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Re: HSC 2012 Marathon :) Progress Thats just my progress on Step 3, but I keep running into that barrier, I want to be able to prove that v^{(k)}=0 \ \ \ OR \ \ \ C \ \ \ $(constant C)$ But I cant because I just end up with: v^{(k)}=\sum_{i=0}^{k-1}u^{(k-1)} Am I missing something here?

Have a go at this, very well in 2U limits

I understand what the +C is there for very well indeed, yet I have forgotten it before (rare occurrence now). Its mostly due to rushing the paper and time issues that people forget to put +C. Id like to see even the best mathematician be flawless 24/7

List of silly mistakes I have made: - Forgetting +C - Not dotting line when graphing inequalities (i.e. y > x needs inequalities) - Confusing the exact ratios - Misreading f^-1 (x) for f'(x) and actually finding f'(x) even though question asked for f-1(x) (yes this has truly happened to...

Now say in the 2U test tommorow we are given the generic maxima minima with something that is a pain to differentiate due to multiple chain rules etc. Say that differential is: u=vw But because it may contain square roots and stuff, I will square u u^2=v^2 w^2 Now u and u^2...

You never know, teacher put it in a quiz of ours to test who actually knew the difference between them.

Another question I thought up $Define the function $ f(x)=1+x+x^2+...+x^n \ \ \ \ $for $ \ \ -1<x<1\\ \\ $By using differentiation show that $ \sum_{r=1}^{n}rx^r=\frac{x-nx^n+x^{n+1}(n-1)}{(1-x)^2}

If it has a degrees sign: i.e. 2\cos{\frac{\pi}{5}^{\circ}} \ \rightarrow $ DEGREES$ 2\cos{\frac{\pi}{5}} \ \rightarrow $ RADIANS$ So basically if they give you the number without the degrees sign IT IS ALWAYS RADIANS If it has the degrees sign it is always DEGREES

Alright then lol. I will see if I can post the interest question that I got wrong.

The only reason why that condition is given is because it makes the series converge to something. In the case of 0 however it just well, stays put it doesnt do anything, so I assumed that it isnt a limiting sum

Yeah that happens to me with the ambiguous wording of the annuity questions :/ Sometimes I cant tell whether A_1= 1000 or A_1=1000(interest) Good idea with the number line though

Oh I see what I did, when I counted on my fingers from 1991, I started with t=1 instead of t=0 I read this good 'formula' for counting: \sum_{q}^{p} u The number of terms are p-q+1 Does this work with EVERYthing that I have to count such as years, number of times interest is added etc.?

If you have 2U fitzpatrick they have a nice formula sheet at the back

Well I have problems in general with finding the amount of things such as number of years/interest etc. I got a question wrong in my CSSA trial but I cant find it so I cant post it, but I got this question 'wrong' from HSC 2004 Part iii, I know exactly how to do everything, the problem is...

Doing 2U trials and the HSCs, I sometimes lose marks over the annuity and loan repayment questions (the more complex ones). I usually lose marks because I miscount the number of times interest is added and such and it frustrates me quite a bit. So I am going to ask, how do people approach...

Im feeling fine about the 2U paper, its just that I cant be bothered studying for 2u since the papers are boring compared to 3U :/

Re: HSC 2012 Marathon :) Alright nice I will see if I can complete this tomorrow (inb4 detroyed)

Re: HSC 2012 Marathon :) Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol $Define the polynomial $ u=a_0 x^n+a_1 x^{n-1}+...+a_n $Step 1 $: n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ $for all n in sequence$ \\...

Re: HSC 2012 Marathon :) When you define u' does it mean this: u=(u_1,u_2,u_3...) \\ \\ u'=(u_1',u_2',u_3'...) If it means this then I will have a go at the question first thing tomorrow.

Re: HSC 2012 Marathon :) $Using the definition of e$ \\ \\ e=\lim_{n \to \infty}(1+\frac{1}{n})^n \\ \\ $Prove from first principles $ \frac{d}{dx}(\ln x)=\frac{1}{x} \\ \\ $Use the substitution $ u=\frac{h}{x} \ \ \ $for $ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}