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4.1: For n=2, obvious. Suppose true for n=k. Then consider n=k+1. Consider the graph of P(x). Suppose x=m (m constant) is a multiple zero for n=k+1. (1) Then that means that x=m is a zero for n=k. (2) So: From (1), 1+m-m^2/2......+(-1)^k.(m^k/k!)+(-1)^(k+1).(m^(k+1)/(k+1)!)=0 (A) From...

sorry :-) I specialise in not making sense...

I usually answer only the hard q's... but if there's an easy q lying around that no one has answered, why not do that? :-) For reference, usually I log on say once a day, and by the time I see a q someone has answered it 90% of the time... So the latter I guess..

Part (b): I'm lazy, so assuming xayma is right... x^2+x=y-y^2 So x^2+x+y^2-y=0 So (x+0.5)^2+(y-0.5)^2=0.5 So locus is circle on argand diagram with centre (-0.5+0.5i) with radius 1/root(2).

there's a reason why bashing it doesn't work... it's called RSA encryption... the reason keypad is doing it is because he doesn't realise how short these holidays are...

Short answer: no. Long answer: No. It all depends on how you define convergent series... there is, after all, no good way to well-define series as there is numbers. (eg you can well-order rationals, but not reals...) I suppose defining a new series b_n from a_n a convergent series with...

Soln follows, look away ***SPOILER WARNING*** if you want to do it yourself. Case bash. Consider n objects, and no. of ways of choosing k of them. Let one of them arbitrarily be called A. Case 1: A is chosen. Then we have to choose k-1 from the other n-1, hence n-1C(k-1). Case 2: A isn't...

sorry, my definition of maths is relatively narrow... as in pure maths, not applied... :-P (edit) or stats for that matter...ugh

Probably the best alternative method is spice girl's... Basically, "stretch" the plane in the direction of the y-axis. As you do that, the x-values don't change, so OT and ON remain constant. As a result, as the minor axis of the ellipse lies on the y-axis, you can "stretch" the plane such that...

Umm... you don't :-)

Umm, spice girl is spicing it up :-) Probably the other one is easier to understand and apply generally... unless you have an affinity for geometric transformations, ha ha...

see kimmeh... It's to do with limits as x -> infinity. y=(x^2)/(x^2-x+2)=1/(1+(1/x)-(2/x^2)) on dividing by the highest factor. Now observe that 1/x-2/x^2 has roots 0,2. Also, from that it is less than 0 for x < -2 and > 0 for x > 2. As x goes to infinity, the 2/x^2 term becomes less...

lol, no... I saw the film at preview sessions.. :-)

lose.

try 2 years if you're talking about maniacguy..

500. And no, doing 40 papers is not worth 500 dollars... do the maths :-P

you're wrong. Well... depends on who for I guess.

Just in case there are those who didn't follow my comment above, note that generators k have property of k^((p-1)/2) cong to -1 mod p

For p congruent to 1 mod 4, k being a generator means that -k is a generator. Hence the result. Yet to figure out for 3 mod 4, but it's getting late, more later...

Been away for a while.. and usually by the time I get to a post someone else has got a proof for it... gah. Hmm... didn't work. I don't think it's going to be easy... I'll figure out something.