Rom

half life?

\(y = A e^{-2x}+Be^{-3x} + 3x^2 - 4x\\ \dfrac{dy}{dx} = -2A e^{-2x} -3Be^{-3x} + 6x-4\\ \left . \dfrac{dy}{dx}\right|_{x=0} = -2A -3B - 4 = 7\\ y(0) = A + B = 4\\ -2A -3B = 11\\ A+B = 4\\ \text{I leave you to solve for $A,~B$}\)

In this problem the elementary matrix would be

\(e=\begin{pmatrix}1&0&0&0\\0&1&0&0\\2&0&-1&0\\0&0&0&1\end{pmatrix}\)

\(f(f(x)) = x\\ f(x) = f^{-1}(x)\\~\\ y = \dfrac{ax+b}{cx+d}\\ cxy+dy = ax + b\\ (cy-a)x = b-dy\\ x = \dfrac{b-dy}{cy-a}\\ f^{-1}(x) = \dfrac{b-dx}{cx-a}\)

\(\dfrac{ax+b}{cx+d} = \dfrac{b-dx}{cx-a}\\ acx^2 +(bc-a^2)x-ab = -cd x^2 +(b c - d^2)x + bd\\ ac = -cd\\ bc-a^2 = bc - d^2\\ -ab = bd\\ a = -d\\~\\ a+d = 0\)

\(\text{Subtract 3 times the first equation from the second}\\ -16y = 48\\ y=-3\\ x+3(-3)=-4\\ x = 5\\~\\ (x,y)=(5,-3)\)

here

whoever is teaching this notation needs to be flogged.

whatever happened to good old elementary matrices?

The problem makes sense.  It's easy enough to visualize a sequence of length 2018 whose elements are in {1,2,3,4,5,6,10},

and whose terms sum to 3860.  It's an integer partition problem.

How sensible the bound on the cardinality of the set of such sequences is will take some figuring.

\(\text{Suppose $c\neq 0$}\\ y = f(c x)\\ x = \dfrac{f^{-1}(y)}{c}\\ \text{In other words for all values of $c\neq 0, g(x)$ is invertible}\\ \text{The sum of $0$ is $0$}\)

\((2-i)z+(2+i)\bar{z} = 20\\ z = x + i y\\ (2-i)(x+iy) + (2+i)(x-i y) = 20\\ (2x+y) - i(x-2y)+(2x+y)+i(x-2y)=20\\ 2x+y=10\\ y=-2x+10\\ y=0 \Rightarrow x=5\\ \text{Intersection is at $z=5$}\)