norm – Quadratic matrix bounds

Let A be a singular matrix with a simple (non-repeated) zero-eigenvalue.
Dose the following inequality hold?

$$|Ax|^2geqsigma_2|x|^2, qquad forall xnotin Null(A)$$

where $sigma_2$ is the smallest nonzero singular value of the matrix $A$. If it is true, where can I find a proof??

algebra precalculus – Application quadratic functions help!

Kevin wants to fence a rectangular garden using 14 rails of
8-foot rail, which cannot cut. What are the dimensions of the rectangle that will maximize the fenced area?

So the number of rails in each dimension the rectangle could be either $2$ in the width by $5$ in the length (area $2(8)*5(8)=640ft^2$) or, $3$ in the width by $4$ in the length (area $3(8)*4(8)=1,152ft^2$), $1$ in the width and $6$ in the length (area $1(8)*6(8)=384ft^2$).

Therefore the dimensions that maximize the fenced area are 24 feet by 32 feet. But I did that by trial and error. Not using a quadratic function.

This is what I did using quadratic function.

Let be $w$ the number of rails in the width and $l$ the number of rails in the length.

The perimeter would be:


The area would be:

$$A=wcdot l$$
$$A=(7-l)cdot l$$

$$lapprox 4$$

Shall I round this to $4$? Since I can’t cut the rail?


So the dimension of the length is $4(8)=32ft$ and the width is $3(8)=24ft$

Is this right?

number theory – On quadratic form $L^2+27M^2$

If $4q=L^2+27M^2$, where $qequiv 1 (mod 3)$ is a power of $p$. How can I show that there is only one pair of integers (L,M) satisfying the condition where $Lequiv 1 (mod 3)$ and M up to sign.

I have proved by elementary number theory that there is at most one pair of POSITIVE integers. But what if $4q=L’^2=L^2+27M^2$, where $Lequiv L’equiv 1$?

stochastic processes – Applyin Itô’s formula in function of quadratic variation

I am learning some basic stochastic calculus and came across the following exercise:

Consider a local martingale $M$ with continuous trajectories. Let $Z_t = exp(M_t −0.5(M)_t)$. Show that Z satisfies the equation $$Z_t = Z_0 +int_0^tZ_sdMs$$ Is $Z$ a local martingale? Compute $Z$ for the case where $M_t = σB_t$ for a standard Brownian motion $B$.

What confuses me, is the presence of quadratic variation in the function of a martingale. I’ve tried to apply standard Ito’s formula:

$f(X) = f(X_0)+int f^prime(X),dX + frac{1}{2}int f^{primeprime}(X),d(X)$

But I am not sure how to treat the $(M)_t$ element in the function. Could someone help? Thanks!

systems of equations – Numerically find the half-iterate of a quadratic

Let’s say we have a function $f(x)=n+mx+lx^2$, and want to find another function $h(x) = a+bx+cx^2 $ such that $h(h(x)) = f(x)$. Here’s what you get when composing h on itself:

$$h(h(x)) = a+ab+b^{2}x+bcx^{2}+a^{2}c+2abcx+2ac^{2}x^{2}+b^{2}cx^{2}+2bc^{2}x^{3}+c^{3}x^{4}$$

If we ignore the $x^3$ and onwards terms and gather the constants, $x$ terms, and $x^2$ terms, we can get a system of three equations which ignores enough terms to not be overly specified, but including enough terms to be a half-decent approximation:

$$a+ab+a^{2}c = n,$$
$$b^{2} + 2abc = m,$$
$$bc+2ac^{2}+b^{2}c = l$$

My original plan was to get $c$ isolated in each of these like so:
$$c = frac{(n – a (b + 1))}{a^2},$$
$$c = frac{(m – b^2)}{2 a b},$$
$$c = frac{l}{(b^2 + b)}$$

This way I could treat each of these as 3D functions and find their intersection (much like how you can solve a system of 2 equations by treating the equations as 2D functions). Using the identity that $2x-y-z=0$ if $x=y=z$, I could reduce the intersection problem further down into one equation (which can also be treated as a 3D function) which I need to find the zero/root of:
$$2cdot{frac{(n – a (b + 1))}{a^2}} -frac{(m – b^2)}{2 a b} – frac{l}{(b^2 + b)} = 0$$

The problem is, I don’t know of any root-finding algorithms supporting 2-argument functions, since for example, Netwon’s method requires a derivative which we can’t do here. So, is my approach wrong, is there a root-finding method I don’t know of, or is there no solution?

quadratic forms – Do we include operators before the coefficient when applying algebraic formulas to equations?

I do some simple retraining on college algebra after a long time, but I have come to find some inconsistencies in my understanding of how to apply the formulas, do we include operators before the coefficient when applying algebraic formulas to equations?
I fight in particular with these two formulas:

Quadratic formula, operators included here:
2x²+14x-196 = 0, when applied to the formula (-b±√(b²- 4ac))/2a, gives (-14±√((14)²-4(2)(-196)))/2(2). The answer here is x = -14, x = 7 checked.

Note that the coefficient "c" must take the negative operator to give the answer.

If we are consistent and say yes, operators must be included in formulas:
When using the formula (what is this algebraic formula called btw?) (a-b)² = a²-2ab+b², when applied to (2-3)², gives: 2²-2(2)(-3)+(-3)² = 4+12+9=25. Which is clearly the wrong answer.

I'm struggling with this inconsistency, can anyone help me?

Find the nearest quadratic Bézier curve

Given a set of quadratic Bézier curves in three dimensions.

I am looking for an analytical solution to find the curve closest to an arbitrary point in space.


I already have a brute force solution with the calculation of the nearest point on each curve and selecting one with the minimum distance. But this algorithm requires a lot of computing resources.

quadratic – Find the values ​​of m that x ^ 2 – 4mx +20 = 0: a) no solutions and b) 2 solutions

I understand how to solve this problem, but my method / logic seems to be a roundabout way of doing it. Advice on how to resolve this issue faster / easier would be appreciated 🙂

Solve for part a) no solution:

Find the discriminant: 16m ^ 2-80

For no solution: 16m ^ 2 – 80 <0

When 16m ^ 2 -80 = 0, m = 5, -5. The interval between 5 and -5 is either below the x-axis (less than 0, i.e. no solutions) or above the x-axis x (more than 0, 2 solutions).

Substituting a random number (0) between 5 and -5 for the discriminant gives -80 (less than 0) so that we can confirm that -5 <m <5 is the range of m required to give no solution.

For part b) I apply the same logic, so 2 solutions obtained when m < -root5 and m > root 5

This is from the Cambridge Methods Unit 1 manual.

Quadratic and double hash sounding

What is the time complexity of quadratic probing and double hashing in the hash table?

Signature of a quadratic form?

I think it's pretty clear that if we have $ f (x) $ = $ x ^ 2 $ , then the function is defined positive, as for each value other than 0, $ x ^ 2 $ assumes a positive value. However, if I have
$ f (x, y, z, t) $ = $ x ^ 2 $ , is it always positive defined or is it only positive? Because if I calculate the associated matrix, I have three eigenvalues ​​= 0, and the manual says that a matrix is ​​definitely positive only if all its eigenvalues ​​are positive. But what is the difference? According to the definition, a function is definitely positive if, regardless of the input outside of 0, the output is positive. Isn't that always the case?
Thank you