6 ms·

Formerly good at math, but I've lost my skills. Any advice to grow them back?

75 points by gulbrandr 16y ago

chegra 16y ago
Oddly enough that he should mention this, but last week or so I was thinking, "my how I have lost some of my mathematical prowess". There was a room of 4 males and 10 females, now there was 5 other guys who were to be paired with someone in that room. What is the probability that the pairs would be as follows (MM,MM,MM,MF,MF) where M=Males and F=Females. This was a real life problem I create for myself to determine the bias by the selector. I struggled to work it out by pen and paper. This was a problem 8 years ago it wouldn't take a minute to do, most of the work would be in punching numbers in the calculator. Eventually, I said when I go home I will program it up. Between the time it took for a friend to restart her computer, I had whipped up a simulation. This is something I wouldn't have been able to do 8 years ago. So in a sense, I call it a skills exchange. I lose some skills and gain some skills. I'm sure with practice I would be back up to par, the same goes for this guy.
- light3 16y ago
  Is the answer: 4x3x2x10x9 / (14x13x12x11x10) * 5! Is there some easier way to solve this?
  - piaskal 16y ago
    Is there any difference between x and * operations?
    - light3 16y ago
      No, if I use * with no space it doesn't show.
  - chalst 16y ago
    The complexity of the formula seems to match the complexity of the problem: three lists of decreasing integers matching the three groups, multiplied/divided as appropriate. The hard bit is having the intuition to see that this is right, which is just what Bill Dubuque's answer goes on about in the linked article.
  - malbertife 16y ago
    You need to compute the ratio between the number of successful experiments and the number of total experiments. If what you want to know is the probability of having the 5 guys paired with 3 men and 2 women (which is what I think the grandfather meant), the answer is C(4,3) x C(10,2) / C(14,5) = about 9% (number of ways to pick 3 men out of 4 x number of ways to pick 2 women out of 10 / number of ways to pick 5 people out of 14, /regardless of the order/) If you want to know the probability of having the pairs made exactly in the order the grandfather mentioned, it's P(4,3) x P(10,2) / P(14,5) = about 0.9% (as above, but considering the order) where C(M,N) and P(M,N) are the number of combinations and permutations of N items out of M, respectively. (edit to change multiplication symbol, asterisk doesn't show)
  - deleted 16y ago
    [deleted]
  - ohmygodel 16y ago
    You need to divide by 3! and 2!. One way to look at it is that every selection of five from the 4M and 10F is equally likely. There are \binom{14}{5} (i.e. 14 choose 5) such selections. Of these, \binom{4}{3} * \binom{10}{2} have exactly 3M and 2F. Therefore, the probability is \binom{4}{3} * \binom{10}{2} / \binom{14}{5}, or (4 * 3 * 2 * 10 * 9) 5! / (3! 2! (14 * 13 * 12 * 11 * 10)) = 10 * 9/(13 * 11 * 7) = 0.08991...
    - dcosson 16y ago
      This is how I'd solve it too. Plus, with a username like that and a LaTex formatted answer, I'm already inclined to trust your math skills :)
    - light3 16y ago
      Ahh indeed, I thought there were 5!=120 ways to order 3M and 2F, but apparently there are only 5!/(2!3!) = 10 ways: {MMMFF,MMFFM,MFFMM,FFMMM,FMMMF,MFMMF,MMFMF,FMMFM,MFMFM,FMFMM}. I made the mistake of thinking you can distinguish between individual males/females.
wheels 16y ago
I'd felt exactly the same a couple years back, but the root cause was rather amusing. When I started working on stuff for Directed Edge I dove into a lot of academic research. I found myself struggling with a lot of it. I really believed that my math abilities had slipped significantly since my glory days and continued with this perception for several months... ...until I picked up some stuff that I'd thought was hard in college. I was dumbfounded at how easy it was. What had changed was not my math abilities, but my perception of what constituted hard math.
- tjarratt 16y ago
  My experience with Haskell reflects this pretty closely. A year ago I was using it primarily for scripting and daily exercise but I couldn't get past monads as a concept. A few days ago I brought out Learn You A Haskell and took a look at the monad section and suddenly it all made sense. Sometimes your unconscious needs time to digest concepts and internalize them before you can master them and move on to hairier problems.
- nostrademons 16y ago
  I've noticed the same thing. I'll routinely struggle to understand an academic paper with a notation-heavy theoretical bent, but when I found I had to dig out high school math for a couple 2D-graphics problems recently, it all came back easily. When you're in high school, you don't realize how much of a gulf there is between Ph.D/MS-level (or even professional-level) math and what you learn in school. The public school system teaches you the barest foundations; there's a gigantic world out there beyond it.
jhlyerte 16y ago
Click on our website: == ( http://cha.la/bkef http://cha.la/bkef ) == Christan Audigier bikini $23 Ed Hardy Bikini $23 Smful short_t-shirt_woman $15 ed hardy short_tank_woman $16 Sandal $32 christian louboutin $80 Sunglass $15 COACH_Necklace $27 handbag $33 AF tank woman $17 puma slipper woman $30 == ( http://cha.la/bkef http://cha.la/bkef ) == ~ ¤ ╭⌒╮ ╭⌒╮ ╭⌒╭⌒╮╭⌒╮～╭⌒╮ ,））））,''）~~ ,''~） ╱◥█◣ ╱◥█◣ ｜田｜田｜｜田｜田｜ ╬╬╬╬╬╬╬╬╬╬╬╬╬╬
cletus 16y ago
I have a similar problem actually. In high school and even the first year of university I was pretty much a natural at math. I'd been doing algebra and trigonometry since I was 10, calculus since I was 13 and started on differential equations at 15. Not that this made me any kind of child prodigy mind you but it was a fairly isolating experience because my first 18 years took places in isolated towns with <15,000 people so I just didn't come across other people like me or with similar interests. But I digress... The second year of university maths bored me senseless and it sucked all the fun out of maths for me. Three hour proofs, numerical methods (in particular) and so on. So I stopped doing it and just did CS instead (originally I'd been intending some kind of double major equivalent). Having not used calculus (for example) for years I can still remember how to derive/integrate, etc. So I retain a decent part of what I knew (or so it seems). But the one area I suffer in, largely because I never got that far, is in deciphering academic papers. This applies to CS too, which is where it's a pain. I like dealing with actual code rather than abstract ideas put in text (and the vaguest of algorithms that you can't just take an implement, or at least I can't). So my situation isn't identical to yours but I really would like to know how I could improve my maths and CS in this area.
- tel 16y ago
  There's an incredibly common roadblock in math education when you make the jump from mechanical maths to creative maths. On one side, math makes sense as a set of powerful rules which help to guide you to understand and analyze the world. On the other side, math is more like a penpal friend while you're living in a foreign country. It offers advice, but largely you're on your own in a confusing land. Academic papers are written from the far side of that roadblock. They're often best considered anecdotal pointers to how to survive abroad rather than comprehensive guides like textbooks (or review papers) offer. Reading them is thus both an exercise in deciphering something distant from your comfort zone and learning to apply its vague knowledge to your own situation. In order to understand academic papers then, you've kind of got to become a traveler yourself. At that point you appreciate the vague note passing because even though it's difficult, there's a real opportunity there to see something few ever have before. --- So practically, academic papers are never easy to decipher unless you're basically part of the same communities as the authors (or they are really dedicated and skillful communicators who have an uncommonly deep understanding of the topics they're writing about). If you're doing work that cannot be solved by what exists and is common knowledge today, however, they're your best bet at finding a guide and no matter how difficult they are you want to read them through.
sudont 16y ago
Similarly burned-out, I'm having trouble caring about design: learning about how it interacts with other disciplines and practicing the skill on these issues, rather than just practicing design itself, is essential. Design, like math, is a problem-solving application and there's definitely a "feel" that can be lost. These solutions can definitely be abstracted out to other areas, but really--what part of math can't?
codelion 16y ago
The best way to study math is to fight it ... http://abstrusegoose.com/353 http://abstrusegoose.com/353
deskamess 16y ago
I wanted to brush up on my math with a goal of shoring up fundamentals and understanding probability and statistics (not just formulaic solving). I was not sure where to start, so I have decided to start by practicing the fundamentals. Addition and so on. Khan Academy has that covered. As of today they have 'practices' that go from basic addition all the way through simple differentiation. Most important, if you run into problems with an exercise set, they have video links to the topic you are practicing. Once you have watched the video and understand the concepts, you can come back to the practice set. (RUPPPL - read understand practice practice practice loop) You should be able to run through the set without much effort, but I think there is value in doing these simpler exercises. It establishes routine and steeps the mind for greater challenges. I will likely add some permutation & combination stuff to my 'course work' before I go on to my eventual goal of understanding prob & stats. Getting exercise for these higher level topics will be a challenge.
- tricky 16y ago
  This is EXACTLY what I'm doing and it seems to be working out great. Finding exercise for the higher level topics is a bit more challenging, but there is a lot out there. The thinkstats book is great if you're a programmer: http://www.greenteapress.com/thinkstats/html/index.html http://www.greenteapress.com/thinkstats/html/index.html I've also been studying a berkely stats course on itunes u and working the exams/notes from this class: https://netfiles.uiuc.edu/fenghong/www/STAT100/ https://netfiles.uiuc.edu/fenghong/www/STAT100/ I'm also reading a lot about financial analysis and risk assessment to fiddle around with real-world applications of the things I'm learning. it's a hell of a journey.
- tel 16y ago
  Fundamentals are nice, but rote learning of them is directly on the route to formulaic solving. You have to have something to force you to use math like an artist and not like a computer. Which is to say, I highly suggest doing this stuff as long as you get value out of it. It'll be a boon to your comprehension speed later. Eventually, however, go find a hard textbook and work through every single line of it. Really. Understand that it'll possibly take years and involve finding many lesser sources to fill in the gaps, but without a hard motivating problem you're going to get stuck in an affective loop on formulaic knowledge. Easy, satisfying, but somewhat flat.
cheez 16y ago
Teach your kids (or someone else's)
- abbasmehdi 16y ago
  Exactly! Teach anyone and explain the concepts to them. Go your local community college and volunteer at the math clinic, even better go to a local high school that's not been doing very well. Public school systems can use all the help they can get. I went from sucky math guy to an awesome math guy in just 1 year - just because I was tutoring.
michaelochurch 16y ago
I don't think you lose "good at math" if you're actually good at math. Being actually good at math, as I'd define it, means you understand problem solving and logical reasoning. It's not about computational techniques. Most math professors have to review linear algebra and the calculus sequence before they can teach their first undergraduate class. It's like this: if you study a language intensely for a year, you'll have a good understanding of it after that year. If you stop using it, the grammar will remain with you but your vocabulary will fade over time. If, however, you travel back to that country, it'll only take you a couple weeks to regain took you a year the first time around. Math is a language, and what you lose is vocabulary, not the language itself. I'm 27 and have been out of school for 5 years. I couldn't do integration by parts without consulting a textbook, and I'd probably do poorly on a Calc-3 final if I had to take it closed-book, right now, in the alloted time. But I also know where to look when I need to review calculus or linear algebra. There's a huge difference between "rusty" and lost skills. For "rusty", you just need to study again. It starts out hard and gets easier as you regain the skills. It's actually fun, at least for me, to revisit familiar friends like set theory. And I'm still learning new things (although not at as fast a rate as when I was in school) even though I've been out of school for half a decade.
- deleted 16y ago
  [deleted]
nextparadigms 16y ago
Try www.khanacademy.org
32ftpersecond 16y ago
see: G H Hardy's "A Mathematician's Apology"
CallMeV 16y ago
Like all things at which you have to work hard to gain ability, the original poster has but one course: the hardest one. To delve into those tiresome, brain-aching problems at which he once balked in school - the ones where he found that he had to put in a real effort. And he will have to accept lots of failures before he can start to claim lots of successes. With nobody to mark his efforts down for not showing sufficient workings, however, admitting to failure can come a lot more easily when you only have your own self as teacher and student.
ciopte7 16y ago
Check out this blog for a lot of free resources http://hbpms.blogspot.com/ http://hbpms.blogspot.com/
pitdesi 16y ago
I'd recommend going through Khan academy and starting from the basics: http://www.khanacademy.org/#browse http://www.khanacademy.org/#browse
parfe 16y ago
Grab a Calculus book and solve some random problems. It can get fairly entertaining. Solving how much rocket fuel you need to reach orbit or how fast a conical object will drain or the angle you need to fire a canon ball into a castle wall or something. I haven't found much use for Calculus though in regular life. On a side note, we've been arguing about 48÷2(9+3) at the office today. There is a correct answer, of course, but also an argument (that some think is reasonable) against that answer.
- e03179 16y ago
  48÷2(9+3) = 2
  - nlco 16y ago
    not quite, 48÷(2(9+3)) = 2, but 48÷2(9+3) = 288 to clarify: PEMDAS is the order of operations. so parenthesis first: 48÷2(9+3) = 48÷2(12) then multiplication and division (they have the same precedence): 48÷2(12) = 24*12 = 288 for another reference: http://www.google.com/search?q=48%C3%B72(9%2B3) http://www.google.com/search?q=48%C3%B72(9%2B3)
    - dcosson 16y ago
      multiplication and division have the same precedence, and division (like most operations) is left-associative. For example, exponentiation is right-associative, so 2^3^2 = 2^9 = 512, not 64.
      - wging 16y ago
        While you're technically correct, of the standard four half (+, *) are both left- and right-associative. / and - are the ones that are only left-associative.
    - sga 16y ago
      Best bet is to replace x÷y by x*(1/y) as division is not commutative while multiplication is commutative. Therefore 48÷2(9+3) => 48(1/(2(9+3)) => 2
projectileboy 16y ago
This topic comes up from time to time on HN. Here's what I did to dive back into mathematics: http://news.ycombinator.com/item?id=108924 http://news.ycombinator.com/item?id=108924

75 points by gulbrandr 16y ago

Oddly enough that he should mention this, but last week or so I was thinking, "my how I have lost some of my mathematical prowess". There was a room of 4 males and 10 females, now there was 5 other guys who were to be paired with someone in that room. What is the probability that the pairs would be as follows (MM,MM,MM,MF,MF) where M=Males and F=Females. This was a real life problem I create for myself to determine the bias by the selector. I struggled to work it out by pen and paper. This was a problem 8 years ago it wouldn't take a minute to do, most of the work would be in punching numbers in the calculator. Eventually, I said when I go home I will program it up. Between the time it took for a friend to restart her computer, I had whipped up a simulation. This is something I wouldn't have been able to do 8 years ago. So in a sense, I call it a skills exchange. I lose some skills and gain some skills. I'm sure with practice I would be back up to par, the same goes for this guy.

light3 16y ago
Is the answer: 4x3x2x10x9 / (14x13x12x11x10) * 5! Is there some easier way to solve this?
- piaskal 16y ago
  Is there any difference between x and * operations?
  - light3 16y ago
    No, if I use * with no space it doesn't show.
- chalst 16y ago
  The complexity of the formula seems to match the complexity of the problem: three lists of decreasing integers matching the three groups, multiplied/divided as appropriate. The hard bit is having the intuition to see that this is right, which is just what Bill Dubuque's answer goes on about in the linked article.
- malbertife 16y ago
  You need to compute the ratio between the number of successful experiments and the number of total experiments. If what you want to know is the probability of having the 5 guys paired with 3 men and 2 women (which is what I think the grandfather meant), the answer is C(4,3) x C(10,2) / C(14,5) = about 9% (number of ways to pick 3 men out of 4 x number of ways to pick 2 women out of 10 / number of ways to pick 5 people out of 14, /regardless of the order/) If you want to know the probability of having the pairs made exactly in the order the grandfather mentioned, it's P(4,3) x P(10,2) / P(14,5) = about 0.9% (as above, but considering the order) where C(M,N) and P(M,N) are the number of combinations and permutations of N items out of M, respectively. (edit to change multiplication symbol, asterisk doesn't show)
- deleted 16y ago
  [deleted]
- ohmygodel 16y ago
  You need to divide by 3! and 2!. One way to look at it is that every selection of five from the 4M and 10F is equally likely. There are \binom{14}{5} (i.e. 14 choose 5) such selections. Of these, \binom{4}{3} * \binom{10}{2} have exactly 3M and 2F. Therefore, the probability is \binom{4}{3} * \binom{10}{2} / \binom{14}{5}, or (4 * 3 * 2 * 10 * 9) 5! / (3! 2! (14 * 13 * 12 * 11 * 10)) = 10 * 9/(13 * 11 * 7) = 0.08991...
  - dcosson 16y ago
    This is how I'd solve it too. Plus, with a username like that and a LaTex formatted answer, I'm already inclined to trust your math skills :)
  - light3 16y ago
    Ahh indeed, I thought there were 5!=120 ways to order 3M and 2F, but apparently there are only 5!/(2!3!) = 10 ways: {MMMFF,MMFFM,MFFMM,FFMMM,FMMMF,MFMMF,MMFMF,FMMFM,MFMFM,FMFMM}. I made the mistake of thinking you can distinguish between individual males/females.

wheels 16y ago

I'd felt exactly the same a couple years back, but the root cause was rather amusing. When I started working on stuff for Directed Edge I dove into a lot of academic research. I found myself struggling with a lot of it. I really believed that my math abilities had slipped significantly since my glory days and continued with this perception for several months... ...until I picked up some stuff that I'd thought was hard in college. I was dumbfounded at how easy it was. What had changed was not my math abilities, but my perception of what constituted hard math.

tjarratt 16y ago
My experience with Haskell reflects this pretty closely. A year ago I was using it primarily for scripting and daily exercise but I couldn't get past monads as a concept. A few days ago I brought out Learn You A Haskell and took a look at the monad section and suddenly it all made sense. Sometimes your unconscious needs time to digest concepts and internalize them before you can master them and move on to hairier problems.
nostrademons 16y ago
I've noticed the same thing. I'll routinely struggle to understand an academic paper with a notation-heavy theoretical bent, but when I found I had to dig out high school math for a couple 2D-graphics problems recently, it all came back easily. When you're in high school, you don't realize how much of a gulf there is between Ph.D/MS-level (or even professional-level) math and what you learn in school. The public school system teaches you the barest foundations; there's a gigantic world out there beyond it.

jhlyerte 16y ago

Click on our website: == ( http://cha.la/bkef http://cha.la/bkef ) == Christan Audigier bikini $23 Ed Hardy Bikini $23 Smful short_t-shirt_woman $15 ed hardy short_tank_woman $16 Sandal $32 christian louboutin $80 Sunglass $15 COACH_Necklace $27 handbag $33 AF tank woman $17 puma slipper woman $30 == ( http://cha.la/bkef http://cha.la/bkef ) == ~ ¤ ╭⌒╮ ╭⌒╮ ╭⌒╭⌒╮╭⌒╮～╭⌒╮ ,））））,''）~~ ,''~） ╱◥█◣ ╱◥█◣ ｜田｜田｜｜田｜田｜ ╬╬╬╬╬╬╬╬╬╬╬╬╬╬

cletus 16y ago

I have a similar problem actually. In high school and even the first year of university I was pretty much a natural at math. I'd been doing algebra and trigonometry since I was 10, calculus since I was 13 and started on differential equations at 15. Not that this made me any kind of child prodigy mind you but it was a fairly isolating experience because my first 18 years took places in isolated towns with <15,000 people so I just didn't come across other people like me or with similar interests. But I digress... The second year of university maths bored me senseless and it sucked all the fun out of maths for me. Three hour proofs, numerical methods (in particular) and so on. So I stopped doing it and just did CS instead (originally I'd been intending some kind of double major equivalent). Having not used calculus (for example) for years I can still remember how to derive/integrate, etc. So I retain a decent part of what I knew (or so it seems). But the one area I suffer in, largely because I never got that far, is in deciphering academic papers. This applies to CS too, which is where it's a pain. I like dealing with actual code rather than abstract ideas put in text (and the vaguest of algorithms that you can't just take an implement, or at least I can't). So my situation isn't identical to yours but I really would like to know how I could improve my maths and CS in this area.

tel 16y ago
There's an incredibly common roadblock in math education when you make the jump from mechanical maths to creative maths. On one side, math makes sense as a set of powerful rules which help to guide you to understand and analyze the world. On the other side, math is more like a penpal friend while you're living in a foreign country. It offers advice, but largely you're on your own in a confusing land. Academic papers are written from the far side of that roadblock. They're often best considered anecdotal pointers to how to survive abroad rather than comprehensive guides like textbooks (or review papers) offer. Reading them is thus both an exercise in deciphering something distant from your comfort zone and learning to apply its vague knowledge to your own situation. In order to understand academic papers then, you've kind of got to become a traveler yourself. At that point you appreciate the vague note passing because even though it's difficult, there's a real opportunity there to see something few ever have before. --- So practically, academic papers are never easy to decipher unless you're basically part of the same communities as the authors (or they are really dedicated and skillful communicators who have an uncommonly deep understanding of the topics they're writing about). If you're doing work that cannot be solved by what exists and is common knowledge today, however, they're your best bet at finding a guide and no matter how difficult they are you want to read them through.

sudont 16y ago

Similarly burned-out, I'm having trouble caring about design: learning about how it interacts with other disciplines and practicing the skill on these issues, rather than just practicing design itself, is essential. Design, like math, is a problem-solving application and there's definitely a "feel" that can be lost. These solutions can definitely be abstracted out to other areas, but really--what part of math can't?

codelion 16y ago

The best way to study math is to fight it ... http://abstrusegoose.com/353 http://abstrusegoose.com/353

deskamess 16y ago

I wanted to brush up on my math with a goal of shoring up fundamentals and understanding probability and statistics (not just formulaic solving). I was not sure where to start, so I have decided to start by practicing the fundamentals. Addition and so on. Khan Academy has that covered. As of today they have 'practices' that go from basic addition all the way through simple differentiation. Most important, if you run into problems with an exercise set, they have video links to the topic you are practicing. Once you have watched the video and understand the concepts, you can come back to the practice set. (RUPPPL - read understand practice practice practice loop) You should be able to run through the set without much effort, but I think there is value in doing these simpler exercises. It establishes routine and steeps the mind for greater challenges. I will likely add some permutation & combination stuff to my 'course work' before I go on to my eventual goal of understanding prob & stats. Getting exercise for these higher level topics will be a challenge.

tricky 16y ago
This is EXACTLY what I'm doing and it seems to be working out great. Finding exercise for the higher level topics is a bit more challenging, but there is a lot out there. The thinkstats book is great if you're a programmer: http://www.greenteapress.com/thinkstats/html/index.html http://www.greenteapress.com/thinkstats/html/index.html I've also been studying a berkely stats course on itunes u and working the exams/notes from this class: https://netfiles.uiuc.edu/fenghong/www/STAT100/ https://netfiles.uiuc.edu/fenghong/www/STAT100/ I'm also reading a lot about financial analysis and risk assessment to fiddle around with real-world applications of the things I'm learning. it's a hell of a journey.
tel 16y ago
Fundamentals are nice, but rote learning of them is directly on the route to formulaic solving. You have to have something to force you to use math like an artist and not like a computer. Which is to say, I highly suggest doing this stuff as long as you get value out of it. It'll be a boon to your comprehension speed later. Eventually, however, go find a hard textbook and work through every single line of it. Really. Understand that it'll possibly take years and involve finding many lesser sources to fill in the gaps, but without a hard motivating problem you're going to get stuck in an affective loop on formulaic knowledge. Easy, satisfying, but somewhat flat.

cheez 16y ago

Teach your kids (or someone else's)

abbasmehdi 16y ago
Exactly! Teach anyone and explain the concepts to them. Go your local community college and volunteer at the math clinic, even better go to a local high school that's not been doing very well. Public school systems can use all the help they can get. I went from sucky math guy to an awesome math guy in just 1 year - just because I was tutoring.

michaelochurch 16y ago

I don't think you lose "good at math" if you're actually good at math. Being actually good at math, as I'd define it, means you understand problem solving and logical reasoning. It's not about computational techniques. Most math professors have to review linear algebra and the calculus sequence before they can teach their first undergraduate class. It's like this: if you study a language intensely for a year, you'll have a good understanding of it after that year. If you stop using it, the grammar will remain with you but your vocabulary will fade over time. If, however, you travel back to that country, it'll only take you a couple weeks to regain took you a year the first time around. Math is a language, and what you lose is vocabulary, not the language itself. I'm 27 and have been out of school for 5 years. I couldn't do integration by parts without consulting a textbook, and I'd probably do poorly on a Calc-3 final if I had to take it closed-book, right now, in the alloted time. But I also know where to look when I need to review calculus or linear algebra. There's a huge difference between "rusty" and lost skills. For "rusty", you just need to study again. It starts out hard and gets easier as you regain the skills. It's actually fun, at least for me, to revisit familiar friends like set theory. And I'm still learning new things (although not at as fast a rate as when I was in school) even though I've been out of school for half a decade.

deleted 16y ago
[deleted]

nextparadigms 16y ago

Try www.khanacademy.org

32ftpersecond 16y ago

see: G H Hardy's "A Mathematician's Apology"

CallMeV 16y ago

Like all things at which you have to work hard to gain ability, the original poster has but one course: the hardest one. To delve into those tiresome, brain-aching problems at which he once balked in school - the ones where he found that he had to put in a real effort. And he will have to accept lots of failures before he can start to claim lots of successes. With nobody to mark his efforts down for not showing sufficient workings, however, admitting to failure can come a lot more easily when you only have your own self as teacher and student.

ciopte7 16y ago

Check out this blog for a lot of free resources http://hbpms.blogspot.com/ http://hbpms.blogspot.com/

pitdesi 16y ago

I'd recommend going through Khan academy and starting from the basics: http://www.khanacademy.org/#browse http://www.khanacademy.org/#browse

parfe 16y ago

Grab a Calculus book and solve some random problems. It can get fairly entertaining. Solving how much rocket fuel you need to reach orbit or how fast a conical object will drain or the angle you need to fire a canon ball into a castle wall or something. I haven't found much use for Calculus though in regular life. On a side note, we've been arguing about 48÷2(9+3) at the office today. There is a correct answer, of course, but also an argument (that some think is reasonable) against that answer.

e03179 16y ago
48÷2(9+3) = 2
- nlco 16y ago
  not quite, 48÷(2(9+3)) = 2, but 48÷2(9+3) = 288 to clarify: PEMDAS is the order of operations. so parenthesis first: 48÷2(9+3) = 48÷2(12) then multiplication and division (they have the same precedence): 48÷2(12) = 24*12 = 288 for another reference: http://www.google.com/search?q=48%C3%B72(9%2B3) http://www.google.com/search?q=48%C3%B72(9%2B3)
  - dcosson 16y ago
    multiplication and division have the same precedence, and division (like most operations) is left-associative. For example, exponentiation is right-associative, so 2^3^2 = 2^9 = 512, not 64.
    - wging 16y ago
      While you're technically correct, of the standard four half (+, *) are both left- and right-associative. / and - are the ones that are only left-associative.
  - sga 16y ago
    Best bet is to replace x÷y by x*(1/y) as division is not commutative while multiplication is commutative. Therefore 48÷2(9+3) => 48(1/(2(9+3)) => 2

projectileboy 16y ago

This topic comes up from time to time on HN. Here's what I did to dive back into mathematics: http://news.ycombinator.com/item?id=108924 http://news.ycombinator.com/item?id=108924