Ian
@Math5Tutor
Maths teacher working hard to ensure everybody is better than average.
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Having looked at #ALevelMaths Paper 1 (Edexcel), here are my "best guess" topics video checklist for Paper 2. Word of caution - this is just a "Best guess", any topic that appeared on Paper 1 could still appear! mrmanmaths.wordpress.com/2025/06/04/a-l…
Each prime (except 2 and 3) is next to a multiple of 6.
ICYMI - great for revision for your L2 FM classes this weekend!
⭐️Bumper Practice Paper for AQA Level 2 Further Maths Paper 2 now available⭐️ PDF version here: drfrost.org/downloadables.… Worksheet to practice/set as task here: drfrost.org/worksheets.php… (Intended to be used as a revision tool, not a prediction of paper content!)
Ideas, thoughts and views on which topics or activities make a good and interesting #ALevel #Maths transition lesson #Year12
Michael Collins, the astronaut who took this photo in 1969, is the only human, alive or dead, not captured in the frame of this picture.
The sum of first n odd numbers is equal to n². Classic. Here's a very necessary 🧵that consists of 7 different proofs of this fact. Let's get into it! [1/9]
![Mathinity_'s tweet image. The sum of first n odd numbers is equal to n². Classic.
Here's a very necessary 🧵that consists of 7 different proofs of this fact. Let's get into it! [1/9]](https://pbs.twimg.com/media/F2h1Wv6WIAAJlyu.jpg)
PROOF of the Birthday Paradox The probability that all 23 individuals have different birthdays is: P_diff = (365/365)*(364/365)*(363/365)*...*(343/365) Thus the probability that at least 2 of them share a birthday is: P_shared = 1 - P_diff ≈ 0.5
In the fall of 1972, President Nixon announced that "the rate of increase of inflation was decreasing". Probably the first time a sitting president used the 3rd derivative to increase the chances of reelection.
This identity is called the Sophomore's dream - in contrast to the name "Freshman's dream" which is given to the incorrect identity (x+y)ⁿ=xⁿ+yⁿ, this one has a similar too-good-to-be-true feel, but is true!
A short Python program to estimate π: import random def estimate_pi(n): m = 0 for i in range(n): x = random.uniform(0, 1) y = random.uniform(0, 1) if x*x + y*y <= 1: m += 1 return 4 * m / n
Here's a 1,400 years old approximation to the sine function (0 ≤ x ≤ π) by Mahabhaskariya of Bhaskara I
A visual representation of the first 100,000 digits of π
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