Browsing category algebra

Summing with Generating Functions

A nice challenge puzzle via Reddit: Find $\sum_{n=1}^{\infty} \frac{n2^n}{(n+2)!}$ There was a video attached to it that I didn’t watch, something about telescoping sums, but the moment I saw this, I thought: generating functions! Why would I think such a thing? The thing that jumped out at me was the

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A GCSE surprise

Some time ago, I was surprised to see the following question in a predicted GCSE paper: Solve for $x$: $ \frac{2x}{3x+2} \leq \frac{3}{4x+1}$ Give your answers to two decimal places (3 marks) Why surprised? Surprised because the techniques you need to solve it correctly are Further Maths A-level, rather than

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Heads, Tails and Bumpsdaisy, revisited

At a recent East Dorset Mathsjam , the puzzle of two heads resurfaced: if you repeatedly flip a fair coin, how long (on average) do you have to wait until you get two heads in a row? Two fine answers are available here. However, the estimable Barney Maunder-Taylor went down

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A Textbook Error?

In class, a student asked to work through a question: Let $f(x) = \frac{5(x-1)}{(x+1)(x-4)} - \frac{3}{x-4}$. (a) Show that $f(x)$ can be written as $\frac{2}{x+1}$. (b)Hence find $f^{-1}(x)$, stating its domain. The answer they gave was outrageous1. Part (a) Part (a) was fine: combine it all into a single fraction

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A moment of neatness

Working through an FP2 question on telescoping sums (one of my favourite topics - although FP2 is full of those), we determined that $r^2 = \frac{\br{2r+1}^3-\br{2r-1}^3-2}{24}$. Adding these up for $r=1$ to $r=n$ gave the fairly neat result that $24\sum_{r=1}^{n} r^2 = \br{2n+1}^3 - 1 - 2n$. Now, there are

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Can you find a centre and angle of rotation without any construction?

Some time ago, I had a message from someone who - somewhat oddly - wanted to find a centre of rotation (with an unknown angle) without constructing any bisectors. (Obviously, if it was a right-angle rotation, they could use the set-square trick; if it was a half-turn, the centre of

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Why the factor and remainder theorem work

So there I was, merrily teaching the factor and remainder theorems, and my student asked me one of my favourite questions: "I accept that the method works, but why does it?" (I like that kind of question because it makes me think on my feet in class, and that makes

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$\cos(72º)$, revisited again: De Moivre’s Theorem

In previous articles, I've looked at how to find $\cos(72º)$ using some nasty algebra and some comparatively nice geometry. In this one, inspired by @ImMisterAl, I try some nicer - although quite literally complex - geometry. De Moivre's Theorem I'm going to assume you're ok with complex numbers. If you're

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Ask Uncle Colin: A Fractional Kerfuffle

Dear Uncle Colin, I was trying to work out $\frac{\frac{3}{7+h}-\frac{3}{7}}{h}$, and I got it down to $\frac{\frac{3}{h}}{h}$ - but that's not the answer in the book! What have I done wrong? - Likely I've Mistreated It Terribly Hi, LIMIT, and thank you for your message! I'm afraid you're right, you

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The Involution of Polynomials

Last time out, I looked at a problem unearthed by @mathsjem - to find the cube root of a degree-six polynomial. This led (unsurprisingly) to a quadratic: $3 + 4x - 2x^2$. When checking whether this was indeed the answer, I hit a problem: is there a simple way to

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