That wouldn't be too hard.
But the product is for k=1 to n, so n
Oh never mind, I mis-interpreted K as a fixed value. Didn't see it as the dummy variable.But the product is for k=1 to n, so nk
I know I didn't say this, but it was implied from the question ok
Yea that's right but I did it in a different way without using Riemann Sums=\frac{1}{n}\sum_{k=1}^n\log\left(1+\frac{k}{n}\right).\\ \textrm{ which is a Riemann sum.} \\ \\ \Rightarrow \lim_{n\rightarrow\infty} \log(S_n) = \int_1^2 \log(t)\, dt = 2\log(2)-1.\\ \\ \textrm{Exponentiate both sides to end up with }4e^{-1}.)
I'm not sure if this is what you're looking for:With MX2 a week away, I will post one or two difficult questions each day, hopefully covering all topics (polynomials, integration, etc). As usual, this is more to stretch good students and hone their problemsolving skill for Q16s than to provide exercises at doing the routine stuff quickly and accurately...practice and mindset is all there is to the latter.
First, polynomials. A result similar in flavour to the conjugate root theorem:
$ be a polynomial with rational coefficients such that $p(A^{1/3}+A^{-1/3})=0.$ Find all other roots (with proof) that any such $p$ must have. $)
Cool, so you are mostly done. What you have done is constructed a particular rational polynomial (your cubic) with alpha as a root. You have then found the two other roots of this poly (which look like the two roots I am looking for, although I haven't double-checked).I'm not sure if this is what you're looking for:
By cubing, then subbing x back in
Considering the graph of this cubic, we find that the maximum turning point at x=-1 is
Since,
Forming a quadratic equation
Subbing that in, in terms of alpha
Those are the other 2 roots that must be there in order to create a rational polynomial p(x) with those roots.
So for a general polynomial Q(x), with a root
So close, but why "must" we have Q=pR? Why must any Q be divisible by p? This needs proof but otherwise good.So for a general polynomial Q(x), with a root, in order to 'ensure rationality', then
R(x) being a rational polynomial, therefore any rational polynomial Q(x) that has
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I have a feeling that I haven't actually proven anything here lol
im not quite sure if this is right?? i got the expression I = k^n n!/(m+1)(m+n+1)(m+2n+1)...(m+kn+1)
^n < \frac{1}{n!} \int_0^1 x^m (1-x^k)^n \ dx < \left( \frac{k}{m+1} \right)^n )
For I, I'm getting:
To have pk points make k polygons, then there are p points per polygon, in order for there to be no polygons with no common area, then all points for each polygon must be 'grouped together', that is if I have A1 A2 A3 belonging to one polygon and B1 B2 B3 belonging to another, in cyclic oritentation:Consider a circle withequally spaced points around it's circumference.
I chooseof these points where
I then makeamount of polygons using my chosen
and
Find the probability that NO polygons share a common area.
I don't think that's correct. Consider ifTo have pk points make k polygons, then there are p points per polygon, in order for there to be no polygons with no common area, then all points for each polygon must be 'grouped together', that is if I have A1 A2 A3 belonging to one polygon and B1 B2 B3 belonging to another, in cyclic oritentation:
A1 B1 A2 A3 B2 B3, will be such that they overlap
The only way that they dont overalp is if they are grouped together, i.e.
A1 A2 A3 B1 B2 B3
Therefore, for the k groups of p points, we are simply going to find the number of ways to arrange pk points into k groups of p points, and divide by total ways to pick pk points from pn points
Number of ways to arrange k groups of p points around a circle is:
Hopefully that is correct
That's true, I think my calculation is wrong, I am still intent on the grouping idea.I don't think that's correct. Consider ifso I choose all the points on the circle. Then the denominator is 1 since pn=pk. So the probability becomes:
Which will give you a probability > 1
(correct me if I have misinterpreted you)