User talk:The Buzzfizzler

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God Jadinkos
On Saradomin's_blessing/Obtaining I outline the steps to find the proper combination to attract God jadinkos, but it seems too long the way I have it written. When you have a few minutes could you review it please? And tell me what I need to fix. Thanks man.--Degenret01 02:04, April 19, 2011 (UTC)

Gaz Lloyd,

 * I want to learn how to write/edit articles on the Wiki here (though it's likely pointless cuz it's almost all up to date.)


 * I did the homework for formatting, and saw you were a teacher. WIll you check it out for me? I did everything about Fletching. The page is [HERE]]

Amos0206 01:26, April 20, 2011 (UTC)Amos0206 Set that out as text to display text to display 20:40, June 11, 2011 (UTC)

UoL
Thanks for the pass. For the bullets, I was trying to stack them to get them indented, but it just put 2 bullets... Test:
 * This is what I did
 * This is what I tried to do

You have to put 1 bullet before doing a 2nd, I think... 20:41, June 11, 2011 (UTC)

Trouble with Templates
Gaz Lloyd,
 * Disregard this. I asked you some questions about the Template lesson in UoL, but I mustered through it and figured most things out. :)

Thanks, Amos0206 01:29, April 21, 2011 (UTC)Amos0206

hi
mr events team guy how does this look w:c:iiiiiii:Events Team -- 01:36, April 23, 2011 (UTC)
 * How about switching the events and the summary? 01:38, April 23, 2011 (UTC)

CiteForum Problems
Hey Gaz! Per this, I changed all of the citation templates to 180 days. They all (at least seemingly) work, but I just looked at CiteForum and it was having some problems, so I reverted my edit to that one. Could you take a look at it and possibly fix it? I'm not sure what I did. It seemed simple enough, I guess it wasn't. 14:56, April 24, 2011 (UTC)
 * Thank you!  18:02, April 24, 2011 (UTC)

Requesting a Block/Ban
I have to request that the IP 87.212.22.245 is banned, because I have had to rollback two of his/her edits, and also undo one. Check the contribs if you need. I came to you because I figured you would get the job done. 16:15, April 25, 2011 (UTC)

fun fun fun
so i heard u liek fun?? 17:44, April 25, 2011 (UTC)

Rank change
Gareth, can you please make Dtm a deputy owner when you get the chance? Thanks. 02:37, April 27, 2011 (UTC)
 * can u pls maek me an awesum person??? 03:16, April 27, 2011 (UTC)

I Wanna be the Scapeguy
Make me the official wiki scapegoat! Do it! 05:36, April 27, 2011 (UTC)

Signature
Gaz, How can I make myself a cool-looking signature for when I post on a talk page?

Amos0206 00:31, April 28, 2011 (UTC)Amos0206

Dirty mind
http://www.youtube.com/watch?v=TywmpMQYojs

14:27, April 29, 2011 (UTC)

Effigy Calculator
Hello Gaz, I have been using this wiki for a long time, but have just created an account on it in order to become a more active member of the rs wiki community. I have attended some of the events so far, such as the Saradomin GWD a couple weeks ago (I was in your group, i was Sylvannyn.) Anyhoo, I had an idea for a calculator on this wiki, and i'd write it myself, but I have less than no clue how to write them in JavaScript (or whatever theyre written in.) The calc would calculate the percent chance of getting an Ancient Effigy drop from killing a specific number of monsters, and their combat level, both inputted by the user. I already have a formula, I just need someone to turn it into a calc. The formula is as follows, given n, the number of monsters killed, and l, their combat level:

Percent chance: 100(1-(1-1/(11018254*l^-1.9469123))^n)

Example: Probability of 1 effigy from 100 steel dragons:

100(1-(1-1/(11018254*246^-1.9469123))^100)=100(1-(1-1/243.8775)^100)=100(.99958995)^100=

100*0.66306324=66.306324%.

This means that by killing 100 steel dragons, you have around a 66% chance of receiving an Ancient effigy. FYI: the formula for effigy drop rate is taken from this wiki's page on Ancient Effigies.<

D4RK3N R4HL 03:25, May 1, 2011 (UTC)

I'm back!
Hi, Gaz. You probably don't remember me, but I'm Ameobea10 and have been away form the wiki for a long time and just returned. You helped me a lot in the past, from editing to my dignature, but most recently with parser functions and wiki calculators, etc. Anway, I saw that you finally got that guide up and running, and its great! Just wanted to say thanks and that I am glad to see the wiki still doing great. Also, if you want, you can drop in on my userpage, its new and improved!

Thanks for everything, 23:37, May 3, 2011 (UTC)

re: cannot revive FACTS
ow dude not you too ... its a fact.

I can understand not everyone knows but that why I am posting it on here.

Because it is a fact and I am for sure ( I am 2452 total) and know a lot of upper level players

And all of my sources say only Castle wars has it.

The only real real proof you can have is a post on HIGH LEVEL RS FORUM

I was able to acces it thanks to one of my friends and jagex mods directly talked to Castle wars saying: "We will have something specially for you." But you know. if a bunch of level 2100s wanna play wannabeknowitall, fine with me.

Shinigamidaio 21:08, May 5, 2011 (UTC)

Grabbafabbadabbahabbatrabba?
Ignoring the idiot above this post who thinks levels mean power on the wiki...I figured you'd like this:

http://www.youtube.com/watch?v=XGwcrAzYcUE&feature=channel_video_title

Event
Can you put me as a defender? No guarantee I'll make it, but I'll try. 13:39, May 14, 2011 (UTC) 000cena 10:58, August 23, 2011 (UTC)

Question
Enigma, hey, long time no see. Um, I just wanted to ask a question: The Gravestones article mentions that due an update, should your membership be terminated, your Gravestone (such as the Angel of Death) will be reverted to a non-member one. Is this actually true? Ralnon (talk) 00:20, May 15, 2011 (UTC)

I can do this?
RS:CALC 17:47, May 16, 2011 (UTC)

Rank change
Hi Gareth, when you have a chance, please change ENlGMATlC to an admin. He's User:9the Enigma9. I made him a general temporarily. Thanks. 17:09, May 22, 2011 (UTC)

Hai
It's monday 21:44, May 23, 2011 (UTC)
 * I'm guessing by the number of people asking you for their ranks I'm meant to come to you, mind giving my bananas in the clan chat pl0x? I've yet to get around to asking for them as I've been fairly inactive since the new clan system came out, as I'm more active now I could probably get a use out of my rank, thanks  15:03, May 25, 2011 (UTC)
 * Never mind liquidhelm got to it first 16:28, May 25, 2011 (UTC)

troll ladder
http://i747.photobucket.com/albums/xx119/JimtehMage/trollladder.jpg 15:22, June 5, 2011 (UTC)

Mods are asleep...
http://img204.imageshack.us/img204/5833/modsareasleep.jpg

Your calculator guide
Could you possibly make examples on it?

Say....

For example, a template needs 3 inputs. If you call them a, b, and c, to use the template you'll use. If you called them 1, 2, and 3, to use them you'll use  (or you could use  )


 * would become:

, or

Like, show a bit what would happen with the things (kind of like the signature page.) 20:56, June 11, 2011 (UTC)

Actually, no. You weren't even specifying templates, so that wouldn't work. My mistake. :P 20:57, June 11, 2011 (UTC)

back
The koalas are back... If you see any possibility, and see this before really coming online, could you please come online asap? Thanks, 11:45, June 15, 2011 (UTC)

Lulzhammer
Ban back to 2 weeks. Gogogogogo.

Thank you
Thanks for deleting page Coppercab as of how it was inappropriate to be posted. At the same time I was trying to delete it but did not know how. If it is possible, could you explain to me on how to delete pages? Thanks!

Firebolt4848  18:40, June 25, 2011 (UTC)

The 90s
You nostalgia you lose. 08:11, June 26, 2011 (UTC)

Sadface.jpeg
Sergeants can't kick guests, thought I'd let you know. 06:22, July 2, 2011 (UTC)

Rank
Horsehead needs a bronze key. 13:20, July 10, 2011 (UTC)

Clan
Hey Gareth, How do I join the wiki clan?!? :P? I saw that you were the owner.

--G12Fighter 22:03, July 10, 2011 (UTC)

citadel stuffs
lol u trollin bro? everyone in the cc is waitin for u and u don't show up http://images4.wikia.nocookie.net/__cb20110725191006/callofduty/images/thumb/7/73/Trollface.png/50px-Trollface.png 10:29, July 26, 2011 (UTC)
 * YA MAEKIN ME CRY BRO  10:43, July 26, 2011 (UTC)

I regret to inform you...
http://i537.photobucket.com/albums/ff331/Soldier_1033/ScreenShot2011-07-27at30431AM.png

luv u bai 08:17, July 27, 2011 (UTC)


 * BLASPHEMY!!!!! I will not commit such treason! 12:54, August 2, 2011 (UTC)

Re:Banned from Chat
Ah right sorry about that, I didn't realise. I've been having problems with disconnecting from the chat today, I don't know why. 21:12, July 31, 2011 (UTC)

Duplicate citations
I made a change to Template:DropsLine to accommodate items that have value but cannot be traded, such as Seren statuette. Problem is the citation shows up multiple times if the parameter is specified on more than one item. Help? 03:22, August 4, 2011 (UTC)
 * Nevermind, Cook Me Plox set me straight :P 04:18, August 4, 2011 (UTC)

Stubborn anon
Hi Gaz, this *insert long line on insult* anon (75.177.129.139), has been making my nipples hard by reverting Blink and adding a false fact, I've RS:UCS and ignored RS:3RR but he simply will not quit, any help on this please. 02:00, August 5, 2011 (UTC)

Altered comment?
Hey Gaz, what's this all about? <.< 19:46, August 6, 2011 (UTC)

Seriously Gaz, I don't want to make a big deal out of this, but you've now not only altered my comment, you've ignored my question too. Is there some bigger picture here? O_o 10:59, August 8, 2011 (UTC)

Oh, sorry then. I didn't know it was only an edit conflict, and in the difference between revisions log my comment appeared to have been altered as opposed to removed, and then when you didn't answer for a while I just started to get worked up over it. Sorry for the trouble, 14:59, August 8, 2011 (UTC)

Calculators?
I read about the calculator guides (well, the start of it), but do you have any guides on making calculators for our own website? would be great...

Advice noob.jpeg
http://i747.photobucket.com/albums/xx119/JimtehMage/Advicenoob.jpg 04:14, August 12, 2011 (UTC)

Click if you dare
Click me

You can't browse Recent changes is the only thing I found so far if you click it. 13:24, August 12, 2011 (UTC)

99 Summon
Hey gaz i'm lewa username is c5s

I was wondering if it is possible for me to get 99 summon untrimmed and how you went about doing it, like your story

i just recently started thinking about it and the math looks right

i have 2.9m summon in bank in charms

you can find me on highscores

20:31, August 14, 2011 (UTC)

I know its possible with Bork (37 summon exp/hp exp) and effigies and cannoning cave crawlers and penguins and troll invasion and genie lamps and kbd killing (3 summon exp/hp exp) and double exp weekend pouch making (1.1x multiplier)

Back in the day I got 68-88 summon on double exp weekend (the first)

I dont think ill do it though, it seems too much of a hassle and too much of a limit on my rs time

Thanks for the help 05:28, August 15, 2011 (UTC)

CC Invitation
Gaz, you're afk right now and a guest is looking for an invite. I'm not sure if you're doing something on the wiki so you might not see this on time, but he's waiting. 12:09, August 15, 2011 (UTC)

Cannon
we both forgot about cannon, rofl 02:06, August 18, 2011 (UTC)

check out this thread http://forums.zybez.net/topic/1327001-mu7c-ronaldos-98s-with-his-untrimmed-slayer-cape/

Getting list of all pages in the "Exchange:" namespace?
Is this possible somehow? If not, is it then possible to get a list of the "Exchange:"-names of the all GE-tradeable items? 16:44, August 21, 2011 (UTC)
 * Yeah, but like more a plain list so I can use it in a Template. 17:16, August 21, 2011 (UTC)
 * How do I with DPL, list all the pages in "Exchange:"-namespace ending with "pyre logs" or "Pyre logs"? Can I parse that list to a template? 18:11, August 21, 2011 (UTC)
 * Thank you very much! 19:27, August 21, 2011 (UTC)
 * One more thing: What does the commas in the format field do? 21:23, August 21, 2011 (UTC)

Charms to experience calculator
What's wrong with the calculator from the link I provided?

Adding a calculator to the calculator page
Hey Gaz, I made my first guide which calculates the cost per exp of burning logs then picking up the ashes. If you think it's worth it could you help me and show me how to add it thanks.

Hoppingmad9 09:45, August 23, 2011 (UTC)

Pi
Problem: Prove that pi exists.

Well, the first problem is understanding what the problem is. Indeed, even the compendium [L. Berggren, J. Borwein, P. Borwein, Pi: A Source Book, Springer Verlag, New York 1997] fails to provide a proof of pi's existence! Basically, you need to figure out what the exact definition of pi is, and then rigorously prove that this defines a unique real number. In effect, this problem is an exercise in mathematical rigor.

Sub Problem 1: What is the correct definition of pi?

Some people try to get around the technical difficulties encountered in the proof below by defining pi in unusual ways (e.g., Apostol in what is therefore his deficient Calculus text defines pi as the area of a circle of radius 1), but you really can't get around the following:

Definition: Pi is the ratio of the circumference of a circle to its diameter.

Now that the definition is settled, it's time to grapple with the more subtle issue which is simply:

Sub Problem 2: Why does the above definition require any kind of proof?

Now is the time to think deep thoughts.

Deep Thought Break.

Well, the first deep point shouldn't be that hard to figure out: The definition of pi is for any circle, so you have to prove the (obvious) fact that the ratio of circumference to diameter is always the same for all circles.

Lemma 1: Define pi(C) to be the ratio of the circumference of circle C to its diameter. Then for any two circles C_1, C_2 one has pi(C_1) = pi(C_2).

That is not too hard to prove, as will be seen below. But before doing that, the time has come to face the more difficult issue, which is that even Lemma 1 is not sufficient to prove pi's existence (and even Lemma 1 fails to work at this point). Deep thought time again.

Deep Thought Break.

The problem is that the circumference of a circle is not a straight line, and it may be possible that its length is infinite. If that were the case, then pi would certainly not be defined, since it would not be a real number (and Lemma 1 would be meaningless). In other words, one must provide a completely rigorous argument which shows that the circumference of a circle is bounded above.

Lemma 2: Let C be any circle and let pi(C) be defined as above, then pi(C) <= 4.

Finally, Lemmas 2 and 1 imply the existence of the real number pi proving the main result.

Proof of Lemma 2: I will give an outline of a proof and state at every point what assumptions I am making and what proofs I am omitting or leaving to the reader. These gaps can be filled in by refering to any geometry text. A complete proof of this result from a set of axioms and postulates (what is the difference between these two?) appears in the book [E. Moise, Elementary Geometry from an Advanced Viewpoint, Addison-Wesley, Reading MA, 1974]. Note that I have not included any diagrams. You will have to use your imagination!

Consider a circle C of radius r (and diameter 2r). Note that a circle is defined to be the set of points equidistant from a point (this common distance is the definition of the radius). Before proving the upper bound, I will first prove the lower bound pi(C) > 3. This will be useful to understand the subtleties involved.

To prove the lower bound, just inscribe a regular hexagon. Each side of the regular hexagon has length r (explain why), so the total perimeter of the hexagon is 6r. But the length of the perimeter of the circle is greater than the perimeter of the hexagon. The reason is that each arc subtending a side of the hexagon is longer than the side because it is an axiom that a line is the shortest distance between two points.

It follows that

pi(C) = (perimeter of C)/(diameter of C)= (perimeter of C)/(2 r) > (6r)(2r) = 3,

which is the lower bound.

That was not too difficult, but showing that pi(C) <= 4 is much harder, as will now be seen. The basic idea is just as simple, you circumscribe a square of side 2r and since the square has a longer perimeter, the result should follow.

Unfortunately, it is no longer an axiom that the perimeter of the square is longer than the perimeter of the circle. Of course, you can make it an axiom, which is exactly what Archimedes did (the main point is that Archimedes recognized that this needs to be based on some unproved assumption). Archimedes' axiom says [Archimedes, On the Sphere and Cylinder I, Assumption 2, in ``The Works of Archimedes,'' translated by T.L. Heath, Dover, New York, 1953]:

``Of other lines in a plane and having the same extremities, [any two] such are unequal whenever both are concave in the same direction and one of them is either wholly included between the other and the straight line which has the same extremities with it, or is partly included by, and is partly common with, the other; and that [line] which is included in the lesser [of the two].''

This definition says that if two concave curves are such that one is inside the other except that their ends meet, then the outside one is longer. In fact, it is not that hard to prove Archimedes' axiom,

Self Study Problem: Prove Archimedes' axiom from first principles. Since the circle and the square are both concave and the four pieces of the square that touch the circle satisfy the assumption, it follows that the square has a longer perimeter than the circle, proving the lemma.

The appeal to a new axiom is not too elegant, and should be avoided if possible. Of course, one can simply prove the axiom directly, but that is a bit of work. One is spared this effort in this case, because the two curves involved in the lemma, the circle and the square are much simpler than general concave curves, and it is possible in this case to prove Archimedes' axiom directly. Before you can do this, though, you have to define what you mean by length. One definition can go as follows:

Definition. Consider a curve in the plane given by a map f:[0, 1] -> R^2 (explain why this assumption does not really lose any generality). Then the length of the curve from f(0) to f(1) will be the least upper bound of the numbers


 * f(a_0) - f(a_1)| + |f(a_1) - f(a_2)| +... + |f(a_{n-1}) - f(a_n)|,

where 0=a_0 < a_1 < a_2 <... < a_n = 1 ranges over all partitions of [0, 1].

In other words, you approximate the curve by little segments, and add up the lengths of all the little segments and look at the values this takes. It is actually a theorem that must be proved that no matter how you decide to partition [0, 1], as long as you have |a_i - a_{i+1}| -> 0 (i.e., your partitions become finer and finer), the sum will approach the same number. Note that this definition requires the least upper bound axiom of the real numbers, i.e., that every bounded set of real numbers has a least upper bound that is a real number.

In the case of a circle, the definition of length is much simpler:

Definition: The circumference of a circle is the least upper bound of the perimeters of all inscribed polygons.

Given this definition, in order to show that pi(C) <= 4, one has to show that given any partition of the circle, i.e., an inscribed polygon, its perimeter is less than the circumbscribed square

The proof of this is based on a simple lemma

Lemma 3. Consider an isosceles triangle ABC with AB = AC. If AB and AC are extended to AD and AE, then DE >= BC.

Before proving this lemma, let's see how it proves the result. Given an inscribed polygon, take any side meeting the circle at P and at Q. If the circle has center O, then OP=OQ. Now Let OP meet the square at R and OQ meet the square at S, then by the lemma RS > PQ. Doing this for every side of the polygon shows that the perimeter of the polygon is smaller than the perimeter of the square.

It follows that any inscribed polygon has perimeter smaller than the square, so the least upper bound of the perimeters is smaller than or equal to the perimeter of the square which is equal to 8r. Since the least upper bound of the perimeters is equal to the length of the perimeter of C, one has

pi(C) = (perimeter of C)/(diameter of C) = (perimeter of C)(2r) <= (8r)(2r) = 4.

It follows that pi(C) <= 4, so you need to do even more work to show that pi(C) < 4. The main point, however, is that you have at least shown that pi(C) is finite.

Proof of Lemma 3: Assume that AE >= AD (the other case is similar) then let F be on AE such that AF=AD.

The triangles ABC and ADF are similar so it follows that DF >=BC (explain) and it is sufficient to prove that DE \ge DF. To do this, note the angle AFD is less than 90 degrees. This is shown by appealing to the elementary theorem that the base angles of an isosceles triangle are equal. Since the angle sum of a triangle is 180 degrees (why?), this implies that the base angles are each less than 90 degrees. Since angle AFD is less than 90 degrees, angle EFD is greater than 90 degrees (why?), so that the other angles in triangle EFD are less than 90 degrees (why?) so that angle EFD is the largest angle in this triangle. The result now follows from

Proposition. If two sides of a triangle are not congruent, then the angles opposite them are not congruent, and the larger side is opposite the larger angle.

Proof: Let the triangle be PQR, where PQ != PR.

Assume that PR > PQ (the other case is similar), and let S be on PR such that PS=PQ. It follows from the isosceles triangle theorem that angle PQS is equal to angle PSQ. Since S is in the interior of PR, it follows that angle PQS is less than angle PQR (though obvious, this must be proved from the axioms). Another elementary fact (though obvious this must be proved from the axioms) is that S lying in the interior of PR implies that angle PSQ is greater than the angle PRQ. Combining all these facts shows that angle PQR is greater than angle PRQ proving the proposition. Q.E.D.

Proof of part Lemma 1: To show that pi(C) is the same for all circles, one uses the method of exhaustion invented by Eudoxus (ca. 408B.C.-355B.C.) and used extensively by Archimedes to prove his results on areas and volumes of circles, spheres, etc. You will note that this method is essentially the method of limits used in calculus (so the ancient Greeks already understood limits).

This method uses a proof by contradiction. So assume that there are two circles C_1 and C_2 such that pi(C_1) != pi (C_2). One can assume that pi(C_1) > pi(C_2) since the other case works the same way.

The first step is to move circle C_1 so it has the same center as circle C_2, i.e., they are concentric. You have to prove that this does not change the length of the perimeter of circle C_2 nor its radius. This is a good exercise in seeing if you have understood the definitions.

Given that you have proved that this does not change pi(C_2), I will arrive at a contradiction. Now, since pi(C_1)> pi(C_2), the definition of least upper bound implies that there is a partition P_1 of the circle C_1 such that the length of P_1, i.e., the perimeter of the inscribed polygon given by P_1, divided by the diameter of C_1 is greater than pi(C_2).

Now consider the partition P_2 of C_2 generated by P_1 as follows: For each point p_1 of P_1 find the corresponding point p_2 on C_2 by drawing a radius through p_1 and letting p_2 be the point on C_2 meeting this radius (on the same side of the common center).

Now for any two adjacent points p_2,p_2' of P_2, consider the ratio |p_2 - p_2'|/r_2, where r_2 is the radius of C_2. By similar triangles and the construction of P_2, this is equal to |p_1 - p_1'|/r_1, where r_1 is the radius of C_1. It follows that

1/r_1 sum |p_1 - p_1'| = 1/r_2 \sum|p_2 - p_2'|

where the summation is over all adjacent points in each partition. Dividing this by 2 gives

1/d_1 \sum |p_1 - p_1'| = 1/d_2 \sum|p_2 - p_2'|

where d_1,d_2 are the diameters of C_1,C_2, respectively. But the assumption that the partition P_1 gave a ratio greater than pi(C_2) and the above equation leads to

1/d_2 \sum|p_2 - p_2'| > pi(C_2)

which contradicts that pi(C_2) is the least upper bound for such ratios (why is it the least upper bound of such ratios?). The result follows by contradiction.

Huh?
Why didn't you delete the ogre chatheads? There were animations uploaded, which are better. 14:47, September 2, 2011 (UTC)

Wyverns and Mauls
Just for your information, the maul IS the most effective weapon against Skeletal Wyverns. You may take massive damage, but with Soul Split and a Unicorn stallion, one may heal back the damage just as quickly. Shinray1kuo 23:12, September 5, 2011 (UTC)

Occupation
Is that really your occupation? 01:36, September 6, 2011 (UTC)

== helo i wanted to know how i can get the word out in runescape wikia about world 51 house party in yanille for PURES under 30def we been alive since 2006 august 20th.We have some vids out on youtube have a check if u would like :)

Thanks for taking ==

helo i wanted to know how i can get the word out in runescape wikia about

world 51 house party in yanille for PURES under 30def we been alive since 2006 august 20th.We have some vids out on youtube have a check if u would like :)

Thanks for taking ur time to read this.