor) is kV1. Hence, the starting current of the motor with autotransformer starting is given by,

$$\mathrm{𝐼_{𝑠𝑡𝑚} =\frac{𝑘𝑉_{1}}{𝑍_{𝑒10}}… (2)}$$

If the no-load current of the transformer is neglected, then, the ratio of currents in a transformer is inversely proportional to the voltage ratio, i.e.,

$$\mathrm{\frac{𝐼_{1}}{𝐼_{2}}=\frac{𝑉_{2}}{𝑉_{1}}}$$

$$\mathrm{\Rightarrow\:𝑉_{1}𝐼_{1} = 𝑉_{2}𝐼_{2}}$$

Thus,

$$\mathrm{𝑉_{1}{𝐼^{′}_{𝑠𝑡𝑙}} = 𝑘𝑉_{1}𝐼_{𝑠𝑡𝑚}}$$

Where, ${𝐼^{′}_{𝑠𝑡𝑙}}$ is the line current drawn from the supply by the autotransformer.

$$\mathrm{\Rightarrow 𝐼^{′}_{𝑠𝑡𝑙} = 𝑘𝐼_{𝑠𝑡𝑚} … (3)}$$

Substituting the value of 𝐼𝑠𝑡𝑚 from eqn. (2) in eq. (3), we get,

$$\mathrm{ 𝐼^{′}_{𝑠𝑡𝑙}=𝑘(\frac{𝑘𝑉_{1}}{𝑍_{𝑒10}})=\frac{𝑘^{2}𝑉_{1}}{𝑍_{𝑒10}}… (4)}$$

Therefore,

$$\mathrm{\frac{Starting\:current\:with \:autotransformer\:starter\:(𝐼′_{𝑠𝑡𝑙} )}{Starting\:current\:with\:direct\:switching \:(𝐼_{𝑠𝑡𝑙 })}=\frac{\frac{𝑘^{2}𝑉_{1}}{𝑍_{𝑒10}}}{\frac{𝑉_{1}}{𝑍_{𝑒10}}}= 𝑘^{2}}$$

$$\mathrm{\:therefore;\: Starting \:current\: with \:\:autotransformer\: starter= 𝑘^2\: × \:Starting \:current \:with \:direct \:switching … (5)}$$

Since the torque developed in an induction motor is directly proportional to the square of the applied voltage. Thus, the torque with direct switching of the motor is,

$$\mathrm{\tau_{𝑠𝑡𝑑} \propto 𝑉_{1}^{2} … (6)}$$

And, the torque developed with the autotransformer starter is,

$$\mathrm{\tau_{𝑠𝑡𝑎} \propto 𝑘^2 𝑉_{1}^{2} … (7)}$$

Hence,

$$\mathrm{\frac{Starting\:torque \:with\: autotransformer \:starter (\tau_{𝑠𝑡𝑎})}{Starting \:torque \:with \:direct \:switching (𝐼_{𝑠𝑡𝑑})}=\frac{𝑘^2𝑉_{1}^{2}}{𝑉_{1}^{2}} = 𝑘^2}$$

$$\mathrm{\therefore\:Starting\: torque\: with \:autotransformer\: starter\:=\:𝑘^2 × Starting \:torque \:with \:direct \:switching … (8)}$$

From the eqns. (5) and (8), it is clear that with an autotransformer starter, the starting current of the motor drawn from the main supply line and the starting torque are reduced to k2 times of their corresponding values with the direct switching starting of the motor

Published on 23-Aug-2021 14:29:09


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# Power

#### What is Power?

In general, the term power is applied to provide an indication of how much work (energy conversion) can be accomplished in a specified amount of time; that is, power is a rate of doing work.
For instance, a large motor has more power than a smaller motor because it has the ability to convert more electrical energy into mechanical energy in the same period of time. Since energy is measured in joules (J) and time in seconds (s), power is measured in joules/second (J/s).
The electrical unit of measurement for power is the watt (W), defined by
$$\text{1 watt (W)} = \text{1 joule/sec}$$ In equation form, power is determined by $$P = {W \over t}$$ with the energy (W) measured in joules and the time t in seconds.
James Watt
The unit of measurement - the watt - is derived from the surname of James Watt. Who was instrumental in establishing the standards for power measurements. He introduced the horsepower (hp) as a measure of the average power of a strong dray horse over a full working day. It is approximately 50% more than can be expected from the average horse. The horsepower and watt are related in the following manner: $$\text{1 horsepower} = \text{746 watts}$$ The power delivered to, or absorbed by, an electrical device or system can be found in terms of the current and voltage by, $$P = {W \over t} = {QV \over t} = V {Q \over t}$$ but $$I = Q/t$$ so that, $$\bbox[5px,border:1px solid red] {\color{blue}{P = VI}}$$ By direct substitution of Ohm's law, the equation for power can be obtained in two other forms: $$P = VI = V ({V \over R})$$ $$\bbox[5px,border:1px solid red] {\color{blue}{P = {V^2 \over R}}} \text{ (watts,W)}$$ Or $$P = VI = (IR) I$$ $$\bbox[5px,border:1px solid red] {\color{blue}{P = I^2 R }} \text{ (watts,W)}$$
##### Power Related Questions
Energy Previous | Next Efficiency


text:
overview - Maple Help

# Online Help

###### All Products    Maple    MapleSim

Pages Related to Algorithms Used by dsolve for Exact Solutions of ODEs

Description

 • The following links point to help pages containing details of the implementation of specific ODE solving algorithms:

 See Also


text:
# $\max_{i\in \{1,…,M\}} (a_i+ r_i)$ strictly convex?

For any $r\equiv(r_1,...,r_M)\in \mathbb{R}$, consider the function $$a\equiv(a_1,...,a_M) \in \mathbb{R}^M\mapsto G(a)\equiv \max_{i\in \{1,...,M\}} (a_i+ r_i)$$

Is $G$ strictly convex or convex? Could you help me to show it?

• it's definitely convex, I don't think it's strictly convex though. – TSF May 25 '18 at 11:43
• Any proof for non strict convexity @TonyS.F.? – TEX May 25 '18 at 11:53

The maximum of convex functions is convex.

See here.

Since $\phi_i(a)=a_i+r_i$ is convex for every $i$, then $G(a)=\max \{\phi_i(a)\}$ is convex.

Consider $M=1$ and $r=0$, since $G(a)=a$, $G$ is not strictly convex.

• Thank you very much. Two clarifications: 1) $\phi_i$ is linear and hence convex and concave, correct? 2) $G(a)=a$ implies $G$ not strictly convex because it is a linear function (hence convex and concave), correct? – TEX May 25 '18 at 12:00
• $\phi_i$ is affine, but that does not change the conclusion. You are correct. – nicomezi May 25 '18 at 12:04


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# Is the variance of an eigenfunction of a finite state space aperiodic irreducible markov chain starting at a single state always non-decreasing?

I am reposting a previous question due to incorrect initial formulation.

Given an ergodic (aperiodic and irreducible) finite state space Markov chain $P$. Let $f$ be an eigenfunction, i.e., $P_t f = \lambda^t f$. Let the chain start at a single state $x \in \Omega$. Is it true that $\mathbb{E}_x (P_tf- \mathbb{E}_x( P_t f))^2$ is nondecreasing in t? As pointed out by James Martin, it's important to start with the point mass distribution at a single state, since otherwise one could choose an initial distribution that has an f-variance larger than the stationary $f$-variance.

Wilson's method gives a way to bound the variance of eigenfunctions in t in a way that's reminiscent of the Martingale difference method, but since the variance at time $\infty$ is usually easy to calculate, if we know the above monotonicity result, we could bound the variance at finite time easily.

-


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# Perceptual Similarity

##### 🌏 Source

Downloadable at: Open Access - CVPR 2018. Source code is available at: GitHub - richzhang/PerceptualSimilarity.

## Brief Introduction

The paper argues that widely used image quality metrics like SSIM and PSNR are simple and shallow functions that may fail to account for many nuances of human perception. The paper introduces a new dataset of human perceptual similarity judgments to systematically evaluate deep features across different architectures and tasks and compare them with classic metrics.

Findings of this paper suggests that perceptual similarity is an emergent property shared across deep visual representations.

## Main contributions

In this paper, the author provides a hypothesis that perceptual similarity is not a special function all of its own, but rather a consequence of visual representations tuned to be predictive about important structure in the world.

• To testify this theory, the paper introduces a large scale, highly varied perceptual similarity dataset containing 484k human judgments.
• The paper shows that deep features trained on supervised, self-supervised, and unsupervised objectives alike, model low-level perceptual similarity surprisingly well, outperforming previous, widely-used metrics.
• The paper also demonstrates that network architecture alone doesn't account for the performance: untrained networks achieve much lower performance.

The paper suggests that with this data, we can improve performance by calibrating feature responses from a pre-trained network.

## Methodology

### The perceptual similarity dataset

This content is less related to my interests. I'll cover them briefly.

• Traditional distortions: photometric distortions, random noise, blurring, spatial shifts, corruptions.
• CNN-based distortions: input corruptions (white noise, color removal, downsampling), generator networks, discriminators, loss/learning.
• Distorted image patches.
• Superresolution.
• Frame interpolation.
• Video deblurring.
• Colorization.

### Similarity measures

• 2AFC similarity judgments. Two-alternative forced choice (2AFC) is a method for measuring the subjective experience of a person or animal through their pattern of choices and response times. The subject is presented with two alternative options, only one of which contains the target stimulus, and is forced to choose which one was the correct option. Both options can be presented concurrently or sequentially in two intervals (also known as two-interval forced choice, 2IFC).
• Just noticeable differences. Just-noticeable difference or JND is the amount something must be changed in order for a difference to be noticeable, detectable at least half the time (absolute threshold). This limen is also known as the difference limen, difference threshold, or least perceptible difference.

## Deep Feature Spaces

### Network activations to distance

The distance between reference and distorted patches $x$ and $x_0$ is calculated using this workflow and the equation below with a network $\mathcal{F}$. The paper extract feature stack from L layers and unit-normalize in the channel dimension. Then the paper scales the activations channel-wise and computes the $\ell_2$ distance.

$d(x,x_0)=\sum_l\frac{1}{H_lW_l}\sum_{h,w}\|w_l\odot(\hat{y}^l_{hw}-\hat{y}^l_{0hw})\|_2^2$

### Training on this data

The paper considers the following variants:

• lin: the paper keep pre-trained network weights fixed and learn linear weights $w$ on top.
• tune: the paper initializes from a pre-trained classification model and allow all the weights for network $\mathcal{F}$ to be fine-tuned.
• scratch: the paper initializes the network from random Gaussian weights and train it entirely on the author's judgments.

Finally, the paper refer to these as variants of the proposed Learned Perceptual Image Patch Similarity (LPIPS).

## Experiments

### Performance of low-level metrics and classification networks

Figure 4 shows the performance of various low-level metrics (in red), deep networks, and human ceiling (in black).

### Metrics correlate across different perceptual tasks

The 2AFC distortion preference test has high correlation to JND: $\rho = .928$ when averaging the results across distortion types. This indicates that 2AFC generalizes to another perceptual test and is giving us signal regarding human judgments.

### Where do deep metrics and low-level metrics disagree?

Pairs which BiGAN perceives to be far but SSIM to be close generally contain some blur. BiGAN tends to perceive correlated noise patterns to be a smaller distortion than SSIM.

## Conclusions

The stronger a feature set is at classification and detection, the stronger it is as a model of perceptual similarity judgments.

Features that are good at semantic tasks, are also good at self-supervised and unsupervised tasks, and also provide good models of both human perceptual behavior and macaque neural activity.


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+0

# Vieta's formula problem

0
246
2

The roots of the polynomial x^3 - 52x^2 + 581x - k are distinct prime numbers. Find k.

Aug 11, 2019

#1
+6180
+1

$$((x-a)(x-b)(x-c) =x^3 + x^2 (-a-b-c)+x (a b+a c+b c)-a b c \\ 52=a+b+c\\ 581 = ab + ac + bc \\ k = abc$$

$$\text{Solving this is a pain but conceptually straightforward.}\\ \text{Start by knowing one of them must be 2, as the sum of 3 odds is odd and 52 is even}\\ \text{The roots are 2, ~13,~37}\\ k = 2 \cdot 13 \cdot 37 = 962$$

.
Aug 11, 2019
#2
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You could also just use the rational root theorem

Aug 11, 2019


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# Render To Texture - depthStencilView

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Hello,

I recently stumbled over a problem with my render-to-texture mechanism. I was setting

 d3d11DevCon->OMSetRenderTargets(1, &renderTargetView, 0);
, what obviously will let me render to the renderTargetView, but without using a depthStencilView.

Now, when I try to pass my depthStencilView as the last parameter, nothing gets drawn anymore. Is there a specific parameter I have to set in the depthStencilView to allow it to be used when I draw to a texture?

Again, without the depthStencilView it works fine, but of course without depth.

If you need some more code on how I initialise stuff, let me know.

Thank you very much!

3D    562

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Nik02    4348

Does the depth stencil view have the same dimensions and sample count as the color target?

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PKLoki    1492

Could be a few reasons. Check your depth-stencil buffer is cleared correctly unless you're reusing it from a previous pass - depth initially set to 1 typically, so things with less depth pass the depth test. Stencil typically set to 0. And check your depth stencil tests are set as you expect, i.e. you haven't inadvertently left a stencil test set inappropriately from a previous operation, or got the depth test set the wrong way round. You could try temporarily turning both tests off completely to see if anything is drawn then.

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Found the problem. I forgot to ZeroMemory the descriptionbuffer of the depthstencilview, not sure why this cause such an akward problem though. Well, now it's running fine.


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HomeEnglishClass 11PhysicsChapterFluid Mechanics

A cylindrical tank has a hole ...

# A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered wooden cylindrical block of diameter 4r, height h and density rho//3.

 Situation I: Initially, the tank is filled with water of density rho to a height such that the height of water above the top of the block is h_1 (measured from the top of the block).
 Situation II: The water is removed from the tank to a height h_2 (measured from the bottom of the block), as shown in the figure. The height h_2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.
 In situation 2, if h_2 is further decreased, then

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cylinder will not move up and remains at its original positionfor h_(2)=h//3 cylinder again starts moving up for h_(2)=h//4 cylinder again starts moving up for h_(2)=h//5 cylinder again starts moving up

When the height h_(2) of water level is further decreased, then the upward force acting on the wooden block decreases. The total force downward remains the same. This difference will be compensated by the normal reaction by the tank wall on the wooden block. Thus, the blcok does not move up return to its original position.