User:Jasonw56k

Jason Wan's math page.

For Kevin
$$\begin{align} &ln (1-10x)^{1/x}=\frac{ln(1-10x)}{x}\\ &\lim_{x \to 0} \frac{ln(1-10x)}{x}=\lim_{x \to 0} \frac{\frac{d}{dx} ln (1-10x)}{\frac{dx}{dx}}= - \frac{10}{1-10x} \mbox{(By LHospitals rule)}\\ &\lim_{x \to 0} (1-10x)^{1/x} = \lim_{x \to 0} e^{ln (1-10x)^{1/x}} = \lim_{x \to 0} e^{- \frac{10}{1-10x}} = e^{-10} = \frac {1}{e^{10}}\\ \end{align}$$

$$\begin{align} &f(x)=\frac{2+x^2}{1+x^2}=\frac{1+1+x^2}{1+x^2}=\frac{1}{1+x^2}+\frac{1+x^2}{1+x^2}=\frac{1}{1+x^2}+1\\ &\int f(x) dx=\int \frac{1}{1+x^2} dx +\int 1 dx=\tan^{-1}x+x+C\\ \end{align}$$

$$\begin{align} &\mbox{first problem:}\\ &\mbox{By product rule,}\\ &f'(x)\\&=g(x) \frac{d}{dx} x^2 \cdot x^2 g'(x)\\ &=2xg(x)+x^2g'(x)\\\\ &\mbox{balloon problem:}\\ &\frac{dS}{dr}=4\pi\cdot2r =8\pi r\\ &\frac{dS}{dr} \bigg | _{r=1}=8\pi~\mbox{ft}^2 / \mbox{ft}\\\\ &\mbox{Last problem:}\\ &\mbox{Let u = } sin^{-1} (2x) \mbox{, let v = } 2x \mbox{, by the chain rule:}\\ &y'=\frac{d}{du} u^2 \cdot \frac{d}{dv} sin^{-1} v \cdot \frac{d}{dx} 2x\\ &=2u \cdot \frac{1}{\sqrt{1-v^2}} \cdot 2\\ &=2(sin^{-1} (2x)) \cdot \frac{1}{\sqrt{1-(2x)^2}} \cdot 2\\ &=\frac{4sin^{-1}(2x)}{\sqrt{1-4x^2}}\\ \end{align}$$

For Sharon
$$\begin{align} &\frac{2^{n+1}}{2^{n-1}}\\ &=\frac{2^n\cdot2}{\frac{2^n}{2}}\\ &=\frac{2^n\cdot2\cdot2}{2^n}\\ &=2\cdot2\\ &=4 \end{align}$$

$$\begin{align} &a^{m+n}\\ &=a_1\cdot a_2\cdot a_3\cdot ...\cdot a_{m+n}\\ &=(a_1\cdot a_2\cdot a_3\cdot ...\cdot a_m)\cdot(a_1\cdot a_2\cdot a_3\cdot ...\cdot a_n)\\ &=a^m\cdot a^n \end{align}$$

For 創辦人
$$\begin{align} &\mbox{p. 163 2.5(a)(i)}\\ &\mbox{Let }S_m=\sum_{k=1}^nk^m\mbox{ where }m,n\in\mathbb{N}\\ &\mbox{Prove that, for all }m\in\mathbb{N}\\ &S_m=\frac{1}{m+1}\left[(n+1)^{m+1}-1-\sum_{r=0}^{m-1}C_r^{m+1}S_r\right]\\\\ &\frac{1}{m+1}\left[(n+1)^{m+1}-1-\sum_{r=0}^{m-1}C_r^{m+1}S_r\right]\\ &=\frac{1}{m+1}\left[(n+1)^{m+1}-1-\sum_{r=0}^{m+1}C_r^{m+1}S_r+(m+1)S_m+S_{m+1}\right]\\ &=\frac{1}{m+1}\left[(n+1)^{m+1}-1-\sum_{k=1}^{n}\sum_{r=0}^{m+1}C_r^{m+1}k^r+(m+1)S_m+S_{m+1}\right]\\ &=\frac{1}{m+1}\left[(n+1)^{m+1}-1-\sum_{r=0}^{m+1}C_r^{m+1}n^r-\sum_{k=1}^{n-1}\sum_{r=0}^{m+1}C_r^{m+1}k^r+(m+1)S_m+S_{m+1}\right]\\ &=\frac{1}{m+1}\left[(n+1)^{m+1}-1-(n+1)^{m+1}-\sum_{k=1}^{n-1}(k+1)^{m+1}+(m+1)S_m+S_{m+1}\right]\\ &=\frac{1}{m+1}\left[-1^{m+1}-\sum_{k=1}^{n-1}(k+1)^{m+1}+(m+1)S_m+S_{m+1}\right]\\ &=\frac{1}{m+1}\left[-1^{m+1}-\sum_{k=2}^{n}k^{m+1}+(m+1)S_m+S_{m+1}\right]\\ &=\frac{1}{m+1}[(m+1)S_m]\\ &=S_m\\ \end{align}$$

$$\begin{align} &\mbox{p. 163 2.6(a)}\\ &\mbox{Let }m\mbox{ be a non-negative integer and for each }n\in\mathbb{N}\\ &S_m(n)=\sum_{k=1}^{n}k^m\\ &\mbox{Prove that }S_{m+1}(n)=(n+1)S_m(n)-\sum_{r=1}^nS_m(r)\\\\

&S_{m+1}(n)=\sum_{k=1}^{n}k(k^{m})\\ &=\sum_{k=1}^{n}(n+1)(k^{m})-\sum_{k=1}^{n}(n+1-k)(k^{m})\\ &=(n+1)S_{m}(n)-\sum_{k=1}^{n}(n+1-k)(k^{m})\\ &S_{m+1}(n)=(n+1)S_m(n)-\sum_{r=1}^nS_m(r)\\\\ \end{align}$$

$$\begin{align} &\mbox{p. 164 2.6(b)}\\ &\mbox{Using (a), deduce that}\\ &S_1(n)=\frac{1}{2}n(n+1)\\ &S_2(n)=\frac{1}{6}n(n+1)(2n+1)\\ &S_3(n)=\frac{1}{4}n^2(n+1)^2\\\\

&S_2(n)=(n+1)S_1(n)-\sum_{r=1}^n\frac{1}{2}r(r+1)\\ &S_2(n)=(n+1)S_1(n)-\sum_{r=1}^n\frac{r}{2}-\sum_{r=1}^n\frac{r^2}{2}\\ &\frac{3}{2}S_2(n)=\frac{n(n+1)^2}{2}-\frac{n(n+1)}{4}\\ &S_2(n)=\frac{1}{6}n(n+1)(2n+1)\\\\

&S_3(n)=(n+1)S_2(n)-\sum_{r=1}^n\frac{1}{6}r(r+1)(2r+1)\\ &S_3(n)=(n+1)S_2(n)-\sum_{r=1}^n\frac{2}{6}r^3-\sum_{r=1}^n\frac{3}{6}r^2-\sum_{r=1}^n\frac{1}{6}r\\ &\frac{4}{3}S_3(n)=\frac{n(n+1)^2(2n+1)}{6}-\frac{\frac{1}{6}n(n+1)(2n+1)}{2}-\frac{\frac{1}{2}n(n+1)}{6}\\\\ &..........................\\\\ \end{align}$$

Random Math Stuff
$$\begin{align} &\mbox{Prove that }\sqrt{6}\mbox{ is an irrational number}\\\\

&\mbox{Assume that }\sqrt{6}\mbox{ is a rational number}\\ &\mbox{Hence }\sqrt{6}=\frac{n}{m}\mbox{, where }m,n\in\mathbb{Z}\mbox{ and have no common factors}\\\\ &\therefore n^2=6m^2\cdot(1)\\ &\therefore n=6k,k\in\mathbb{Z}\\\\ &\mbox{Substitute into (1)}\\ &36k^2=6m^2\\ &m^2=6k^2\\ &\therefore m=6l, l\in\mathbb{Z}\\ \end{align}$$

$$\begin{align} &1+(1+2)+(1+2+3)+...+(1+2+3+...+n) \\ &=\sum_{m=1}^n \sum_{k=1}^m k \\ &=\sum_{m=1}^n \frac{1}{2} m(m+1) \\ &=\frac{1}{2} \sum_{m=1}^n (m^2+m) \\ &=\frac{1}{2} \sum_{m=1}^n m^2 + \frac{1}{2} \sum_{m=1}^n m \\ &=\frac{1}{2} \left[\frac{1}{6}n(n+1)(2n+1)\right] + \frac{1}{2}\left[\frac{1}{2}n(n+1)\right] \\ &=\frac{n(n+1)(2n+1)+3n(n+1)}{12} \\ &=\frac{n(n+1)(2n+1+3)}{12} \\ &=\frac{2n(n+1)(n+2)}{12} \\ &=\frac{n(n+1)(n+2)}{6} \end{align}$$

$$\begin{align} &1^3-2^3+3^3-4^3+\dots+(2n+1)^3\\ &=[1^3+3^3+5^3+\dots+(2n+1)^3]-[2^3+4^3+6^3+\dots+(2n)^3]\\ &=\sum_{k=1}^{n+1}(2k-1)^3-\sum_{k=1}^n(2k)^3\\ &=\sum_{k=1}^{n+1}[(2k)^3-3(2k)^2+3(2k)-1] -\sum_{k=1}^n(2k)^3\\ &=8\sum_{k=1}^{n+1}k^3 - 12\sum_{k=1}^{n+1}k^2 + 6\sum_{k=1}^{n+1}k   - (n+1) -  8\sum_{k=1}^{n}k^3\\ &=2(n+1)^2(n+2)^2 - 2(n+1)(n+2)(2n+3) + 3(n+1)(n+2) - 2n^2(n+1)^2 -n-1\\ &=2(n^4+6n^3+13n^2+12n+4)-2(2n^3+9n^2+13n+6)+3(n^2+3n+2)-2(n^4+2n^3+n^2)-n-1\\ &=2n^4+12n^3+26n^2+24n+8-4n^3-18n^2-26n-12+3n^2+9n+6-2n^4-4n^3-2n^2-n-1\\ &=4n^3+9n^2+6n+1 \end{align}$$