EPT
Latin, Math, and some probability theory.
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Recognizable
Netherlands1552 Posts
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lcms
Netherlands27 Posts
There are 4*3*2*1 = 24 different arrangements of pairs of which one is the correct one, so 1/24th i'd say :-) | ||
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Recognizable
Netherlands1552 Posts
On January 17 2013 23:19 lcms wrote: Concerning the chances: There are 4*3*2*1 = 24 different arrangements of pairs of which one is the correct one, so 1/24th i'd say :-) Ah, thanks. That makes a lot more sense than what I did. | ||
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Smancer
United States379 Posts
If you really understand algebra, and you still find that you are making mistakes, try improving your handwriting. When you write neater, you force yourself to slow down. This just may help to avoid 1 or two mistakes that you make. Its out of the box advice, but I have seen it work before. | ||
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duckett
United States589 Posts
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Recognizable
Netherlands1552 Posts
On January 17 2013 23:34 Smancer wrote: I have a couple of degrees in Mathematics and I have taught High school level (everything from Algebra to Calculus). If you really understand algebra, and you still find that you are making mistakes, try improving your handwriting. When you write neater, you force yourself to slow down. This just may help to avoid 1 or two mistakes that you make. Its out of the box advice, but I have seen it work before. This is something I noticed as well but I've only been trying to improve it for about a month. My handwriting is absolutely terrible. I remember messing up a test a year ago because my P looks exactly like an 8 for some reason. Teacher told me he had to fail me. It's a slow process tho, and sometimes laziness kicks in and I go back to my terrible handwriting. I don't think I really understand Algebra. I can apply all the rules. But truly understanding? I don't really. It's never explained sadly, they only show that. Yes, it really does work with numbers. For some things I have an intuitive understanding, but for others I have not. It's such a pain to improve on this. I can just remember it and solve ten problems. Or I can try to get some kind of understanding for these rules but I won't get any practice. making computation mistakes is a good problem to have - its much harder to solve problems with building conceptual understanding. Yes, I'm happy this is the biggest problem as well. Still, my biggest fear is that someday I can't understand it conceptually anymore because it has gotten too hard. Hope that never happens. It's intruiging to see my Chemistry class where half the people just don't understand these things conceptually and fails hard, and the other half understands everything and gets amazing grades because the tests are made a bit easier because of the other half failing hard. I wonder why they don't understand it conceptually. I have no doubts that they can, they just aren't being taught right/thinking right. I was trying to explain something to somebody sitting next to me and she never thought of trying to visualize, for example, the titration itself whilst doing problems on titration. The other thing I have noticed is that they don't really want to learn. They have this mindset that they can't ever understand, which is sad. | ||
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duckett
United States589 Posts
On January 17 2013 23:44 Recognizable wrote: The other thing I have noticed is that they don't really want to learn. They have this mindset that they can't ever understand, which is sad. very insightful - it's often the case that the biggest barrier to learning concepts in math and science is the perception that it is too hard or I am too dumb etc. overcome that and you're well on your way to becoming a strong mathematical thinker | ||
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Recognizable
Netherlands1552 Posts
On January 17 2013 23:53 duckett wrote: very insightful - it's often the case that the biggest barrier to learning concepts in math and science is the perception that it is too hard or I am too dumb etc. overcome that and you're well on your way to becoming a strong mathematical thinker This is why Math/Physics is so great. I look at the stuff I had to do two/three years ago and I'm stunned I had ever any trouble understanding any of that. The only thing I am struggling with is geometric proofs. Many people are. It's a different way of problem solving, but I'm definitely getting a lot better and it was only a year ago that I thought I can never understand this. | ||
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Smancer
United States379 Posts
On January 17 2013 23:44 Recognizable wrote: don't think I really understand Algebra. I can apply all the rules. But truly understanding? I don't really. It's never explained sadly, they only show that. Yes, it really does work with numbers. For some things I have an intuitive understanding, but for others I have not. Haha, there are parts of algebra that I don't really understand either. Serge Lang is my arch nemesis. He treats tensor products about as elegantly as an elephant on a balance beam. Anyway, I am curious to see if I can help. If there are things you really never had explained and don't understand, I can try. If not good luck in school! | ||
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Recognizable
Netherlands1552 Posts
On January 18 2013 00:06 Smancer wrote: Haha, there are parts of algebra that I don't really understand either. Serge Lang is my arch nemesis. He treats tensor products about as elegantly as an elephant on a balance beam. Anyway, I am curious to see if I can help. If there are things you really never had explained and don't understand, I can try. If not good luck in school! Primitives. I mean, I understand all that jazz with rectangles for primitives but. Why is:F(x)=x ln(x) - x + c -->f(x)=ln(x) --> f'(x)=1/x. The great thing is. They show you that x ln(x) - x + c really is the primitive of ln(x) because if you differentiate x ln(x) - x you get ln(x) again! Haha. I also don't really have an understanding of the primitive function. You get a function, but what does this function represent? When you put in an X, what do you get, what is F(x)? I guess it's the surface between 0 and X for the original function, but again, why does this function represent the surface? | ||
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Nehsb
United States380 Posts
On January 18 2013 00:06 Smancer wrote: Haha, there are parts of algebra that I don't really understand either. Serge Lang is my arch nemesis. He treats tensor products about as elegantly as an elephant on a balance beam. Anyway, I am curious to see if I can help. If there are things you really never had explained and don't understand, I can try. If not good luck in school! If you don't already know it, maybe you should try reading some algebraic geometry (in particular some scheme theory.) So much of algebra becomes so much clearer when worded in terms of algebraic geometry... Tensor products of rings become fibered products of schemes, tensor product of a ring and a module becomes pullback of a sheaf, and tensor product of modules is some notion of "twisting a sheaf by another sheaf." It's unfortunate that a lot of the algebra is clarified by scheme-theoretic algebraic geometry, but scheme-theoretic algebraic geometry requires too many prerequisites to be used to actually show the intuition behind the algebra... On January 18 2013 00:14 Recognizable wrote: Primitives. I mean, I understand all that jazz with rectangles for primitives but. Why is:F(x)=x ln(x) - x + c -->f(x)=ln(x) --> f'(x)=1/x. The great thing is. They show you that x ln(x) - x + c really is the primitive of ln(x) because if you differentiate x ln(x) - x you get ln(x) again! Haha. I also don't really have an understanding of the primitive function. You get a function, but what does this function represent? When you put in an X, what do you get, what is F(x)? I guess it's the surface between 0 and X for the original function, but again, why does this function represent the surface? Hmm, so I'll try to clarify some things: As to why this primitive represents the surface: I think the right way to do this is to start from the function that is the area of the surface, and not to start from the primitive. If you have a function f, then the Riemann integral of f is not defined as the primitive (which unfortunately some calculus textbooks do, and IMO it's a terrible way to do things if you want to really understand what's going on.) The Riemann integral of f is defined as a certain Riemann sum which represents the area under f. However, in practice, this explicit Riemann sum is often annoying to calculate. But we have the very convenient fact that if g is the integral from 0 to x of f, then the derivative of g is f. The reason for this is that if you slightly increase x, the area should change as fast as f(x). (This should be clearer with a good diagram, but I don't have one.) This means that you have the following computational technique to find this function g: Find some function h such that the derivative of h is f. Then, the derivative of g-h is 0, but it is a theorem in calculus (usually proven with Mean Value Theorem) that the only functions with derivative 0 are constant functions. So g must be a constant off of h, and you just need to find the constant. So, in my opinion, it's not so much "a primitive represents the area of the surface" but "the area of the surface should be a primitive, and all primitives differ from each other by a constant." I'm not exactly very good at explaining things, so I'm sorry if this doesn't make sense. | ||
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duckett
United States589 Posts
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King Geedorah
190 Posts
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Recognizable
Netherlands1552 Posts
As to why this primitive represents the surface: I think the right way to do this is to start from the function that is the area of the surface, and not to start from the primitive. If you have a function f, then the Riemann integral of f is not defined as the primitive (which unfortunately some calculus textbooks do, and IMO it's a terrible way to do things if you want to really understand what's going on.) The Riemann integral of f is defined as a certain Riemann sum which represents the area under f. However, in practice, this explicit Riemann sum is often annoying to calculate. But we have the very convenient fact that if g is the integral from 0 to x of f, then the derivative of g is f. The reason for this is that if you slightly increase x, the area should change as fast as f(x). (This should be clearer with a good diagram, but I don't have one.) This means that you have the following computational technique to find this function g: Find some function h such that the derivative of h is f. Then, the derivative of g-h is 0, but it is a theorem in calculus (usually proven with Mean Value Theorem) that the only functions with derivative 0 are constant functions. So g must be a constant off of h, and you just need to find the constant. So, in my opinion, it's not so much "a primitive represents the area of the surface" but "the area of the surface should be a primitive, and all primitives differ from each other by a constant." Took me half an hour or so. But I believe I understand ^.^ The bolded part is indeed how I used to think about integrals. Basically, x times f(x) is the the F(x) and the area changes as ''fast'' as f(x) so therefore the derative of F(x) is f(x). Right? Name thief, how dare you. Sir, I believe it's the other way around. | ||
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hypercube
Hungary2735 Posts
Also try to figure out what kind of mistakes are you usually making, so you'll know where you need to pay extra attention. Most of high school math as well as a lot of university math for engineering and science comes down to a few key concepts and lots of algorithmic computation. Maybe it's necessary maybe it isn't, but it is what it is. The way to learn algorithms is through solving exercises and eliminating typical mistakes one by one. | ||
BigFan
TLADT24920 Posts
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sam!zdat
United States5559 Posts
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Recognizable
Netherlands1552 Posts
On January 18 2013 06:00 sam!zdat wrote: What are you reading in Latin? Seneca, not very exciting. But it's pretty easy to translate compared to other things I've had or maybe I'm just getting better at translating. After a while you really get used to his style and his metaphors are nice. De Ira, Ad Helviam Matrem, De Tranquilitate Animi, De Constantia Sapientis, Epistulae Morales and De Clementia I believe. We translate a selection of these works and we have to understand the content/philosophy. Which is where I usually get all my points from :p, However this test was only about translation. The more I learn Latin the more I appreciate it actually. I don't really understand why I've never exerted myself for this class, lotsa shitty teachers, but right now I have a very good one who is also invested in helping me improve which is great. For that matter I haven't exerted myself a lot the past 5 years. I don't say I regret it, a lot, because my grades are still good. However if I had been as motivated as I am right now I could've been way ahead at everything, but it's really hard to challenge yourself when you can get a passing grade without effort. I see it with a lot of my classmates. The Dutch education system really fails in this aspect. There are literally zero incentives whatsoever to get a grade beyond one that is passible. How do you think they 'fixed' this issue? Yes, by raising the minimum requirements. The US does this a lot better, but it fails were the Dutch system excels at and that is giving everyone equal chances and creative thought is encouraged, which asia seems to be struggling with. For my exam program for History they state explicitly that they don't want to encourage regurgitation. | ||
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sam!zdat
United States5559 Posts
On January 18 2013 06:21 Recognizable wrote: The more I learn Latin the more I appreciate it actually. You will find this to be a robust trend. Keep studying Latin! Beautiful language. | ||
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Recognizable
Netherlands1552 Posts
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