5 ms·

Don't you have to have math knowledge ? Any good resource you'd recommend ?

by activee 11y ago

cheez 11y ago
I just realized my version of accessible is in fact... Inaccessible. Sorry, yes the math needed is moderate.
taneq 11y ago
There's "math knowledge" and then there's "math knowledge". I'm still learning the beginner stuff but so far all that it seems to need is basic calculus (ie. "do you understand what a gradient is and how to calculate one numerically?") and some linear algebra ("do you know what matrices are and how to multiply them together?")
- unixhero 11y ago
  And then there's math knowledge. Social scientist here, and I have no clue of what you're talking about. Calculus destroyed my engineering aspirations.
  - 50CNT 11y ago
    Basic calculus doesn't seem that hard though. You understand asymptotes, right? When something gets closer to a certain value, but doesn't quite touch it, like 1/x with 0. That leads quite naturally to limits, where you say "ok, at a close enough value, we'll just ignore that little bit and say it touches". Then come derivatives. So you have a function graph, and you're trying to determine the slope at a certain point. It's really simple for say y = x, where you just take any two points, but what if the graph is curvy? You'd take the point and then a point close to it, and calculate the slope for that. If the function is fairly smooth, that's fairly close to the actual slope. So you push the other point closer to the original point, and the slope you calculate gets closer to the real slope. You edge closer and closer, and suddenly it sounds a lot like a limit, where your initial point is x, your second point x + h, and h becomes smaller and smaller, so h -> 0. Here's what's cool though. Since you didn't use 5, or 7.23, but x, you can put any point into this, so you've got a function that maps x onto the slope at x of the original function (at least if I remember correctly). You play around with a couple of different kinds of functions and arrive at various rules for differentiating things without having to do the whole limit thing, like x^2 -> 2x, x^3 -> 3x^2, e^x -> e^x, etc. And derivatives are really handy. Say in kinematics, velocity is the derivative over time of position, acceleration is the derivative over time of velocity, and suddenly a whole bunch of things make a lot of sense. Integrals are pretty neat too. So you want to estimate the area under a function from point a to point b. so you draw a box, and there's an area you're missing, or an area you're including but shouldn't, but it's close to the actual area underneath the function. So you think...hrm, I could use two boxes of half the width, that'd be closer to what I want to find. Then three, ten, a hundred, with the sum of the area of boxes getting closer to the actual area. Said more mathy, it'd be a to b, with a width of b-a, and each box gets a width b-a/h, and h gets infinitely large. Then we make that little limit jump again, and get a function. We do that for a couple of functions, deduce some rules, so we don't need to go through the entire process whilst integrating simple functions. a -> x + C, x -> 1/2x^2 + C for example. With some understanding of the FTC (fundamental theorem of calculus), which roughly states that derivatives and integrals are inverse operations of each other (isn't that neat?), it again helps us make sense of things. For example, look at the equations in physics for constant accelerations: v = v_i + at, x = x_i + vt + 1/2at^2. See how v is the integral of a, and x is the integral of v? For the first one, "v_i" is the constant factor C, and "at" is what comes from the factor a. For the second one, "x_i" is the constant factor C, "vt" is "v_i" integrated, and "1/2at^2" is the integral of "at". Calculus is super neat.
    - MawNicker 11y ago
      Thanks for writing all of this. I enjoyed it. It makes no sense to me how calculus makes so much sense. It all just fits together. It's an eerie high point in the "unreasonable" effectiveness of mathematics.
    - trampi 11y ago
      I enjoyed your writeup too. If I had to choose between being able to low level do everything without having a high level understanding of how they fit together and on the other hand having this high level understanding with no low level applying skills I would definitely choose the high level view. It is IMHO far more valuable to know how they fit together.
    - ssebastianj 11y ago
      Very nice. I know it's pretty dated by now (mid '80s, I guess) but the "The Mechanical Universe...and Beyond" [0] videos really opened my eyes on the matter. [0] http://www.its.caltech.edu/~tmu/ http://www.its.caltech.edu/~tmu/
- cshenton 11y ago
  In theory this is true. But it's very hard to find good instruction that is wilfully light on math. Andrew Ng's course for one requires you to be very comfortable with lin alg and convex optimisation.
- aje404 11y ago
  Go show someone how to calculate a gradient vector (assuming they can calculate basic derivatives) and multiply two matrices, and then try to teach them how gradient descent works or the SVM derivation. That's not going to end even slightly well for your average person. There really is no silver bullet for this stuff. You either need some mathematical maturity, or you need to put in an exorbitant amount of time to understand things.
- deleted 11y ago
  [deleted]
Retric 11y ago
Not really, you can go far playing around with off the shelf components. IMO, the math is less helpful in NN than other areas because your trying to create a useful aproximation not get a specific correct answer. Thus knowing why something works is only so helpful.
PeterisP 11y ago
In general, if you're an user of ML (as in, apply already known ML methods to your particular problem instead of researching better conceptual ML methods) then you'd be able to work on a higher abstraction levels where all the math is inside the libraries and you're not really touching it, much less implementing the formulae.

Don't you have to have math knowledge ? Any good resource you'd recommend ?

by activee 11y ago

Don't you have to have math knowledge ? Any good resource you'd recommend ?

cheez 11y ago

I just realized my version of accessible is in fact... Inaccessible. Sorry, yes the math needed is moderate.

taneq 11y ago

There's "math knowledge" and then there's "math knowledge". I'm still learning the beginner stuff but so far all that it seems to need is basic calculus (ie. "do you understand what a gradient is and how to calculate one numerically?") and some linear algebra ("do you know what matrices are and how to multiply them together?")

unixhero 11y ago
And then there's math knowledge. Social scientist here, and I have no clue of what you're talking about. Calculus destroyed my engineering aspirations.
- 50CNT 11y ago
  Basic calculus doesn't seem that hard though. You understand asymptotes, right? When something gets closer to a certain value, but doesn't quite touch it, like 1/x with 0. That leads quite naturally to limits, where you say "ok, at a close enough value, we'll just ignore that little bit and say it touches". Then come derivatives. So you have a function graph, and you're trying to determine the slope at a certain point. It's really simple for say y = x, where you just take any two points, but what if the graph is curvy? You'd take the point and then a point close to it, and calculate the slope for that. If the function is fairly smooth, that's fairly close to the actual slope. So you push the other point closer to the original point, and the slope you calculate gets closer to the real slope. You edge closer and closer, and suddenly it sounds a lot like a limit, where your initial point is x, your second point x + h, and h becomes smaller and smaller, so h -> 0. Here's what's cool though. Since you didn't use 5, or 7.23, but x, you can put any point into this, so you've got a function that maps x onto the slope at x of the original function (at least if I remember correctly). You play around with a couple of different kinds of functions and arrive at various rules for differentiating things without having to do the whole limit thing, like x^2 -> 2x, x^3 -> 3x^2, e^x -> e^x, etc. And derivatives are really handy. Say in kinematics, velocity is the derivative over time of position, acceleration is the derivative over time of velocity, and suddenly a whole bunch of things make a lot of sense. Integrals are pretty neat too. So you want to estimate the area under a function from point a to point b. so you draw a box, and there's an area you're missing, or an area you're including but shouldn't, but it's close to the actual area underneath the function. So you think...hrm, I could use two boxes of half the width, that'd be closer to what I want to find. Then three, ten, a hundred, with the sum of the area of boxes getting closer to the actual area. Said more mathy, it'd be a to b, with a width of b-a, and each box gets a width b-a/h, and h gets infinitely large. Then we make that little limit jump again, and get a function. We do that for a couple of functions, deduce some rules, so we don't need to go through the entire process whilst integrating simple functions. a -> x + C, x -> 1/2x^2 + C for example. With some understanding of the FTC (fundamental theorem of calculus), which roughly states that derivatives and integrals are inverse operations of each other (isn't that neat?), it again helps us make sense of things. For example, look at the equations in physics for constant accelerations: v = v_i + at, x = x_i + vt + 1/2at^2. See how v is the integral of a, and x is the integral of v? For the first one, "v_i" is the constant factor C, and "at" is what comes from the factor a. For the second one, "x_i" is the constant factor C, "vt" is "v_i" integrated, and "1/2at^2" is the integral of "at". Calculus is super neat.
  - MawNicker 11y ago
    Thanks for writing all of this. I enjoyed it. It makes no sense to me how calculus makes so much sense. It all just fits together. It's an eerie high point in the "unreasonable" effectiveness of mathematics.
  - trampi 11y ago
    I enjoyed your writeup too. If I had to choose between being able to low level do everything without having a high level understanding of how they fit together and on the other hand having this high level understanding with no low level applying skills I would definitely choose the high level view. It is IMHO far more valuable to know how they fit together.
  - ssebastianj 11y ago
    Very nice. I know it's pretty dated by now (mid '80s, I guess) but the "The Mechanical Universe...and Beyond" [0] videos really opened my eyes on the matter. [0] http://www.its.caltech.edu/~tmu/ http://www.its.caltech.edu/~tmu/
cshenton 11y ago
In theory this is true. But it's very hard to find good instruction that is wilfully light on math. Andrew Ng's course for one requires you to be very comfortable with lin alg and convex optimisation.
aje404 11y ago
Go show someone how to calculate a gradient vector (assuming they can calculate basic derivatives) and multiply two matrices, and then try to teach them how gradient descent works or the SVM derivation. That's not going to end even slightly well for your average person. There really is no silver bullet for this stuff. You either need some mathematical maturity, or you need to put in an exorbitant amount of time to understand things.
deleted 11y ago
[deleted]

Retric 11y ago

Not really, you can go far playing around with off the shelf components. IMO, the math is less helpful in NN than other areas because your trying to create a useful aproximation not get a specific correct answer. Thus knowing why something works is only so helpful.

PeterisP 11y ago

In general, if you're an user of ML (as in, apply already known ML methods to your particular problem instead of researching better conceptual ML methods) then you'd be able to work on a higher abstraction levels where all the math is inside the libraries and you're not really touching it, much less implementing the formulae.