The only thing that is clear (based on adjusting the coefficients in the desmos plot) is that the shift forms a continuous function and is very sensitive to where the root initially is, but I will need to check in more detail to determine what that function look like
@Semiclassical I was asked to use the equioscillation theorem to show that the best approximation of an even function on $[-d,d]$ by a degree-$n$ polynomial w.r.t. the $L^\infty$ norm is also even
i.e. let $f \in C^0[-d,d]$ even and $p_n^\ast$ be a deg <= n polynomial that best approximates $f$ in the sense that $\|f-p_n^\ast\|_\infty$ is minimized
"Let $f:[a,b]\to \mathbb{R}$ be a continuous function . Among all the polynomials of degree at most $n$, the polynomial $g$ minimizes $\|f-g\|_{\infty }$ if and only if there are $n+2$ points $a\leq x_{0}<x_{1}<\cdots <x_{n+1}\leq b$ such that $f(x_{i})-g(x_{i})=\sigma (-1)^{i}\|f-g\|_{\infty }$ where $\sigma =\pm 1$."
just to have a common reference (that's from the wiki page on equioscillation)
So evidently you want to show that, if those conditions are fulfilled and $f$ is even, then $g$ must also be even.
If I have two polynomials completely factored into linear polynomials over some field, what is their gcd? Isn't it the product of the common factors raised to the larger power?
@LeakyNun Ah, okay. Thanks. In my case I actually have the polynomials factored into linear polynomials, but thanks for informing me about the general case.
Note that you can't run this backwards. If you've got n+2 points in [0,d], then you automatically get 2n+2 points on [-d,d] which is certainly sufficient
But if you had n+2 points in [-d,d], then you'd presumably lose some in going to [0,d]
i.e. can a function have two best-approximations on a given interval?
my guess is that, if you do that, you end up having two polynomials which are equal in enough places (relative to their degrees) that they have to be identical
i.e. the difference between the two polynomials would itself be a degree $\leq n$ polynomial, and this difference would have more than $n$ zeros
So I think it all hangs together.
actually, something seems incomplete here
Suppose $p_n^\ast$ on $[0,d]$ were an odd polynomial
Then there's no way to extend that to an even polynomial on [-d,d]
a man stands on the bank of a frozen river. he has to reach the opposite point on the river in the shortest time possible.The river has friction coefficient 0, and the grass on both sides of the frozen river has friction coefficient $\mu$. What will be the minimum time required for the man to cross the river
my guess would be that you'd walk along the river for some distance, then walk in a straight line towards the other point. in that case the arc length is just pythagorean
How can we fix a? that is basically the distance along the river that the man finally travels, and that will depend on the curve f(x) that we choose right?
once you've walked along the river, you're just trying to find the path which gets you from point a to point b in the fastest time while going at a constant velocity
that's just a matter of finding the shortest path. that's a straight line
I am looking for advice on what would be a reasonable or useful generalization of vertex- and edge-connectivity to the graphs with 0 and 1 vertices (null graph and singleton graph).
Motivation: Creating a software package that computes these quantities, as well as checks connectivity and biconne...
Suppose we have the following: All the complete graphs are connected. There are simple graphs that are not connected. Therefore, there are simple graphs that are not complete.
I know that the propositions are: $$\begin{array}{l} p(x)=\text{A graph is complete}.\\ q(x)=\text{A graph is connected}.\\ r(x)=\text{A graph is simple}. \end{array}$$
And the reasoning is as follows: $$\begin{array}{cc} \forall x:&p(x)\implies q(x)\\ \exists x:&r(x)\wedge\neg q(x)\\\hline \exists x:&r(x)\wedge\neg p(x) \end{array}$$
My question is, is it correct to say that when using $\exists$, an implicator should NEVER appear, but should ALWAYS be a conjunction? Because it has already happened to me several times that when I work with the existential and the sentence seems to use an implicator, in reality the premise is with a conjunction.
In the example, why $\exists x(r(x)\wedge\neg q(x))$ and not $\exists x(r(x)\implies\neg q(x))$?
@TedShifrin but read the first sentence: "All the complete graphs are connected". It says are and an implicator is used. Read the second one: " There are simple graphs that are not connected". It says are, but in this case a conjunction is used. Why?
@manooooh: It has nothing to do with the quantifier in general. Certain conjunctive sentences can be rewritten as implications; certain other ones cannot.
It really is a question about what the statement itself is.
It really is a question, intuitively, of causality, which is what an implication entails. "If it rains, then the ground is wet." Well, I could say, "the ground is wet and it is raining." That's a different statement.
It could be that the ground is wet because someone spilled water there, nothing to do with rain.
Some guests are engineers. Some engineers give classes in the faculty. Therefore, some guests teach at the faculty.
We have: $$\begin{array}{l} p(x)=\text{A person is a guest}.\\ q(x)=\text{A person is an engineer}.\\ r(x)=\text{An engineer teaches in the faculty}. \end{array}$$
And the reasoning is: $$\begin{array}{cc} \exists x:&p(x)\wedge q(x)\\ \exists x:&q(x)\wedge r(x)\\\hline \exists x:&p(x)\wedge r(x).\end{array}$$
See: "Some guests are engineers" uses "are" but in the formalization you use a conjunction, meanwhile in the other example we used $\Longrightarrow$. Why?
Typically "$\implies$" requires "if ... then" (i.e., some sort of causality, as I keep saying). You could say that if you're an engineer, then you teach classes. Then the "some" still would apply. It's a matter of the English (or whatever language) phrasing.
But you could say it that way, as I just indicated. It is a matter of what logical statement you're trying to make. That is the context. It's not about rules.
In "All the complete graphs are connected" we use an implicator, but in "Some guests are engineers" we use a conjunction. Both have the same structure, but differ in the quantifier.
I think you're looking for absolute rules and there aren't any.
No, there is no one correct way. It's a matter of what the sentence is interpreted to mean. You're making too big a deal. You're trying to have a rule.
@TedShifrin What I am seeing is that for two same structures that differ in a quantifier the logical operator is changed completely. Why does that happen? And I answer "Well for the use of the quantifier". It is right?
@TedShifrin but!!! You do not see that "All the complete graphs are connected" and "Some guests are engineers" have the same structure of sentence, differing in a quantifier?
Then, like a teacher, when a student makes a mistake you must say why it is wrong. If I formalize "Some guests are engineers" as $\text{If a person is a guest then is an engineer}$ and it's wrong, why is it wrong?
I agree that "all" suggests that it becomes an "if then" rule. But that might be false.
Every day, if the ground is wet, then it has been raining. That's a false statement, because it's phrased with "every." Whereas, every day, if it is raining, then the ground is wet seems to be a true statement. They are both implications. If I want to make the first statement true, I have to change the logical structure of the statement.
@Semiclassical Well... so please tell me why in one sentence we use "for all" with an implicator, and in other sentence with the same structure differing by a quantifier, we change it to "and"
What you can try is to look at the power series: Series[Integrate[Log[ArcSin[Sinh[x]]], x], {x, 0, 12}] $$x (\log (x)-1)+\frac{x^3}{9}+\frac{x^5}{45}+\frac{233 x^7}{19845}+\frac{194 x^9}{25515}+\frac{1406 x^{11}}{245025}+O\left(x^{13}\right)$$
