• Posted by Konstantin 23.11.2016 4 Comments

    Basic linear algebra, introductory statistics and some familiarity with core machine learning concepts (such as PCA and linear models) are the prerequisites of this post. Otherwise it will probably make no sense. An abridged version of this text is also posted on Quora.

    Most textbooks on statistics cover covariance right in their first chapters. It is defined as a useful "measure of dependency" between two random variables:

        \[\mathrm{cov}(X,Y) = E[(X - E[X])(Y - E[Y])].\]

    The textbook would usually provide some intuition on why it is defined as it is, prove a couple of properties, such as bilinearity, define the covariance matrix for multiple variables as {\bf\Sigma}_{i,j} = \mathrm{cov}(X_i, X_j), and stop there. Later on the covariance matrix would pop up here and there in seeminly random ways. In one place you would have to take its inverse, in another - compute the eigenvectors, or multiply a vector by it, or do something else for no apparent reason apart from "that's the solution we came up with by solving an optimization task".

    In reality, though, there are some very good and quite intuitive reasons for why the covariance matrix appears in various techniques in one or another way. This post aims to show that, illustrating some curious corners of linear algebra in the process.

    Meet the Normal Distribution

    The best way to truly understand the covariance matrix is to forget the textbook definitions completely and depart from a different point instead. Namely, from the the definition of the multivariate Gaussian distribution:

    We say that the vector \bf x has a normal (or Gaussian) distribution with mean \bf \mu and covariance \bf \Sigma if:

        \[\Pr({\bf x}) =|2\pi{\bf\Sigma}|^{-1/2} \exp\left(-\frac{1}{2}({\bf x} - {\bf\mu})^T{\bf\Sigma}^{-1}({\bf x} - {\bf \mu})\right).\]

    To simplify the math a bit, we will limit ourselves to the centered distribution (i.e. {\bf\mu} = {\bf 0}) and refrain from writing out the normalizing constant |2\pi{\bf\Sigma}|^{-1/2}. Now, the definition of the (centered) multivariate Gaussian looks as follows:

        \[\Pr({\bf x}) \propto \exp\left(-0.5{\bf x}^T{\bf\Sigma}^{-1}{\bf x}\right).\]

    Much simpler, isn't it? Finally, let us define the covariance matrix as nothing else but the parameter of the Gaussian distribution. That's it. You will see where it will lead us in a moment.

    Transforming the Symmetric Gaussian

    Consider a symmetric Gaussian distribution, i.e. the one with {\bf \Sigma = \bf I} (the identity matrix). Let us take a sample from it, which will of course be a symmetric, round cloud of points:

    We know from above that the likelihood of each point in this sample is

    (1)   \[P({\bf x}) \propto \exp(-0.5 {\bf x}^T {\bf x}).\]

    Now let us apply a linear transformation {\bf A} to the points, i.e. let {\bf y} ={\bf Ax}. Suppose that, for the sake of this example, {\bf A} scales the vertical axis by 0.5 and then rotates everything by 30 degrees. We will get the following new cloud of points {\bf y}:

    What is the distribution of {\bf y}? Just substitute {\bf x}={\bf A}^{-1}{\bf y} into (1), to get:

    (2)   \begin{align*} P({\bf y}) &\propto \exp(-0.5 ({\bf A}^{-1}{\bf y})^T({\bf A}^{-1}{\bf y}))\\ &=\exp(-0.5{\bf y}^T({\bf AA}^T)^{-1}{\bf y}) \end{align*}

    This is exactly the Gaussian distribution with covariance {\bf \Sigma} = {\bf AA}^T. The logic works both ways: if we have a Gaussian distribution with covariance \bf \Sigma, we can regard it as a distribution which was obtained by transforming the symmetric Gaussian by some {\bf A}, and we are given {\bf AA}^T.

    More generally, if we have any data, then, when we compute its covariance to be \bf\Sigma, we can say that if our data were Gaussian, then it could have been obtained from a symmetric cloud using some transformation \bf A, and we just estimated the matrix {\bf AA}^T, corresponding to this transformation.

    Note that we do not know the actual \bf A, and it is mathematically totally fair. There can be many different transformations of the symmetric Gaussian which result in the same distribution shape. For example, if \bf A is just a rotation by some angle, the transformation does not affect the shape of the distribution at all. Correspondingly, {\bf AA}^T = {\bf I} for all rotation matrices. When we see a unit covariance matrix we really do not know, whether it is the “originally symmetric” distribution, or a “rotated symmetric distribution”. And we should not really care - those two are identical.

    There is a theorem in linear algebra, which says that any symmetric matrix \bf \Sigma can be represented as:

    (3)   \[{\bf \Sigma} = {\bf VDV}^T,\]

    where {\bf V} is orthogonal (i.e. a rotation) and {\bf D} is diagonal (i.e. a coordinate-wise scaling). If we rewrite it slightly, we will get:

    (4)   \[{\bf \Sigma} = ({\bf VD}^{1/2})({\bf VD}^{1/2})^T = {\bf AA}^T,\]

    where {\bf A} = {\bf VD}^{1/2}. This, in simple words, means that any covariance matrix \bf \Sigma could have been the result of transforming the data using a coordinate-wise scaling {\bf D}^{1/2} followed by a rotation \bf V. Just like in our example with \bf x and \bf y above.

    Principal Component Analysis

    Given the above intuition, PCA already becomes a very obvious technique. Suppose we are given some data. Let us assume (or “pretend”) it came from a normal distribution, and let us ask the following questions:

    1. What could have been the rotation \bf V and scaling {\bf D}^{1/2}, which produced our data from a symmetric cloud?
    2. What were the original, “symmetric-cloud” coordinates \bf x before this transformation was applied.
    3. Which original coordinates were scaled the most by \bf D and thus contribute most to the spread of the data now. Can we only leave those and throw the rest out?

    All of those questions can be answered in a straightforward manner if we just decompose \bf \Sigma into \bf V and \bf D according to (3). But (3) is exactly the eigenvalue decomposition of \bf\Sigma. I’ll leave you to think for just a bit and you’ll see how this observation lets you derive everything there is about PCA and more.

    The Metric Tensor

    Bear me for just a bit more. One way to summarize the observations above is to say that we can (and should) regard {\bf\Sigma}^{-1} as a metric tensor. A metric tensor is just a fancy formal name for a matrix, which summarizes the deformation of space. However, rather than claiming that it in some sense determines a particular transformation \bf A (which it does not, as we saw), we shall say that it affects the way we compute angles and distances in our transformed space.

    Namely, let us redefine, for any two vectors \bf v and \bf w, their inner product as:

    (5)   \[\langle {\bf v}, {\bf w}\rangle_{\Sigma^{-1}} = {\bf v}^T{\bf \Sigma}^{-1}{\bf w}.\]

    To stay consistent we will also need to redefine the norm of any vector as

    (6)   \[|{\bf v}|_{\Sigma^{-1}} = \sqrt{{\bf v}^T{\bf \Sigma}^{-1}{\bf v}},\]

    and the distance between any two vectors as

    (7)   \[|{\bf v}-{\bf w}|_{\Sigma^{-1}} = \sqrt{({\bf v}-{\bf w})^T{\bf \Sigma}^{-1}({\bf v}-{\bf w})}.\]

    Those definitions now describe a kind of a “skewed world” of points. For example, a unit circle (a set of points with “skewed distance” 1 to the center) in this world might look as follows:

    And here is an example of two vectors, which are considered “orthogonal”, a.k.a. “perpendicular” in this strange world:

    Although it may look weird at first, note that the new inner product we defined is actually just the dot product of the “untransformed” originals of the vectors:

    (8)   \[{\bf v}^T{\bf \Sigma}^{-1}{\bf w} = {\bf v}^T({\bf AA}^T)^{-1}{\bf w}=({\bf A}^{-1}{\bf v})^T({\bf A}^{-1}{\bf w}),\]

    The following illustration might shed light on what is actually happening in this \Sigma-“skewed” world. Somehow “deep down inside”, the ellipse thinks of itself as a circle and the two vectors behave as if they were (2,2) and (-2,2).

    Getting back to our example with the transformed points, we could now say that the point-cloud \bf y is actually a perfectly round and symmetric cloud “deep down inside”, it just happens to live in a skewed space. The deformation of this space is described by the tensor {\bf\Sigma}^{-1} (which is, as we know, equal to ({\bf AA}^T)^{-1}. The PCA now becomes a method for analyzing the deformation of space, how cool is that.

    The Dual Space

    We are not done yet. There’s one interesting property of “skewed” spaces worth knowing about. Namely, the elements of their dual space have a particular form. No worries, I’ll explain in a second.

    Let us forget the whole skewed space story for a moment, and get back to the usual inner product {\bf w}^T{\bf v}. Think of this inner product as a function f_{\bf w}({\bf v}), which takes a vector \bf v and maps it to a real number, the dot product of \bf v and \bf w. Regard the \bf w here as the parameter (“weight vector”) of the function. If you have done any machine learning at all, you have certainly come across such linear functionals over and over, sometimes in disguise. Now, the set of all possible linear functionals f_{\bf w} is known as the dual space to your “data space”.

    Note that each linear functional is determined uniquely by the parameter vector \bf w, which has the same dimensionality as \bf v, so apparently the dual space is in some sense equivalent to your data space - just the interpretation is different. An element \bf v of your “data space” denotes, well, a data point. An element \bf w of the dual space denotes a function f_{\bf w}, which projects your data points on the direction \bf w (recall that if \bf w is unit-length, {\bf w}^T{\bf v} is exactly the length of the perpendicular projection of \bf v upon the direction \bf w). So, in some sense, if \bf v-s are “vectors”, \bf w-s are “directions, perpendicular to these vectors”. Another way to understand the difference is to note that if, say, the elements of your data points numerically correspond to amounts in kilograms, the elements of \bf w would have to correspond to “units per kilogram”. Still with me?

    Let us now get back to the skewed space. If \bf v are elements of a skewed Euclidean space with the metric tensor {\bf\Sigma}^{-1}, is a function f_{\bf w}({\bf v}) = {\bf w}^T{\bf v} an element of a dual space? Yes, it is, because, after all, it is a linear functional. However, the parameterization of this function is inconvenient, because, due to the skewed tensor, we cannot interpret it as projecting vectors upon \bf w nor can we say that \bf w is an “orthogonal direction” (to a separating hyperplane of a classifier, for example). Because, remember, in the skewed space it is not true that orthogonal vectors satisfy {\bf w}^T{\bf v}=0. Instead, they satisfy {\bf w}^T{\bf \Sigma}^{-1}{\bf v} = 0. Things would therefore look much better if we parameterized our dual space differently. Namely, by considering linear functionals of the form f^{\Sigma^{-1}}_{\bf z}({\bf v}) = {\bf z}^T{\bf \Sigma}^{-1}{\bf v}. The new parameters \bf z could now indeed be interpreted as an “orthogonal direction” and things overall would make more sense.

    However when we work with actual machine learning models, we still prefer to have our functions in the simple form of a dot product, i.e. f_{\bf w}, without any ugly \bf\Sigma-s inside. What happens if we turn a “skewed space” linear functional from its natural representation into a simple inner product?

    (9)   \[f^{\Sigma^{-1}}_{\bf z}({\bf v}) = {\bf z}^T{\bf\Sigma}^{-1}{\bf v} = ({\bf \Sigma}^{-1}{\bf z})^T{\bf v} = f_{\bf w}({\bf v}),\]

    where {\bf w} = {\bf \Sigma}^{-1}{\bf z}. (Note that we can lose the transpose because \bf \Sigma is symmetric).

    What it means, in simple terms, is that when you fit linear models in a skewed space, your resulting weight vectors will always be of the form {\bf \Sigma}^{-1}{\bf z}. Or, in other words, {\bf\Sigma}^{-1} is a transformation, which maps from “skewed perpendiculars” to “true perpendiculars”. Let me show you what this means visually.

    Consider again the two “orthogonal” vectors from the skewed world example above:

    Let us interpret the blue vector as an element of the dual space. That is, it is the \bf z vector in a linear functional {\bf z}^T{\bf\Sigma}^{-1}{\bf v}. The red vector is an element of the “data space”, which would be mapped to 0 by this functional (because the two vectors are “orthogonal”, remember).

    For example, if the blue vector was meant to be a linear classifier, it would have its separating line along the red vector, just like that:

    If we now wanted to use this classifier, we could, of course, work in the “skewed space” and use the expression {\bf z}^T{\bf\Sigma}^{-1}{\bf v} to evaluate the functional. However, why don’t we find the actual normal \bf w to that red separating line so that we wouldn’t need to do an extra matrix multiplication every time we use the function?

    It is not too hard to see that {\bf w}={\bf\Sigma}^{-1}{\bf z} will give us that normal. Here it is, the black arrow:

    Therefore, next time, whenever you see expressions like {\bf w}^T{\bf\Sigma}^{-1}{\bf v} or ({\bf v}-{\bf w})^T{\bf\Sigma}^{-1}({\bf v}-{\bf w}), remember that those are simply inner products and (squared) distances in a skewed space, while {\bf \Sigma}^{-1}{\bf z} is a conversion from a skewed normal to a true normal. Also remember that the “skew” was estimated by pretending that the data were normally-distributed.

    Once you see it, the role of the covariance matrix in some methods like the Fisher’s discriminant or Canonical correlation analysis might become much more obvious.

    The Dual Space Metric Tensor

    “But wait”, you should say here. “You have been talking about expressions like {\bf w}^T{\bf\Sigma}^{-1}{\bf v} all the time, while things like {\bf w}^T{\bf\Sigma}{\bf v} are also quite common in practice. What about those?”

    Hopefully you know enough now to suspect that {\bf w}^T{\bf\Sigma}{\bf v} is again an inner product or a squared norm in some deformed space, just not the “internal data metric space”, that we considered so far. Which space is it? It turns out it is the “internal dual metric space”. That is, whilst the expression {\bf w}^T{\bf\Sigma}^{-1}{\bf v} denoted the “new inner product” between the points, the expression {\bf w}^T{\bf\Sigma}{\bf v} denotes the “new inner product” between the parameter vectors. Let us see why it is so.

    Consider an example again. Suppose that our space transformation \bf A scaled all points by 2 along the x axis. The point (1,0) became (2,0), the point (3, 1) became (6, 1), etc. Think of it as changing the units of measurement - before we measured the x axis in kilograms, and now we measure it in pounds. Consequently, the norm of the point (2,0) according to the new metric, |(2,0)|_{\Sigma^{-1}} will be 1, because 2 pounds is still just 1 kilogram “deep down inside”.

    What should happen to the parameter ("direction") vectors due to this transformation? Can we say that the parameter vector (1,0) also got scaled to (2,0) and that the norm of the parameter vector (2,0) is now therefore also 1? No! Recall that if our initial data denoted kilograms, our dual vectors must have denoted “units per kilogram”. After the transformation they will be denoting “units per pound”, correspondingly. To stay consistent we must therefore convert the parameter vector (”1 unit per kilogram”, 0) to its equivalent (“0.5 units per pound”,0). Consequently, the norm of the parameter vector (0.5,0) in the new metric will be 1 and, by the same logic, the norm of the dual vector (2,0) in the new metric must be 4. You see, the “importance of a parameter/direction” gets scaled inversely to the “importance of data” along that parameter or direction.

    More formally, if {\bf x}'={\bf Ax}, then

    (10)   \begin{align*} f_{\bf w}({\bf x}) &= {\bf w}^T{\bf x} = {\bf w}^T{\bf A}^{-1}{\bf x}'\\ & =(({\bf A}^{-1})^T{\bf w})^T{\bf x}'=f_{({\bf A}^{-1})^T{\bf w}}({\bf x}'). \end{align*}

    This means, that the transformation \bf A of the data points implies the transformation {\bf B}:=({\bf A}^{-1})^T of the dual vectors. The metric tensor for the dual space must thus be:

    (11)   \[({\bf BB}^T)^{-1}=(({\bf A}^{-1})^T{\bf A}^{-1})^{-1}={\bf AA}^T={\bf \Sigma}.\]

    Remember the illustration of the “unit circle” in the {\bf \Sigma}^{-1} metric? This is how the unit circle looks in the corresponding \bf\Sigma metric. It is rotated by the same angle, but it is stretched in the direction where it was squished before.

    Intuitively, the norm (“importance”) of the dual vectors along the directions in which the data was stretched by \bf A becomes proportionally larger (note that the “unit circle” is, on the contrary, “squished” along those directions).

    But the “stretch” of the space deformation in any direction can be measured by the variance of the data. It is therefore not a coincidence that {\bf w}^T{\bf \Sigma}{\bf w} is exactly the variance of the data along the direction \bf w (assuming |{\bf w}|=1).

    The Covariance Estimate

    Once we start viewing the covariance matrix as a transformation-driven metric tensor, many things become clearer, but one thing becomes extremely puzzling: why is the inverse covariance of the data a good estimate for that metric tensor? After all, it is not obvious that {\bf X}^T{\bf X}/n (where \bf X is the data matrix) must be related to the \bf\Sigma in the distribution equation \exp(-0.5{\bf x}^T{\bf\Sigma}^{-1}{\bf x}).

    Here is one possible way to see the connection. Firstly, let us take it for granted that if \bf X is sampled from a symmetric Gaussian, then {\bf X}^T{\bf X}/n is, on average, a unit matrix. This has nothing to do with transformations, but just a consequence of pairwise independence of variables in the symmetric Gaussian.

    Now, consider the transformed data, {\bf Y}={\bf XA}^T (vectors in the data matrix are row-wise, hence the multiplication on the right with a transpose). What is the covariance estimate of \bf Y?

    (12)   \[{\bf Y}^T{\bf Y}/n=({\bf XA}^T)^T{\bf XA}^T/n={\bf A}({\bf X}^T{\bf X}){\bf A}^T/n\approx {\bf AA}^T,\]

    the familiar tensor.

    This is a place where one could see that a covariance matrix may make sense outside the context of a Gaussian distribution, after all. Indeed, if you assume that your data was generated from any distribution P with uncorrelated variables of unit variance and then transformed using some matrix \bf A, the expression {\bf X}^T{\bf X}/n will still be an estimate of {\bf AA}^T, the metric tensor for the corresponding (dual) space deformation.

    However, note that out of all possible initial distributions P, the normal distribution is exactly the one with the maximum entropy, i.e. the “most generic”. Thus, if you base your analysis on the mean and the covariance matrix (which is what you do with PCA, for example), you could just as well assume your data to be normally distributed. In fact, a good rule of thumb is to remember, that whenever you even mention the word "covariance matrix", you are implicitly fitting a Gaussian distribution to your data.

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  • Posted by Konstantin 17.11.2016 1 Comment

    Mass on a spring

    Imagine a weight hanging on a spring. Let us pull the weight a bit and release it into motion. What will its motion look like? If you remember some of your high-school physics, you should probably answer that the resulting motion is a simple harmonic oscillation, best described by a sinewave. Although this is a fair answer, it actually misses an interesting property of real-life springs. A property most people don't think much about, because it goes a bit beyond the high school curriculum. This property is best illustrated by

    The Slinky Drop

    The "slinky drop" is a fun little experiment which has got its share of internet fame.

    The Slinky Drop

    The Slinky Drop

    When the top end of a suspended slinky is released, the bottom seems to patiently wait for the top to arrive before starting to fall as well. This looks rather unexpected. After all, we know that things fall down according to a parabola, and we know that springs collapse according to a sinewave, however neither of the two rules seem to apply here. If you browse around, you will see lots of awesome videos demonstrating or explaining this effect. There are news articles, forum discussions, blog posts and even research papers dedicated to the magical slinky. However, most of them are either too sketchy or too complex, and none seem to mention the important general implications, so let me give a shot at another explanation here.

    The Slinky Drop Explained Once More

    Let us start with the classical, "high school" model of a spring. The spring has some length L in the relaxed state, and if we stretch it, making it longer by \Delta L, the two ends of the spring exert a contracting force of k\Delta L. Assume we hold the top of the spring at the vertical coordinate y_{\mathrm{top}}=0 and have it balance out. The lower end will then position at the coordinate y_{\mathrm{bot}} = -(L+mg/k), where the gravity force mg is balanced out exactly by the spring force.

    How would the two ends of the spring behave if we let go off the top now? Here's how:

    The falling spring, version 1

    The horozontal axis here denotes the time, the vertical axis - is the vertical position. The blue curve is the trajectory of the top end of the spring, the green curve - trajectory of the bottom end. The dotted blue line is offset from the blue line by exactly L - the length of the spring in relaxed state.

    Observe that the lower end (the green curve), similarly to the slinky, "waits" for quite a long time for the top to approach before starting to move with discernible velocity. Why is it the case? The trajectory of the lower point can be decomposed in two separate movements. Firstly, the point is trying to fall down due to gravity, following a parabola. Secondly, the point is being affected by string tension and thus follows a cosine trajectory. Here's how the two trajectories look like separately:

    They are surprisingly similar at the start, aren't they? And indeed, the cosine function does resemble a parabola up to o(x^3). Recall the corresponding Taylor expansion:

        \[\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \dots \approx 1 - \frac{x^2}{2}.\]

    If we align the two curves above, we can see how well they match up at the beginning:

    Consequently, the two forces happen to "cancel" each other long enough to leave an impression that the lower end "waits" for the upper for some time. This effect is, however, much more pronounced in the slinky. Why so?

    Because, of course, a single spring is not a good model for the slinky. It is more correct to regard a slinky as a chain of strings. Observe what happens if we model the slinky as a chain of just three simple springs:

    Each curve here is the trajectory of one of the nodes inbetween the three individual springs. We can see that the top two curves behave just like a single spring did - the green node waits a bit for the blue and then starts moving. The red one, however, has to wait longer, until the green node moves sufficiently far away. The bottom, in turn, will only start moving observably when the red node approaches it close enough, which means it has to wait even longer yet - by that time the top has already arrived. If we consider a more detailed model, the movement  of a slinky composed of, say, 9 basic springs, the effect of intermediate nodes "waiting" becomes even more pronounced:

    To make a "mathematically perfect" model of a slinky we have to go to the limit of having infinitely many infinitely small springs. Let's briefly take a look at how that solution looks like.

    The Continuous Slinky

    Let x denote the coordinate of a point on a "relaxed" slinky. For example, in the two discrete models above the slinky had 4 and 10 points, numbered 1,\dots, 4 and 1,\dots, 10 respectively. The continuous slinky will have infinitely many points numbered [0,1].

    Let h(x,t) denote the vertical coordinate of a point x at time t. The acceleration of point x at time t is then, by definition \frac{\partial^2 h(x,t)}{\partial^2 t}, and there are two components affecting it: the gravitational pull -g and the force of the spring.

    The spring force acting on a point x is proportional to the stretch of the spring at that point \frac{\partial h(x,t)}{\partial x}. As each point is affected by the stretch from above and below, we have to consider a difference of the "top" and "bottom" stretches, which is thus the derivative of the stretch, i.e. \frac{\partial^2 h(x,t)}{\partial^2 x}. Consequently, the dynamics of the slinky can be described by the equation:

        \[\frac{\partial^2 h(x,t)}{\partial^2 t} = a\frac{\partial^2 h(x,t)}{\partial^2 x} - g.\]

    where a is some positive constant. Let us denote the second derivatives by h_{tt} and h_{xx}, replace a with v^2 and rearrange to get:

    (1)   \[h_{tt} - v^2 h_{xx} = -g,\]

    which is known as the wave equation. The name stems from the fact that solutions to this equation always resemble "waves" propagating at a constant speed v through some medium. In our case the medium will be the slinky itself. Now it becomes apparent that, indeed, the lower end of the slinky should not move before the wave of disturbance, unleashed by releasing the top end, reaches it. Most of the explanations of the slinky drop seem to refer to that fact. However when it is stated alone, without the wave-equation-model context, it is at best a rather incomplete explanation.

    Given how famous the equation is, it is not too hard to solve it. We'll need to do it twice - first to find the initial configuration of a suspended slinky, then to compute its dynamics when the top is released.

    In the beginning the slinky must satisfy h_t(x, t) = 0 (because it is not moving at all), h(0, t) = 0 (because the top end is located at coordinate 0), and h_x(1, t) = 0 (because there is no stretch at the bottom). Combining this with (1) and searching for a polynomial solution, we get:

        \[h(x, t) = h_0(x) = \frac{g}{2v^2}x(x-2).\]

    Next, we release the slinky, hence the conditions h_t(x,t)=0 and h(0,t)=0 disappear and we may use the d'Alembert's formula with reflected boundaries to get the solution:

        \[h(x,t) = \frac{1}{2}(\phi(x-vt) + \phi(x+vt)) - \frac{gt^2}{2},\]

        \[\text{ where }\phi(x) = h_0(\mathrm{mod}(x, 2)).\]

    Here's how the solution looks like visually:

    Notice how the part of the slinky to which the wave has not arrived yet, stays completely fixed in place. Here are the trajectories of 4 equally-spaced points on the slinky:

    Note how, quite surprisingly, all points of the slinky are actually moving with a constant speed, changing it abruptly at certain moments. Somewhat magically, the mean of all these piecewise-linear trajectories (i.e. the trajectory of the center of mass of the slinky) is still a smooth parabola, just as it should be:

    The Secret of Spring Motion

    Now let us come back to where we started. Imagine a weight on a spring. What will its motion be like? Obviously, any real-life spring is, just like the slinky, best modeled not as a Hooke's simple spring, but rather via the wave equation. Which means that when you let go off the weight, the weight will send a deformation wave, which will move along the spring back and forth, affecting the pure sinewave movement you might be expecting from the simple Hooke's law. Watch closely:

    Here is how the movement of the individual nodes looks like:

    The fat red line is the trajectory of the weight, and it is certainly not a sinewave. It is a curve inbetween the piecewise-linear "sawtooth" (which is the limit case when the weight is zero) and the true sinusoid (which is the limit case when the mass of the spring is zero). Here's how the zero-weight case looks like:

    And this is the other extreme - the massless spring:

    These observations can be summarized into the following obviously-sounding conclusion: the basic Hooke's law applies exactly only to the the massless spring. Any real spring has a mass and thus forms an oscillation wave traveling back and forth along its length, which will interfere with the weight's simple harmonic oscillation, making it "less simple and harmonic". Luckily, if the mass of the weight is large enough, this interference is negligible.

    And that is, in my opinion, one of the interesting, yet often overlooked aspects of spring motion.

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  • Posted by Konstantin 11.11.2016 2 Comments

    A question on Quora reminded me that I wanted to post this explanation here every time I got a chance to teach SVMs and Kernel methods, but I never found the time. The post expects basic knowledge of those topics from the reader.

    Introductory Background

    The concept of kernel methods is probably one of the coolest tricks in machine learning. With most machine learning research nowadays being centered around neural networks, they have gone somewhat out of fashion recently, but I suspect they will strike back one day in some way or another.

    The idea of a kernel method starts with the curious observation that if you take a dot product of two vectors, x^T y, and square it, the result can be regarded as a dot product of two "feature vectors", where the features are all pairwise products of the original inputs:

        \begin{align*} &k(x,y) = (x^Ty)^2 = (\sum_i x_iy_i)^2 = \sum_{i,j} x_ix_jy_iy_j \\ & = \langle (x_1x_1, x_1x_2,\dots,x_ix_j,\dots,x_nx_n), (y_1y_1,\dots,y_iy_j,\dots,y_ny_n)\rangle \end{align*}

    Analogously, if you raise x^Ty to the third power, you are essentially computing a dot product within a space of all possible three-way products of your inputs, and so on, without ever actually having to see those features explicitly.

    If you now take any linear model (e.g. linear regression, linear classification, PCA, etc) it turns out you can replace the "real" dot product in its formulation model with such a kernel function, and this will magically convert your model into a linear model with nonlinear features (e.g. pairwise or triple products). As those features are never explicitly computed, there is no problem if there were millions or billions of them.

    Consider, for example, plain old linear regression: f(x) = w^Tx + b. We can "kernelize" it by first representing w as a linear combination of the data points (this is called a dual representation):

        \[f(x) = \left(\sum_i \alpha_i x_i\right)^T x + b = \sum_i \alpha_i (x_i^T x) + b,\]

    and then swapping all the dot products x_i^T x with a custom kernel function:

        \[f(x) = \sum_i \alpha_i k(x_i,x) + b.\]

    If we now substitute k(x,y) = (x^T y)^2 here, our model becomes a second degree polynomial regression. If k(x,y) = (x^T y)^5 it is the fifth degree polynomial regression, etc. It's like magic, you plug in different functions and things just work.

    It turns out that there are lots of valid choices for the kernel function k, and, of course, the Gaussian function is one of these choices:

        \[k(x, y) = \exp\left(-\frac{|x - y|^2}{2\sigma^2}\right).\]

    It is not too surprising - the Gaussian function tends to pop up everywhere, after all, but it is not obvious what "implicit features" it should represent when viewed as a kernel function. Most textbooks do not seem to cover this question in sufficient detail, usually, so let me do it here.

    The Gaussian Kernel

    To see the meaning of the Gaussian kernel we need to understand the couple of ways in which any kernel functions can be combined. We saw before that raising a linear kernel to the power i makes a kernel with a feature space, which includes all i-wise products. Now let us examine what happens if we add two or more kernel functions. Consider k(x,y) = x^Ty + (x^Ty)^2, for example. It is not hard to see that it corresponds to an inner product of feature vectors of the form

        \[(x_1, x_2, \dots, x_n, \quad x_1x_1, x_1x_2,\dots,x_ix_j,\dots, x_n x_n),\]

    i.e. the concatenation of degree-1 (untransformed) features, and degree-2 (pairwise product) features.

    Multiplying a kernel function with a constant c is also meaningful. It corresponds to scaling the corresponding features by \sqrt{c}. For example, k(x,y) = 4x^Ty = (2x)^T(2y).

    Still with me? Great, now let us combine the tricks above and consider the following kernel:

        \[k(x,y) = 1 + x^Ty + \frac{(x^Ty)^2}{2} + \frac{(x^Ty)^3}{6}.\]

    Apparently, it is a kernel which corresponds to a feature mapping, which concatenates a constant feature, all original features, all pairwise products scaled down by \sqrt{2} and all triple products scaled down by \sqrt{6}.

    Looks impressive, right? Let us continue and add more members to this kernel, so that it would contain all four-wise, five-wise, and so on up to infinity-wise products of input features. We shall choose the scaling coefficients for each term carefully, so that the resulting infinite sum would resemble a familiar expression:

        \[\sum_{i=0}^\infty \frac{(x^Ty)^i}{i!} = \exp(x^Ty).\]

    We can conclude here that k(x,y) = \exp(x^Ty) is a valid kernel function, which corresponds to a feature space, which includes products of input features of any degree, up to infinity.

    But we are not done yet. Suppose that we decide to normalize the inputs before applying our linear model. That is, we want to convert each vector x to \frac{x}{|x|} before feeding it to the model. This is quite often a smart idea, which improves generalization. It turns out we can do this “data normalization” without really touching the data points themselves, but by only tuning the kernel instead.

    Consider again the linear kernel k(x,y) = x^Ty. If we normalize the vectors before taking their inner product, we get

        \[\left(\frac{x}{|x|}\right)^T\left(\frac{y}{|y|}\right) = \frac{x^Ty}{|x||y|} = \frac{k(x,y)}{\sqrt{k(x,x)k(y,y)}}.\]

    With some reflection you will see that the latter expression would normalize the features for any kernel.

    Let us see what happens if we apply this kernel normalization to the “infinite polynomial” (i.e. exponential) kernel we just derived:

        \begin{align*} \frac{\exp(x^Ty)}{\sqrt{\exp(x^Tx)\exp(y^Ty)}} &= \frac{\exp(x^Ty)}{\exp(|x|^2/2)\exp(|y|^2/2)}  \\ &= \exp\left(-\frac{1}{2}(|x|^2+|y|^2-2x^Ty)\right) \\ &= \exp\left(-\frac{|x-y|^2}{2}\right). \end{align*}

    Voilà, the Gaussian kernel. Well, it still lacks \sigma^2 in the denominator but by now you hopefully see that adding it is equivalent to scaling the inputs by 1/\sigma

    To conclude: the Gaussian kernel is a normalized polynomial kernel of infinite degree (where feature products if i-th degree are scaled down by \sqrt{i!} before normalization). Simple, right?

    An Example

    The derivations above may look somewhat theoretic if not "magical", so let us work through a couple of numeric examples. Suppose our original vectors are one-dimensional (that is, real numbers), and let x = 1, y = 2. The value of the Gaussian kernel k(x, y) for these inputs is:

        \[k(x, y) = \exp(-0.5|1-2|^2) \approx 0.6065306597...\]

    Let us see whether we can obtain the same value as a simple dot product of normalized polynomial feature vectors of a high degree. For that, we first need to compute the corresponding unnormalized feature representation:

        \[\phi'(x) = \left(1, x, \frac{x^2}{\sqrt{2!}}, \frac{x^3}{\sqrt{3!}}, \frac{x^4}{\sqrt{4!}}, \frac{x^5}{\sqrt{5!}}\dots\right).\]

    As our inputs are rather small in magnitude, we can hope that the feature sequence quickly approaches zero, so we don't really have to work with infinite vectors. Indeed, here is how the feature sequences look like:

    ----\phi'(1) = (1, 1, 0.707, 0.408, 0.204, 0.091, 0.037, 0.014, 0.005, 0.002, 0.001, 0.000, 0.000, ...)

    ----\phi'(2) = (1, 2, 2.828, 3.266, 3.266, 2.921, 2.385, 1.803, 1.275, 0.850, 0.538, 0.324, 0.187, 0.104, 0.055, 0.029, 0.014, 0.007, 0.003, 0.002, 0.001, ...)

    Let us limit ourselves to just these first 21 features. To obtain the final Gaussian kernel feature representations \phi we just need to normalize:

    ----\phi(1) =\frac{\phi'(1)}{|\phi'(1)|} = \frac{\phi'(1)}{1.649} = (0.607, 0.607, 0.429, 0.248, 0.124, 0.055, 0.023, 0.009, 0.003, 0.001, 0.000, ...)

    ----\phi(2) =\frac{\phi'(2)}{|\phi'(2)|} = \frac{\phi'(2)}{7.389} = (0.135, 0.271, 0.383, 0.442, 0.442, 0.395, 0.323, 0.244, 0.173, 0.115, 0.073, 0.044, 0.025, 0.014, 0.008, 0.004, 0.002, 0.001, 0.000, ...)

    Finally, we compute the simple dot product of these two vectors:

        \[\scriptstyle\phi(1)^T\phi(2) = 0.607\cdot 0.135 + 0.607\cdot 0.271 + \dots = {\bf 0.6065306}602....\]

    In boldface are the decimal digits, which match the value of \exp(-0.5|1-2|^2). The discrepancy is probably more due to lack of floating-point precision rather than to our approximation.

    A 2D Example

    The one-dimensional example might have seemed somewhat too simplistic, so let us also go through a two-dimensional case. Here our unnormalized feature representation is the following:

        \begin{align*} \scriptscriptstyle\phi'(&\scriptscriptstyle x_1, x_2) = \left(1, x_1, \frac{x_1x_1}{\sqrt{2!}}, \frac{x_1x_2}{\sqrt{2!}}, \frac{x_2x_1}{\sqrt{2!}}, \frac{x_2x_2}{\sqrt{2!}}, \right. \\ &\scriptscriptstyle \left.\frac{x_1x_1x_1}{\sqrt{3!}}, \frac{x_1x_1x_2}{\sqrt{3!}}, \frac{x_1x_2x_1}{\sqrt{3!}}, \frac{x_1x_2x_2}{\sqrt{3!}}, \frac{x_2x_1x_2}{\sqrt{3!}},  \frac{x_2x_2x_1}{\sqrt{3!}}, \dots\right).\\ \end{align*}

    This looks pretty heavy, and we didn't even finish writing out the third degree products. If we wanted to continue all the way up to degree 20, we would end up with a vector with 2097151 elements!

    Note that many products are repeated, however (e.g. x_1x_2 = x_2x_1), hence these are not really all different features. Let us try to pack them more efficiently. As you'll see in a moment, this will open up a much nicer perspective on the feature vector in general.

    Basic combinatorics will tell us, that each feature of the form \frac{x_1^a x_2^b}{\sqrt{n!}} must be repeated exactly \frac{n!}{a!b!} times in our current feature vector. Thus, instead of repeating it, we could replace it with a single feature, scaled by \sqrt{\frac{n!}{a!b!}}. "Why the square root?" you might ask here. Because when combining a repeated feature we must preserve the overall vector norm. Consider a vector (v, v, v), for example. Its norm is \sqrt{v^2 + v^2 + v^2} = \sqrt{3}|v|, exactly the same as the norm of the single-element vector (\sqrt{3}v).

    As we do this scaling, each feature gets converted to a nice symmetric form:

        \[\sqrt{\frac{n!}{a!b!}}\frac{x_1^a x_2^b}{\sqrt{n!}} = \frac{x_1^a x_2^b}{\sqrt{a!b!}} = \frac{x^a}{\sqrt{a!}}\frac{x^b}{\sqrt{b!}}.\]

    This means that we can compute the 2-dimensional feature vector by first expanding each parameter into a vector of powers, like we did in the previous example, and then taking all their pairwise products. This way, if we wanted to limit ourselves with maximum degree 20, we would only have to deal with 21^2 = 231 features instead of 2097151. Nice!

    Here is a new view of the unnormalized feature vector up to degree 3:

        \[\phi'_3(x_1, x_2) = \scriptstyle\left(1, x_1, x_2, \frac{x_1^2}{\sqrt{2!}}, \frac{x_1x_2}{\sqrt{1!1!}}, \frac{x^2}{\sqrt{2!}}, \frac{x_1^3}{\sqrt{3!}}, \frac{x_1^2x_2}{\sqrt{2!1!}}, \frac{x_1x_2^2}{\sqrt{1!2!}}, \frac{x_2^3}{\sqrt{3!}}\right).\]

    Let us limit ourselves to this degree-3 example and let x = (0.7, 0.2), y = (0.1, 0.4) (if we picked larger values, we would need to expand our feature vectors to a higher degree to get a reasonable approximation of the Gaussian kernel). Now:

    ----\phi'_3(0.7, 0.2) = (1, 0.7, 0.2, 0.346, 0.140, 0.028, 0.140, 0.069, 0.020, 0.003),

    ----\phi'_3(0.1, 0.4) = (1, 0.1, 0.4, 0.007, 0.040, 0.113, 0.000, 0.003, 0.011, 0.026).

    After normalization:

    ----\phi_3(0.7, 0.2) = (0.768, 0.538, 0.154, 0.266, 0.108, 0.022, 0.108, 0.053, 0.015, 0.003),

    ----\phi_3(0.1, 0.4) = (0.919, 0.092, 0.367, 0.006, 0.037, 0.104, 0.000, 0.003, 0.010, 0.024).

    The dot product of these vectors is 0.8196\dots, what about the exact Gaussian kernel value?

        \[\exp(-0.5|x-y|^2) = 0.{\bf 81}87\dots.\]

    Close enough. Of course, this could be done for any number of dimensions, so let us conclude the post with the new observation we made:

    The features of the unnormalized d-dimensional Gaussian kernel are:

        \[\phi({\bf x})_{\bf a} = \prod_{i = 1}^d \frac{x_i^{a_i}}{\sqrt{a_i!}},\]

    where {\bf a} = (a_1, \dots, a_d) \in \mathbb{N}_0^d.

    The Gaussian kernel is just the normalized version of that, and we know that the norm to divide by is \sqrt{\exp({\bf x}^T{\bf x})}. Thus we may also state the following:

    The features of the d-dimensional Gaussian kernel are:

        \[\phi({\bf x})_{\bf a} = \exp(-0.5|{\bf x}|^2)\prod_{i = 1}^d \frac{x_i^{a_i}}{\sqrt{a_i!}},\]

    where {\bf a} = (a_1, \dots, a_d) \in \mathbb{N}_0^d.

    That's it, now you have seen the soul of the Gaussian kernel.

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