Because it's for implicit equations as opposed to explicit equations of the form y=f(x)
12.11.2025 11:52 β π 0 π 0 π¬ 1 π 0@vmnalex.bsky.social
#mtbos #iteachmath | Engelmann | Free notes & resources @ http://vicmathsnotes.weebly.com | Author for OUP Maths | CL @ Ochre
Because it's for implicit equations as opposed to explicit equations of the form y=f(x)
12.11.2025 11:52 β π 0 π 0 π¬ 1 π 0Great etymology video from @robwords.bsky.social
09.11.2025 22:27 β π 2 π 0 π¬ 0 π 0Inflation hitting so hard, it's affecting the percentages
09.11.2025 22:25 β π 3 π 0 π¬ 0 π 0Literally just came across this the other day
youtu.be/oIhdrMh3UJw
The main other ones I'd drawn attention to are probably sin(0)=tan(0)=0 and cos(0)=1 for y-intercepts and because they show up in integrals regularly, and I want to make sure students don't mistake e^0=cos(0)=1 with being 0 which happens frequently.
29.10.2025 20:43 β π 1 π 0 π¬ 0 π 0We usually have already got that sin(45)=cos(45)=1β/2 by osmosis, but making that explicit (useful for intersections too)
We use the fact that tan is increasing from 1/β3 to 1 to β3 for 30 to 45 to 60Β° to help differentiate those.
That basically leaves sin(60)=cos(30)=β3/2 and the multiples of 90
After they've been getting familiar with those and we're trying to improve efficiency, we start memorising the rational ones in tri's: sin(30)=cos(60)=Β½, tan(45)=1 (because rational tends to feel easier to remember)
29.10.2025 20:43 β π 1 π 0 π¬ 1 π 0I'm in the same boat. I call them the half-square and half-equilateral triangle to help Ss remember how to construct the tri's if they forget.
I also use the unit circle definitions (x, y coordinates for cosine and sine and gradient of radius or length of tangent to x-axis) for the multiples of 90
We only do x^2, log x, and 1/x, so it's a fairly short ladder. But it was /so/ much easier to remember how to label the circle.
Yes, I can do it by considering similarities to the respective graphs, but this subject doesn't go through that.
circle of transformations with possible transforms listed. signs indicate the sign of the correlation after transformation (without compensating for the reversal of 1/x)
the bulging rule where the ladder is labelled as a scale across each axis
ladders going from x^-3 to x^-1 to x^0 treated as log(x) to x^1/2 x^1 then through x^5
Got to test something new in #mathstoday. Normally, to decide which transformation to linearise a data set, I'd just show a circle of transformations where each quadrant has the options listed. I recently stumbled upon Tukey and Mosteller's bulging rule and ladder of powers and it's /so/ much easier
22.10.2025 09:01 β π 2 π 0 π¬ 1 π 0I think I love it even more than discorectangle
21.10.2025 13:14 β π 1 π 1 π¬ 2 π 0I've discovered some people call it a convex angle, which makes some sense, although it suggests a reflex angle should be called concave.
Incidentally, did you know that angles that add uo to 360Β° are called explementary?
Explementary is such a handy word
21.10.2025 10:17 β π 1 π 0 π¬ 1 π 0Giving a child a calculator too early doesnβt help them learn maths; it helps them avoid learning maths.
Great piece from @amandavande.bsky.social on why fluency must come before function. springmath.org/fluency-befo...
easy change
18.10.2025 06:36 β π 4 π 0 π¬ 1 π 0differentiating x^2, x^3, and x^4 by factorising (x+h)^n - x^n using the factorisation of a^n - b^n and letting h β 0 without expanding the second factor.
differentiating x^n by factorising (x+h)^n - x^n using the factorisation of a^n - b^n and letting h β 0 without expanding the second factor, where there are n terms of x^(n-1)
the (hopefully) fully corrected version and x^n
18.10.2025 05:58 β π 3 π 2 π¬ 1 π 0First principles, yeah, heaps. By factorising rather than expanding? Surprisingly, no!
18.10.2025 04:12 β π 1 π 0 π¬ 0 π 0This was wrong too, lol #OnARoll
18.10.2025 03:57 β π 1 π 0 π¬ 1 π 0differentiating x^2, x^3, and x^4 by factorising (x+h)^n - x^n using the factorisation of a^n - b^n and letting h β 0 without expanding the second factor. *corrected x^4 working
18.10.2025 03:23 β π 1 π 0 π¬ 1 π 0Doesn't surprise me
18.10.2025 03:22 β π 2 π 0 π¬ 1 π 0differentiating x^2, x^3, and x^4 by factorising (x+h)^n - x^n using the factorisation of a^n - b^n and letting h β 0 without expanding the second factor.
Can't say I'd thought to do this before. #mathstoday
18.10.2025 02:23 β π 18 π 1 π¬ 6 π 0Would appreciate any reposts to try and get to the bottom of this
13.10.2025 08:52 β π 0 π 0 π¬ 0 π 016 principles of task design with Nathan Day
Loved discussing some excellent maths tasks in @nathanday.bsky.socialβs session. Great to have the opportunity to dig into what makes them so good and how they could be adapted. #MathsConf39
11.10.2025 15:20 β π 9 π 2 π¬ 0 π 0Queries:
1. Are the two methods actually distinct and have been inadvertently conflated somewhere along the line?
2. Which (if either) should be favoured as an introductory/straight-forward/trend-agnostic method?
average percentage method applied to data over 2 years to determine seasonal indices and deseaonalised sales; sales in millions of dollars
simple average method applied to data over 2 years to determine seasonal indices and deseaonalised sales; sales in millions of dollars
differences in seasonal indices: -0.0093950283 0.0278223623 -0.0145401985 -0.0038871355 differences in deseasonalised values (in millions of dollars) 0.032382259 -0.04141642 0.025299911 0.017857768 0.019279611 -0.035355481 0.015528388 0.011324438
Additional example over 2 years, quarterly.
They're out JUST enough that you could get marked incorrectly from an assumption of one method or the other because it looks like a rounding error
seasonally adjusted time series showing original data, and deseasonalised values using average percentage and simple averages methods which are generally equivalent but have deviations
same graph without the original data to exaggerate discrepancies
differences between seasonal indices 0.001 0 0.001 0 -0.001 -0.002 -0.002 -0.001 0 0.002 0.002 0.002
Here's the catch: they're not equivalent (nuisance 2). They produce VERY similar results, but not exactly the same (algebraically distinct). And the difference is enough that it could look like a rounding error to 2-4 decimal places occasionally.
11.10.2025 04:11 β π 0 π 0 π¬ 1 π 0text refers to average percent method: data as percentages of the average for the year which are then averaged
text refers to simple averages method: calculates average for each quarter across all years, then written as a percentage of the overall average
text lists average percentage and simple average as the same
Some texts describe the "average percentage" method but call it "simple average", but "average percentage" is pretty consistently as above (feels very self-referential tbh).
11.10.2025 04:11 β π 0 π 0 π¬ 1 π 0When I've looked into other methods for seasonal adjustment to see what else is out there, different texts tend to refer to them in place of each other or as being equivalent.
11.10.2025 04:11 β π 0 π 0 π¬ 1 π 0Simple averages is much more efficient calculation wise, even if you have to first average the seasonal values for each season. For reference, 8 years of monthly data goes from 116 calculations for average percent to 25 calculations for simple averages to get the averaged seasonal indices.
11.10.2025 04:11 β π 0 π 0 π¬ 1 π 0simple averages method applied to the same data over 8 years, monthly; conveniently the average of the seasonal averages is equal to the average of all the data values
Past exam question that refers to long-term monthly average heating costs
past exam question referring to long-term quarterly average visitor numbers
Some questions we've gotten give the average values for each season (assuming across years) and ask to compute the seasonal indices from those. From what I've found, this is the "simple averages" method.
11.10.2025 04:11 β π 0 π 0 π¬ 1 π 0