[QUOTE=Cosa8888;50370922]Not really, what about 1/n? It converges to 0, but all it's terms are different, no matter how high n is.
Is this ok?
Define N such that for every e>0, n>N -> |Un-l|<e
Then |Un-l| is always less that inf(e>0)=0, because it is a lower bound. Then, |Un-l|<=0. But the term it's always positive or equal to zero, so |Un-l|=0, and that clearly implies that Un=l.
With this, I'm saying that there exists an N such that for all n>N, Un=l, and that implies that there exists a finite number of different terms.
But why can't I apply this to the normal definition of limit? thereust be something wrong with this proof, in that case.[/QUOTE]
Yeah what I wrote was nonsense, I somehow read that your sequence attains its limit point. When you say definition of 'limit', do you mean the definition of a convergent sequence? In that case, what you have given isn't the definition of a convergent sequence, note that the quantifiers 'for all e > 0' and 'there exists N' have been swapped around.
Also, didn't mean to rate your original post as funny.
Anyone ever worked with secant or bisection method on very high dimensional spaces? like 256 dimensions or more
so I forgot a negative in a question for differential equations and lost 1/4 of the tests marks (7), then made a process error in another question worth probably somewhere around 2-3 marks out of 8 at most, and got 0 for the whole question. I lost 15 marks for 2 mistakes and it's completely messed me up going into the exam. I mean yeah technically what i wrote was wrong but holy crap thats a dickish way to mark
That really is a shitty, lazy way to grade.
My favorite professor used to write -epsilon next to questions which had a mistake but it was too small to be worth taking points off.
[QUOTE=JohnnyMo1;50383128]That really is a shitty, lazy way to grade.
My favorite professor used to write -epsilon next to questions which had a mistake but it was too small to be worth taking points off.[/QUOTE]
Yeah the worst part is I've seen him do some error carried forward marking before for most people including myself, but I think my paper was at the bottom of the bunch so he just started getting lazy. I mean there aren't even any corrections which is why I'm pretty sure the rest of my questions were right, it's just a fat 0 no crosses or anything
[editline]24th May 2016[/editline]
the whole maths department is like this to some extent, heavily demotivates me from doing maths when I know like 90% of the work but get 40% in the test because I made a few addition or algebra errors
Can you talk to your professor about it? Maybe he could get you some points back.
yeah I'm gonna try tomorrow, the problem is when I got my test back they had already recorded the marks in the system for the exams office so I don't know if he can change them. Even worse of a clusterfuck with numerical methods 2, we switched lecturers between our first and second test and the first lecturer was so disorganized with his marking and ish that we didn't get our first test back or any feedback and it simply didn't count towards our mark. No discussion just an email informing us
When I'm marking I usually take off a small amount of marks for a small mistake. The problem is when the small mistake utterly destroys the question, and the student ends up skipping a lot of the "moves" we expected them to make. In those cases the student will get very few marks for their work, which is disheartening.
[QUOTE=Wunce;50385485]When I'm marking I usually take off a small amount of marks for a small mistake. The problem is when the small mistake utterly destroys the question, and the student ends up skipping a lot of the "moves" we expected them to make. In those cases the student will get very few marks for their work, which is disheartening.[/QUOTE]
Well that at least makes sense, but I've seen cases as a tutor of kids making a sign error and having the answer be correct up to a sign and losing half or all credit for the problem.
got a b+ in calc2! didn't pass data structs tho ):
You don't specify what the end goal of what you're doing is. Do you have to do something special to do what?
I'll have to think about that for a bit. A priori, no. The generators of the Lorentz group don't commute with each other in general, but they may in this case.
If you guys want to play with complex functions :)
[url]http://www.science-to-touch.com/DGD/CindyJS/complexFunctions/04_Explorer.html[/url]
This'll be noob talk (compared to the stuff you guys are talking about) but I'm half way through exam period and C1 and C2 (edexcel A level maths) were the easiest papers I've ever done in the last 9 years worth of papers.
Like I legitimately think they're trying to make mathematics (at least in the UK) more accessible to the general public but that means concepts are massively under-developed - might make the jump to uni a bit much. What about you guys?
Can someone give an example of this property of determinants
[img]http://i.imgur.com/ax2KVCb.png[/img]
[QUOTE=proboardslol;50498097]Can someone give an example of this property of determinants
[img]http://i.imgur.com/ax2KVCb.png[/img][/QUOTE]
An example?
e.g. suppose you want to calculate
[;\left| \begin{array}{ccc} 3 & 1 & 2 \\2 & 0 & 1 \\1 & 2 & 4 \end{array} \right| ;]
Then you can linearly break up the middle column to show
[;=\left| \begin{array}{ccc} 3 & 1 & 2 \\2 & 0 & 1 \\1 & 0 & 4 \end{array} \right| + 2 \left| \begin{array}{ccc} 3 & 0 & 2 \\2 & 0 & 1 \\1 & 1 & 4 \end{array} \right|;]
This is the basis of doing column ops to simplify determinant calculation.
Guys I have the Further Pure IGCSE Maths Exam today and I'm stressin out please help me I don't get this question on vectors
[IMG]http://i.imgur.com/2bjRMYG.png[/IMG]
The answer key says that OA should be 2(a+e) but I don't get it
[QUOTE=MyBumBum;50529330]Guys I have the Further Pure IGCSE Maths Exam today and I'm stressin out please help me I don't get this question on vectors
The answer key says that OA should be 2(a+e) but I don't get it[/QUOTE]
Start by drawing in the line OC, and then continue to fill up the hexagon with equilateral triangles of length OA. That should become a lot more obvious...
[QUOTE=Matthew0505;50532138]Assuming you meant AB is 2(a+e), let a = |OA|[cos pi/3, sin pi/3] and e = |OA|[cos pi/3, -sin pi/3] and given that cos pi/3 = 1/2, you should be able to figure it from there.[/QUOTE]
My exam is over but I'm still interested in finding the answer. Franky I'm quite terrible at vectors, so I'd appreciate any guidance. Here's the answer key
[t]http://i.imgur.com/BO4I70M.png[/t]
[editline]17th June 2016[/editline]
[QUOTE=Joey90;50531551]Start by drawing in the line OC, and then continue to fill up the hexagon with equilateral triangles of length OA. That should become a lot more obvious...[/QUOTE]
OOOOOHHHHHHHHHHHH
Thanks a lot!
EDIT:
Teacher just sent us this years' paper, I would like you guys' help in solving some of the questions I didn't know how to solve on the exam. Thanks a lot guy!
[t]http://i.imgur.com/VmRKr54.png[/t]
This question from (b) onwards and
[t]http://i.imgur.com/RDdpoJb.png[/t]
Part (c) of this question, I knew how to solve the logarithms and draw the correct curves but I've completely forgotten how solving an equation by drawing lines works. I just solved the equation normally and put the answer but I know i'll lose marks for it.
Much appreciated guys!
Just finished my last lower division math courses, heading into Intro to Number Theory and an Advanced Problem Solving course in the fall
Not sure if I posted it here a few years ago - but there is a problem that I still haven't solved and it runs through my head every now and again.
It went something like,
There is a series of circles; the first circle touches the x-axis, y-axis, and the function y = e^(-x), the rest of them subsequently touch the previous circle as well as the x-axis, and y = e^(-x).
What is the sum of the area of those circles?
Example, but the circles continue on:
[t]http://i.imgur.com/66QsCtf.png[/t]
The area under the curve is 1, so the sum of the areas of the circles should be less than that. The whole problem gives me a headache.
Can somebody explain to me the difference between adiabatic quantum computing and the other kind?
[QUOTE=Bradyns;50535754]Not sure if I posted it here a few years ago - but there is a problem that I still haven't solved and it runs through my head every now and again.
It went something like,
There is a series of circles; the first circle touches the x-axis, y-axis, and the function y = e^(-x), the rest of them subsequently touch the previous circle as well as the x-axis, and y = e^(-x).
What is the sum of the area of those circles?
Example, but the circles continue on:
[t]http://i.imgur.com/66QsCtf.png[/t]
The area under the curve is 1, so the sum of the areas of the circles should be less than that. The whole problem gives me a headache.[/QUOTE]
Are you sure there's a neat answer? Unless there's a very clever trick, I can't see it dropping out...
(Obviously it's [i]about[/i] pi/4)
[QUOTE=Bradyns;50535754]Not sure if I posted it here a few years ago - but there is a problem that I still haven't solved and it runs through my head every now and again.
It went something like,
There is a series of circles; the first circle touches the x-axis, y-axis, and the function y = e^(-x), the rest of them subsequently touch the previous circle as well as the x-axis, and y = e^(-x).
What is the sum of the area of those circles?
Example, but the circles continue on:
[t]http://i.imgur.com/66QsCtf.png[/t]
The area under the curve is 1, so the sum of the areas of the circles should be less than that. The whole problem gives me a headache.[/QUOTE]
So it boils down to finding the radii of each inscribed circle? Perhaps use Lagrange multipliers to find the maximum radius given those constraints? Finding the radius of the first circle already seems quite hard...
[QUOTE=Joey90;50537614]Are you sure there's a neat answer? Unless there's a very clever trick, I can't see it dropping out...
(Obviously it's [i]about[/i] pi/4)[/QUOTE]
I threw out some weird fraction multiplied by pi years ago, but I can't recall it.
[QUOTE=Number-41;50539456]So it boils down to finding the radii of each inscribed circle? Perhaps use Lagrange multipliers to find the maximum radius given those constraints? Finding the radius of the first circle already seems quite hard...[/QUOTE]
I was hoping there would be an interesting ratio between the radii as they got smaller.
You'd think that it would be tied somehow to the function e^(-x).
[QUOTE=Bradyns;50535754]Not sure if I posted it here a few years ago - but there is a problem that I still haven't solved and it runs through my head every now and again.
It went something like,
There is a series of circles; the first circle touches the x-axis, y-axis, and the function y = e^(-x), the rest of them subsequently touch the previous circle as well as the x-axis, and y = e^(-x).
What is the sum of the area of those circles?
Example, but the circles continue on:
[t]http://i.imgur.com/66QsCtf.png[/t]
The area under the curve is 1, so the sum of the areas of the circles should be less than that. The whole problem gives me a headache.[/QUOTE]
Hmmm unless I am completely wrong, I started off with an assumption:
*The circle and the function have the same derivative in the point that they touch.
Once you find where they meet, then ( radius of the first circle) r = 0.5*( e(-2*x) - x)
Again I might be completely wrong, but it seems like a good place to start.
I was reading Algebra, Chapter 0 last night before bed and I had an algebra dream. :|
Also:
[img]http://i.imgur.com/gh7mrpA.png[/img]
sorry if this isn't the right place to post this, but i didn't want to make my own thread for a single question. i'm having some trouble calculating something for a MOO i'm working on.
12:46 PM - ▽ Dead: delay = toint(player.stats_current[2][5] / 5.0);
12:46 PM - ▽ Dead: this is the current expression to calculate time it takes to move from one room to the next
12:46 PM - ▽ Dead: i'm trying to make it so that higher agility = faster movement
12:46 PM - ▽ Dead: but, it does this
12:47 PM - ▽ Dead: if agility were say, 5
12:47 PM - ▽ Dead: 5.0 / 5.0
12:47 PM - ▽ Dead: = 1 second
12:47 PM - ▽ Dead: if agility were 10
12:47 PM - ▽ Dead: 10.0 / 5.0
12:47 PM - ▽ Dead: = 2 seconds
12:47 PM - ▽ Dead: = higher agility but slower movement
12:47 PM - ▽ Dead: can you think of a solution to this?
in short;
delay = toint(agility / 5.0);
the above line is what i'm using to calculate the player's agility, which is used in my game to determine how long it takes to move from room to room. the problem is that with the current expression, it takes longer if you have higher agility (5.0 / 5.0 = 1 second, 10.0 / 5.0 = 2 seconds, and so on), which is completely backwards for what i'm trying to do. can anyone suggest a different expression to achieve the desired result?
[editline]11th August 2016[/editline]
nevermind, i got it figured out. reversing the variables achieved the desired effect.
[QUOTE=Matthew0505;50399492]If a velocity has components in more than one axis, can I use a rotated coordinate system such that V = [Vx, Vy, Vz] becomes [|V|, 0, 0], do a one-dimensional Lorentz boost, then reverse the rotation, or do I have to do something special to the time coordinate?
i.e.
[img]https://i.imgur.com/IAXFxyt.png[/img]
1. Convert event to green coordinate system without modifying time
2. Perform x-boost on green event
3. Convert x-boosted green event to a red event to get the event as viewed by an observer with velocity v relative to the original observer of the event[/QUOTE]
This is going to be a tough one..And for good reason. Disclaimer: I don't even have a high school diploma. I have a GED and have passed a college course or two, I've just done a lot of independent study over my life and self-stimulation.
Alright, so from my PoV what your trying to say in lehmans terms: Is that while these 3 axis all have a distinct velocity corresponding equivalently so with eachother, what you're trying to do is eliminate 2 of the axis and correspondingly extend/boost the one dimension you have left, to a certain point in space-time.
I'm not sure if geodesic is the right term but what your trying to convey is a one dimensional segment through space, correct? Well, keep in line, space and time are connected-There's no denying that. So conversely, there will be an equal reaction to your elimination of the 2 dimensions with their respective velocities before hand. Now, when you extend that one dimension, this stretches that reaction, along with, I assume, other things. This has to do with Question #1:converting the spatial event into a polar opposite system which time will not apply to it. There is a fallacy and a truth in that, with respect to duality.
And that's that you basically have to circumvent the loop here. The only possible way that I personally could think of to do that is to say that when your doing this, you are in fact re-inventing the wheel. And that my friend is not a bad thing. There's an assumption that relativity is absolute because it is absolute, it's just no one can perceive it as Einstein had been able to perceive it. That's why we can't disprove it nor prove it any more so. He understood all these concepts like we know how to connect two and two, and I suppose he withheld those techniques of understanding for good reason! That's my personal opinion but I really hope I have helped because I feel like this has gone on to a rambling conjecture. Uh, let me know if you'd like any more input!
[QUOTE=x2yzh9;50876398]That's my personal opinion but I really hope I have helped because I feel like this has gone on to a rambling conjecture.[/QUOTE]
Sorry, I think I have to be a little harsh here for a second: that was all ramble.
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