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Re: 2012 HSC MX2 Marathon Perhaps time for a hint...the derivatives of this polynomial look quite similar to it! A sign that some sort of induction will work.

Re: HSC 2012 Marathon :) Doesn't make it any more legal unfortunately...

Answer: 16x-x^3. Since it is odd, it must have roots -4,0,4. So p(x)=kx(x^2-16). The remainder theorem tells us p(3)=21 which lets us find k=-1.

Re: HSC 2012 Marathon :) This is the problem with your argument Sy123, you cannot treat such a series "term by term". There isn't really a proof of this equality that a 3U student could be expected to produce...I will post a sandwich argument a bit later.

Re: HSC 2012 Marathon :) That is all this question needs :), but it is sort of uni analysis-y.

Re: HSC 2012 Marathon :) A classic fallacious argument in analysis. The number of terms in this series is not fixed, so we cannot simply take the limits of each term individually and add them. If things like this were legit then we would have for example lim (1+1/n)^n = lim 1^n = 1, as each...

Re: HSC 2012 Marathon :) Equation too long haha.

You can't find "the equation" of that curve, many different equations will have identical looking sketches.

Some graphing software will 'simplify' the expression before graphing it, this can cause some domain problems.

There should be an open circle at x=0. The expression 0/0 is undefined, so the function we are sketching is not defined at x=0. That the limit of f(x) is 1 as x approaches 0 is irrelevant. Eg, the graph of y=x/x is the constant line y=1 with an open circle at the point (0,1). Look up the...

emery <3. good taste bro.

grats! (and nice subject choices.)

Re: 2012 HSC MX2 Marathon Not quite, you are adding the two terms x^{2n}/(2n)! and x^{2n-1}/(2n-1)!. You would need the SUM of these to be positive for negative x to make your argument work. (Or equivalently you could show that their difference is positive for positive x). And in fact this is...

Re: 2012 HSC MX2 Marathon It's not what I did but it might lead somewhere :).

and http://www.youtube.com/watch?v=kxrIBCwyk80

http://www.youtube.com/watch?v=Psk09J-ZqXU

For maths mostly. The weighting is very heavily towards exams, and how well you do in a course is essentially determined by how good you are at taking exams. Every other subject I have done is assessed very differently. The reason for this is presumably because it takes a lot less effort to...

this.

Re: 2012 HSC MX2 Marathon Nope, at least not the way I did it. (And I don't see how that fact would be helpful. Remember, for a fixed n, this polynomial will resemble e^x less and less the further you get from x=0, so the properties of the function e^x have little bearing here.)

Re: 2012 HSC MX2 Marathon Also, the sum of the squares of the reciprocals of the \beta_i's being negative does not imply that every \beta_i has nonzero imaginary part, it only implies that SOME of the \beta_i's have nonzero imaginary part. Eg (i*sqrt(2))^2+1^2=-1<0 even though 1 is real.