Taylor Polynomials from Orthogonal Projections
One of the standard applications of orthogonal projections to function spaces comes in the form of Fourier series, where we use 
In particular, projecting a function onto the span of the first 2n+1 functions in the set 
where the coefficients 
A natural follow-up question is then whether or not we similarly get Taylor polynomials when we project a function down onto the span of the set of functions 
![\mathcal{C}[-1,1]](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BC%7D%5B-1%2C1%5D&bg=ffffff&fg=000&s=0&c=20201002)
and consider the orthogonal projection ![P : \mathcal{C}[-1,1] \rightarrow \mathcal{C}[-1,1]](https://s0.wp.com/latex.php?latex=P+%3A+%5Cmathcal%7BC%7D%5B-1%2C1%5D+%5Crightarrow+%5Cmathcal%7BC%7D%5B-1%2C1%5D&bg=ffffff&fg=000&s=0&c=20201002)
![\mathcal{P}_n[-1,1]](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BP%7D_n%5B-1%2C1%5D&bg=ffffff&fg=000&s=0&c=20201002)
It does not take long to work through an example to see that, no, we do not get Taylor polynomials when we do this. For example, if we choose n = 2 then projecting the function ![\mathcal{P}_2[-1,1]](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BP%7D_2%5B-1%2C1%5D&bg=ffffff&fg=000&s=0&c=20201002)
These plots illustrate why the orthogonal projection of
However, something interesting happens if we change the interval that the orthogonal projection acts on. In particular, if we let 
![P_c : \mathcal{C}[-c,c] \rightarrow \mathcal{C}[-c,c]](https://s0.wp.com/latex.php?latex=P_c+%3A+%5Cmathcal%7BC%7D%5B-c%2Cc%5D+%5Crightarrow+%5Cmathcal%7BC%7D%5B-c%2Cc%5D&bg=ffffff&fg=000&s=0&c=20201002)
![\mathcal{P}_2[-c,c]](https://s0.wp.com/latex.php?latex=%5Cmathcal%7BP%7D_2%5B-c%2Cc%5D&bg=ffffff&fg=000&s=0&c=20201002)
While this by itself is an ugly mess, something interesting happens if we take the limit as c approaches 0:
which is exactly the Taylor polynomial of 
![[-c,c]](https://s0.wp.com/latex.php?latex=%5B-c%2Cc%5D&bg=ffffff&fg=000&s=0&c=20201002)
Here is an interactive Desmos plot that you can play around with to see the orthogonal projection approach the Taylor polynomial as 
.
.
is the positive real number such that 
(and we take 
if 
, and vice-versa). The idea is then to maximize over 
and 
one at a time, alternately.
and fixing a randomly-chosen vector 
, scaled so that 
.
fixed, and let 
be the vector attaining this maximum. By 
. Furthermore, the equality condition of Hölder’s inequality tells us that the vector 
, and whose magnitudes are such that 
is a multiple of 
(here the notation 
means we take the absolute value and the q-th power of every entry of the vector).
be the vector attaining this maximum. By an argument almost identical to that of step 2, this maximum is equal to 
, where 
is the positive real number such that 
. Furthermore, the vector 
, and whose magnitudes are such that 
is a multiple of 
.
by 1 and return to step 2. Repeat until negligible gains are made after each iteration.
.
) as well. First, for a matrix 
, we define its 
only acts on the diagonal entries of 
and just zeros out the off-diagonal entries. However, it seems likely that this norm’s computation is also difficult even in the special cases p = q = 1 and p = q = ∞ (however, it is straightforward to compute when p = q = 2).
, one at a time, alternately, just as before:
, scaled so that 
.
fixed, and let 
be the matrix attaining this maximum. By the 
. Furthermore, the matrix 
, and whose singular values are such that there is a constant 
so that 
for all 
(i.e., the vector of singular values of 
power).
be the matrix attaining this maximum. By essentially the same argument as in step 2, we know that this maximum value is exactly equal to 
, where 
is the map that is dual to 
, and whose singular values are such that there is a constant 
for all 
that comes up from time to time when dealing with quantum entanglement:
and 
and 
, but perhaps not its exact value. If you want to take this algorithm out for a spin, it has been implemented in QETLAB as the 
where we maximize on the left and right by pure states with 
and 
are 
-valued measurement settings, then the following inequality holds in classical physics (where 
denotes expectation):
(this is 
.
are vector spaces then
does hold, so it’s very tempting to think “inclusion-exclusion, yadda yadda, it’s simple enough to prove that it’s not worth writing down or working through the details”. However, it’s not true: a counterexample is provided by 3 distinct lines through the origin in 
.
for all n, and I am happy to announce that Ben Chaffin has proved this conjecture when n = 5. More specifically, he showed that minimal superpermutations in the n = 5 case have length 153, and there are exactly 8 such superpermutations (previously, it was only known that minimal superpermutations have either length 152 or 153 in this case, and there are at least 2 superpermutations of length 153).
can be written in the form
is a real scalar and the sets 
and 
form orthonormal bases of 
.
‘s are singular values of some matrix and the sets 
: it is entangled if and only if it has more than one non-zero 
is a mixed quantum state then it can be written in its operator-Schmidt decomposition:
and 
form orthonormal bases of Hermitian matrices in 
(under the 
).
is separable or entangled.
in the operator-Schmidt decomposition of 
then 
. To see that 
is indeed an entanglement witness, we must show that 
for all 
and 
. Well, some algebra shows that
. To see this notice that
(and similarly for 
).
). This is straightforward to show by making use of the fact that the sets 
by 
and extending by linearity, and let 
denote the trace norm (i.e., the sum of the singular values), then 
implies that 
are exactly the coefficients 
be a symmetric gauge function (i.e., a norm that is invariant under permutations and sign changes of the 
entries of the input vector). If we can conclude that 
then it must be the case that 
so that 
. We first prove two facts about 
for all mixed states 
(which itself is kind of a pain to prove: it follows from the fact that the 
Schatten norm of the realignment map 
, but if anyone knows of a more direct and/or simpler way to prove this, I’d love to see it). If we assume without loss of generality that 
then
![[\frac{1}{d},1]](https://s0.wp.com/latex.php?latex=%5B%5Cfrac%7B1%7D%7Bd%7D%2C1%5D&bg=ffffff&fg=000&s=0&c=20201002)
. To see why this is the case, first notice that there exists a separable state 
and 
for all 
: the state 
is one such example. Similarly, there is a separable state with 
and 
is one such example. Furthermore, it is straightforward to interpolate between these two extremes to find separable states (even 
for all ![\gamma_1 \in[\frac{1}{d},1]](https://s0.wp.com/latex.php?latex=%5Cgamma_1+%5Cin%5B%5Cfrac%7B1%7D%7Bd%7D%2C1%5D&bg=ffffff&fg=000&s=0&c=20201002)
. For such states we have
. However, in this case we have
with the property that we can use the value of 
to conclude that 
then 
for some 
. Since 
and 
and then we are done.
are also separable, as was shown in [6].
for some 
. Throughout this proof, we use the notation 
, and so on.
and 
are both positive semidefinite operators. It is then straightforward to verify that
for some 
(in order to do this, we actually need the partial traces of 
for all eigenvalues 
of 
and 
of 
. If we absorb scalars between 
then this implies that 
for all 
and 
are both positive semidefinite. Furthermore, a straightforward calculation shows that
for some 
(similar to before, we note that there is a standard continuity argument that can be used to handle the fact that 
, which is non-negative as a result of 
has operator-Schmidt rank 
open. Very recently, it was shown in [7] that a mixed state in 
with operator-Schmidt rank 
.
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