alternative solution to nsb 2025 Q16ciii (1 Viewer)

tywebb · 2025-08-27T16:26:58+1000

when question says "or otherwise" u can do the oos methods.

so in nsb 2025 trial Q16ciii u can use hypergeometric function to solve much more quickly like this

$\text{Leading coefficient in polynomial expansion of }\cos(n\theta)$

$=\lim_{x\rightarrow\infty}\frac{{}_2F_1(-n,n;\frac{1}{2};\frac{1-x}{2})}{x^n}=2^{n-1}$

killer queen · 2025-08-27T19:15:20+1000

what is a hypergeometric function and what are its use cases?

coolcat6778 · 2025-08-27T19:33:42+1000

killer queen said:
what is a hypergeometric function and what are its use cases?

something not in the syllabus, therefore teachers wont understand and will probably give you 0

tywebb · 2025-08-27T20:14:23+1000

coolcat6778 said:
something not in the syllabus, therefore teachers wont understand and will probably give you 0

well if they are that stupid they shouldn't put "or otherwise" in their exams.

but they do and that allows for all the oos under the sun

killer queen said:
what is a hypergeometric function and what are its use cases?

${}_2F_1(a,b;c;z)\text{ is the analytic continuation of }\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$

$\text{along any path in the complex plane that avoids the branch points 1 and infinity}$

$\text{where }(q)_n\text{ is the rising Pochhammer symbol}$

it is related to Chebyshev polynomials via

$\textstyle\cos(ax)={}_2F_1(-a,a;\frac{1}{2};\frac{1-\cos x}{2})$

and this is how it is related to the nsb question

by way of example with n=8

$\textstyle{}_2F_1(-8,8;\frac{1}{2};\frac{1-x}{2})=128x^8-256x^6+160x^4-32x^2+1$

hence without having to do a long calculation involving de Moivre's theorem we can deduce instantly that

$\cos(8\theta)=128\cos^8\theta-256\cos^6\theta+160\cos^4\theta-32\cos^2\theta+1$

$\text{now if u notice the leading coefficient }128=2^7=2^{8-1}$

$\text{one may wonder how one might extract this from the polynomial$

$128x^8-256x^6+160x^4-32x^2+1$

$\text{simply divide the whole thing by }x^8\text{ and take limit as }x\rightarrow\infty$

$\lim_{x\rightarrow\infty}\textstyle\frac{128x^8-256x^6+160x^4-32x^2+1}{x^8}=128$

now apply this method to the nsb question with n instead of 8 and u solve the whole question in 1 line like this:

$\lim_{x\rightarrow\infty}\frac{{}_2F_1(-n,n;\frac{1}{2};\frac{1-x}{2})}{x^n}=2^{n-1}$

the fact that this higher level of abstraction results in much more efficient solution is very common in maths

killer queen · 2025-08-27T20:22:38+1000

I'm going to be completely honest this makes zero sense to me but I'm not surprised haha

what is an analytic continuation? what are a, b and c? I presume there's some sort of series going on, but how does the 2F1 fit in? in the n=8 example, is there math behind the scenes that gets from LHS to RHS, or is it some sort of 'formula' or 'process' one needs to remember in order to use this technique?

atp I'm just curious haha thank you for indulging me I'm very unlikely to understand with my basic 4U math

tywebb · 2025-08-27T20:32:41+1000

killer queen said:
I'm going to be completely honest this makes zero sense to me but I'm not surprised haha

what is an analytic continuation? what are a, b and c? I presume there's some sort of series going on, but how does the 2F1 fit in? in the n=8 example, is there math behind the scenes that gets from LHS to RHS, or is it some sort of 'formula' or 'process' one needs to remember in order to use this technique?

atp I'm just curious haha thank you for indulging me I'm very unlikely to understand with my basic 4U math

there are more details on hypergeometric functions at https://en.wikipedia.org/wiki/Hypergeometric_function

but do u at least understand this:

$\lim_{x\rightarrow\infty}\frac{128x^8-256x^6+160x^4-32x^2+1}{x^8}$

$=\lim_{x\rightarrow\infty}\left(128-\frac{256}{x^2}+\frac{160}{x^4}-\frac{32}{x^6}+\frac{1}{x^8}\right)$

$=128-0+0-0+0$

$=128$

$=2^7$

$=2^{8-1}$

?

that literally is all i did, just generalised with n instead of 8

coolcat6778 · 2025-08-27T20:39:53+1000

tywebb said:
well if they are that stupid they shouldn't put "or otherwise" in their exams.

but they do and that allows for all the oos under the sun

${}_2F_1(a,b;c;z)\text{ is the analytic continuation of }\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$

$\text{along any path in the complex plane that avoids the branch points 1 and infinity}$

$\text{where }(q)_n\text{ is the rising Pochhammer symbol}$

it is related to Chebyshev polynomials via

$\textstyle\cos(ax)={}_2F_1(-a,a;\frac{1}{2};\frac{1-\cos x}{2})$

and this is how it is related to the nsb question

by way of example with n=8

$\textstyle{}_2F_1(-8,8;\frac{1}{2};\frac{1-x}{2})=128x^8-256x^6+160x^4-32x^2+1$

hence without having to do a long calculation involving de Moivre's theorem we can deduce instantly that

$\cos(8\theta)=128\cos^8\theta-256\cos^6\theta+160\cos^4\theta-32\cos^2\theta+1$

$\text{now if u notice the leading coefficient }128=2^7=2^{8-1}$

$\text{one may wonder how one might extract this from the polynomial$

$128x^8-256x^6+160x^4-32x^2+1$

$\text{simply divide the whole thing by }x^8\text{ and take limit as }x\rightarrow\infty$

$\lim_{x\rightarrow\infty}\textstyle\frac{128x^8-256x^6+160x^4-32x^2+1}{x^8}=128$

now apply this method to the nsb question with n instead of 8 and u solve the whole question in 1 line like this:

$\lim_{x\rightarrow\infty}\frac{{}_2F_1(-n,n;\frac{1}{2};\frac{1-x}{2})}{x^n}=2^{n-1}$

the fact that this higher level of abstraction results in much more efficient solution is very common in maths

unless the markers of hsc math extension 2 papers are uni professors with masters in maths i don't think they would understand or remember this method

tywebb · 2025-08-28T17:34:00+1000

here is yet another way to do it, again using hypergeomeric function, but this time using n-th derivative instead of limit

$\textstyle\frac{1}{n!}\cdot\frac{d^n}{dx^n}\left({}_2F_1(-n,n;\frac{1}{2};\frac{1-x}{2})\right)=2^{n-1}$

tywebb · 2025-08-28T17:49:48+1000

wanna know how this works?

again although it looks fancy the idea is again quite simple

so again i go back to the example how to extract the 128 from function

$128x^8-256x^6+160x^4-32x^2+1$

8-th derivative is

$8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot128x^0$

so leading coefficient is

$\textstyle\frac{1}{8!}\cdot\frac{d^8}{dx^8}(128x^8-256x^6+160x^4-32x^2+1)=128$

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alternative solution to nsb 2025 Q16ciii (1 Viewer)

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