About lloydstagg

Extensive experience as a Victorian Certificate of Education mathematics teacher. Enjoys combining maths and technology via CAS calculators and dynamic geometry software. Available for face-to-face tutoring in Mildura or via Skype.

Two Up

Posted on 2019/04/26 by lloydstagg

Yesterday being ANZAC Day prompted news stories of the game Two Up, where two coins are flipped and punters bet on either two heads or two tails coming up. Neither punter wins when one of each occurs.

I happened to be talking with someone who had played a bit of Two Up. To my comment that it is the fairest game to bet on, he replied; “Except at the casino. If one of each comes up four times in a row the house takes all current bets.”

The chance of one of each occurring four times in a row is binomial with probability one in sixteen, but the astute gambler and wants to know the expected number of throws before this occur four times successively.

Matrix Row Reduction

Posted on 2018/12/12 by lloydstagg

Thinking about how matrix row reduction works I realised that it’s basically scaled addition of ordinates.

RowReduction

MASA conference 2018

Posted on 2018/04/30 by lloydstagg

Being a recent arrival in South Australia I was surprised to find the annual, two-day MASA conference is routinely scheduled during the Autumn school holidays and on this occasion adjacent to the ANZAC Day holiday.

Post-conference reflections:-

Presenters, mostly active classroom teachers, were passionate, enthusiastic and genuinely want to share their mathematical achievements and interests.
Delegates were enthusiastic and attentive. In one workshop, when the presenter asked, mid-presentation, if the audience wanted to see the proof behind his assertion there was an enthusiastic “yes” followed by applause when he had done so.
Delegates were keen to engage in mathematical conversation. Even at happy hour, when I mentioned to someone I recognised from one of the sessions, that I was unfamiliar with a component of the presentation, he happily got out some paper and explained it to me.
A smaller attendance allows for a compact venue where registration,workshops, trade display and meals are all in easy-to-find proximity to each other.

Overall, a very satisfying experience.

Randomness

Posted on 2018/04/26 by lloydstagg

Galton Board using sand

Fascinated with the Galton Board: a vertical triangle of pins. Marbles or sand are poured in through a gap at the apex. Each object bounces randomly of the pins as it descends to the base. The spread of objects invariably resembles the bell-shaped normal distribution curve. “Overall, the pattern is utterly predictable: it always forms a bell-shaped distribution – even though it is impossible to predict where any given ball will end up“; Steven Strogatz, The Joy of X, p 176.

Memories of a Bairnsdale High School physics prac

Posted on 2017/10/29 by lloydstagg

Recent news items about the oil slick on the River Torrens reminded me of my Bairnsdale High School physics prac to measure the dimensions of a molecule.

Count the number of oil drops needed to measure 1 cc of a low viscosity oil; divide the 1 cc (or 1000 cubic millimetres) by the number of oil drops to calculate the average volume of one oil drop; sprinkle talcum powder over the surface of a shallow ripple tank full of water; place one drop of the oil in the center of the powdered surface. The oil spreads in a roughly circular manner pushing the talc out to surround its perimeter. Several estimates of the dispersed oil drop’s diameter are used to calculate its area. Assuming the oil slick has spread until it is one molecule thick, (and keeping units the same) divide the volume of one oil drop by the oil slick’s area to calculate the thickness of the oil slick and thus the dimension/s of an oil molecule. A most memorable experiment.

Complex Regions

Posted on 2015/03/04 by lloydstagg

$\begin{array}{l}\left\{ {z:\left| {z+1-i} \right|+\left| {z+1-i} \right|=4} \right\}\\\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}}}+\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}=4\\\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}}}=4-\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\{{\left( {\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}}}} \right)}^{2}}={{\left( {4-\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}} \right)}^{2}}\\{{\left( {x+1} \right)}^{2}}+{{\left( {y-1} \right)}^{2}}=16-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}+{{\left( {x+1} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}\\{{\left( {y-1} \right)}^{2}}-{{\left( {y+1} \right)}^{2}}-16=-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\{{y}^{2}}-2y+1-\left( {{{y}^{2}}+2y+1} \right)-16=-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\-4y-16=-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\{{\left( {-y-4} \right)}^{2}}={{\left( {-2\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}} \right)}^{2}}\\{{y}^{2}}+8y+16=4\left( {{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}} \right)\\=4{{\left( {x+1} \right)}^{2}}+4{{y}^{2}}+8y+4\\\frac{{12=4{{{\left( {x+1} \right)}}^{2}}+4{{y}^{2}}}}{{12}}\\\frac{{{{{\left( {x+1} \right)}}^{2}}}}{3}+{{y}^{2}}=1\end{array}$

Complex solve using complete the square variation.

Posted on 2015/02/27 by lloydstagg

$\displaystyle \begin{array}{l}{{z}^{4}}-2{{z}^{2}}+4=0,z\in C\\\left( {{{z}^{4}}-2{{z}^{2}}-2{{z}^{2}}+4} \right)+2{{z}^{2}}=0\\\left( {{{z}^{2}}-2} \right)+2{{z}^{2}}=0\\{{\left( {{{z}^{2}}-2} \right)}^{2}}-{{\left( {i\sqrt{2}z} \right)}^{2}}=0\\\left( {{{z}^{2}}-i\sqrt{2}z-2} \right)\left( {{{z}^{2}}+i\sqrt{2}z-2} \right)=0\\{{z}^{2}}-i\sqrt{2}z-2=0\\a=1,b=-i\sqrt{2},c=-2\\z=\frac{{-\left( {-i\sqrt{2}} \right)\pm \sqrt{{{{{\left( {-i\sqrt{2}} \right)}}^{2}}-4\left( 1 \right)\left( {-2} \right)}}}}{{2\left( 1 \right)}}\\=\frac{{i\sqrt{2}\pm \sqrt{6}}}{2}\\=\frac{{\pm \sqrt{6}}}{2}+\frac{{i\sqrt{2}}}{2}\\{{z}^{2}}+i\sqrt{2}z-2=0\\a=1,b=i\sqrt{2},c=-2\\z=\frac{{-\left( {i\sqrt{2}} \right)\pm \sqrt{{{{{\left( {i\sqrt{2}} \right)}}^{2}}-4\left( 1 \right)\left( {-2} \right)}}}}{{2\left( 1 \right)}}\\=\frac{{-i\sqrt{2}\pm \sqrt{6}}}{2}\\=\frac{{\pm \sqrt{6}}}{2}-\frac{{i\sqrt{2}}}{2}\\z=\left( {\frac{{\pm \sqrt{6}}}{2}+\frac{{i\sqrt{2}}}{2}} \right),\left( {\frac{{\pm \sqrt{6}}}{2}-\frac{{i\sqrt{2}}}{2}} \right)\end{array}$

Complex Solve 1

Posted on 2015/02/27 by lloydstagg

$\begin{array}{l}{{z}^{4}}-2{{z}^{2}}+4=0,z\in C\\\text{Let }w={{z}^{2}}\Rightarrow \\{{w}^{2}}-2w+4=0\\\left( {{{w}^{2}}-2w+1} \right)+4-1=0\\{{\left( {w-1} \right)}^{2}}-3=0\\{{\left( {w-1} \right)}^{2}}-\left( {i\sqrt{3}} \right)=0\\\left( {w-1-i\sqrt{3}} \right)\left( {w-1+i\sqrt{3}} \right)=0\\w-1-i\sqrt{3}=0\Rightarrow \\{{z}^{2}}=1+i\sqrt{3}=2cis\left( {\frac{{\pi \left( {6n+1} \right)}}{3}} \right)\\z=\sqrt{2}cis\left( {\frac{{\pi \left( {6n+1} \right)}}{6}} \right)\\n=0,{{z}_{1}}=\sqrt{2}cis\left( {\frac{{\pi \left( 1 \right)}}{6}} \right)=\sqrt{2}\left( {\frac{{\sqrt{3}}}{2}+i\frac{1}{2}} \right)\\n=1,{{z}_{2}}=\sqrt{2}cis\left( {\frac{{\pi \left( 7 \right)}}{6}} \right)=\sqrt{2}\left( {-\frac{{\sqrt{3}}}{2}-i\frac{1}{2}} \right)\\w-1+i\sqrt{3}=0\Rightarrow \\{{z}^{2}}=1-i\sqrt{3}=2cis\left( {\frac{{\pi \left( {6n-1} \right)}}{3}} \right)\\z=\sqrt{2}cis\left( {\frac{{\pi \left( {6n-1} \right)}}{6}} \right)\\n=0,{{z}_{1}}=\sqrt{2}cis\left( {\frac{{\pi \left( {-1} \right)}}{6}} \right)=\sqrt{2}\left( {\frac{{\sqrt{3}}}{2}-i\frac{1}{2}} \right)\\n=1,{{z}_{2}}=\sqrt{2}cis\left( {\frac{{\pi \left( 5 \right)}}{6}} \right)=\sqrt{2}\left( {-\frac{{\sqrt{3}}}{2}+i\frac{1}{2}} \right)\end{array}$

Triumph of the City

Posted on 2014/12/31 by lloydstagg

Reading “Triumph of the city” by Edward Glaeser. Struck by his claim (page 7): “There is a near-perfect correlation between urbanisation and prosperity across nations”.

Practicing with mathtype.

Posted on 2014/10/26 by lloydstagg

My first practice using mathtype with wordpress.
$\displaystyle {{s}_{{N-1}}}=\sqrt{{\frac{1}{{N-1}}\sum\limits_{{i=1}}^{N}{{({{x}_{i}}-\bar{x}}}{{)}^{2}}}}$

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