Hard Questions (2 Viewers)

leehuan · Apr 12, 2016

KINGOM 885 said:
Why?
I am an extension 1 student

You're an Ext 1 student and you posted here. Are you saying you want hard Ext 1 maths questions or questions that are actually hard for real mathematicians

Paradoxica · Apr 12, 2016

KINGOM 885 said:
Why?
I am an extension 1 student

leehuan said:
You're an Ext 1 student and you posted here. Are you saying you want hard Ext 1 maths questions or questions that are actually hard for real mathematicians

Prove or disprove the existence of a universal method to determine whether or not a given equation that defines elliptic curves over ℚ, has finitely, or infinitely many solutions in ℚ.

seanieg89 · Apr 12, 2016

Paradoxica said:
Prove or disprove the existence of a universal method to determine whether or not a given equation that defines elliptic curves over ℚ, has finitely, or infinitely many solutions in ℚ.

Went to an amazing lecture on this by Venkatesh last year. Fascinating problem.

InteGrand · Apr 12, 2016

leehuan said:
You're an Ext 1 student and you posted here. Are you saying you want hard Ext 1 maths questions or questions that are actually hard for real mathematicians

I think it was originally posted elsewhere (in Maths Extension 1 forum if I recall correctly), but got moved to this area by a Moderator.

Paradoxica · Apr 12, 2016

seanieg89 said:
Went to an amazing lecture on this by Venkatesh last year. Fascinating problem.

I know. I've already read up on a similar problem for general diophantine equations and was disappointed (but it didn't go against my expectations) that the answer was a negative.

I won't be surprised if the answer to that problem is a negative though.

tywebb · May 12, 2023

tywebb said:
Here are some more papers from 1956-1962 containing some hard questions:
http://4unitmaths.com/lc1956-1962.pdf

We can see from the 1957 leaving certificate paper that

We can also see from the 2014 HSC Extension 2 exam that

These can be combined to produce a new formula for π in terms of binomial coefficients which was discovered in 2007 by J.C. Toloza:

as follows:

kev@year1223 · May 20, 2023

Hello @tywebb , would like to buy Cambridge worked out solutions for Maths Extension 1 from you. Could you please advise?

tywebb · Jun 10, 2023

tywebb said:
discovered in 2007 by J.C. Toloza:

$\textstyle2+{2\choose2}^{-1}+{3\choose2}^{-1}-{4\choose2}^{-1}-{5\choose2}^{-1}+{6\choose2}^{-1}+{7\choose2}^{-1}-\cdots=\pi$

$\text{Toloza's formula essentially relates } \pi \text{ to one diagonal of Pascal's triangle via a pairwise$
$\text{alternating sum of reciprocals of binomial coefficients.}$

$\text{But Toloza's formula can be generalised to relate }\pi\text{ to every second diagonal of Pascal's triangle.}$

$\text{Suppose }T_p:=\textstyle\frac{2p}{2p-1}+\sum_{n=1}^\infty\sum_{r=0}^{2p-1}(-1)^{n-1}{2pn+r\choose 2p}^{-1}\text{ for positive integers }p.$

$\text{ Then Toloza's formula is now }T_1=\pi.$

$\text{ See if you can prove that for }p>1,$

$T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}(\csc\frac{\pi i}{2p}-1).$

$\text{For example, with } p=2\text{ we find the 4-fold alternating sum }$

$\scriptsize{\textstyle\frac{4}{3}+{4\choose4}^{-1}+{5\choose4}^{-1}+{6\choose4}^{-1}+{7\choose4}^{-1}-{8\choose4}^{-1}-{9\choose4}^{-1}-{10\choose4}^{-1}-{11\choose4}^{-1}+\cdots=2(\sqrt2-1)\pi}$

$\text{or with }p=3,\text{ the 6-fold alternating sum$

${\textstyle\frac{6}{5}+{6\choose6}^{-1}+{7\choose6}^{-1}+{8\choose6}^{-1}+{9\choose6}^{-1}+{10\choose6}^{-1}+{11\choose6}^{-1}-{12\choose6}^{-1}-{13\choose6}^{-1}-{14\choose6}^{-1}-{15\choose6}^{-1}-{16\choose6}^{-1}-{17\choose6}^{-1}+\dots=2(5-\frac{8}{3}\sqrt3)\pi}$

$\text{etc.}$

tywebb · Jul 8, 2023

tywebb said:
$\text{Toloza's formula essentially relates } \pi \text{ to one diagonal of Pascal's triangle via a pairwise$
$\text{alternating sum of reciprocals of binomial coefficients.}$

$\text{But Toloza's formula can be generalised to relate }\pi\text{ to every second diagonal of Pascal's triangle.}$

$\text{Suppose }T_p:=\textstyle\frac{2p}{2p-1}+\sum_{n=1}^\infty\sum_{r=0}^{2p-1}(-1)^{n-1}{2pn+r\choose 2p}^{-1}\text{ for positive integers }p.$

$\text{ Then Toloza's formula is now }T_1=\pi.$

$\text{ See if you can prove that for }p>1,$

$T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}(\csc\frac{\pi i}{2p}-1).$

$\text{For example, with } p=2\text{ we find the 4-fold alternating sum }$

$\scriptsize{\textstyle\frac{4}{3}+{4\choose4}^{-1}+{5\choose4}^{-1}+{6\choose4}^{-1}+{7\choose4}^{-1}-{8\choose4}^{-1}-{9\choose4}^{-1}-{10\choose4}^{-1}-{11\choose4}^{-1}+\cdots=2(\sqrt2-1)\pi}$

$\text{or with }p=3,\text{ the 6-fold alternating sum$

${\textstyle\frac{6}{5}+{6\choose6}^{-1}+{7\choose6}^{-1}+{8\choose6}^{-1}+{9\choose6}^{-1}+{10\choose6}^{-1}+{11\choose6}^{-1}-{12\choose6}^{-1}-{13\choose6}^{-1}-{14\choose6}^{-1}-{15\choose6}^{-1}-{16\choose6}^{-1}-{17\choose6}^{-1}+\dots=2(5-\frac{8}{3}\sqrt3)\pi}$

$\text{etc.}$

I found a proof which is attached. There might be an easier way to do it, or at least a simplification of the attached proof.

tywebb · Feb 21, 2024

tywebb said:
Toloza's formula can be generalised to relate π to every second diagonal of Pascal's triangle

When you learned trig did you ever use these archaic trig functions?

Versine: versin(θ)=1-cos(θ)
Vercosine: vercosin(θ)=1+cos(θ)
Coversine: coversin(θ)=1-sin(θ)
Covercosine: covercosine(θ)=1+sin(θ)
Haversine: haversin(θ)=versin(θ)/2
Havercosine: havercosin(θ)=vercosin(θ)/2
Hacoversine: hacoversin(θ)=coversin(θ)/2
Hacovercosine: hacovercosin(θ)=covercosin(θ)/2
Exsecant: exsec(θ)=sec(θ)-1
Excosecant: excsc(θ)=csc(θ)-1

It means the formula for $T_p$ is actually a sum of excosecants.

So it could also be written as

$T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}\text{excsc}\frac{\pi i}{2p}$

Hard Questions (2 Viewers)

leehuan

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Paradoxica

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