Installing OpenCV on Ubuntu was painful for me in the past. There are lots of dependencies to be sorted out first and there’s no comprehensive installation tutorial shipped with the OpenCV distribution.
This script has done a great job assembling all necessary “apt-get-install’s” into an executable script. Some personal notes about this:
- It seems to me ” sudo apt-get install libopencv-dev” for dependencies is not necessary – this will be superseded by the subsequent installation of opencv from source. At the end of the scripting installation, the package seems to be in some cache place and be not in effect:
- It seems ” sudo apt-get install libavcodec-dev libavformat-dev libswscale-dev” is also unnecessary. These libraries are to be provided by the next fresh installation of ffmpeg follows. Checking their location seems to also confirm my guess:
- About version of ffmpeg: ffmepg is under active development as always. The choice of its version is sometimes tricky, especially in running together with OpenCV. I remember the 1.x series was not compatible with OpenCV 2.3 in my previous trial (compilation errors were thrown due to some problem with ffmpeg libraries). The current 2.4.3 version seems to be at least fine with ffmpeg 0.11.2 in compilation.
[Update] From test by Mr. Alok Singh Mahor (comments below), installing Ubuntu package “libopencv-dev” would already get things work, though the linking order with g++ has to be taken care of as discussed (also comments below and here) . Of course, it’s not bad idea to install from source that always guarantees to bring in the newest feature of actively developed OpenCV. [Dec 28 2012]
.
over 
and apply an integral lower bound:![\displaystyle \begin{array}{rcl} \sum_{j=1}^k \arctan \frac{1}{j} &\geq & \int_{1}^{k+1} \arctan 1/x \; dx = \left. x\arctan 1/x + \frac{1}{2} \ln \left(1+x^2\right) \right|_{x = 1}^{k+1} \\ & = & \frac{1}{2}\ln \left[\left(k+1\right)^2 +1\right] + \left(k+1\right) \arctan \frac{1}{k+1} - \frac{\pi}{4} - \frac{1}{2} \ln 2 \\ & \geq & \frac{1}{2}\ln \left[\left(k+1\right)^2 +1\right] + \left(k+1\right) \left(\frac{1}{k+1} - \frac{1}{3} \frac{1}{\left(k+1\right)^3}\right) - 1.132 \\ & = & \frac{1}{2}\ln \left[\left(k+1\right)^2 +1\right] - 0.132 - \frac{1}{3} \frac{1}{\left(k+1\right)^2}, \end{array}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Barray%7D%7Brcl%7D+%5Csum_%7Bj%3D1%7D%5Ek+%5Carctan+%5Cfrac%7B1%7D%7Bj%7D+%26%5Cgeq+%26+%5Cint_%7B1%7D%5E%7Bk%2B1%7D+%5Carctan+1%2Fx+%5C%3B+dx+%3D+%5Cleft.+x%5Carctan+1%2Fx+%2B+%5Cfrac%7B1%7D%7B2%7D+%5Cln+%5Cleft%281%2Bx%5E2%5Cright%29+%5Cright%7C_%7Bx+%3D+1%7D%5E%7Bk%2B1%7D+%5C%5C+%26+%3D+%26+%5Cfrac%7B1%7D%7B2%7D%5Cln+%5Cleft%5B%5Cleft%28k%2B1%5Cright%29%5E2+%2B1%5Cright%5D+%2B+%5Cleft%28k%2B1%5Cright%29+%5Carctan+%5Cfrac%7B1%7D%7Bk%2B1%7D+-+%5Cfrac%7B%5Cpi%7D%7B4%7D+-+%5Cfrac%7B1%7D%7B2%7D+%5Cln+2+%5C%5C+%26+%5Cgeq+%26+%5Cfrac%7B1%7D%7B2%7D%5Cln+%5Cleft%5B%5Cleft%28k%2B1%5Cright%29%5E2+%2B1%5Cright%5D+%2B+%5Cleft%28k%2B1%5Cright%29+%5Cleft%28%5Cfrac%7B1%7D%7Bk%2B1%7D+-+%5Cfrac%7B1%7D%7B3%7D+%5Cfrac%7B1%7D%7B%5Cleft%28k%2B1%5Cright%29%5E3%7D%5Cright%29+-+1.132+%5C%5C+%26+%3D+%26+%5Cfrac%7B1%7D%7B2%7D%5Cln+%5Cleft%5B%5Cleft%28k%2B1%5Cright%29%5E2+%2B1%5Cright%5D+-+0.132+-+%5Cfrac%7B1%7D%7B3%7D+%5Cfrac%7B1%7D%7B%5Cleft%28k%2B1%5Cright%29%5E2%7D%2C+%5Cend%7Barray%7D+&bg=ffffff&fg=000000&s=0&c=20201002)
. The integral approximation gives very messy terms and a somewhat loose lower bound. We can obtain a much neater one:
as 
. Now suppose the claim holds for 
, i.e., 
, we need to show it holds for 
. It suffices to show 
. Now we consider the series expansions of 
and 
:
, and 
satisfies the condition. 
(Thanks to Dai Liang! See first comment below). Basically it follows from the fact that 
, where the latter can be observed as follows: 
and ![{\left(\arctan x\right)' \geq \left[\log \left(1+x\right)\right]'}](https://s0.wp.com/latex.php?latex=%7B%5Cleft%28%5Carctan+x%5Cright%29%27+%5Cgeq+%5Cleft%5B%5Clog+%5Cleft%281%2Bx%5Cright%29%5Cright%5D%27%7D&bg=ffffff&fg=000000&s=0&c=20201002)
for 
. (03/07/2012)
. Thus we have
; meanwhile I doubt part of the motivation of the current paper is to raise awareness of the community to the dataset)
source to Blogger compatible sources with the support of the powerful 
norm in the spectral domain, reminiscent of the intriguing relationship between the 
(quasi)norm and the 
, its nuclear norm is defined as 
, where 
denotes the 
singular value (henceforth assuming singular values sorted in descending order). Then 
is the best under-estimator (technically the convex envelope) of 
over the domain 
, where 
is the spectral norm or equivalently the largest singular value 
. The last condition on the spectral norm is in fact trivial, as one can always scale the matrix to make 
to problem 
, where 
is the singular value decomposition of 
. Moreover, 
, and unitarily matrices 
and 
(aka. 
, and similar for 
is unitarily invariant if and only if 
for any 
here denotes conjugate transpose over the complex matrix field)
norms defined as 
(assuming 
); the other is the Ky-Fan 
(assuming descending order of singular values). Clearly the nuclear norm is the Schatten 
norm (or Ky-Fan 
norm), and hence unitarily invariant. Similarly the Frobenius norm is the Schatten 
norm and hence also enjoys the nice property.
is convex if and only if its domain 
is a convex set and 
for 
and 
(the description can be considerably simplified with the extended-value extensions); it’s strictly convex if and only if the inequality always holds strictly. Now consider minimizing a convex function 
is the convex domain that’s dictated by constraints. Suppose 
, 
with convex components, 
, 
is strictly convex so long as any component function is strictly convex.
for example. The objective is obviously not strictly convex (in fact, local linearity is a typical kind of antidote for thus), but the minimizer is only the origin (it looks as if it was strictly convex in the 
vicinity of the origin). There are numerous reasons for which such unicity can arise and I won’t push towards enumerating them; instead, I’d like to note that a useful technique to prove such unicty is by perturbation analysis: for a candidate minimizer of 
, it is the unique minimizer if and only if 
where 
. This is justified since for convex functions, any local minimizer is also a global minimizer while the convexity of domain suggests such minimizers must be spatially contiguous.
, for SVD of 
as 
, where operations for 
are taken component-wise.
, for eigenvalue decomposition of 
.
as 
.
is diagonal. The last assertion comes from the facts that for any matrix 
, we have 
(generalized from Lemma 13 in 
, 
for example). So the problem to solve reduces to seeking 
(solved by simple component-wise arguments similar to that of soft-thresholding for the Lasso problem), upon which the optimal variable is 
.
for orthogonal 
), for which both directions are easy by the definition of positive semi-definiteness.
for symmetric 
), one can easily verify that 
. So the objective is equivalent to 
. 
is the SVD of 
and 
for convenience. Furthermore 
.
is reminiscent of the solution to 
is enforced through a constraint or a soft loss term. Besides though 
are strictly convex w.r.t. 
, we can only show that 
is convex w.r.t. 
, we have
,
denotes matrix inner product. Notice that for the last inequality, the equality holds only when 
, or back to our complex form, 
. On the other hand, for the inner product to norm product inequality in line 3 to 4, obviously the equality is taken only when 
for a non-zero constant 
or any taking 
. Altogether, the equality is achieved only when 
. This provides an important hint on when 
is strictly convex:
, 
, implying 
and 
. By the UIP of Schattern norms, it’s easy to see 
, and hence the objective can be written as 
with the substitution 
. For the diagonal (but non-vanishing) 
and a positive 
, it’s not hard to see that 
is minimized only when 
is also diagonal (by the pinching equality for nuclear norm as mentioned above and also property for squared Frobenius norm). The diagonal elements can be uniquely determined from the corresponding scalar optimization problems. For invertible 
. The positive semi-definiteness can be easily established as the diagonal matrix 
, i.e., is component-wise nonnegative.
(for clarity, we switch to 
for a general orthogonal matrix). The form of 
indicates that only those singular vectors corresponding to singular values that are greater than 
matter, i.e., 
, where we have partitioned the matrices according to the threshold. However we must have 
for a certain rotation matrix 
that represents the in-plane rotation of the subspace basis 
and 
. The second way goes as follows. Suppose there is another minimizer in the form 
, we must have the perturbation relationship 
. It follows
. 
(this can be found in 
and 
where 
is the SVD of 
. On the other hand, the Cauchy-Schwarz inequality for unitarily invariant norms tells 
(see, e.g., Ex IV 2.7 in Matrix Analysis by Bhatia).
Vision and around 