Math by Richard Li

Epsilon

rzli — 2017-08-28 15:10:53 UTC

\text{For any } x > 0 \text{ and any } 0 < \epsilon < x^2

\begin{aligned} x-\sqrt{x^2-\epsilon} & > \sqrt{x^2+\epsilon}-x \\ 2x & > \sqrt{x^2-\epsilon} + \sqrt{x^2+\epsilon} \\ 4x^2 & > x^2-\epsilon+x^2+\epsilon+2\sqrt{x^4-\epsilon^2} \\ 2x^2 & > 2\sqrt{x^4-\epsilon^2} \\ 4x^4 & > 4x^4-4\epsilon^2 \\ \epsilon^2 & > 0 \end{aligned}

\text{The last statement is trivial, Q.E.D.}

\text{For any } x > 0 \text{ and any } 0 < \epsilon < \sqrt{x}

\begin{aligned} x-(\sqrt{x}-\epsilon)^2 & < (\sqrt{x}+\epsilon)^2-x \\ 2x & < (\sqrt{x}-\epsilon)^2+(\sqrt{x}+\epsilon)^2 \\ 2x -2(x-\epsilon^2) & < (\sqrt{x}+\epsilon)^2 - 2(\sqrt{x}+\epsilon)(\sqrt{x}-\epsilon) + (\sqrt{x}-\epsilon)^2 \\ 2\epsilon^2 & < ((\sqrt{x}+\epsilon)-(\sqrt{x}-\epsilon))^2 \\ 2\epsilon^2 & < (2\epsilon)^2 \\ 2\epsilon^2 & < 4\epsilon^2 \end{aligned}

\text{The last statement is trivial, Q.E.D.}

Newton's Method

rzli — 2017-10-09 14:10:26 UTC

\text{Here is the graph of }f(x)=x^3-x^2-2x+1\text{.}

\text{I start with }{\color{blue}x_1 = 1}\text{. I take the tangent line of }f(x)\text{ at 1, and find the x-intercept of that line.}

\text{Now }{\color{red}x_2 = 0}\text{. I keep repeating this method of finding the tangent and x-intercept.}

\text{After {\color{green}three} repetitions, I estimate a zero of }f(x)\text{ to be 0.445. It turns out to be 0.444.}

\text{The general equation is } \displaystyle x_{x+1}=x-\frac{f(x)}{f'(x)}\text{.}

https://www.desmos.com/calculator/9xd3gs97sp

WHee

rzli — 2018-05-04 16:42:15 UTC

\text{5. Let }f\text{ be the function satisfying }f'(x) =4x-2f(x)\text{ for all real numbers }x\text{, with }f(0)=5\text{ and }\lim\limits_{x\to\infty}f(x)=2\text{.}

\displaystyle\text{(a) Find the value of }\int_0^\infty(4x-4xf(x))\,dx\text{.}

\displaystyle\int\limits_0^\infty f'(x)\,dx = \lim\limits_{a\to\infty}f(x)\Big\vert_0^a=\lim\limits_{a\to\infty}f(a)-f(0)=2-5=\boxed{-3}

\displaystyle\text{(b) Use Euler's method to approximate }f(-1)\text{, starting at }x=0\text{, with two steps of equal size.}

\begin{aligned} \text{at }x=0\text{: }&f'(0)=(4)(0)-(2)(0)f(0)=0\\ &y-5=0(x-0)\\ \text{at }x=-0.5\text{: }&y-5=(0)(-0.5-0)\implies y=5\\ &f'(0.5)=(4)(-0.5)-(2)(-0.5)(5)=-2+5=3\\ &y-5=3(x+0.5)\\ \text{at }x=-1\text{: }&y-5=3(-1+0.5)\implies y=\boxed{3.5} \end{aligned}

\displaystyle\text{(c) Find the particular solution }y=f(x)\text{ to the differential equation }\frac{dy}{dx}=4x-2xy\text{ with initial condition }f(0)=5\text{.}

\begin{aligned} \frac{dy}{dx}&=4x-2xy=(2x)(2-y)\\ \int\frac{1}{2-y}\,dy&=\int 2x\,dx\\ -\ln\left|2-y\right|&=x^2+C\\ \ln\left|2-y\right|&=-x^2+C\\ y&=\pm Ce^{-x^2}+2\\ \text{Initial cond}&\text{ition of }f(0)=5.\\ 5&=Ce^{-0^2}+2\\ C&=3\\ &\boxed{y=3e^{-x^2}+2} \end{aligned}