Putnam Rush
by djmathman, Nov 27, 2017, 5:58 AM
because this is actually in a week oops
I should probably be doing some harder problems, but meh, my main hope is that I get HM again which doesn't really require going into the back end of the test too much. (Instead it seems it's just about being able to solve a good number of the early numbered questions on a consistent basis, which doesn't feel unreachable.)
On an unrelated note, I have 700+ moons in Mario Odyssey right now so I'd say that game was worth the wait
I should probably be doing some harder problems, but meh, my main hope is that I get HM again which doesn't really require going into the back end of the test too much. (Instead it seems it's just about being able to solve a good number of the early numbered questions on a consistent basis, which doesn't feel unreachable.)
Putnam 2012 A2 wrote:
Let 
be a commutative and associative binary operation on a set 
Assume that for every 
and 
in 
there exists 
in 
such that 
(This may depend on and 
) Show that if 
are in and 
then 
be a commutative and associative binary operation on a set 
Assume that for every 
and 
in 
there exists 
in 
such that 
(This may depend on and 
) Show that if 
are in and 
then 
there exists 
such that 
,![\[a * s = a * (c * r) = (a*c)*r = (b*c) * r = b * (c*r) = b * s.\]](http://latex.artofproblemsolving.com/f/5/5/f55b0375efa410b92fb1ca0fe4eb2b531bbbb2ef.png)
In other words, 
for any 
and 
such that ![\[a * x = b\quad\text{and}\quad b * y = a.\]](http://latex.artofproblemsolving.com/d/6/8/d688693b1768c8933ea69ac4fe213345c5b3617c.png)
Then ![\[a = b * y = (a * x) * y = a * (x * y)\]](http://latex.artofproblemsolving.com/3/d/3/3d3842baf0d4bfa55ea53d5773378f3d8f7e68df.png)
and ![\[b = a * x = (b * y) * x = b * (y * x).\]](http://latex.artofproblemsolving.com/6/2/c/62c2ec44e4044d742cac9432e0339c1bf357946c.png)
Since 
from the above, we must have 
as desired.Putnam 2008 B2 wrote:
Let 
For 
and 
let 
Evaluate 
For 
and 
let 
Evaluate 
denote the 
Harmonic number for 
,
is defined as an empty sum. I claim that ![\[n!F_n(x) = x^n\ln x - x^n\mathcal H_n\]](http://latex.artofproblemsolving.com/d/1/3/d13433213e7f3c45f0bc796b889bf00275ffd2f9.png)
for all 
.
is trivial, since ![\[0!F_0(x) = 1\cdot \ln x = x^0\ln x - x^0\mathcal H_0.\]](http://latex.artofproblemsolving.com/8/b/9/8b9dffcbdf68c10a12de21c2c8d0eb643bf6c61a.png)
Now assume the result hold for some integer 
Now remark that by L'Hopital's Rule, ![\[\lim_{y\to 0}y^{n}\ln y = \lim_{y\to 0}\frac{\ln y}{y^{-n}} = \lim_{y\to 0}\frac{1/y}{-ny^{-n-1}} = \lim_{y\to 0}-\frac1ny^n = 0\]](http://latex.artofproblemsolving.com/3/5/a/35ab7978bbf5364d1c19d0145c470706d8878720.png)
for all 
to clear denominators the given rewrites as ![\[(n+1)!F_{n+1}(x) = x^{n+1}\ln x - \frac{x^{n+1}}{n+1} - \mathcal H_nx^{n+1} = x^{n+1}\ln x - x^{n+1}\mathcal H_{n+1},\]](http://latex.artofproblemsolving.com/2/1/9/219da87471d7c97a2f47613fb82aba2020b730ef.png)
completing the induction.
,![\[\lim_{n\to \infty}\frac{n!F_n(1)}{\ln n} = \lim_{n\to\infty}\frac{-\mathcal H_n}{\ln n} = \boxed{-1}.\]](http://latex.artofproblemsolving.com/4/d/f/4df9a94ee57223b0ff6a0df4d5f48a9cf619c9bf.png)
Putnam 1990 B5 wrote:
Is there an infinite sequence 
of nonzero real numbers such that for 
, the polynomial ![\[p_n(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n\]](//latex.artofproblemsolving.com/1/1/6/116d1f6fd174d34934236bacaa78d98e0651209b.png)
has exactly 
distinct real roots?
of nonzero real numbers such that for 
, the polynomial ![\[p_n(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n\]](http://latex.artofproblemsolving.com/1/1/6/116d1f6fd174d34934236bacaa78d98e0651209b.png)
has exactly 
distinct real roots?
and 
.
has been constructed to have ![\[r_1 < r_2 < \cdots < r_n.\]](http://latex.artofproblemsolving.com/2/e/7/2e7882c91348c18135a32fa64d6ca15a64da7ff8.png)
Note that by the Extremal Value Theorem there exists 
such that 
obtains a maximum in 
at 
(note that the "all roots distinct" condition forces this to be in the open interval). Now choose 
so small that ![\[|a_{n+1}x^{n+1}| < \frac12\min\{|p(c_i)|:1\leq i\leq n-1\}\quad\text{for all }x\in [r_1,r_n].\]](http://latex.artofproblemsolving.com/4/b/7/4b7a3cb9467df7617cb70aa42ae7ffe58f969c93.png)
Then Intermediate Value Theorem on the intervals ![\[(-\infty,c_1),(c_1,c_2),\ldots, (c_{n-1},\infty)\]](http://latex.artofproblemsolving.com/8/f/8/8f8c56e6d8aea63b9e44fc4e3ca64f275c058da6.png)
guarantees the existence of ![\[p_{n+1}(x) := p_n(x) - a_{n+1}x^{n+1}\]](http://latex.artofproblemsolving.com/b/e/2/be2fc24da1f95d97a0d056b8709e54cf191c5f18.png)
has Putnam 2008 A5 wrote:
Let 
be an integer. Let 
and 
be polynomials with real coefficients such that the points 
in 
are the vertices of a regular 
-gon in counterclockwise order. Prove that at least one of and has degree greater than or equal to 
be an integer. Let 
and 
be polynomials with real coefficients such that the points 
in 
are the vertices of a regular 
-gon in counterclockwise order. Prove that at least one of and has degree greater than or equal to 
are the 
and 
are polynomials with degree 
satisfying ![\[f(j) = \cos\frac{2\pi j}n\quad\text{and}\quad g(j) = \sin\frac{2\pi j}n\]](http://latex.artofproblemsolving.com/7/c/a/7cad68e722717ba65632843efdbe25f41f037a8e.png)
for all 
.
to be an ![\[f(j) + ig(j) = \omega^j\]](http://latex.artofproblemsolving.com/4/0/e/40e886c8dcf6ada7666391f180e1ffc6de084f26.png)
for all 
,![\[\sum_{j=1}^n(-1)^j\binom{n-1}j\cos\frac{2\pi j}n = 0\]](http://latex.artofproblemsolving.com/d/c/3/dc37652247949e3a767f39a71c7ccb967c39fa0a.png)
by finite differences. Similarly, the sum with sines also equals zero. But then 
so the real and imaginary parts can't both be zero, contradiction. Hence one of 
.
.
for shorthand. Denote by 
the complex number representing the center of the polygon in question. Then there exists a complex number 
with 
such that the numbers ![\[\omega(z_j - r)\quad(1\leq j\leq n)\]](http://latex.artofproblemsolving.com/d/f/6/df66fb7c6baa35246e9f2300f0b3c31ca9f95240.png)
form the roots of unity polygon as described in the first case. Now writing 
and 
,![\begin{align*}\omega(z_j - r) &= (p+qi)(f(j) + ig(j) - a - bi) \\&= p(f(j) - a) - q(g(j) - b) + \left[q(f(j) - a) + p(g(j) - b)\right]i,\end{align*}](http://latex.artofproblemsolving.com/8/4/3/8432657037dd3f9a9d8c9b00edbdd54e0fb11520.png)
so by the previous part, one of the polynomials ![\[p(f(x) - a) - q(g(x) - b)\quad\text{or}\quad q(f(x) - a) + p(g(x) - b)\]](http://latex.artofproblemsolving.com/2/2/8/22847fc0561041766f5656f9069bd48b24cedd13.png)
must have degree at least 
.
and 
,On an unrelated note, I have 700+ moons in Mario Odyssey right now so I'd say that game was worth the wait
This post has been edited 5 times. Last edited by djmathman, Nov 27, 2017, 6:21 AM
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