new resources for new syllabus (1 Viewer)

coolcat6778 · Jul 30, 2025

tywebb said:
View attachment 48372

i'm glad they put that in

i've been arguing for years they shouldn't be doing the tildes

the tilde is for handwritten vectors, not typed vectors

this is made clear in new syllabus

textbooks and exams for current syllabus have the tilde quite a lot

but typing it with tildes in word or latex is difficult for a reason.

it's not a standard notation

nsw looks rather peculiar with all the typed vectors replete with uncustomary tildes

so to make this change makes it more consistent with what the rest of the world does

tildes look fucking ugly too

SS173 · Jul 30, 2025

They removed "work allowances" topic from Standard Math to reflect the reality in work places

tywebb · Jul 31, 2025

tywebb said:
some chapters of ext 1 cambridge y 11 up now, ch 6, 11, 16, 17, 18

https://issuu.com/cambridge3/docs/cambridgemathsnsw_stage_6_ext1_y11_2ed_uncorrect

maybe they put the rest up later but dunno when

rest is up now

https://issuu.com/cambridge3/docs/cambridgemaths_nsw_stage_6_extension_11_2ed_4pp

this one has all the chapters

Socialism · Aug 3, 2025

tywebb said:
so to make this change makes it more consistent with what the rest of the world does

omg standardisation we

that

tywebb · Aug 7, 2025

https://4unitmaths.com/expected-publication-dates.pdf updated again to include new books by Concept Press called Concept Mathematics

advanced year 11 is expected to come september 19, 2025

they look like this:

tywebb · Sunday at 8:02 PM

with the formal introduction of the cauchy-schwarz inequality as we transition from current syllabus to new one one might reenvision inequalities as in the following example

$\text{Prove that if }a,b,c\in\Bbb R^+,\textstyle\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca$

rephrase it for current syllabus

$\text{For }a,b,c\in\Bbb R^+\text{ using the fact that }a^2+b^2+c^2\ge ab+bc+ca\text{ prove using the AM-GM inequality that }$

$\textstyle\frac{a^3}{b}+ab\ge2a^2\text{ and hence prove that }\textstyle\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca$

$\text{By AM-GM inequality, }\textstyle\frac{\frac{a^3}{b}+ab}{2}\ge\sqrt{\frac{a^3}{b}\times ab}=a^2$

$\text{Now }\textstyle\frac{a^3}{b}+ab\ge2a^2,\frac{b^3}{c}+bc\ge2b^2\text{ and }\frac{c^3}{a}+ca\ge2c^2$

$\begin{aligned}\text{so }\textstyle\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+bc+ca&\ge2(a^2+b^2+c^2)\\&\ge2(ab+bc+ca) \end{aligned}$

$\therefore\textstyle\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca$

rephrase it for new syllabus

$\text{For }a,b,c\in\Bbb R^+\text{ using the fact that }a^2+b^2+c^2\ge ab+bc+ca\text{ prove using the Cauchy-Schwarz inequality that$

$\textstyle\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca$

$\begin{aligned}\textstyle\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)(ab+bc+ca)&\ge(a^2+b^2+c^2)^2\text{ by the Cauchy-Schwarz inequality}\\ &\ge(ab+bc+ca)^2\end{aligned}$

$\therefore\textstyle\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca$

if u looked at my spoilers u would appreciate the inclusion of cauchy-schwarz will help to increse the efficiency of solutions

tywebb · Monday at 1:10 PM

here is a somewhat simpler example related to the previous post

$\text{Prove that for }a,b,c\in\Bbb R^+,a^2+b^2+c^2\ge ab+bc+ca$

rephrased for current syllabus:

$\text{For }a,b,c\in\Bbb R^+\text{ by considering }(a-b)^2+(b-c)^2+(c-a)^2$

$\text{prove that }a^2+b^2+c^2\ge ab+bc+ca$

$(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ca)\ge0$

$\therefore a^2+b^2+c^2\ge ab+bc+ca$

rephrased for new syllabus

$\text{Use the Cauchy-Schwarz inequality to prove that for }a,b,c\in\Bbb R^+,$

$a^2+b^2+c^2\ge ab+bc+ca$

$\text{By the Cauchy-Schwarz inequality,}$

$(a^2+b^2+c^2)(b^2+c^2+a^2)\ge(ab+bc+ca)^2$

$\therefore a^2+b^2+c^2\ge ab+bc+ca$

tywebb · Monday at 5:44 PM

here is a more challenging one

alfred mortimer nesbitt, headmaster of toowoomba grammar school 1882-1888 published in the may 1, 1902 educational times https://www.mediafire.com/file/2uh9o45f3rg1i91/e-times.pdf on pdf page 33, problem 15114 what has become known as nesbitt's inequality

prove it using the Cauchy-Schwarz inequality

$\text{Start by considering }\textstyle((a+b)+(b+c)+(c+a))\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)$

$\begin{aligned}\textstyle((a+b)+(b+c)+(c+a))\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)&\ge(1+1+1)^2\text{ by the Cauchy-Schwarz inequality}\\ \textstyle\frac{2(a+b+c)}{a+b}+\frac{2(a+b+c)}{b+c}+\frac{2(a+b+c)}{c+a}&\ge9\\ \textstyle2+\frac{2c}{a+b}+2+\frac{2a}{b+c}+2+\frac{2b}{c+a}&\ge9\\ \textstyle\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\ge\textstyle\frac{3}{2} \end{aligned}$

coolcat6778 · Monday at 6:48 PM

tywebb said:
here is a more challenging one

alfred mortimer nesbitt, headmaster of toowoomba grammar school 1882-1888 published in the may 1, 1902 educational times https://www.mediafire.com/file/2uh9o45f3rg1i91/e-times.pdf on pdf page 33, problem 15114 what has become known as nesbitt's inequality

View attachment 48687

prove it using the Cauchy-Schwarz inequality

$\begin{aligned}\textstyle((a+b)+(b+c)+(c+a))\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)&\ge(1+1+1)^2\text{ by the Cauchy-Schwarz inequality}\\ \textstyle\frac{2(a+b+c)}{a+b}+\frac{2(a+b+c)}{b+c}+\frac{2(a+b+c)}{c+a}&\ge9\\ \textstyle2+\frac{2c}{a+b}+2+\frac{2a}{b+c}+2+\frac{2b}{c+a}&\ge9\\ \textstyle\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\ge\textstyle\frac{3}{2} \end{aligned}$

how tf were they so smart back then

tywebb · Wednesday at 8:04 AM

it is not immediately clear in the new syllabus what they mean by this

there is a proof in the sample paper but not much on

- use for what?
- why use?
- how to use?
- when to use and when not to use?

there is however hidden in the support document

there is a problem severely restricting applicability to the result for real numbers, namely that it is easily extendable to n dimensions, yet we only have 2 and 3 dimensional vectors in the syllabus

the proof in the sample paper still holds for the n-dimensional case

it may also be investigated how consequences and generalisations can be applied, such as titu's lemma and hölder's inequality

these often result in more efficient solutions than simply applying cauchy-schwarz.

titu's is easy to prove, hölder's is a bit harder to prove, but when you've proved them, the skill developed in using cauchy-schwarz is immediately transferable to using titu's and hölder's.

inclusion of n-dimensional cauchy-schwarz, titu's and hölder's will allow exploration of a wider set of applications.

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new resources for new syllabus (1 Viewer)

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