Given $n+1$ control points, B-spline blending functions are polynomials   of degree $d-1$, $(1&lt;d\leq n+1)$.    This much is easy to comprehend. Now comes the part I am not able to make any sense of.      Each polynomial function is defined over $d$ subintervals of the total   range of $u$. The selected set of subinterval endpoints $u_j$ is referred   to as a knot vector.   What is the purpose of knot vector? What is its physical significance?
122279
A B-spline blending curve formulation is:     $P(u)=\sum_{k=0}^np_k B_{k,d}(u)$      Given $n+1$ control points, B-spline blending functions are polynomials   of degree $d-1$, $(1&lt;d&lt;=n+1)$.    This much is easy to comprehend. Now comes the part I am not able to make any sense of.      Each polynomial function is defined over $d$ subintervals of the total   range of $u$. The selected set of subinterval endpoints $u_j$ is referred   to as a knot vector.   What is the purpose of knot vector? What is its physical significance?
114891
I need to do a polynomial interpolation of a set $N$ of experimental points; the functional form I have to use to interpolate is this:  $$ f(x) = a + bx^2 + cx^4,$$ as you can see the coefficient that I need to find are just 3: $a, b, c$; however the points I have are $\#N&gt;3$ and so it looks like the determination of the coefficients is impossible because is over-determined. Does anyone have an idea of what should be done in such case (supposing it is even possible)?
19743
I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?   In short, what is the benefit or reason to define a function this way?  At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).  Thanks,  Moshe
100412
Consider the graphs of the functions $f_1(x) = |x|$, and $f_2(x) = x$ under the subspace topology of $\mathbb{R}^2$.   Both of these graphs are smooth manifolds, just pick coordinate charts to be $(x, f_i(x)) \leftrightarrow x$.   Moreover, they are diffeomorphic via the map $(x, f_1(x)) \rightarrow (x, f_2(x))$.   This seems to clash with my intuition. For example, the graph of $f_1$ has a corner, so it "shouldn't" be smooth, much less diffeomorphic to $f_2$, which is just a straight line.   Can someone explain what's going on here?   In light of these examples, how should I visualize smooth manifolds and diffeomorphisms?
70
Let $U$ be an open subset of $\mathbf R^n$ and $V:U\to\mathbf R^n$ be a differentiable vector field on $U$. Let $\mathbf p\in U$ be a singular point of $V$, that is, $V(\mathbf p)=\mathbf 0$.     Then the only integral curve which starts at $\mathbf p$ is the constant curve.   I know that one could simply use the theorem of "uniqueness of integral curves" and in fact adapt the same proof for this particular case.  But is anybody aware of a simple argument for this?
22818
The problem:  Consider the smooth quadric $Q=V(X_{0}X_{1}+X_{2}X_{3}+X_{4}^{2})\subset\mathbb{P}^{4}$ and the line $L=V(X_{0},X_{2},X_{4})$ contained in it. Prove that there exists a vector bundle $F$ on $Q$ with a section vanishing exactly at $L$.  The hardest thing is to get the transition matrices. Any help is appreciated.
90653
Some simple vector fields have components that are just a function of their own variable. for example, $$F(x,y,z)=\begin{bmatrix}x^2+1\\\sin y\\3\end{bmatrix}$$ as opposed to: $$F(x,y)=\begin{bmatrix}5x+y^2\\4\sin x\end{bmatrix}$$ They feel somewhat special, and they are easier to work with in general. Is there a formal name for these? How would I refer to such a vector field concisely?
78476
In high school textbooks, the following characterizations are often found:     A function is continuous if its graph can be drawn without lifting the pencil.   and     A function is differentiable if its graph has no sharp bends.   Is there a similarly intuitive characterization for twice differentiable functions based on the shape of the function graph alone? Of course, we can say things like "a function is twice differentiable if the graph of its rate of change has no sharp bends" but that does not really help when trying to recognize the graph of a twice differentiable function.  A notable example for where this could be relevant is , which are widely used in design and architecture and often appear as "perfectly smooth" shapes, although they are in general only twice differentiable at the joints between the individual spline polynomials.
70261
I happened to stumble upon the following matrix: $$ A = \begin{bmatrix}     a       &amp; 1  \\     0       &amp; a  \end{bmatrix} $$  And after trying a bunch of different examples, I noticed the following remarkable pattern. If $P$ is a polynomial, then: $$ P(A)=\begin{bmatrix}     P(a)       &amp; P'(a)  \\     0       &amp; P(a) \end{bmatrix}$$  Where $P'(a)$ is the derivative evaluated at $a$.  Futhermore, I tried extending this to other matrix functions, for example the matrix exponential, and wolfram alpha tells me: $$ \exp(A)=\begin{bmatrix}     e^a       &amp; e^a  \\     0       &amp; e^a \end{bmatrix}$$ and this does in fact follow the pattern since the derivative of $e^x$ is itself!  Furthermore, I decided to look at the function $P(x)=\frac{1}{x}$. If we interpret the reciprocal of a matrix to be its inverse, then we get: $$ P(A)=\begin{bmatrix}     \frac{1}{a}       &amp; -\frac{1}{a^2}  \\     0       &amp; \frac{1}{a} \end{bmatrix}$$ And since $f'(a)=-\frac{1}{a^2}$, the pattern still holds!  After trying a couple more examples, it seems that this pattern holds whenever $P$ is any rational function.  I have two questions:   Why is this happening? Are there any other known matrix functions (which can also be applied to real numbers) for which this property holds?
88707
I'm having trouble understanding this homework problem.   Suppose four polynomials are defined by the following:   $ p_{1}(x) = x^3 - 2x^2 + x + 1 \\ p_{2}(x) = x^2 - x + 2 \\ p_{3}(x) = 2x^3 + 3x + 4 \\ p_{4}(x) = 3x^2 + 2x + 1 \\ $  Does the set $S = $ { ${p_{1}, p_{2}, p_{3}, p_{4}}$ } span $P_{3}$ (the space of all polynomials of degree at most 3)?  So if I start with the polynomial $y = ax^3 + bx^2 + cx + d$ I understand(I think) that in order for $S$ to span $P_{3}$ then $y$ must be a linear combination of $S$ but I'm not sure where to go from there.   EDIT:   $$ A =  \begin{bmatrix} 1 &amp; -2 &amp; 1 &amp; 1\\  0 &amp; 1 &amp; -1 &amp; 2\\  2 &amp; 0 &amp; 3 &amp; 4\\  0 &amp; 3 &amp; 2 &amp; 1 \end{bmatrix}  \sim  \begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0\\  0 &amp; 1 &amp; 0 &amp; 0\\  0 &amp; 0 &amp; 1 &amp; 0\\  0 &amp; 0 &amp; 0 &amp; 1 \end{bmatrix} $$  So $\operatorname{rank}(A) = 4$ which means the vectors in the set are linearly independent because there are only 4 column vectors in the matrix (AND there is only the trivial solution to the matrix) and therefore we span $P_{3}$?
80836
I'm working on an optimization problem on manifolds and I'm having a bit of a conceptual issue with choosing between the Grassmann and Stiefel manifolds. Grassmann(2, 3) is the linear subspace of dimension 2 within the space $\mathbb{R}^3$, so all planes through the origin. So a point on the manifold corresponds to a plane, invariant to linear mixing of support vectors. Stiefel(2, 3) would be all possible planes through the origin that are the span of two orthonormal vectors. So my questions are:   Does that mean that a point on the Stiefel manifold is also a point on the Grassmann manifold (in the sense that all linear subspaces of a given span can be reduced to the same orthonormal components)?  What do I gain by optimizing on the Grassmann vs the Stiefel manifold, if the parametrization of the points is the same?
4930
Suppose that $P = (1,1,7)$ and $Q = (8,6,1)$. Inside parenthesis are x-coordinate value, y-coordinate and z-coordinate. The question is to find vector and parametric equation for a line segment.   Now, I used equation that goes like $(1-t)&lt;1,1,7&gt;+t&lt;8,6,1&gt;, where, 0\leq t\leq 1$.   But, the answer says that I am wrong. So, I am wondering what is going on.   Also, why do we write parametric equations? Are they really different from vector functions anyway?   If vector function is $&lt;2t,3t,1&gt;$, according to my knowledge, parametric equations would be $x(t)=2t, y(t)=3t, z(t)=1$ ....
91168
What would the graph of $(-1)^{x}$ look like?   I know that the value of the function alternates between $1$ and $-1$ when it is defined so I think it would just be points spread over the lines $x=1$ and $x=-1$.   Is this correct? Also, will there be any definite pattern of the points.
93533
I have a function $y=sin(x)$  I sample the points $ x_i=\{0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1\}$ and $$ y_i=sin(x_i) + \epsilon \  \ \epsilon \sim iid\ Normal$$  Now given only the set $\{x_i,y_i\}$ I want to uncover the underlying functional form $f(x)$ given as  $$y=f(x)+\epsilon \ \ \ \ \epsilon \sim iid\ Normal  $$  My question is : ARE $y_i$'s independent ?  My intuition is that $y_1$ and $y_2$ are not independent. This is because given a new ${x}$ say ${x_\star}$ we can use $y_i$'s to learn some information about ${y_\star}$.   Does the fact that the input points come from the same underlying function make them somehow correlated or posses commonalities ? and if not why do most methods (ex: Non-parametric splines, Gaussian process) use the $y_i$'s to get information and learn ${y_\star}$ ? Isn't the fact that if samples are independent then they can't learn new information from each other ?
84909
Given $n+1$ points in the plane, Lagrangian interpolation gives the unique polynomial of degree less or equal to $n$.  From the examples I've seen so far, interpolation polynomials all have degree exactly $n$ but I don't know if there might be a set of points for which the interpolation gives a polynomial of degree strictly less than $n$.  Many thanks for your answers.
95590
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a smooth function such that for every $x \in \mathbb{R}$ there exists $n$ such that $f^{(n)}(x) = 0$, then f is a polynomial.   I'm kind of lost on this one. I know that I have to use Baire's category theorem somewhere here, but I'm not sure exactly how.
104951
In , there's this line under the section describing Smoothing Spline's tuning parameter $\lambda$:       In fitting a smoothing spline, we do not need to select the number or   location of knots - there will be a knot at each training observation,   $x_1,...,x_n$.   But, in my opinion wouldn't that be overfitting?     Reason for such a claim: If there are more points which act as knots in the Spline fit, then the range for each slot for fitting is very small. Thus each change in the data, is recognized and imititated in the fit. Thus fitting the changes, which are random, and do not mean anything. Thus overfitting.    Possibilities: We know that cubic splines and Smoothing splines are different. The basic differences are:     Smoothing splines try to minimize the function $$\sum_{i=1}^n(y_i-g(x_i))^2+\lambda\int g''(t)^2dt$$ Smoothing splines shrink the parameters.   Moreover, Smoothing Splines are basically natural cubic splines, and thus they're smooth too.  So, how can these differences( and properties) save the Smoothing Spline from overfitting? Is there anything else that I'm missing here?