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Wow. Thanks a lot for that insight. You know I still think sometimes that I am a retard and I can't possibly understand stuff. I know it's rather self-defeatin

Wow. Thanks a lot for that insight. You know I still think sometimes that I am a retard and I can't possibly understand stuff. I know it's rather self-defeating, but when I used to see a few teachers handle stuff with amazing dexterity I used to just gape at the whiteboard. Aware that something beautiful was going on, without being able to put my finger on it. Whenever I used to talk about this to people they would say that I don't work hard enough. Or, that I am a fool to expect anything more than just get the marks which are expected of me. Or, that this is not the time or place to do such stuff. When I told my mother right now about this amazing place called HN and this thread she told me to shut up and stop being so impractical (don't judge her on this she's quite amazing). Long story short, in the end I assumed that the problem was with me and that I was deficient in some way or the other. Now, I'm learning that is not so and hence this thread. So, your comment was an eye opener. I am so sorry that I put proofs in there. To me that was the biggest symptom of not seeing stuff, you know. So, I thought that people may understand from that. My bad.

Tichy 16y ago
Nothing against proofs, of course proofs are the basis of maths. What I meant is that typically I couldn't just read a mathematical text, follow every step logically, and then have an understanding of that mathematical subject. Rather, the understanding would sink in after a while of trying to follow the proofs and solving exercises. Could you give examples of the kind of problems that stump you, and the things you try to do? I am sure a lot of people, including me, would be glad to help. To me it just sounds as if you are trying to take too many steps at once, if you can not follow some of the proofs.
- todayiamme 16y ago
  For example a few days ago this teacher wrote this problem in class; (x^2 - 2x + 2^|a|)/(x^2 - a^2) > 0 I had no clue how to do it. When he wrote the solution on the board only then could I follow it. I didn't understand why I couldn't see the solution on my own. It looks so simple later on and I just can't understand why I couldn't get it on my own in the first place. Can you help me out with that? Thank you.
  - Tichy 16y ago
    Here's hoping I won't embarrass myself too much... Since presumably your teacher already wrote the solution onto the board, I'll just talk about how I would arrive at a solution. So first of all, it is a very common problem. We must have solved countless polynomial problems at school. So usually, seeing this kind of problem should already ring several bells. So what I associate with polynomials (from school day) is - zero points - factorization - quadratic equation - "curve discussion" (don't know the english name for it, determining minima and maxima by calculating differentials) Did you have the same associations? If not, then that is simply a matter of experience that would come with practice. Without that experience, naturally arriving at a solution would be hard. It should be immediately obvious that we can factorize this polynomial without problems: simply because we know that we can factorize every second degree polynomial, thanks to the quadratic equation. This knowledge is basically "rote learning", except that you'll probably pick it up without decidedly doing rote learning. The polynomial presented here is already factorized into two second degree polynomials, so no problem there - do you agree so far? (Bonus points for seeing immediately that (x^2-a^2) = (x+a)(x-a) - this is a really common and basic formula that you should probably just "know". However, in principle you can also derive it from the quadratic equation, knowing this formula is just a shortcut). So, are these zero points of use after all? Yes!!! We can simply check some values of the function between the zero points and determine if they are greater or less than zero. Since the function can only "switch sides" at the zero points, we then know it has to have the same sign (negative or positive) everywhere between those zero points (and same for outside of all the zero points). Do you agree so far? If not, where is that you were stuck? With that idea, let's just determine the zero points. From the (x+a)(x-a) part we already know that a and -a are zero points. Using the quadratic formula on x^2 - 2x + a^|a| I get -1-sqrt(1-2^|a|) and -1+sqrt(1-2^|a|) - just ask if that is too fast, also be aware that I am very prone to make errors in such calculations. From there on, I admit it is not much fun. We have to somehow order the zero points, so that we now when the sign could change. So we have to make case distinctions. In fact, I haven't yet thought further than this, and now I am looking for ways to make it easier (I hate case distinctions). Laziness is the main motivator for maths, therefore I am not that much in favor of the rote learning approach... So one step back: we have a products of two polynomials, poly1*poly2 with poly2 = x^2-a^2. That product is > 0 if poly1 > 0 and poly2 > 0 or poly1 < 0 and poly2 < 0 (if this is not obvious, please ask). It is clear that poly2 < 0 for |x| < |a| and > 0 for |x| > |a|. So let's look at poly1 = (x+1-sqrt(1-2^|a|))(x+1+sqrt(1-2^|a|)) (if my calculation was correct). Since 1-sqrt(1-2^|a|) < 1+sqrt(1-2^|a|) we get poly1 < 0 for -(1+sqrt(1-2^|a|)) < x < -(1-sqrt(1-2^|a|)) and > 0 for x < -(1+sqrt(1-2^|a|)) or x > -(1-sqrt(1-2^|a|)) So alltogether, the (intermediate) solution is (-(1+sqrt(1-2^|a|)) < x < -(1-sqrt(1-2^|a|)) and |x| < |a|) or (x < -(1+sqrt(1-2^|a|)) or x > -(1-sqrt(1-2^|a|))) and |x| > |a| I have a feeling this should be simplified further, but I'll post it like this because it becomes a little bit intractable in the small comment box of HN. Maybe I'll find time to finish later. Anyway, I probably made mistakes, and overlooked some much easier approach. The main point is that most of the steps are very common approaches, and I think/hope I demonstrated dividing the problem into smaller sub-problems. One advice I would give for maths is to not be intimidated - you need a little bit of faith... If you fear the equations, you'll be too paralyzed to find a solution. Also, I used to cheat a little bit on the problems in school: usually chances are high that the problem at hand has something to do with a recent thing the teacher has taught us. So it pays to look for ways to use the knowledge from the latest lessons. Also, this problem seems comparatively ugly. Sorry that this got rather complicated. Let me know how I can help better. I'll try to create a shorter summary of the steps involved. (Meanwhile, please post the shorter solution if there is one and you find it).

Tichy 16y ago

Nothing against proofs, of course proofs are the basis of maths. What I meant is that typically I couldn't just read a mathematical text, follow every step logically, and then have an understanding of that mathematical subject. Rather, the understanding would sink in after a while of trying to follow the proofs and solving exercises. Could you give examples of the kind of problems that stump you, and the things you try to do? I am sure a lot of people, including me, would be glad to help. To me it just sounds as if you are trying to take too many steps at once, if you can not follow some of the proofs.

todayiamme 16y ago
For example a few days ago this teacher wrote this problem in class; (x^2 - 2x + 2^|a|)/(x^2 - a^2) > 0 I had no clue how to do it. When he wrote the solution on the board only then could I follow it. I didn't understand why I couldn't see the solution on my own. It looks so simple later on and I just can't understand why I couldn't get it on my own in the first place. Can you help me out with that? Thank you.
- Tichy 16y ago
  Here's hoping I won't embarrass myself too much... Since presumably your teacher already wrote the solution onto the board, I'll just talk about how I would arrive at a solution. So first of all, it is a very common problem. We must have solved countless polynomial problems at school. So usually, seeing this kind of problem should already ring several bells. So what I associate with polynomials (from school day) is - zero points - factorization - quadratic equation - "curve discussion" (don't know the english name for it, determining minima and maxima by calculating differentials) Did you have the same associations? If not, then that is simply a matter of experience that would come with practice. Without that experience, naturally arriving at a solution would be hard. It should be immediately obvious that we can factorize this polynomial without problems: simply because we know that we can factorize every second degree polynomial, thanks to the quadratic equation. This knowledge is basically "rote learning", except that you'll probably pick it up without decidedly doing rote learning. The polynomial presented here is already factorized into two second degree polynomials, so no problem there - do you agree so far? (Bonus points for seeing immediately that (x^2-a^2) = (x+a)(x-a) - this is a really common and basic formula that you should probably just "know". However, in principle you can also derive it from the quadratic equation, knowing this formula is just a shortcut). So, are these zero points of use after all? Yes!!! We can simply check some values of the function between the zero points and determine if they are greater or less than zero. Since the function can only "switch sides" at the zero points, we then know it has to have the same sign (negative or positive) everywhere between those zero points (and same for outside of all the zero points). Do you agree so far? If not, where is that you were stuck? With that idea, let's just determine the zero points. From the (x+a)(x-a) part we already know that a and -a are zero points. Using the quadratic formula on x^2 - 2x + a^|a| I get -1-sqrt(1-2^|a|) and -1+sqrt(1-2^|a|) - just ask if that is too fast, also be aware that I am very prone to make errors in such calculations. From there on, I admit it is not much fun. We have to somehow order the zero points, so that we now when the sign could change. So we have to make case distinctions. In fact, I haven't yet thought further than this, and now I am looking for ways to make it easier (I hate case distinctions). Laziness is the main motivator for maths, therefore I am not that much in favor of the rote learning approach... So one step back: we have a products of two polynomials, poly1*poly2 with poly2 = x^2-a^2. That product is > 0 if poly1 > 0 and poly2 > 0 or poly1 < 0 and poly2 < 0 (if this is not obvious, please ask). It is clear that poly2 < 0 for |x| < |a| and > 0 for |x| > |a|. So let's look at poly1 = (x+1-sqrt(1-2^|a|))(x+1+sqrt(1-2^|a|)) (if my calculation was correct). Since 1-sqrt(1-2^|a|) < 1+sqrt(1-2^|a|) we get poly1 < 0 for -(1+sqrt(1-2^|a|)) < x < -(1-sqrt(1-2^|a|)) and > 0 for x < -(1+sqrt(1-2^|a|)) or x > -(1-sqrt(1-2^|a|)) So alltogether, the (intermediate) solution is (-(1+sqrt(1-2^|a|)) < x < -(1-sqrt(1-2^|a|)) and |x| < |a|) or (x < -(1+sqrt(1-2^|a|)) or x > -(1-sqrt(1-2^|a|))) and |x| > |a| I have a feeling this should be simplified further, but I'll post it like this because it becomes a little bit intractable in the small comment box of HN. Maybe I'll find time to finish later. Anyway, I probably made mistakes, and overlooked some much easier approach. The main point is that most of the steps are very common approaches, and I think/hope I demonstrated dividing the problem into smaller sub-problems. One advice I would give for maths is to not be intimidated - you need a little bit of faith... If you fear the equations, you'll be too paralyzed to find a solution. Also, I used to cheat a little bit on the problems in school: usually chances are high that the problem at hand has something to do with a recent thing the teacher has taught us. So it pays to look for ways to use the knowledge from the latest lessons. Also, this problem seems comparatively ugly. Sorry that this got rather complicated. Let me know how I can help better. I'll try to create a shorter summary of the steps involved. (Meanwhile, please post the shorter solution if there is one and you find it).