The Flying Colours Maths Blog: Latest posts

Ask Uncle Colin: Am I working too hard?

Dear Uncle Colin, I work in a call centre, which is run by the numbers. Every employee has to stay above a certain conversion rate, calculated as the number of successful deals resulting from their calls, divided by the number of times they have a call that doesn't go to

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The Mathematical Ninja And The Cubes

The Mathematical Ninja glanced at the Rubik Cube and paused. "And $45^3$ is..." A reach for the calculator. A flurry of colour. "Ow!" "91,125," said the Mathematical Ninja, catching the cube on the rebound and swizzling it solved. "Only needed 12 that time." The student sighed. "Go on, then. I

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Ask Uncle Colin: A Huge Power Of Two

Dear Uncle Colin, I've been asked to find $2^{64}$ without a calculator, to four significant figures. How would you go about this? -- Large Exponent, Horrific Multiplication, Extremely Repetitive Hi, LEHMER! To get a rough answer, I'd usually start with the rule of thumb that $2^{10} \approx 10^3$. I'd conclude

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Wrong, But Useful: Episode 37

In the October episode of Wrong, But Useful, @icecolbeveridge and @reflectivemaths have their habitual chat. Dave insults mechanics, then gets stroppy about Colin insulting stats The number of the podcast is 16, 40 and 64, because one number isn't enough for Dave Dave visits a fete and gets ripped off

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Ask Uncle Colin: Invariant Lines

Dear Uncle Colin, I've got a matrix, and I'm not afraid to use it. It's $\begin{pmatrix} 3 & -5 \\ -4 & 2\end{pmatrix}$ Apparently, it has invariant lines. Those, I'm afraid of. How do I find them? -- Terrors About Rank, Safely Knowing Inverses Hi, TARSKI! An invariant line of

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A curve-sketching challenge

Via @DrTrapezio, an interesting question: Year 13 curve sketching challenge: y = | | | |x| - 1| - 1| - 1| — Luciano Rila (@DrTrapezio) July 14, 2016 Where do you even start with that mess? The answer to that, my friend, is you start in the middle and

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Ask Uncle Colin: A Vector Line

Dear Uncle Colin, I've got three points: $A$, with a position vector of $(2\bi + 4\bj)$, $B$, with a position vector of $(6\bi + 8\bj)$ and $C$, with a position vector of $(k\bi + 25\bj)$, and they all lie on the same straight line. I have to find $k$, and

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Getting closer to $\pi$

A lovely curiosity came my way via @mikeandallie and @divbyzero: In 1992 Daniel Shanks observed that if p~pi to n digits, then p+sin(p)~pi to 3n digits. For instance, 3.14+sin(3.14)=3.1415926529… — Dave Richeson (@divbyzero) July 15, 2016 Isn't that neat? If I use an estimate $p = 3.142$, then this method

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Ask Uncle Colin: A Stretch, Indeed

Dear Uncle Colin, My research has determined that female adults have a mean overhead reach of 208.5cm, with a standard deviation of 8.6cm, and follows a normal distribution. I wanted to know the probability that the mean overhead reach of 50 female adults would lie between 180cm and 200cm and

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The Echo and a simple answer

Don't get me wrong, The Dorset Echo is one of my favourite local newspapers. They have been kind enough to feed my ego on several occasions, and even if their headlines sometimes don't quite reflect the gist of the story, I appreciate that. This time, though, they've gone too far.1

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