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Lol, how do you know my last name? And so it wouldn't be an issue for you that in order to utilize and study from the textbook, you will always have to carry around two books at the same time instead of one (if you wanna be able to check your work)? Cause I figured option 2 would mean that for...

Hi all, I have a question for which I have included a poll for you to vote on below: Suppose that a textbook is to be written for the extension 2 new syllabus that is set to include very detailed solutions, such that the book would most likely become too big and heavy if everything was all...

An alternative approach for part (i) can be done as follows: \begin{align*} \frac{z_{1}-z_{2}}{z_{1}+z_{2}} &= ki \text{ } (1) \\[0.4cm] \Longrightarrow \overline{(\frac{z_{1}-z_{2}}{z_{1}+z_{2}})} &= -ki \\[0.4cm] \Longrightarrow...

For your solution to the last question of the paper, although not impactful to the proof, you mistakenly represented \overline{\alpha} - \alpha as -2\text{Im}(\alpha) when it really should be -2i\text{Im}(\alpha). The point I'm making is that, by definition, \text{Im}(\alpha)\in\mathbb{R}, so...

If we consider a situation where for every n tosses, we get exactly r heads, then we can calculate P(\text{obtaining a head}) = \frac{\text{total number of heads}}{\text{total number of tosses}} = \frac{r}{n} , which I believe is exactly what you're saying. So yes, you are right. The...

Essentially, the outcome after carrying out the n tosses is a total of r heads (which implies n-r tails). Your task is to determine how many arrangements of the r heads and n-r tails are possible if the first "slot" is taken up by a head, and there is basically no restriction on the remaining...

Basically, you have: 1 + \sum\limits_{k=1}^{n}u_{k}x^{k} + \frac{x^{n+1}R_{n}(x)}{(1-ax)(1-bx)} \equiv -1 + \frac{1}{1-ax} + \frac{1}{1-bx} Use the trick \frac{1}{1-ax} = 1 + \frac{ax}{1-ax} = 1 + ax(1+ \frac{ax}{1-ax}) = 1 + ax + a^{2}x^{2}(1+\frac{ax}{1-ax}) = ... = 1 +...

Yeah I just tried that and you're right. I think that means it's impossible to have an imaginary root that lies halfway between the origin and the point of inflection. I'll try a few other ideas and see where it gets me..

Try k = \pm\frac{\sqrt{7}}{2}.

Okay, I will change the question to this: Suppose that \alpha is a non-real root of P occurring when k\neq\pm1 such that its position in the Argand plane corresponds to the midpoint between the origin and the point of inflection of P in the Cartesian plane. (f) Show that \text{Re}(\alpha) =...

There's one real root of multiplicity 3 when k = \pm 1, i.e there are three real roots - they just all happen to have the same value.

Okay so I just realized I made yet another mistake and my amendment actually is possible, but you end up getting k = \pm\sqrt{1+\sqrt{6}/2} which isn't very 'nice-looking' so I think I'll still try to change it further.

You are actually correct. I just remembered that a while ago when I was trying to come up with polynomials questions, I tried to set one up that had a monic cubic equation with the remaining coefficients being all arbitrary, such that the cubic had a non-real root corresponding to its point of...

Yep, you can also use that same line of reasoning for the proof by contradiction approach.

Lol, I didn't even see that - yes, they are trivial..whoops. You can also do part (a) by contradiction if you wanted to.

It looks good, BUT... when you were showing that k\pm\sqrt{k^{2}-1} both have the same sign for any k (in fact, the same sign as k), there should be a total of four cases to consider. These are: 1) k-\sqrt{k^{2}-1} > 0 \text{ for } k\geq1 2) k+\sqrt{k^{2}-1} < 0 \text{ for } k\leq-1 3)...

Just a fun question I made up.

\begin{align*} \int\frac{x^{5}+3}{(x-5)^{3}(x-1)}dx &= \int\frac{x^{5}-1+4}{(x-5)^{3}(x-1)}dx \\ &= \int\frac{x^{5}-1}{(x-5)^{3}(x-1)}dx + \int\frac{4}{(x-5)^{3}(x-1)}dx \\ &= \int\frac{(x-1)(x^{4}+x^{3}+x^{2}+x+1)}{(x-5)^{3}(x-1)}dx + \int\frac{4}{(x-5)^{3}(x-1)}dx \\ &=...

It will take quite some time for me to type it all up so not sure if I'll do it tonight. Though I should mention that my method still uses partial fractions at one point, but it's a much simpler case that can be done by inspection.

How long exactly is your partial fraction approach? I do have an alternative method but it may be even longer than the way you did it.