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3

Notice that a linear transformation $T$ preserves the norm if and only if it preserves the dot product, we can prove this simply using $$\langle T(u+v),T(u+v)\rangle=\langle u+v,u+v\rangle$$ moreover, the nice characterization by the matrix is: $$A^TA=I_n$$ so we see that $\det T=\pm1$ hence $T$ is invertible.

2

$p(x) = 8 + 44i + x^2(-(1+3i) + x^2(1+2i + x))$ Now, let $x=-2-i$, thus, $x^2 = 4-1-4i = 3+4i$ \begin{align}p(x) &= 8 +44i + (3+4i)(-(1+3i) + (3+4i)(-1+i)))\\ &=8+44i + (3+4i)(-(1+3i) -3+3i-4i-4))\\ &=8+44i +(3+4i)(-1-3i-3+3i-4i-4)\\ &=8+44i + (3+4i)(-8 - 4i)\\ &=8+44i -24 -12i -32i +16\\ &=0 \end{align} Thus $-2-i$ is a root.

1

A simple Monte Carlo approach to estimate the value of $\pi$ is to generate random numbers $(x_n)$ with uniform distribution in $[0,1]$ and to compute four times the proportion of pairs $(x_{2n-1},x_{2n})$ such that $x_{2n-1}^2+x_{2n}^2\leqslant1$. The error when using $2n$ points is $\Theta(1/\sqrt{n})$.

1

Call the cubic B-spline centered at zero $β_3$. Then a linear combination of B-splines is $$f(t)=\sum_{i=1}^nc_i\,β_3(x-i)$$ On any interval $x\in[i,i+1]$, the value of $f$ is given by the polynomial basis functions discussed in http://math.stackexchange.com/a/700183/115115 as $$f(t)=B_0(x-i)c_{i-i}+B_1(x-i)c_{i}+B_2(x-i)c_{i+1}+B_3(x-i)c_{i+2}.$$ One ...

1

In general, a b-spline curve will not pass through any of its control points. There is an example at the bottom of this web page, which explains how repeating knot values will cause a b-spline curve to pass through one of its control points. This technique is typically used with the first and last knots, to force the spline to pass through the first and last ...

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Let's think about what we want: We want a set of polynomials that are $1$ on one $x$-value and $0$ on all the other $x$-values. So call the set of $x$-values $\{x_i\}_{i=1}^n$ and let's write the set of polynomials, for $k \in [1,n]$, $$L_k(x) = \prod_{i=1}^n {}' \frac{x-x_i}{x_k-x_i}$$ where the "prime" means to skip $k$ in the product (since otherwise ...

1

It differs in two key ways. Suppose the function is not differentiable. For multivariate functions, the derivative describes a multivariate quantity, but we want to intuit this to a scalar "amount." Consider the following example. Suppose we're given a system of equations: \begin{align*} ax + by &= 0, \\ cx+dy &= 0.\end{align*} Defining ...

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