text:
# Arch-Glacor

 This article has a strategy guide.All information on mechanics, setups, and tactics is on the subpage.
 This article has a money making guide here.Description: Killing the Arch-Glacor (Hard Mode)Please add tips to the subpage, rather than the article below.
 This article has a money making guide here.Description: Killing the Arch-Glacor (Normal Mode)Please add tips to the subpage, rather than the article below.
 This article has a calculator here.Calculators determine experience and costs based on real-time prices from the Grand Exchange Market Watch.

The Arch-Glacor is a massive glacor who "leads" the glacors sent by Wen to the Elder God Wars Dungeon as part of the Glacor Front.[1]

Like Kerapac, the bound and TzKal-Zuk, the Arch-Glacor possesses both normal and hard mode encounters. In normal mode, the boss provides a customisable fight. The player can add or remove mechanics as desired to make the fight easier or harder, which also affects the quantity and quality of rewards received. By defeating the boss with all mechanics active in a solo instance, the player can unlock its hard mode encounter. In normal mode, deaths are considered safe (including for Hardcore Ironmen[2]); the player is teleported to safety should they fall. There is a rolling damage cap in place with a set amount of mechanics active that prevents players from skipping most mechanics. The boss is considered a glacor for the purposes of a slayer task and ushabti, and the slayer helmet provides its benefits for normal mode.[1][3]

In hard mode, the boss fight becomes similar to that of Telos, the Warden - the encounter can only be entered solo, players cannot teleport out of the fight (meaning victory or death), deaths are unsafe, and there is an enrage/streaking system, with the boss' enrage capping at 4,000%. The boss' mechanics are also adjusted for hard mode. The effects of a slayer helmet will not work on hard mode if on task.[3]

Drops from the boss include the Scripture of Wen, a boss pet and a Leng artefact, along with materials to create a pair of tier 85 melee weapons - the Dark ice shard and Dark ice sliver weapons, made from Glacor remnants and Dark nilas. In hard mode, a new drop, the Frozen Core of Leng, is added, which can be used to upgrade the weapons to tier 95 - the Dark Shard of Leng and Dark Sliver of Leng. The Leng artefact, along with the remnants, allows players to upgrade the nightmare gauntlets, gloves of passage and Kerapac's wrist wraps from tier 85 to tier 90.[2]

Defeating the Arch-Glacor at 500% enrage awards the player with the title [Name] the Iceborn. A bronze version of the title, [Name] the Iceborn, can be unlocked by killing the Arch-Glacor at 1,000% enrage. A silver version of the title, [Name] the Iceborn, can be unlocked by killing the Arch-Glacor at 2,000% enrage. A gold version of the title, [Name] the Iceborn, can be unlocked by killing Arch-Glacor at 4,000% enrage.

Obtaining all of the unique drops and crafting all of the unique rewards from this boss completes the achievement The Cold Never Bothered Me Anyway, unlocking the title, [Name] the Cold-Blooded.

Deathtouched darts do not work on this boss, and attempting to do so in either normal mode or hard mode will give the message "You cannot kill that with a dart, so it would be a waste to throw it".

## History

The Arch-Glacors, like their smaller brethren, invaded Leng during the early stages of the current universe. They are said to be ten to a hundred times larger than the normal glacors. Arch-Glacors possess powers far greater than their smaller brethren, such as the ability to come back after being defeated and becoming stronger in the process.

During Zaros' trip across the universe, he discovered Leng and its Arch-Glacors. He subsequently killed one and transformed its energy into the ice crystal, which he used to create the Nihil and later gave it to Nex to serve as a source of power.[4]

During the Sixth Age, one Arch-Glacor was brought from Leng to the ruins of Senntisten by Wen in her attempt to seize the Elder God eggs stolen by Azzanadra. The Arch-Glacor's powers quickly transformed the area it was attacking from into a frozen wasteland. The elves of the Iorwerth and Cadarn clan attempted to battle the Arch-Glacor, but those who challenged it were quickly overwhelmed and killed by its many powers. This warranted the presence of multiple mages to keep its powers in check - Ariane would prevent it from calling glacyte reinforcements, Charos would prevent it from using a powerful cannon attack, Arianwyn would prevent it from randomly changing combat styles, Akthanakos would prevent it from conjuring pillars of ice, Enakhra would prevent it from popping its core out, while Azzanadra used his new god powers to keep the rest of its powers at bay.

Enakhra later abandoned the other mages, forcing Azzanadra to send them off for their own safety. With most of its powers unchecked, the Arch-Glacor advanced towards the cathedral, but Azzanadra's new god powers were able to keep it at bay. However, by moving away from the aqueduct, the Arch-Glacor left behind four Generating glacors that were spawning many glacytes that nearly overwhelmed him until the World Guardian arrived and eliminated them, allowing Azzanadra to fight without distractions. Azzanadra was later able to send the Arch-Glacor back to Leng.

## Location

Arch-Glacor is located at the aqueduct at the Glacor Front on the southern side of Senntisten.

## Mechanics

The Arch-Glacor fight consists of six main mechanics, five of which can be enabled or disabled before the fight. Mechanics can be toggled either from the user interface or by speaking to the mage responsible for holding back the corresponding mechanic. Hard mode can also be toggled in a similar fashion once the player has unlocked the ability to do so by completing the fight solo with all mechanics enabled, in an instance set to one player. Hard Mode activates all mechanics.[1]

The number of active mechanics influences the rewards obtained from the boss. It is possible, although unlikely, to obtain unique drops with no mechanics enabled.

Mechanic Mage Description
Creeping Ice[m 1] N/A N/A Ice Shards approach slowly from either side and force the player into a specific area of the arena.
Glacyte Minions Ariane Additional glacytes are spawned. As well as attacking the player, the Arch-Glacor will be immune to damage until all its minions are killed. When glacytes are killed, they leave behind Unstable glacyte cores that can be clicked to deal damage to the Arch-Glacor.
<NAME> The Arch-Glacor will switch between two different damage types with high damage in their respective style. The player has increased damage reduction when praying against the correct combat style.
Pillars of I<NAME> The Arch-Glacor forces the player to one side and calls down elemental beams from the sky. The beam follows the player, then splits into two to harass the player from two sides. Being hit by the beam deals damage and saps 8% of max Prayer points.
<NAME> The Arch-Glacor traps the player in a small area and launches a huge surge of energy at them. This attack is unavoidable, requiring the use of other tactics to survive.
Exposed Core Enakhra The Arch-Glacor releases its frozen core into the centre of the arena, causing its body to collapse and become inanimate, while slamming both its arms down on either side of the player.

The frozen core begins to emanate a harsh, icy energy that damages and applies a stacking debuff on the player. The damage grows with every stack. To escape, the player must smash through either of the arms and get out of its proximity in order to escape death.

Hard Mode N/A Azzanadra All mechanics are enabled and enhanced.
1. ^ Creeping Ice is the base mechanic and cannot be disabled.

## Strategy

Arch-Glacor's difficulty is scalable: players are able to choose the fight's difficulty which affects both the strength of the boss and potential reward value. In normal mode, this involves enabling or disabling any of the five mechanics before the fight. On the other hand, when initiating a confrontation with Arch-Glacor in hard mode, the player receives an interface in which they may select a custom enrage level (between 0% and the highest enrage level on which they successfully scored a kill plus 5%). As the boss gains enrage, its life points and damage inflicted increase.

In normal mode, Arch-Glacor's life points scale based on the number of mechanics enabled, equalling ${\displaystyle 5{,}000m^{2}+35{,}000m+65{,}000}$, where ${\displaystyle m}$ is the number of mechanics enabled (not including Creeping Ice), meaning the boss will have between 65,000 and 365,000 life points.

In hard mode, the life points of the Arch Glacor, Icy Arm (right) and (left), Glacytes, and Bolstered Glacyte scale with enrage. All but the life points of the Arch-Glacor remain constant up until 2,500% enrage. Above 2,500% enrage, they too will have increased life points. The Arch-Glacor's life points, ${\displaystyle {\text{LP}}}$, is expressed as a function of enrage, ${\displaystyle e}$, in the following way:

${\displaystyle {{\text{LP}}=370{,}000+3{,}700\cdot \left\lfloor e/5\right\rfloor }}$

## Drops

### Normal mode

In normal mode, three sets of drops are awarded at a time. In solo encounters, the player receives all three drops. In duo encounters, the leftover drop is awarded to the top damaging player. In larger groups, the three top damage dealers receive one drop each. The quantity and quality of drops in normal mode are dependent on how many mechanics were enabled.

#### Charms

Charms are dropped in set of five; at most three sets can be obtained from a kill for a total of 15 charms.

Nothing
15
15
15
15
10–12%7–9%7–9%36–39%34–37%
Represents a 90% confidence range based on a sample of 2,470 kills.
15 items are dropped at a time.
Add data to the log (requires JavaScript).

#### Main drops

1. ^ After a roll of 9/100 on components, drop rate and quantity ranges from 995/1,000 for 5, 993/1,000 for 7, 989/1,000 for 11, 980/1,000 for 20, 965/1,000 for 35 and 925/1,000 for 75 for each mechanic active.
2. ^ After a 1/200 roll on the stronger drop table, 3/8 chance of obtaining this drop.
3. ^ a b c d Drop rate increased by 2% after upgrading to enriched pontifex shadow ring.
4. ^ After a roll of 9/100 on components, drop rate ranges from 5/1,000 at none, 7/1,000 with one, 11/1,000 with two, 20/1,000 with three, 35/1,000 with four and 75/1,000 for five mechanics active respectively.
5. ^ a b Only with all five mechanics active.[5]

#### Secondary drops

It is possible to receive multiple secondary drops for a single primary drop in normal mode.

1. ^ Replaced by tradeable version after the player has upgraded their pontifex shadow ring with 1,000 anima.
3. ^ Requires completion of Azzanadra's Quest.
4. ^ Only obtainable when the pontifex shadow ring's Elder Trove modifier is tier 3. Replaces the T1 Trove.
5. ^ Only obtainable when the pontifex shadow ring's Elder Trove modifier is tier 4. Replaces the T2 Trove.
6. ^ a b c Triskelion fragments are dropped in order.
7. ^ There's a 1% chance an elite clue scroll will be upgraded to a master clue scroll.
8. ^ Has a drop rate of ${\displaystyle 1/(18{,}000-m\times 3{,}000)}$ in normal mode per loot pile, where ${\displaystyle m}$ is the number of active mechanics (between 0 and 5). Effectively between 1/18,000 and 1/3,000 per loot pile, with a threshold of 500.

### Hard mode

In hard mode, the rewards are obtained through a streak/enrage system. After each kill, players have the option to continue the challenge, or claim the rewards. One reward is given at a time, but the quantity of these items increases with higher killstreak and possibly enrage. Likewise, if the player dies during their streak, part of their accumulated rewards is lost based on the amount of killstreak the player had prior to their death. This starts at 5% saved for 1 killstreak and caps at 75% saved at 15+ killstreak, rounded down. This reduction applies by whole numbers of the items held in the chest, not the total value of rewards. With the exception of the glacor core, no charms or other secondary drops are obtainable in hard mode.

It was worked out by the players that the equation for the unique item drop rates could be:

${\displaystyle {\frac {1000+10\times e+25\times s}{1000000}}}$

Where ${\displaystyle e}$ is the enrage of Arch-Glacor and ${\displaystyle s}$ is the number of total kills done in the streak (so Streak+1 if using the Streak value in the loot interface).

This is more easily interpreted as:

${\displaystyle {\frac {1+{\frac {e}{100}}+{\frac {s}{40}}}{1000}}}$

Where it can be seen that the base chance is 1/1000, and increases by 1/1000 for every 100 enrage or 40 kills in streak.

The validity of these formulae remains unconfirmed, but it matches the known drop rates at different streaks and enrages given from the reveal of the drop rates from the Elder God Wars Dungeon.

When the unique item table is rolled, the player has the chance for the following:

The expected number of uniques (exclusively hard mode) to complete the Arch-Glacor boss collection log (2 Frozen core of Leng, 1 Scripture of Wen, 1 Leng artefact, pet ignored) is 10.5.

The quantity of items added after each kill seems to follow the formula ${\displaystyle {\text{Quantity}}=\lfloor {\text{Base Quantity}}\times (1+0.1\times {\text{Streak}})\rfloor +{\text{Luck bonus}}}$

• ${\displaystyle {\text{Quantity}}}$ is the size of the reward added after a kill.
• ${\displaystyle {\text{Base Quantity}}}$ is the amount obtained at 0 streak.
• ${\displaystyle {\text{Streak}}}$ is the number of kills shown in the loot interface. After the first kill, streak is equal to 0.
• ${\displaystyle {\text{Luck bonus}}}$ has a chance to increase the quantity of a single stack of items by 1-4 when tier 4 luck is used. Checking the loot chest with and without tier 4 luck can change the quantity of items.
• The Leng artefact, scripture of Wen, and frozen core of Leng will always be added in a quantity of 1.

When claiming streaks, if the player's inventory and bank is full, any remaining rewards will remain inside the chest. If the player logs out before claiming these drops, they will be lost. They will be withdrawn in noted form if this occurs.

Alternatively, if the player dies, their streak ends, some of their loot is deducted, and they can claim their remaining loot from the frozen chest outside the boss arena.

#### Calculator

This calculator along with documentation can be found at the top of this page.

template  = Calculator:Template/Combat/Arch-Glacor
form      = archglacorCalcForm
result    = archglacorCalcResult
param     = mode|Calculate|Drop rate|buttonselect|Drop rate,Streaking expected number of uniques,Claiming expected number of uniques|Drop rate=enrage,streak,pontifex;Streaking expected number of uniques=startingEnrage,useStreak,averageEnrageGained,pontifex;Claiming expected number of uniques=claimEnrage,claimKills,pontifex|Drop rate calculates the chances of receiving unique/dark nilas for a specific kill at the given enrage and streak.Streaking expected number of uniques calculates the cumulative total number of uniques/dark nilas over the entire streak.Claiming expected number of uniques calculates the cumulative total number of uniques/dark nilas when repeating kills at the same enrage and claiming after every kill.
param     = enrage|Enrage (%)|100|int|0-4000
param     = streak|Streak|1|int|1-200
param     = startingEnrage|Starting enrage (%)|100|int|0-4000
param     = useStreak|Streak|1|int|1-200
param     = averageEnrageGained|Average enrage gained (%)|11.5|number|5-20,,0.1||Average enrage gained is 11.5%. However, any value from 5% to 20% can be set.
param     = claimEnrage|Enrage (%)|100|int|0-4000
param     = claimKills|Kill count|25|int|1-200
param     = pontifex|Enriched pontifex shadow ring|No|check|Yes,No||The enriched pontifex shadow ring provides a 2% increased chance at unique drops.

This text will disappear if the form is loaded properly.
This text will disappear if the form is submitted.

#### Main drops

1. ^ The drop quantity is unaffected by streak.
2. ^ The drop rate is unaffected by enrage and streak.
3. ^ a b c Enrage and streak affect drop rate, see here.

#### Secondary drops

1. ^ Completion of Twilight of the Gods not necessary to receive this drop. It appears in inventory when received, or on the ground under the player if their inventory is full.
2. ^ Has a base drop rate of 1/1,000 in hard mode, with a threshold of 500.

## Update history

This information has been compiled as part of the update history project. Some updates may not be included - see here for how to help out!
• patch 13 June 2022 (Update):
• The drop rates for the Chaos die reroll tokens have been increased. Normal mode Arch-Glacor per pile drop rate: 10/2500. Hard mode Arch-Glacor drop rate: 10/1000
• patch 23 May 2022 (Update):
• Hard Mode Glacor loot that was rerolled with the addition of Resonant Anima has been reverted to the previous loot earned for players who are yet to have claimed it.
• hotfix 16 May 2022:
• Fixed an issue with chest loot being modified as a result of resonant anima of Wen being included in hard mode.
• patch 16 May 2022 (Update):
• Hard mode Arch-Glacor encounters now drop resonant anima of Wen to increase supply of anima for Wen arrows, as it could only be obtained from normal mode prior to this change.
• patch 7 March 2022 (Update):
• The scenery versions of TzKal-Zuk and the Arch-Glacor, outside their combat encounters, have had their Santa hat removed.
• patch 28 February 2022 (Update):
• Players now receive correct loot information in the chat after killing Arch-Glacor and will not receive a chat prompt about an effigy if they have five of them banked.
• hotfix 13 December 2021:
• Fixed an issue with players obtaining less rewards while in the middle of a streak following the drop changes for hard mode.
• update 1 November 2021 (Update):
• Having the ring upgraded with a faction no longer decreases the amount of anima gained from bosses.
• Having the ring upgraded will now continue to drop anima at a 100% drop rate rather than the previous 1/6 chance.
• Voice lines from the Mages have been added to the Arch-Glacor boss fight.
• patch 18 October 2021 (Update):
• Adjusted the clickboxes during the Arch-Glacor fight.
• Glacor Remnants will now be dropped in the correct quantities, depending on how many of the Arch-Glacor's mechanics are active.
• patch 11 October 2021 (Update):
• Minions mechanic has been adjusted:
• Glacytes' health in normal mode is adjusted to number of active mechanics.
• Arch-Glacor's barrier during glacyte mechanic no longer provides complete immunity, instead reducing damage by 98%.
• Glacyte minions now have an additional prefix to their name to indicate their combat style.
• Spawn position moved so minions do not obscure boss.
• Barrier does not activate immediately, instead having a small delay before minions appear when mechanic is initiated.
• The unstable core left behind from glacyte minions has a lower click priority to avoid interference while clicking on glacyte minions.
• Tsunami can now be used on the Arch-Glacor's arms.
• Arch-Glacor now awards proper Reaper points and Marks of War for hard mode:
• Reaper points increased from 15, 18 and 21 to 25, 31 and 37 (the same as Elite Dungeons bosses).
• Marks of War increased from 31 to [40 + streak].
• patch 27 September 2021 (Update):
• Fixed a related issue with unlocking the wing adornments (the last unlock would not log the Arch-Glacor Boss as completed).
• A player on Hard Mode joining another player's Normal Mode instance will now benefit from the safe death.
• Fixed an issue with the Arch-Glacor Hard Mode toggle being affected by other bosses.
• patch 20 September 2021 (Update):
• Reduced the screen shake effects that play when activating Hard mode for the Arch-Glacor fight.
• The Arch-Glacor Boss Collection Log will now track more than 250 Remnants.
• The drop broadcast for the Arch-Glacor pet item now states the correct name.
• The freezing blood debuff from the Arch-Glacor fight now consistently times out.
• The Shard and Sliver of Leng weapons are more consistent regarding keepsaking.
• patch 13 September 2021 (Update):
• Reaper Assignments will now correctly indicate that the Arch-Glacor is the target when using quick-chat.
• Fixed a grammar error on the grouping system UI for the Arch-Glacor fight.
• Arch-Glacor logs are now shown in the Adventurer's Log.
• Rewards from the Arch-Glacor fight are now given to the top three players.
• The Arch-Glacor's drops will no longer spawn below the boss, where they are inaccessible to players.
• Players can now consistently leave the Arch-Glacor arena after defeating the boss in hard mode.
• hotfix 6 September 2021 (Update):
• Practice mode is no longer available for the Arch-Glacor.
• Adjusted the Instance Teleport Coordinate in the Arch-Glacor arena so players can surge straight away upon entering.
• Added drop broadcasts for the Frozen core of Leng and Leng artefact.
• Fixed an issue that meant the Arch-Glacor boss would die due to DoT effects, which could kill the player in certain situations.
• Fixed an issue where the Pontifex Ring was reducing Prayer-based damage resistance.
• Uncut Dragonstone will now drop in noted form during the Arch-Glacor fight.
• Normal mode damage reduction no longer kicks in after you have completed all mechanics during an encounter.
• Rejoining a Hard Mode Arch-Glacor instance after toggling it off will no longer place you in a pseudo-Hard Mode encounter.
• Fixed an issue where sequential attacks of the same type from the same side of the Arch-Glacor boss were not appearing.
• The Bolstered Glacytes no longer set the Arch-Glacors health when they die.
• update 6 September 2021 (Update):
• Added a Bank Preset for the Arch-Glacor.
• The Arch-Glacor boss will no longer drop Bones. Where was it even getting those?!
• Fixed an issue where the incorrect armour values were being applied to the new enhanced T90 glove upgrades introduced with the Glacor Front update.
• Fixed an issue that was preventing the 'Exposed Core' and 'Glacyte Minions' from working as expected at high enrage tiers during the Arch-Glacor boss fight.
• hotfix 1 September 2021 (Update):
• Re-addressed the bug where killing the bolstered glacyte reduced the boss' health to 1,250 again.
• The third beam of the Pillars of Ice mechanic in the Hard Mode Arch-Glacor boss fight will no longer disappear and reappear when the mechanic ends.
• Manuscripts and Scriptures of Wen will now drop from the Arch-Glacor correctly in all scenarios.
• hotfix 1 September 2021:
• Hard Mode common drops (and Dark nilas) have had their quantities increased.
• hotfix 31 August 2021 (Update):
• Triskelion keys dropped from the Arch-Glacor will no longer be multiplied by enrage or mechanic modifiers.
• Players can no longer rejoin the Arch-Glacor's Hard Mode encounter after turning it off, unless they re-activate it.
• The Bolstered Glacytes no longer reduce the boss' health to 1,250 upon being killed.

## Trivia

• The player <NAME> was the first player who defeated the Arch-Glacor on 4000% enrage on 22 September 2021 which was shown in-game with the announcement: News: Evil Lucario has scored the first Arch-Glacor boss kill at 4000% enrage!, roughly nine days shy of its first month released.
• During its reward stream, Mod Orion claimed that players would be unable to clear 4000% enrage within a month.
• In order to be visible from a distance, the Arch-Glacor was added to the non-instanced area as an environment piece rather than non-player character due to shorter draw distance for NPCs in the NXT client. During development it had also been considerably scaled up compared to the size of the model initially created by Mod JT.[3]
• The Arch-Glacor has been designed to be an introduction boss meant to encourage more people to try bossing. The ability to customise the encounter is meant to allow players to learn different types of combat mechanics and how to deal with them.[3]
• On release, the Arch-Glacor suffered from several bugs which affected the fight and loot mechanics, such as the Exposed Core not appearing correctly, or persisting in the arena if Arch-Glacor had been killed with poison while the Frozen Core mechanic was in progress. Additionally, the hard mode fight was affected by an invalid teleport glitch on the exit portal, which forced some players to lose their kill streaks. This glitch is currently still active and happens if the player is pushed by the creeping ice at the same time as the Arch-Glacor is defeated.
• Unlike Telos, whose health is capped at 300% enrage, the Arch-Glacor's health in hard mode is not subjected to such a cap. At 4,000% enrage, the Arch-Glacor has 3,330,000 life points, which is the highest amount of health for a solo boss.
• The Arch-Glacor was first mentioned in The Promised Gift, although there were no links between the glacor race and Wen at the time. The blood warden later hinted to its arrival in the Elder God Wars Dungeon by stating "A cold front is coming.", referring to the glacors' abilities to freeze the air around them. Seren later hinted to the boss' arrival by remarking how it was about to become "dark and very cold".

## References

1. ^ a b c Jagex. "Elder God Wars Dungeon: The Glacor Front." 20 August 2021. RuneScape News.
2. ^ a b Jagex. "Elder God Wars Dungeon: The Glacor Front Survival Guide." 27 August 2021. RuneScape News.
3. ^ a b c d RuneScape. Elder God Wars Dungeon: An introduction to the Glacor Front | RuneScape. YouTube. 20 August 2021.*
4. ^ The Promised Gift, written by Nex, "Nex: Angel of Death", RuneScape. "Finally I created the stone of ice from an arch-glacor on the frozen world of Leng. These creatures, that tower ten or a hundred times taller than their smaller kin, dominate the landscape of that brutal and inhospitable wasteland."
5. ^ Mod Shogun. Elder God Wars Dungeon: Drop Rates!. Reddit. 20 December 2021. (Archived from the original on 23 January 2022.) Mod Shogun: "[Are the book and artifact only obtainable from 5 mechanics...?] Only 5 mechanics."


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My Math Forum First order differential equation

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 November 8th, 2016, 09:04 AM #1 Newbie   Joined: Nov 2016 From: London Posts: 3 Thanks: 0 First order differential equation Hey all, I have a separable differential equation: Can someone please explain how the solution is: Thanks
 November 8th, 2016, 09:25 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Post your attempt first. There's nothing that should cause a differential equations student much trouble, especially if you know how to integrate by partial fractions (writing your constant of integration as $\ln c$) and how to use logarithmic identities to good effect. Perhaps the hardest part is removing the absolute value signs.
 November 8th, 2016, 11:09 AM #3 Newbie   Joined: Nov 2016 From: London Posts: 3 Thanks: 0 I get 1/2ln(2x+1)=ln(t/(t+1))+ln(C), but what should I do then? Last edited by skipjack; November 8th, 2016 at 02:49 PM. Reason: to insert parentheses
 November 8th, 2016, 02:54 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Can you write that equation in the form ln(...) = ln(...)?
 November 8th, 2016, 03:38 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra You are looking for the terms in $t$ to be squared, so multiply by 2 and use properties of the logarithm. Last edited by skipjack; November 8th, 2016 at 03:41 PM.
 November 8th, 2016, 03:42 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Mrkukas, are you given that t > 0?
 November 8th, 2016, 04:42 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra He shouldn't be. It should be a valid result for all $t$ with the possible exception of $t=-1$.
 November 8th, 2016, 06:16 PM #8 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 That depends on what you mean by "valid result" and "possible exception". The original equation doesn't imply that t isn't -1, but the supplied solution does, as it's undefined at t = -1.
November 8th, 2016, 11:34 PM   #9
Newbie

Joined: Nov 2016
From: London

Posts: 3
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Quote:
 Originally Posted by <NAME>, are you given that t > 0?
No, I am not.

Last edited by skipjack; November 8th, 2016 at 11:51 PM.

 November 8th, 2016, 11:53 PM #10 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Can you assume that t is real?

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# INEQUATIONS(Inequalities) Part 1

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INEQUATIONS (INEQUALITIES): Prologue

During GMAT exam, where time is precious, complicated inequalities often create a difficult situation. Purpose of this article is to narrate the paradigms of inequalities and to state the methods to solve the inequalities.

This Article is actually a digital version of my handwritten notes collected from various sources. Most of the input is from Indian School Text Books on Maths(CBSE Board) and remaining is from GMAT Math Strategy Guides of Elite Test Prep Companies and Guides for Indian CAT.

INEQUATIONS : Definition and Characterictics

The term Equation is defined as a statement involving variable(s) and the sign of equality(=)
Similarly, we can define the term Inequation as a statement involving variable(s) and the sign of inequality viz, >, ≥, <, or ≤
An Inequation may contain one or more variables. Also, it may be linear or quadratic or cubic etc.

Some Examples of Inequations

5x – 3 > 0

$$x^2$$ - 5x + 4 ≤ 0

$$ax^2$$ + $$by^2$$ < 1

$$x^3$$ + $$6x^2$$ + 11x + 6 ≤ 0

Linear Inequations with one variable

Let a be a non-zero real number and x be a variable.

Then Inequations of the form ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0 are known as linear Inequations in one variable

Examples : 9x – 15 > 0, 2x – 3 ≤ 0

Linear Inequations with two variables

Let a, b be non-zero real numbers and x, y be a variables.

Then Inequations of the form ax + by < c, ax + by ≤ c, ax + by > c, ax + by ≥ c are known as linear Inequations in two variables x and y

Examples : 2x + 3y ≤ 6, 2x + y ≥ 6

Let a be a non-zero real number.

Then Inequations of the form $$ax^2$$ + bx + c < 0, $$ax^2$$ + bx + c ≤ 0, $$ax^2$$ + bx + c > 0, $$ax^2$$ + bx + c ≥ 0 are known as Quadratic Inequations with one variable.

Examples : $$x^2$$ + x - 6 < 0, $$2x^2$$ + 3x + 1 > 0

The equation $$ax^2$$ + $$by^2$$ < 1 is a sample quadric Inequation with two variables.

Solution of an Inequation

Definition :- A Solution of an Inequation is the value(s) or the range of values of the variable(s) that makes the inequation a true statement

Consider the Inequation $$\frac{(3-2x)}{5}$$ < $$\frac{x}{3}$$ – 4

For x = 9, we have, LHS = $$\frac{(3-2*9)}{5}$$ = -3 and, RHS = $$\frac{9}{3}$$ - 4 = -1. Clearly -3 < -1 that means LHS < RHS, which is true. So x = 9 is a solution of the given inequation

For x = 6, LHS = $$\frac{(-9)}{5}$$ and, RHS = -2 ---------> LHS > RHS ---------> So x = 6 is not a solution of the given Inequation.

We can verify that any real number greater than 7 is a solution of the given Inequation.

Let us now consider the Inequation $$x^2$$ + 1 < 0

We know that $$x^2$$ ≥ 0 for all real values of x

$$x^2$$ +1 ≥ 1 for all real values of x--------------(Added 1 to both sides of ineqation)

So, there is no real value of x which makes the Inequation $$x^2$$ + 1 < 0 a true statement. Hence, it has no solution.

Solving an Inequation

It is the process of obtaining all possible solutions of an Inequation. Substitution by numbers is a lengthy process, and it is an impossible process for complicated quadratic equations. Best way to solve Inequations is shown in the following section.

Solution Set

The set of all possible solutions of an Inequation is known as its solution set.

For Example
The solution set of the Inequation $$x^2$$ + 1 ≥ 0 is the set of all real numbers
The solution set of the Inequation $$x^2$$ + 1 < 0 is empty/null set

Solving Linear Inequation with one variable

As mentioned earlier that solving an Inequation is the process of obtaining all possible solutions of the variable.
In the process of solving an Inequation, we use mathematical simplifications which are governed by the following rules

Rule 1 :- Same Number may be added to (or subtracted from) both sides of an Inequation without changing the sign of inequality. Example: x > 5 ---------> x + 2 > 5 + 2

Rule 2 :- Both sides of an Inequation can be multiplied (or divided) by the same positive real number without changing the sign of inequality. However, the sign of inequality is reversed when both sides of an Inequation are multiplied or divided by a negative number. Examples: x > 5 ----> 3x > 15; x > 1 ----> -x < -1

Rule 3 :- Any term of an Inequation may be taken to the other side with its sign changed without affecting the sign of inequality. Example: $$\frac{x}{3}$$ - 2 < $$\frac{x}{6}$$ -----> $$\frac{x}{3}$$ - $$\frac{x}{6}$$ < 2

Rule 4 :-
In inequalities never multiply for a term unless you know it does not equal 0 AND you know its sign.

Type I :- Algorithm to solve the Inequations of the form ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0

Step I :- Obtain the linear Inequation.
Step II :- Collect all terms involving the variable on one side of the Inequation and the constant terms on other side
Step III :- Simplify both sides of inequality in their simplest forms to reduce the Inequation in the form ax < b, or ax ≤ b, or ax > b, or ax ≥ b
Step IV :- Solve the Inequation obtained in step III by dividing both sides of the Inequation by the coefficient of the variable (of x)
Step V :- Write the solution set obtained in step IV in the form of an interval on the real line.

Example 1 :- Solve the Inequation 2x – 4 ≤ 0

Step 1) 2x – 4 ≤ 0
Step 2) (2x – 4) + 4 ≤ 0 + 4 --------------------------[ Adding 4 on both sides ]
Step 3) 2x ≤ 4
Step 4) $$\frac{2x}{2}$$ ≤ $$\frac{4}{2}$$
Step 5) x ≤ 2

Any real number less than or equal to 2 is a solution of the given Inequation.

Hence the solution set of the given Inequation is (-∞, 2]

NOTE :- The Interval [x, y] stands for the values between x and y, both inclusive, where as The interval (x, y) stands for the values between x and y, both exclusive

Example 2 :- -3x + 12 < 0

Step 1) -3x + 12 < 0

Step 2) -3x < -12 --------------------------[ Transposing 12 on right side ]

Step 3) $$\frac{-3x}{-3}$$ > $$\frac{-12}{-3}$$ --------------------------[ Dividing both sides by -3, and hence the sign of inequality changes ]

Step 4) x > 4

Any real number greater than 4 is a solution of the given Inequation.

Hence the solution set of the given Inequation is (4,∞)

Example 3 :- $$\frac{1}{(x-2)}$$ < 0

$$\frac{1}{(x-2)}$$ < 0 ---------> x-2 < 0 -----------------[ If $$\frac{a}{b}$$ < 0 and a > 0 then b < 0]

x < 2 ----------> Solution set is ( -∞, 2)

Type II :- Algorithm to solve the Inequations of the form $$\frac{(ax+b)}{(cx+d)}$$ < k, $$\frac{(ax+b)}{(cx+d)}$$ ≤ k, $$\frac{(ax+b)}{(cx+d)}$$ > k, $$\frac{(ax+b)}{(cx+d)}$$ ≥ k

Step I :- Obtain the Inequation.

Step II :- Transpose all terms on LHS

Step III :- Simplify LHS of the equation obtained in step II to obtain an Inequation of the form $$\frac{(px+q)}{(rx+s)}$$ < 0, $$\frac{(px+q)}{(rx+s)}$$ ≤ 0, $$\frac{(px+q)}{(rx+s)}$$ > 0, $$\frac{(px+q)}{(rx+s)}$$ ≥ 0

Step IV :- Make Coefficient of x positive in numerator and denominator if they are not.

Step V :- Equate Numerator and denominator separately to zero and obtain the values of x. These values of x are generally called critical points.

Step VI :-
Plot the critical points obtained in step V on real line. These points will divide the real line In three regions.

Step VII :-
In the rightmost region the expression on LHS of the Inequation obtained in step IV will be Positive and in other regions it will be alternatively negative and positive. So, mark positive Sign in the right most region and then mark alternatively negative and positive signs in Other regions.

Step VIII :-
Select Appropriate region on the basis of the sign of the information obtained in step IV. For +ve inequality signs (>, ≥), the range will be the region with +ve signs, i.e., the rightmost segment and leftmost segment. For -ve inequality signs (<, ≤), the range will be the region with -ve signs, i.e., the range between critical points. Write these regions in the form of intervals to obtain the desired solution sets of the given Inequation.

Example 1 :- $$\frac{(x-2)}{(x+5)}$$ > 2

Step I :- $$\frac{(x-2)}{(x+5)}$$ > 2

Step II :- $$\frac{(x-2)}{(x+5)}$$ - 2 > 0

Step III :- $$\frac{(-x-12)}{(x+5)}$$ > 0

Step IV :- $$\frac{(x+12)}{(x+5)}$$ < 0 ---------------[Multiplying by -1 to make coefficient of x positive in the expression in numerator]

Step V :- x+12 < 0 -----> x < -12 and x+5 < 0 --------> x < -5

Therefore x = -12, -5 are the critical points.

Step VI and Step VII :-

Step VIII:- The Real line is divided in to three regions and the signs of LHS of Inequation (step IV) marked. Since the Inequation in (step IV) possesses less than sign which means that LHS of the Inequation is negative. So, the solution set of the given Inequation is the union of the regions containing negative sign.

Hence the solution set of given Inequation is -12 < x < -5

Example 2 :- $$\frac{(x-3)}{(x-5)}$$ > 0

$$\frac{(x-3)}{(x-5)}$$ > 0

x-3 > 0 ------> x > 3 --------------------[ Equating x-3 to zero]

x-5 > 0 ------> x > 5 --------------------[ Equating x-5 to zero]

Upon equating x-3 and x-5 separately to zero, we obtain x = 3, 5 as critical points.

Plot these points on Real Line

The Real line is divided into three regions. Since the Inequation possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions Containing positive sign.

Hence x > 5 or x < 3 -----------> (-∞, 3) U (5, ∞)

Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ - 5 ≥ 0

$$\frac{((2x+4)-5(x-1))}{(x-1)}$$ ≥ 0

$$\frac{(-3x+9)}{(x-1)}$$ ≥ 0

$$\frac{(3x-9)}{(x-1)}$$ ≤ 0 ----------------[ Multiplying both sides by -1]

$$\frac{(3(x-3))}{(x-1)}$$ ≤ 0

$$\frac{(x-3)}{(x-1)}$$ ≤ 0 -----------------[ dividing both sides on Inequation by 3]

x = 3, 1

Sign of original Inequation obtained in Step 4 is negative so the solution set will be union of the regions containing negative sign.

Therefore Solution Set is 1 < x ≤ 3

NOTE :-
When x=1, the Inequation becomes $$\frac{6}{0}$$ ≥ 5 and we know that division by zero is not defined, hence we need to exclude x=1 from solution set.

Example 4 :- $$\frac{(x+3)}{(x-2)}$$ ≤ 2

$$\frac{(x+3)}{(x-2)}$$ ≤ 2

$$\frac{(x+3)}{(x-2)}$$ - 2 ≤ 0

$$\frac{(x+3-2x+4)}{(x-2)}$$ ≤ 0

$$\frac{(-x+7)}{(x-2)}$$ ≤ 0

$$\frac{(x-7)}{(x-2)}$$ ≥ 0 -------------------[Multiplying both sides by -1]

x-7 ≥ 0 ---------> x ≥ 7

x-2 ≥ 0 ---------> x ≥ 2

x = 7 and 2 are the critical points.

The Real line is divided into three regions. Since the Inequation (obtained in step 4) possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions containing positive sign.

Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)

In order this Article not be dense with too much information, we have decided to split the same in two parts. This is the first Part of Article. Second Part contains two remaining concepts which are 1) Modules - Absolute Value 2) Quadratic Inequalities.
Here is the link to Second Part of the Article inequations-inequalities-part-154738.html#p1238488.

Courtesy for the Information

Prof. Dr. <NAME>
B.Sc. (Hons) (Gold Medalist), M.Sc. (Gold Medalist), Ph.D. in Mathematics
Author of CBSE Math Books

Mr. <NAME>
Aluminus – IIM Bangalore

Special Thanks to

BUNUEL, doe007, Zarrolou

Regards,

Narenn

---------- End of First Part ------------
Attachments

INEQUATIONS Part 1.pdf [160.45 KiB]

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Last edited by Bunuel on 02 Jul 2013, 12:20, edited 3 times in total.
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01 Jul 2013, 08:42
Narenn wrote:
Rule 4 :- In inequalities never multiply for a term unless you know it does not equal 0 AND you know its sign.

This is really important and it is a common mistake of test takers, especially in data sufficiency. Here is a really simple problem to demonstrate this:

Is x > 3?
1) xy > 3y
2) y > 1

If a test-taker divides each side of the inequality in statement 1 by y, then the result is x > 3. This appears to be sufficient to answer the question, but it is not because of your rule 4. y may not be a positive number, in which case we would have to switch the sign if we divide both sides by y. Instead, one should perform a bifurcated analysis in working statement 1, as below.

Statement 1:
Case 1: y > 0
This results in x > 3 because we can divide both sides by a non-zero y and we don't have to change the sign because y is positive

Case 2: y < 0
This results in x < 3 because we can divide both sides by a non-zero y and we have to change the sign because y is negative

Since x > 3 and x < 3 are each possible, this is enough to prove the insufficiency of statement 1, so we don't have to try the case of y = 0. In fact, y = 0 is not possible because then we would have 0 > 0

Statement 2 forces y to be a positive number. This is not enough by itself because we don't know x.

When we combine both statements, we are only left with case 1 (y > 0). This results in x > 3. Therefore, the answer is C.

Another way to avoid this trap is to choose numbers, put them into the equation, and then observe the outcome. When working with inequalities, be especially prone to try 0, 1, 2, -1, -2, 1/2, -1/2 as these will cover most outcomes. Choosing y = 1 and y = -1 disproves statement 1 by itself.
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12 Aug 2013, 15:49
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"Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)"

x ≤ 2 should be x < 2, not equal to 2 else denom will become 0.
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21 Jul 2014, 13:17
Narenn wrote:

Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ - 5 ≥ 0

$$\frac{((2x+4)-5(x-1))}{(x-1)}$$ ≥ 0

$$\frac{(-3x+9)}{(x-1)}$$ ≥ 0

$$\frac{(3x-9)}{(x-1)}$$ ≤ 0 ----------------[ Multiplying both sides by -1]

$$\frac{(3(x-3))}{(x-1)}$$ ≤ 0

$$\frac{(x-3)}{(x-1)}$$ ≤ 0 -----------------[ dividing both sides on Inequation by 3]

x = 3, 1

Sign of original Inequation obtained in Step 4 is negative so the solution set will be union of the regions containing negative sign.

Therefore Solution Set is 1 < x ≤ 3

NOTE :-
When x=1, the Inequation becomes $$\frac{6}{0}$$ ≥ 5 and we know that division by zero is not defined, hence we need to exclude x=1 from solution set.

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3
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22 Jul 2014, 00:40
Game wrote:
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.
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08 Nov 2014, 19:36
PiyushK wrote:
Game wrote:
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.

Example 4 :- \frac{(x+3)}{(x-2)} ≤ 2

Why based on the same concept, 2 has not been excluded in example 4? In this example also, 2 in the denominator will result the expression to undefined
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25 Jan 2015, 22:44
PiyushK wrote:
Game wrote:
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.

How come when you multiply the equation by (x-1), you don't eliminate it from the denominator?
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text:
# 20-60A adjustable ~200vdc current mode buck design

J

#### <NAME>

Jan 1, 1970
0
I'm looking to build a power supply for a krypton arc lamp, and need a
design that can deliver an adjustable 20-60 amps into a LOW resistance load
(around 5 ohms). I'm a programmer for a living and this is my first power
electronics project (previous electronics projects have mostly been
microcontroller based stuff)

I've done quite a bit of reading, but am running into a brick wall scaling
up some of the various designs that I've found online. I'm working with
LTC's SWCad3 to test concepts, but I'm having problems with input currents
spiking up way too high since my target output is so close to my current
source (AC 220v 60A single phase rectified to 320vdc), and the 120hz line is
just too slow to keep the input capacitor bank charged under heavy load
without some serious high current spikes to recharge.

Any suggestions to smooth out the current draw from the line would be MUCH
appreciated !!

R

#### <NAME>

Jan 1, 1970
0
John said:
I'm looking to build a power supply for a krypton arc lamp, and need a
design that can deliver an adjustable 20-60 amps into a LOW resistance load
(around 5 ohms). I'm a programmer for a living and this is my first power
electronics project (previous electronics projects have mostly been
microcontroller based stuff)

I've done quite a bit of reading, but am running into a brick wall scaling
up some of the various designs that I've found online. I'm working with
LTC's SWCad3 to test concepts, but I'm having problems with input currents
spiking up way too high since my target output is so close to my current
source (AC 220v 60A single phase rectified to 320vdc), and the 120hz line is
just too slow to keep the input capacitor bank charged under heavy load
without some serious high current spikes to recharge.

Any suggestions to smooth out the current draw from the line would be MUCH
appreciated !!

John,
such a power supply is rather tricky. The power
level is rather high, and the nonlinear negative
dynamic impedance load doen't make it simpler.
You'll have to use a switching approach.
Not recommended.

Rene

S

#### [email protected]

Jan 1, 1970
0
John,
such a power supply is rather tricky. The power
level is rather high, and the nonlinear negative
dynamic impedance load doen't make it simpler.
You'll have to use a switching approach.
Not recommended.

Rene

Must agree here.
At this power level , I'd recommend you get a professional power
supply designer
to tackle this.
DNA , you around :0)
If you still decide to tackle this yourself , I recommend some eye
protetion of some sort
and prepair yourself for lots of bangs and smoke :0(
Rob

P

#### [email protected]

Jan 1, 1970
0
and the 120hz line is
just too slow to keep the input capacitor bank charged under heavy load
without some serious high current spikes to recharge.

The old-fashioned
way:

----------- ~ + ------ L ------------ +
|
bridge ===
---
|
----------- ~ - --------------------- -

'L' smoothes the charge peaks, needs an air gap, and will
basically be a very heavy big piece of iron with copper.

The output of this circuit is somewhat lower then 220V DC,
so more current.
See:
http://tpub.com/neets/book7/27f.htm

J

#### <NAME>

Jan 1, 1970
0
Ohh I'm quite prepared for a bit of mess Even got a few spare arc tubes
just in case And a welding helmet for eye protection

As stated... I'm planning a current mode buck regulator, though it has been
suggested that I'll need a boost circuit in front of it to smooth out the
line input current surges (to push up the voltage so that charge is
transfered into the caps more effciently ??) . I've already got a spice
model created that is close (need to work on the control loops a little
more, and a better soft start circuit., and beat these 120+ amp surges that
I'm getting)

"nonlinear negative dynamic impedance load" -- I'm not an EE so I could use
a bit of explanation I'll probably need to understand that as I'd like to
create a spice model of the arc tube for the later stages of testing.
(resistance decreases as current increases ?? am I even close ??)

(cummon guys !!! so far everyone who has responded, here and in alt.lasers
[because this is for a KW output ND:YAG laser] has had nothing to hand out
but warnings !! I'm committed, it will get built !! )

B

Jan 1, 1970
0
IPM - International Perforation Management
high-tech engineering China-Germany-Thailand
fax : 0049 - (0) 1212-5-375-17-531
http://www.microperforation.com
http://www.microperforation.com.cn
http://www.deguodaguan.com/ipm/
http://www.dk3qv.de
Email : [email protected] - [email protected]

We are using AC/DC converters and special DC constanter with 300 V/DC
30 up to 60 Amps for nano or micro perforation machines since more
then 20 years.

There are several companies very well equiped - have a look to
http://www.kfactor.it or http://www.irem.it

Such AC/DC STABLIZED power supplies are to purchase by around 4000 -
6000 bucks by one year guarantee so that it makes not sense to invest
several 10,000 bucks in developments for NEW primary AC switching and
secondary DC regulated units.

Let me know if you need more or specific details - I'm glad to help
and support you.

J

#### <NAME>

Jan 1, 1970
0
IPM - International Perforation Management
high-tech engineering China-Germany-Thailand
fax : 0049 - (0) 1212-5-375-17-531
http://www.microperforation.com
http://www.microperforation.com.cn
http://www.deguodaguan.com/ipm/
http://www.dk3qv.de
Email : [email protected] - [email protected]

We are using AC/DC converters and special DC constanter with 300 V/DC
30 up to 60 Amps for nano or micro perforation machines since more
then 20 years.

There are several companies very well equiped - have a look to
http://www.kfactor.it or http://www.irem.it

Such AC/DC STABLIZED power supplies are to purchase by around 4000 -
6000 bucks by one year guarantee so that it makes not sense to invest
several 10,000 bucks in developments for NEW primary AC switching and
secondary DC regulated units.

Let me know if you need more or specific details - I'm glad to help
and support you.

The only detail that could possibly help me at those kinds of prices is "do
you give free samples ??"

I makes very little sense to invest that kind of money when I've got less
than $500 into this laser so far, and I dont think its going to cost me more than about$1000 more to bring it online. I'm not being funded by a
corporation with deep pockets In any case, for that kind of money, I
could afford to pump this rod with 808nm laser diode bars and skip the
massive current requirements that the krypton arcs are forcing me to deal
with. (\$6,000 times the 3 supplies I need pays for a LOT of laser diode bars
!!)

And I seriously doubt its going to cost me 10K or better to to design my
own -- even if I count my time invested -- I'm pretty close to something
workable now with less than a week invested, and between the folks here and
what I'm finding online, working out the rest of the issues shouldnt take
all that long. And besides -- I get to learn something new in the process !!
<BigGrin>

[Flame On]
So save the commercial hype for someone that doesnt want to be bothered with
the nitty gritty details !!
[Flame Off]

O

#### [email protected]

Jan 1, 1970
0
dude, your insane trying to build a 6Kw. lamp supply from scratch,
those lamps blow if ya sneeze.
I know I just blew one trying to build a 2 kW version. I have one
left.

contact bruce @ contact @ second source lasers do com, he has two
eski lamp supplies in stock. BTW the series 6 pins and two connector
bodies you asked for shipped yesterday.

<NAME>

J

#### <NAME>

Jan 1, 1970
0
I'm looking to build a power supply for a krypton arc lamp, and need a
design that can deliver an adjustable 20-60 amps into a LOW resistance load
(around 5 ohms). I'm a programmer for a living and this is my first power
electronics project (previous electronics projects have mostly been
microcontroller based stuff)

I've done quite a bit of reading, but am running into a brick wall scaling
up some of the various designs that I've found online. I'm working with
LTC's SWCad3 to test concepts, but I'm having problems with input currents
spiking up way too high since my target output is so close to my current
source (AC 220v 60A single phase rectified to 320vdc), and the 120hz line is
just too slow to keep the input capacitor bank charged under heavy load
without some serious high current spikes to recharge.

Any suggestions to smooth out the current draw from the line would be MUCH
appreciated !!

Three phase AC!

But what you're doing is like deciding it would be cool to climb a
mountain for the first time, and buying a ticket to Nepal.

John

D

#### D from BC

Jan 1, 1970
0
Three phase AC!

But what you're doing is like deciding it would be cool to climb a
mountain for the first time, and buying a ticket to Nepal.

John

It's a mountain all right..
I'm very new in smps..
From near scratch, my little 130Watt (90% eff) converter project has
gone on for > 2 months.
mosfet switching losses
trace inductances
fast power diodes
feedback control theory
control topologies
smps simulation
paralleling power devices
reducing EMI
creepage
ground planes
capacitor losses
magnetics design
line filtering and ov protection
snubbers

But, I got it working.
Just have to fix some ringing....
Score:
smps: 1 quality of life: 0

D from BC

R

#### <NAME>

Jan 1, 1970
0
John said:
Ohh I'm quite prepared for a bit of mess Even got a few spare arc tubes
just in case And a welding helmet for eye protection

As stated... I'm planning a current mode buck regulator, though it has been
suggested that I'll need a boost circuit in front of it to smooth out the
line input current surges (to push up the voltage so that charge is
transfered into the caps more effciently ??) . I've already got a spice
model created that is close (need to work on the control loops a little
more, and a better soft start circuit., and beat these 120+ amp surges that
I'm getting)

"nonlinear negative dynamic impedance load" -- I'm not an EE so I could use
a bit of explanation I'll probably need to understand that as I'd like to
create a spice model of the arc tube for the later stages of testing.
(resistance decreases as current increases ?? am I even close ??)

Exactly.

Rene

W

#### <NAME>

Jan 1, 1970
0
D said:
<NAME>
Yep.

It's a mountain all right..
I'm very new in smps..
From near scratch, my little 130Watt (90% eff) converter project
has gone on for > 2 months.
mosfet switching losses
trace inductances
fast power diodes
feedback control theory
control topologies
smps simulation
paralleling power devices
reducing EMI
creepage
ground planes
capacitor losses
magnetics design
line filtering and ov protection
snubbers

But, I got it working.
Just have to fix some ringing....

That's a mere foothill, and a small one at that.
But certainly keep on climbing.

J

#### <NAME>

Jan 1, 1970
0
dude, your insane trying to build a 6Kw. lamp supply from scratch,
those lamps blow if ya sneeze.
I know I just blew one trying to build a 2 kW version. I have one
left.

contact bruce @ contact @ second source lasers do com, he has two
eski lamp supplies in stock. BTW the series 6 pins and two connector
bodies you asked for shipped yesterday.

<NAME>

Thanks for getting those out, Steve !!

and OF COURSE I'm insane.. even more so than you think because I need 8KW !!
(7mm ID bore needing 195-205 volts !!)

However -- the design is progressing -- I've got a model up and running that
(without current regulation) stabilizes at a very smooth 48 amps into the
tube (presuming 5 ohms tube resistance) after startup... its got a bit more
ripple voltage than I would like, and its spiking the AC input hard, but at
least I got the soft start and main current path worked out

having a real problem with the current sense -- trying to use the switch
model to trigger when the ISense voltage passes a preset threshold, but I
must have it modeled wrong some how because it is not switching when the
tube current spikes. (I know -- its a kludge, but I'm too lazy to add in the
full pwm driver until I got everything else close to working as intended)
I'll prolly see what I can do about putting at least a comparator in and see
if that helps

J

#### <NAME>

Jan 1, 1970
0
John said:
Thanks for getting those out, Steve !!

and OF COURSE I'm insane.. even more so than you think because I need 8KW !!
(7mm ID bore needing 195-205 volts !!)

However -- the design is progressing -- I've got a model up and running that
(without current regulation) stabilizes at a very smooth 48 amps into the
tube (presuming 5 ohms tube resistance) after startup... its got a bit more
ripple voltage than I would like, and its spiking the AC input hard, but at
least I got the soft start and main current path worked out

At this high output current, I would probably thinking about
a 2 phase parallel current mode output stage (or pair of
stages. This is a fairly simple way to divide the current
load equally between two parallel paths, while having
something like 80% or 90% of the current ripple cancel. Toy
can work with one of the halves, till you pretty much get it
working at half current and then build a second one. The
trick is to come up with the dual complementary triangle
wave to drive the error amplifiers, so that the output
pulses are always almost precisely 180 degrees out of phase.
I have made fairly high current, low ripple current mode
laser supplies this way, though the voltage was much lower,
having a real problem with the current sense -- trying to use the switch
model to trigger when the ISense voltage passes a preset threshold, but I
must have it modeled wrong some how because it is not switching when the
tube current spikes.

I think you may need a more continuous current sense, rather
than a peak trip.

J

#### <NAME>

Jan 1, 1970
0
<NAME> said:
At this high output current, I would probably thinking about a 2 phase
parallel current mode output stage (or pair of stages. This is a fairly
simple way to divide the current load equally between two parallel paths,
while having something like 80% or 90% of the current ripple cancel. Toy
can work with one of the halves, till you pretty much get it working at
half current and then build a second one. The trick is to come up with
the dual complementary triangle wave to drive the error amplifiers, so
that the output pulses are always almost precisely 180 degrees out of
phase. I have made fairly high current, low ripple current mode laser
supplies this way, though the voltage was much lower, and the load better
behaved.

I think you may need a more continuous current sense, rather than a peak
trip.

I know I need continuous feedback -- but I'm still trying to get a handle on
the main current path before I add complexity -- just wanted something
simple in there to provide SOME regulation so I'd have an idea if I was
getting close.

I have made a deciscion over the last 24 hours -- I'm going with a 2 phase
boost front end for active PFC and an internal 400v DC bus. One phase will
be a non-inverting boost for the positive halves of the 2 line inputs, the
other will be inverting for the negative halves, eliminating all the
problems that I'm having because my negative input to the buck is a half
wave pulse instead of a solid ground.

I'm gonna spend today modeling the non-inverting boost supply.. see if I can
get that working -- nice thing is -- it doesnt have to be strongly
regulated... -- the not-so-nice thing is I'm going to have to do the full
PWM setup because crude regulation methods wont work to drive the boost
switch.

hehehe at this point I would kill for a PWM controller spice model generic
enough to be used for this Gonna have to search around and see whats out
there !!

G

#### Genome

Jan 1, 1970
0
<NAME> said:
hehehe at this point I would kill for a PWM controller spice model generic
enough to be used for this Gonna have to search around and see whats
out there !!

John was talking about interleaving the buck stage but interleaving the
boost stage wouldn't hurt either.

Here you go....

http://www.genomerics.org/stuff/pfcfullb.asc

It's supposed to model a UCC3817....

http://focus.ti.com/docs/prod/folders/print/ucc3817.html

Actually it's not really that super brilliant but you'll get the idea.

DNA

J

#### <NAME>

Jan 1, 1970
0
Genome said:
John was talking about interleaving the buck stage but interleaving the
boost stage wouldn't hurt either.

Here you go....

http://www.genomerics.org/stuff/pfcfullb.asc

It's supposed to model a UCC3817....

http://focus.ti.com/docs/prod/folders/print/ucc3817.html

Actually it's not really that super brilliant but you'll get the idea.

DNA

Thank YOU !! Very much appreciated !!

I was getting a little put out with half a dozen people telling me

Thanks again for what I came here for -- a little education on SMPS design

G

#### Genome

Jan 1, 1970
0
<NAME> said:
Thank YOU !! Very much appreciated !!

I was getting a little put out with half a dozen people telling me

Thanks again for what I came here for -- a little education on SMPS design

Oh..... OK, fiddle fiddle fiddle. Try one of these,

http://www.genomerics.org/laser/pfcpar.asc

That's two stages operated in parallel with the clock waveforms shifted. The
idea is sort of that the current error amplifiers receive the same demand
signal from a single or master voltage error amplifier and program the same
current in their respective power stages. You have to figure out how to tie
the available parts in the ICs together to do the job.

One thing you are forced to do is use current transformers to recreate the
inductor current. At high powers this would be done anyway but at low powers
you would use a sense resistor in the return lead to the input rectifier.
Unfotunately when you try to operate things in parallel you can't do that
any more.... they just won't connect together proper like.

I've added some source sense resistors to the 'mosfets' but not implemented
a current limit with them. Problem is that the drain current transformer is
operated at high duty cycles and might saturate. You lose the signal and
things go bang. Resistors don't, lose signals, so it's a kind of 'oh shit!'
protection.

The peak limit is simplified a bit. It really needs slope compensation but
the IC does not make that easy to add, that's why those transistors were in
the previous one. Here I've just thrown in some appropriate current sources.

I've diddled it to do 1KW per section from a 220V line..... I suppose you
will need 8 of them because you are some sort of cross dressing freak. There
are ways of synchronising these ICs by shoving a pulse up their oscillator.
I'm sure if you read some data sheets you'll get some ideas.

In the old days a 4017 Johnson Counter clocked at N (stages) times the
desired base frequency with the reset pin on the appropriate tap did the
job. These day I suppose you would use a PIC.

Really all you are getting is something to practice your sums on. You will
probably find some of the values I've left in there are sub-optimal. You can
try putting in 'real' components to get some guesses about power losses. I'd
recommend you grab hold of some SiC diodes for the main boost diodes.

Have Fun and try not to go too fruit pie.

DNA

J

#### <NAME>

Jan 1, 1970
0
Oh..... OK, fiddle fiddle fiddle. Try one of these,

http://www.genomerics.org/laser/pfcpar.asc

That's two stages operated in parallel with the clock waveforms shifted.
The idea is sort of that the current error amplifiers receive the same
demand signal from a single or master voltage error amplifier and program
the same current in their respective power stages. You have to figure out
how to tie the available parts in the ICs together to do the job.

One thing you are forced to do is use current transformers to recreate the
inductor current. At high powers this would be done anyway but at low
powers you would use a sense resistor in the return lead to the input
rectifier. Unfotunately when you try to operate things in parallel you
can't do that any more.... they just won't connect together proper like.

I've added some source sense resistors to the 'mosfets' but not
implemented a current limit with them. Problem is that the drain current
transformer is operated at high duty cycles and might saturate. You lose
the signal and things go bang. Resistors don't, lose signals, so it's a
kind of 'oh shit!' protection.

The peak limit is simplified a bit. It really needs slope compensation but
the IC does not make that easy to add, that's why those transistors were
in the previous one. Here I've just thrown in some appropriate current
sources.

I've diddled it to do 1KW per section from a 220V line..... I suppose you
will need 8 of them because you are some sort of cross dressing freak.
There are ways of synchronising these ICs by shoving a pulse up their
oscillator. I'm sure if you read some data sheets you'll get some ideas.

In the old days a 4017 Johnson Counter clocked at N (stages) times the
desired base frequency with the reset pin on the appropriate tap did the
job. These day I suppose you would use a PIC.

Really all you are getting is something to practice your sums on. You will
probably find some of the values I've left in there are sub-optimal. You
can try putting in 'real' components to get some guesses about power
losses. I'd recommend you grab hold of some SiC diodes for the main boost
diodes.

Have Fun and try not to go too fruit pie.

DNA

I've already located inexpensive isolated current sensors with models for 50
or 100 amps and 3000v isolation, so I can easily do high side current
sensing -- just gotta do a spice model for it, which looking at your models
has helped me figure out (a voltage source defined via a formula referencing

I'm seriously looking at the PFC chip you mentioned, and some others (I've
already tried to model out one using the LT1248 but its not working right
yet)

You want fruit pie ?? here is fruit pie -- what would it take if I wanted a
true earth ground on the output of this mess -- I've considered running
seperate boosts on the poitive and negative outputs from the bridge, using
the isolated current sensor to get EVERYTHING out of the ground path.. one
of them inverting to get a positive voltage (add more in pairs if 2 wont get
me enough current at reasonable ripple)

is that fruit pie enough for you ?? or is there a better way ??

G

#### Genome

Jan 1, 1970
0
<NAME> said:
I've already located inexpensive isolated current sensors with models for
50 or 100 amps and 3000v isolation, so I can easily do high side current
sensing -- just gotta do a spice model for it, which looking at your
models has helped me figure out (a voltage source defined via a formula

I'm seriously looking at the PFC chip you mentioned, and some others (I've
already tried to model out one using the LT1248 but its not working right
yet)

You want fruit pie ?? here is fruit pie -- what would it take if I wanted
a true earth ground on the output of this mess -- I've considered running
seperate boosts on the poitive and negative outputs from the bridge, using
the isolated current sensor to get EVERYTHING out of the ground path.. one
of them inverting to get a positive voltage (add more in pairs if 2 wont
get me enough current at reasonable ripple)

is that fruit pie enough for you ?? or is there a better way ??

Ermmmmm...... Current sense transformers, something like

http://www.cd4power.com/data/magnetics/kmp_5600.pdf

Each PFC section needs to control its own inductor current otherwise they
won't share the total load. It's tied up with the volt-second balance,
inductors integrate the voltage placed accross them as the current through
them. If there is a mismatch in the drive to the inductors then one of them
will hog the current.

The local current error amplifiers force them to share it but you need
separate sense signals for each stage so one honking big current sensing
thing on the input won't do. You also need a reasonable frequency response
from the devices if you want to compensate the current loops correctly.

Ahaaaa, LT1248, shows you how crap I am for not checking if they did one or
if LTSpice had the model. I knew they did but there is something wrong with
my head. I suppose I'm fixated on TI because I grew up with Unitrode. It's
all much of a muchness and once you get happy with one you can see that a
lot of the internals are the same in all the others.

If you really really need to ground the output of your beast....... Uhm,
right, your side of the pond has strange electricity with something like
110V balanced either side of ground. My head hurts but I think you might
only get away with it if you use half wave rectification which is going to
be dirty and waste half of the capacity of your supply.

Since you seem to enjoy pain you might as well go the whole banana and
dangle a transformer isolated coverter of the end of your PFC stages.
Something like a half or full bridge. That will make any (well most)

DNA


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Zbl 1157.93519
<NAME>; <NAME>; <NAME>
$H_{\infty }$ model reduction of Markovian jump linear systems.
(English)
[J] Syst. Control Lett. 50, No. 2, 103-118 (2003). ISSN 0167-6911

Summary: The $H_{\infty }$ model reduction problem for linear systems that possess randomly jumping parameters is studied. The development includes both the continuous and discrete cases. It is shown that the reduced order models exist if a set of matrix inequalities is feasible. An effective iterative algorithm involving linear matrix inequalities is suggested to solve the matrix inequalities characterizing the model reduction solutions. Using the numerical solutions of the matrix inequalities, the reduced order models can be obtained. An example is given to illustrate the proposed model reduction method.
MSC 2000:
*93E03 General theory of stochastic systems
93B36 $H^\infty$-control

Keywords: $H_{\infty }$ theory; Model reduction; Markovian jump systems

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# Benchmarks and Visualization

## Quality

To evaluate the performance of our implementation we calculate the mean squared error on the unknown pixels of the benchmark images of [RRW+09].

## Visualization

The following videos show the iterates of the different methods. Note that the videos are timewarped.

 CF KNN LKM RW

## Performance

We compare the computational runtime of our solver with other solvers: pyAMG, UMFPAC, AMGCL, MUMPS, Eigen and SuperLU. Figure 3 shows that our implemented conjugate gradients method in combination with the incomplete Cholesky decomposition preconditioner outperforms the other methods by a large margin. For the iterative solver we used an absolute tolerance of $$10^{-7}$$, which we scaled with the number of known pixels, i.e. pixels that are either marked as foreground or background in the trimap.


text:
# NLopt Python Reference

The NLopt includes an interface callable from the Python programming language.

The main purpose of this section is to document the syntax and unique features of the Python API; for more detail on the underlying features, please refer to the C documentation in the NLopt Reference.

## Using the NLopt Python API

To use NLopt in Python, your Python program should include the lines:

import nlopt
from numpy import *

which imports the nlopt module, and also imports the numpy (NumPy) that defines the array data types used for communicating with NLopt.

## The nlopt.opt class

The NLopt API revolves around an object of type nlopt.opt. Via methods of this object, all of the parameters of the optimization are specified (dimensions, algorithm, stopping criteria, constraints, objective function, etcetera), and then one finally calls the opt.optimize method in order to perform the optimization. The object should normally be created via the constructor:

opt = nlopt.opt(algorithm, n)

given an algorithm (see NLopt Algorithms for possible values) and the dimensionality of the problem (n, the number of optimization parameters). Whereas the C algorithms are specified by nlopt_algorithm constants of the form NLOPT_LD_MMA, NLOPT_LN_COBYLA, etcetera, the Python algorithm values are of the form nlopt.LD_MMA, nlopt.LN_COBYLA, etcetera (with the NLOPT_ prefix replaced by the nlopt. namespace).

There are also a copy constructor nlopt.opt(opt) to make a copy of a given object (equivalent to nlopt_copy in the C API).

If there is an error in the constructor (or copy constructor, or assignment), a MemoryError exception is thrown.

The algorithm and dimension parameters of the object are immutable (cannot be changed without constructing a new object), but you can query them for a given object by the methods:

opt.get_algorithm()
opt.get_dimension()

You can get a string description of the algorithm via:

opt.get_algorithm_name()

## Objective function

The objective function is specified by calling one of the methods:

opt.set_min_objective(f)
opt.set_max_objective(f)

depending on whether one wishes to minimize or maximize the objective function f, respectively. The function f should be of the form:

def f(x, grad):
return ...value of f(x)...

The return value should be the value of the function at the point x, where x is a NumPy array of length n of the optimization parameters (the same as the dimension passed to the constructor).

In addition, if the argument grad is not empty [i.e. grad.size>0], then grad is a NumPy array of length n which should (upon return) be set to the gradient of the function with respect to the optimization parameters at x. That is, grad[i] should upon return contain the partial derivative $\partial f / \partial x_i$, for $0 \leq i < n$, if grad is non-empty. Not all of the optimization algorithms (below) use the gradient information: for algorithms listed as "derivative-free," the grad argument will always be empty and need never be computed. (For algorithms that do use gradient information, however, grad may still be empty for some calls.)

Note that grad must be modified in-place by your function f. Generally, this means using indexing operations grad[...] = ... to overwrite the contents of grad, as described below.

### Assigning results in-place

Your objective and constraint functions must overwrite the contents of the grad (gradient) argument in-place (although of course you can allocate whatever additional storage you might need, in addition to overwriting grad). However, typical Python assignment operations do not do this. For example:

grad = 2*x

might seem like the gradient of the function sum(x**2), but it will not work with NLopt because this expression actually allocates a new array to store 2*x and re-assigns grad to point to it, rather than overwriting the old contents of grad. Instead, you should do:

grad[:] = 2*x

Assigning any slice or view grad[...] of the array will overwrite the contents, which is what NLopt needs you to do. So, you should generally use indexing expressions grad[...] = ... to assign the gradient result.

In specific cases, there are a few other NumPy and SciPy functions that are documented to operate in-place on their arguments, and you can also use such functions to modify grad if you want. If a function is not explicitly documented to modify its arguments in-place, however, you should assume that it does not.

## Bound constraints

The bound constraints can be specified by calling the methods:

opt.set_lower_bounds(lb)
opt.set_upper_bounds(ub)

where lb and ub are arrays (NumPy arrays or Python lists) of length n (the same as the dimension passed to the nlopt.opt constructor). For convenience, these are overloaded with functions that take a single number as arguments, in order to set the lower/upper bounds for all optimization parameters to a single constant.

To retrieve the values of the lower/upper bounds, you can call one of:

opt.get_lower_bounds()
opt.get_upper_bounds()

both of which return NumPy arrays.

To specify an unbounded dimension, you can use ±float('inf') (or ±numpy.inf) in Python to specify ±∞.

## Nonlinear constraints

Just as for nonlinear constraints in C, you can specify nonlinear inequality and equality constraints by the methods:

opt.add_inequality_constraint(fc, tol=0)

where the arguments fc and h have the same form as the objective function above. The optional tol arguments specify a tolerance in judging feasibility for the purposes of stopping the optimization, as in C.

To remove all of the inequality and/or equality constraints from a given problem, you can call the following methods:

opt.remove_inequality_constraints()
opt.remove_equality_constraints()

### Vector-valued constraints

Just as for nonlinear constraints in C, you can specify vector-valued nonlinear inequality and equality constraints by the methods

opt.add_inequality_mconstraint(c, tol)

Here, tol is an array (NumPy array or Python list) of the tolerances in each constraint dimension; the dimensionality m of the constraint is determined by tol.size. The constraint function c must be of the form:

def c(result, x, grad):
result[0] = ...value of c0(x)...
result[1] = ...value of c1(x)...
...

result is a NumPy array whose length equals the dimensionality m of the constraint (same as the length of tol above), which upon return should be set in-place (see above) to the constraint results at the point x (a NumPy array whose length n is the same as the dimension passed to the constructor). Any return value of the function is ignored.

In addition, if the argument grad is not empty [i.e. grad.size>0], then grad is a 2d NumPy array of size m×n which should (upon return) be set in-place (see above) to the gradient of the function with respect to the optimization parameters at x. That is, grad[i,j] should upon return contain the partial derivative $\partial c_i / \partial x_j$ if grad is non-empty. Not all of the optimization algorithms (below) use the gradient information: for algorithms listed as "derivative-free," the grad argument will always be empty and need never be computed. (For algorithms that do use gradient information, however, grad may still be empty for some calls.)

An inequality constraint corresponds to $c_i \le 0$ for $0 \le i < m$, and an equality constraint corresponds to ci = 0, in both cases with tolerance tol[i] for purposes of termination criteria.

(You can add multiple vector-valued constraints and/or scalar constraints in the same problem.)

## Stopping criteria

As explained in the C API Reference and the Introduction), you have multiple options for different stopping criteria that you can specify. (Unspecified stopping criteria are disabled; i.e., they have innocuous defaults.)

For each stopping criteria, there are (at least) two methods: a set method to specify the stopping criterion, and a get method to retrieve the current value for that criterion. The meanings of each criterion are exactly the same as in the C API.

opt.set_stopval(stopval)
opt.get_stopval()

Stop when an objective value of at least stopval is found.

opt.set_ftol_rel(tol)
opt.get_ftol_rel()

Set relative tolerance on function value.

opt.set_ftol_abs(tol)
opt.get_ftol_abs()

Set absolute tolerance on function value.

opt.set_xtol_rel(tol)
opt.get_xtol_rel()

Set relative tolerance on optimization parameters.

opt.set_xtol_abs(tol)
opt.get_xtol_abs()

Set absolute tolerances on optimization parameters. The tol input must be an array (NumPy array or Python list) of length n (the dimension specified in the nlopt.opt constructor); alternatively, you can pass a single number in order to set the same tolerance for all optimization parameters. get_xtol_abs() returns the tolerances as a NumPy array.

opt.set_maxeval(maxeval)
opt.get_maxeval()

Stop when the number of function evaluations exceeds maxeval. (0 or negative for no limit.)

opt.set_maxtime(maxtime)
opt.get_maxtime()

Stop when the optimization time (in seconds) exceeds maxtime. (0 or negative for no limit.)

### Forced termination

In certain cases, the caller may wish to force the optimization to halt, for some reason unknown to NLopt. For example, if the user presses Ctrl-C, or there is an error of some sort in the objective function. You can do this by raise any exception inside your objective/constraint functions:the optimization will be halted gracefully, and the same exception will be raised to the caller. See Exceptions, below. The Python equivalent of nlopt_forced_stop from the C API is to throw an nlopt.ForcedStop exception.

## Performing the optimization

Once all of the desired optimization parameters have been specified in a given object opt, you can perform the optimization by calling:

xopt = opt.optimize(x)

On input, x is an array (NumPy array or Python list) of length n (the dimension of the problem from the nlopt.opt constructor) giving an initial guess for the optimization parameters. The return value xopt is a NumPy array containing the optimized values of the optimization parameters.

You can call the following methods to retrieve the optimized objective function value from the last optimize call, and also the return code (including negative/failure return values) from the last optimize call:

opt_val = opt.last_optimum_value()
result = opt.last_optimize_result()

The return code (see below) is positive on success, indicating the reason for termination. On failure (negative return codes), optimize() throws an exception (see Exceptions, below).

### Return values

The possible return values are the same as the return values in the C API, except that the NLOPT_ prefix is replaced with the nlopt. namespace. That is, NLOPT_SUCCESS becomes nlopt.SUCCESS, etcetera.

## Exceptions

The Error codes (negative return values) in the C API are replaced in the Python API by thrown exceptions. The following exceptions are thrown by the various routines:

RunTimeError
Generic failure, equivalent to NLOPT_FAILURE.
ValueError
Invalid arguments (e.g. lower bounds are bigger than upper bounds, an unknown algorithm was specified, etcetera), equivalent to NLOPT_INVALID_ARGS.
MemoryError
Ran out of memory (a memory allocation failed), equivalent to NLOPT_OUT_OF_MEMORY.
nlopt.RoundoffLimited (subclass of Exception)
Halted because roundoff errors limited progress, equivalent to NLOPT_ROUNDOFF_LIMITED.
nlopt.ForcedStop (subclass of Exception)
Halted because of a forced termination: the user called opt.force_stop() from the user’s objective function or threw an nlopt.ForcedStop exception. Equivalent to NLOPT_FORCED_STOP.

If your objective/constraint functions throw any exception during the execution of opt.optimize, it will be caught by NLopt and the optimization will be halted gracefully, and opt.optimize will re-throw the same exception to its caller.

## Local/subsidiary optimization algorithm

Some of the algorithms, especially MLSL and AUGLAG, use a different optimization algorithm as a subroutine, typically for local optimization. You can change the local search algorithm and its tolerances by calling:

opt.set_local_optimizer(local_opt)

Here, local_opt is another nlopt.opt object whose parameters are used to determine the local search algorithm, its stopping criteria, and other algorithm parameters. (However, the objective function, bounds, and nonlinear-constraint parameters of local_opt are ignored.) The dimension n of local_opt must match that of opt.

This function makes a copy of the local_opt object, so you can freely change your original local_opt afterwards without affecting opt.

## Initial step size

Just as in the C API, you can get and set the initial step sizes for derivative-free optimization algorithms. The Python equivalents of the C functions are the following methods:

opt.set_initial_step(dx)
opt.get_initial_step(x)

Here, dx is an array (NumPy array or Python list) of the (nonzero) initial steps for each dimension, or a single number if you wish to use the same initial steps for all dimensions. opt.get_initial_step(x) returns the initial step that will be used for a starting guess of x in opt.optimize(x).

## Stochastic population

Just as in the C API, you can get and set the initial population for stochastic optimization algorithms, by the methods:

opt.set_population(pop)
opt.get_population()

(A pop of zero implies that the heuristic default will be used.)

## Pseudorandom numbers

For stochastic optimization algorithms, we use pseudorandom numbers generated by the Mersenne Twister algorithm, based on code from <NAME>. By default, the seed for the random numbers is generated from the system time, so that you will get a different sequence of pseudorandom numbers each time you run your program. If you want to use a "deterministic" sequence of pseudorandom numbers, i.e. the same sequence from run to run, you can set the seed by calling:

nlopt.srand(seed)

where seed is an integer. To reset the seed based on the system time, you can call:

nlopt.srand_time()

(Normally, you don't need to call this as it is called automatically. However, it might be useful if you want to "re-randomize" the pseudorandom numbers after calling nlopt.srand to set a deterministic seed.)

## Vector storage for limited-memory quasi-Newton algorithms

Just as in the C API, you can get and set the number M of stored vectors for limited-memory quasi-Newton algorithms, via the methods:

opt.set_vector_storage(M)
opt.get_vector_storage()

(The default is M=0, in which case NLopt uses a heuristic nonzero value.)

## Version number

To determine the version number of NLopt at runtime, you can call:

nlopt.version_major()
nlopt.version_minor()
nlopt.version_bugfix()

For example, NLopt version 3.1.4 would return major=3, minor=1, and bugfix=4.


text:
# How do you find the value of the discriminant and state the type of solutions given -2x^2-8x-14=-6?

Determinant $= 0$ & the roots are real & equal

#### Explanation:

$- 2 {x}^{2} - 8 x - 14 = - 6$

${x}^{2} + 4 x + 4 = 0$

Comparing the above equation with the standard form of quadratic equation: $a {x}^{2} + b x + c = 0$, we get

$a = 1 , b = 4 , c = 4$

hence the determinant is given as

${b}^{2} - 4 a c = {4}^{2} - 4 \left(1\right) \left(4\right) = 0$

Since ${b}^{2} - 4 a c = 0$ the roots are real & equal.


text:
# I Some questions from a numerical analysis book

1. Jun 15, 2017

### awholenumber

Last edited by a moderator: Jun 15, 2017
2. Jun 15, 2017

### BvU

Can't find what this is about ....

3. Jun 15, 2017

### Staff: Mentor

This is a 300 page book. You will have to specify what you are talking about.

4. Jun 15, 2017

### awholenumber

Are we trying to find the roots with all these types of numerical methods ? or is it called finding something else ?

5. Jun 15, 2017

### Staff: Mentor

Generally speaking, a numerical algorithm that allows you to solve a problem of the type $\mathbf{F}(\mathbf{x}) = 0$ is indeed called a root-finding algorithm.

6. Jun 15, 2017

### awholenumber

So what are we trying to do when we use numerical differentiation ? I don't think its called root finding ? is it ?

7. Jun 15, 2017

### Staff: Mentor

No, it is called numerical differentiation.

8. Jun 15, 2017

### awholenumber

OK , thanks

9. Jun 15, 2017

### Staff: Mentor

A book on Numerical Analysis typically discusses techniques for solving equations, finding derivatives numerically, and calculating integrals numerically. It might also discuss techniques for finding matrix inverses, finding eigenvalues and eigenvectors of matrices, and other topics.

10. Jun 15, 2017

### awholenumber

Thanks

11. Jun 15, 2017

### Staff: Mentor

Really, though, @rosiekidcute, if you're struggling with basic concepts like fractions, factoring, and trig (as in recent threads), it's not very likely that you will understand topics in numerical analysis or partial differential equation.

12. Jun 15, 2017

### awholenumber

I keep thinking of naturally occurring things whenever i think of derivatives , differential equations and partial differential equations .Like electricity , magnetic field .
I am always like if there is a derivative in a differential equation , how could have that happened ? what could have made it like that . Maybe i should stop thinking about it that way .Its just some rules , Maybe its my lack of experience .

13. Jun 15, 2017

### jbriggs444

No need to invoke electricity or magnetism. Consider Newton's second law for an object with a fixed mass: F=ma. Force equals mass times acceleration.

That acceleration is a the first derivative of velocity. $F=m \frac{dv}{dt}$.
That velocity is the first derivative of position. $F = m \frac{d^2x}{dt^2}$.
If we are dealing with a mass on a spring, that force is also given by Hooke's Law: F = -kx.
So we have $-kx = m \frac{d^2x}{dt^2}$

That's a differential equation. (A second order homogeneous linear differential equation. There is a straightforward crank-and-grind approach to solving those).

Last edited: Jun 15, 2017
14. Jun 15, 2017

### awholenumber

Thanks ,

Same things right ?

15. Jun 17, 2017

### Staff: Mentor

No, these are different quantities: displacement, velocity, acceleration, and jerk.

16. Jun 17, 2017

### awholenumber

Right , sorry about that .


text:
# Deleting Numbers

## 题目描述

This is an interactive problem. There is an unknown integer $x$ ( $1\le x\le n$ ). You want to find $x$ . At first, you have a set of integers $\{1, 2, \ldots, n\}$ . You can perform the following operations no more than $10000$ times: - A $a$ : find how many numbers are multiples of $a$ in the current set. - B $a$ : find how many numbers are multiples of $a$ in this set, and then delete all multiples of $a$ , but $x$ will never be deleted (even if it is a multiple of $a$ ). In this operation, $a$ must be greater than $1$ . - C $a$ : it means that you know that $x=a$ . This operation can be only performed once. Remember that in the operation of type B $a>1$ must hold. Write a program, that will find the value of $x$ .

## 输入输出格式

### 输入格式

The first line contains one integer $n$ ( $1\le n\le 10^5$ ). The remaining parts of the input will be given throughout the interaction process.

### 输出格式

In each round, your program needs to print a line containing one uppercase letter A, B or C and an integer $a$ ( $1\le a\le n$ for operations A and C, $2\le a\le n$ for operation B). This line desribes operation you make. If your operation has type C your program should terminate immediately. Else your program should read one line containing a single integer, which is the answer to your operation. After outputting each line, don't forget to flush the output. To do it use: - fflush(stdout) in C/C++; - System.out.flush() in Java; - sys.stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages. It is guaranteed, that the number $x$ is fixed and won't change during the interaction process. Hacks: To make a hack, use such input format: The only line should contain two integers $n$ , $x$ ( $1 \leq x \leq n \leq 10^5$ ).

## 输入输出样例

### 输入样例 #1

10

2

4

0

### 输出样例 #1

B 4

A 2

A 8

C 4

## 说明

Note that to make the sample more clear, we added extra empty lines. You shouldn't print any extra empty lines during the interaction process. In the first test $n=10$ and $x=4$ . Initially the set is: $\{1,2,3,4,5,6,7,8,9,10\}$ . In the first operation, you ask how many numbers are multiples of $4$ and delete them. The answer is $2$ because there are two numbers divisible by $4$ : $\{4,8\}$ . $8$ will be deleted but $4$ won't, because the number $x$ will never be deleted. Now the set is $\{1,2,3,4,5,6,7,9,10\}$ . In the second operation, you ask how many numbers are multiples of $2$ . The answer is $4$ because there are four numbers divisible by $2$ : $\{2,4,6,10\}$ . In the third operation, you ask how many numbers are multiples of $8$ . The answer is $0$ because there isn't any number divisible by $8$ in the current set. In the fourth operation, you know that $x=4$ , which is the right answer.