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Re: HSC 2013 4U Marathon Nice work guys, I will post another question soon. (unless someone else could contribute heh, instead of me just giving questions out)

Re: HSC 2013 4U Marathon Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved: $The positive integers in this sequence are as follows$ \ \ \ \ (1) , (2,3) , (4, 5, 6) , (7, 8, 9, 10) , ... \\ \\ $Prove that the sum of the integers in the nth...

Re: HSC 2013 4U Marathon Oh, the taylor series representations for sin x and cos x and e^x, which is what I was implying when I talked about 'Euler's path' But yes there are too many rigour issues according to the earlier post, that complex exponentiation cannot be accounted for and there is...

Re: HSC 2013 4U Marathon This thing

Re: HSC 2013 4U Marathon Could we prove the Taylor Series formula from the generic power series representation: f(x) = \sum_{k=0}^{\infty} c_k x^k And then evaluating the constants in terms of the derivatives of f. This does of course assume that power series exists and that it converges...

Re: HSC 2013 4U Marathon Ooh, one of the better proofs I have seen (other than the original, which was more a deduction) EDIT: Can we assume that for the logarithm though? I read somewhere that someone once derived \ln(\cos x + i\sin x ) = ix but due to the periodicity of trig functions the...

Re: HSC 2013 4U Marathon Nice proof, I would of just done: \frac{dy}{dx} = y \\ \frac{dx}{dy}= \frac{1}{y} \\ x = \ln y + C \\ x-C = \ln y \ \ \ y=e^{x-C} = e^x \div e^C = ke^x \ \ \ \ \ $as C is constant$ EDIT: Happy to have 1000th post in the Maths section =)

Re: HSC 2013 4U Marathon Oh alright I see. Ah well, I can't seem to find a rigorous way to introduce Euler's formula into 4U

Re: HSC 2013 4U Marathon Heh, well that failed then. Am I still allowed to assume a function is sin x or cos x if the power series satisfies the definition y'' + y = 0 ? If not, how can we derive the taylor series for sin and cos using only the derivative definition? And for ke^x, am I...

Re: HSC 2013 4U Marathon I will use my rote learned formula: ^1 C _ 1 = \frac{1!}{(1-1)! \cdot 1!} = \frac{1}{1 \cdot 1} = 1

Re: HSC 2013 4U Marathon Nice work :) $i) Show that $ \ \ y=e^x \ \ $is the only function which satisfies$ \ \ \ y'=y \ \ \ \ \ \fbox{1} $ii) Hence show that $ \\ \\ $If$ \ \ \ \ E(x) = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{x^n}{n!} \\ \\ E(x) = e^x \ \ \ \ \ \fbox{2} $iii) You...

Ah I see, alright. d/dx \sin x = \sin x (\lim_{n \to \infty} \sqrt{n^2-1}-n) + \cos x Using the same conjugate technique I arrive at: d/dx \sin x = \sin x (\lim_{n \to \infty} \frac{-1}{\sqrt{n^2-1}+n}) + \cos x = \cos x Not sure why I didn't see that earlier lol

By multiplying by 1, where we pick the 1 to be the conjugate of the expression, we arrive at: \lim_{n \to \infty} \frac{-n}{\sqrt{n^2+n}+n} \\ \\ =\lim_{n \to \infty} \frac{-\sqrt{n}}{\sqrt{n-1}+\sqrt{n}} = -\frac{1}{2} Yes?

Re: HSC 2013 4U Marathon Nice work both of you: As for a hint to the second part if anyone wants it, I have it in white here: Consider E(alpha) where alpha is the stationary point of E(x) (this is how I proved it anyway)

I this is a contradiction to my assumption or something to consider in my proof? Because I can't see how to use that.

Nice proof there

Nevermind lol: Cant I just say that \lim_{n \to \infty} \sqrt{n^2-1} = \lim_{n \to \infty}n And hence they cancel out?

Re: HSC 2013 4U Marathon No problem man. E(x)=\sum_{r=0}^{n} \frac{x^{r}}{r!} $i) Show that E(x) does not have a double root$ \\ \\ $ii) Show that E(x) does not have any real roots if n is even$

Re: HSC 2013 4U Marathon $Show that$ \\ \\ \sum_{k=1}^{\infty} \frac{1}{k^2} = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int \frac{1-x^n}{1-x} \ dx \right) \ dx

re: HSC Chemistry Marathon Archive Yeah this is what I was hoping people would do but it doesn't really matter.