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Clearly 1,\frac{1}{a},\frac{1}{a^2},\frac{1}{a^3},\frac{1}{a^4} are the 5 distinct roots of x^5-1=0 (sub them in to see) so the denominator is just the product of all factors of x^5-1

^^ dw, Shadowdude's "method" is needlessly complicated for a question that could be more easily done by high school calculus.

You have: I_n = \frac{x}{(1+x^2)^n} + 2nI_n - 2nI_{n+1} Just make I_{n+1} the subject and replace n with n-1 in your final equation

This: \int \frac{1}{(1+x^2)^n} dx &= \frac{x}{(1+x^2)^n} + 2n \int \frac{x^2}{(1+x^2)^{n+1}} dx\\&=\frac{x}{(1+x^2)^n} + 2n (\int \frac{1}{(1+x^2)^{n}}dx - \int \frac{1}{(1+x^2)^{n+1}}dx) Should be easy from here, just replace n with n-1

Not quite right... try using the fact cot(\theta)=tan(\frac{\pi}{2}-\theta) and an appropriate substitution

Use latex for maths summaries if you are doing so electronically. Just search and download miktex which contains packages you need and find a latex editor of your choice. Use google/wiki to learn latex. For graphing, use what mitchy_boy suggested.

A common misconception is that maths complements actuarial studies. In reality, actuarial studies is more like statistics than like maths. So if you were going to do a major purely to complement actuarial studies, then do a stats major if you want to complement actuarial. If you want to learn...

You really should know the whole topic before uni. As Drongoski pointed out, it's not very hard. It's bread and butter stuff for most actuarial courses you do.

For maths, you need at least 70 WAM, while for actuarial, generally those who do it have above 80 WAM. However, it really depends on whether you can find a supervisor who is willing to take you on.

B

Irregardless on whether its in the exam or not, you really should know it well especially since you want to do actuarial studies in uni.

You have to do it in maths. You can do actuarial honours in the year after that.

Consider each case separately: 1) Three digit numbers - First digit must be a five or a six, and last digit must be 2,4,6 (even number). So deal with the restrictions first. Suppose the first digit is 5, then there are 3 choices for the last digit (2,4,6), and hence for the middle digit, there...

we have lim_{t\rightarrow \infty} t^n e^{-t} =0 for any n. This is because the rate at which e^{-t} tends towards 0 is much quicker than the rate at which t^n tends towards infinity. Thus e^{-t} dominates t^n and hence the limit is 0. More formally, using L'Hopital Rule for integer...

Upper bound would be infinity \int_0^{\infty} \frac{du}{4(\frac{3}{4}+u^2)} &= [\frac{2}{4\sqrt{3}} tan^{-1} (\frac{2u}{\sqrt{3}})]_0^{\infty} Note lim_{n\rightarrow \infty} tan^{-1} (n) = \frac{\pi}{2} So the result should be: \frac{2}{4\sqrt{3}}*\frac{\pi}{2} = \frac{\pi}{4\sqrt{3}} Someone...

Question 1 \int \frac{1}{1+sin^2(x)} dx = \int \frac{sec^2(x)}{sec^2(x)+tan^2(x)} dx Then let u=tan(x) So: \int \frac{sec^2(x)}{sec^2(x)+tan^2(x)} dx &= \int \frac{du}{1+2u^2} which should be easy to solve

The question suggests that for each car, there is a probability of 3/100 that it is faulty (and hence a probability of 97/100 that it is not faulty). You are not picking cars out of a group of 100. For 1 faulty, there are three cases: (Faulty, Ok, Ok), (Ok, Faulty, Ok) and (Ok, Ok, Faulty)...

I'm pretty sure you mean \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}=x In that case: x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}=1+x Solving gives x=\frac{1+\sqrt{5}}{2} (Discard negative root of quadratic)

[tex.]\cos^{-1}x-\sin^{-1}x = k[/tex.] but without the dots As for the question \cos^{-1}x-\sin^{-1}x &= k\\ \cos^{-1}x-(\frac{\pi}{2}-\cos^{-1}x) &= k\\ \cos^{-1}x &= \frac{k}{2}+\frac{\pi}{4} Result follows

Yes you can do it by trig as well eg. substitute x=tan\theta