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I think that there's another little trick in this question, the force is only positive for the first three seconds so the particle is going to turn around. a=3/2-t/2 v=(3/2)t-(1/4)t^2 x=(3/4)t^2-(1/12)t^3 The particle is going its fastest in the positive direction at t=3 (before force...

basically use F=ma so a=F/m=(6-2t)/4=3/2-t/2 You should then be able to integrate up to find v=int(a)dt and x=int(v)dt putting in the appropriate constants so that v(0)=0 and x(0)=0. Then put v=19.6 to find the time, t_max, when it reaches this speed, then x(t_max) is the distance in the...

Yeah, nice one ngai. To show that Mathematica's answer is equivalent we need to use the complex logarithm: if z=r*e^(it) Log(z) = ln(r) + it (Note that this has most of the features of the real log, like exp(log(z)) = z, but is defined for all complex numbers). Then -1-i =...

Actually after reading over it again its pretty easy. from above, the integral ≤ x^(2n+2)/(2n+2) if x=1, x^(2n+2)/(2n+2) = 1/(2n+2) -> 0 as n->∞

Yeah, you're right Slide Rule, I didn't think about x=1. My answer doesn't cover it and I'm not sure how you could show that the integral goes to 0. Edit:see next message However maybe we don't need to show that. At x = 1 you get log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + .... Put 10...

0≤t≤1 1≤t+1≤2 1≥1/(t+1)≥1/2 so 1/(t+1)≤1 therefore t^(2n+1)/(t+1) ≤ t^(2n+1) therefore integral from 0 to x of LHS ≤ integral of RHS = x^(2n+2)/(2n+2) and x^(2n+2)/(2n+2) ≤ x^(2n+2) since n is positive and x^(2n+2) -> 0 as n->∞ because 0≤x≤1 since the integral we are interested in...

The integrator http://integrals.wolfram.com/ (which uses mathematica) gives integral(Log[Log[x]]) = xLog[Log[x] - LogIntegral[x] A bit of research shows that LogIntegral[x] = integral from 0 to x of dt/logt. So it seems that it isn't possible to do in terms of elementary functions. This...

I still don't understand your argument. In your original post you have a d₀ but when you expand out your result you get md₀. Also, your proof seems to work for real numbers as well as complex, so you have proved that x²+x+1=0 has 2 real solutions, which isn't...

I don't really follow this whole bit. Where does m come from? Its obviously meant to be something to do with the root but you seem to have pulled it out of the air. Also note (x+m)(x^k+d_k-1x^k-1+d_k-2x^k-2 +...+ d₁x+d₀) =...

Fundamental Theorem of Algebra: Every polynomial equation having complex coefficients and degree at least 1 has at least one complex root. This theorem was first proven by Gauss. The proof isn't in the HSC but is well within the grasp of a university course. I've actually seen proofs (that I...

let the roots be 1,a,a,b,b then a²b² = 1/16 (product of roots) and 2a + 2b + 1 = 0 (sum of roots) so b=-1/2-a we know that one of a and b is +ve and one -ve because they are cos(2Pi/5) and cos(4Pi/5) so ab is -ve. therefore ab = -1/4 so a(-1/2-a) = -1/4...

Yeah it looks right to me, 2cos(pi/5) = 1.618... then 1.618...^4 - 3*1.618...^2 + 1 = 0 (it might come up as something * 10^-5 but that's just rounding errors). The way I did it in the end was that used in another thread on this board. Using De Moivres theorem cos5t + isin5t = (cost +...

Actually, if cos5@ = 1 then 5@=0, 2Pi, 4Pi, 6Pi, 8Pi, 10Pi.... (Note that cosPi = -1). So then @=0, 2Pi/5, 4Pi/5, .... But the hard part of this question is solving the original polynomial (16x^5 - 20x^3 + 5x - 1 = 0) in terms of surds. (x - 1) is a factor but on dividing out I get 16x^4 +...

Yeah thats right, z=± sqrt((3 ± √ 5)/2) Then using a calculator we can see sqrt((3 + √ 5)/2) = 2cos(Pi/5) and similar results. But why is this? Surely there is a proof.

z^4 - 3z^2 + 1 = 0 Show that the roots are 2cos(Pi/5), 2cos(2Pi/5), 2cos(3Pi/5) and 2cos(4Pi/5). I've got a solution but its pretty long and indirect. I wonder if anybody can come up with a nice way of showing it.

Alright I've written up a long division method. If you remember standard long division it isn't too bad. First open the picture then read the description write x-2)x^2-x (with a line over the top) look at the highest powers in the divisor (x - 2) and the thing being divided (x^2 -x). We...

Yeah what you're thinking of is a way of writing it as a polynomial plus a ratio of polynomials with the top one being of smaller degree than the bottom. This way isn't long division but you might like it better. Basically you try and write the top so you can split it up and then cancel with the...

To be perfectly clear, none of the following is relevant in the NSW 4U maths course. In first year uni we only did a week or two of complex numbers but Euler's formula was used a lot. I wondered why we didn't learn it in high school. I guess its difficult to give an honest proof and its very...

All right, I don't think this result is especially trivial. Be nice to your maths teachers, when you haven't done complex analysis for years its easy to forget little tricks like these. Basically to understand this question you just have to know Euler's formula e^ix = cosx + isinx so i =...

I'm another random that got 1st in 3 unit maths (back in 2002) and at the time I wondered about this issue as well. I could definitely see how I got 100%, just answered all the questions and got them right, but I couldn't see why nobody else got full marks. Out of all the people who I'm sure...