![$f,g \in C^1([0,1])$](http://latex.artofproblemsolving.com/6/1/7/617c5be50035205b21074842d7266911aa1dc4d8.png)
.
?
be a ring with unity such that for every 
there exist 
such that 
.
then 
then the result from a) may no longer hold.
,![\[
\sum_{i=1}^{n} \tan^{-1} r_i = \tan^{-1} \frac{a_1 - a_3 + a_5 - \cdots}{a_0 - a_2 + a_4 - \cdots}
\]](http://latex.artofproblemsolving.com/8/1/2/812c7c690b2237f45bc7453ff2b4160fb72c7b0d.png)
and![\[
\sum_{i=1}^{n} \tanh^{-1} r_i = \tanh^{-1} \frac{a_1 + a_3 + a_5 + \cdots}{a_0 + a_2 + a_4 + \cdots}.
\]](http://latex.artofproblemsolving.com/6/1/9/61962993a2c34e1c0ce961d7057127ae75876ac2.png)
![$f \in C([0,1];[0,1])$](http://latex.artofproblemsolving.com/4/7/5/475518b7db0b787dd90b93f602c7b912c663fba3.png)
bijective, 
and ![$(a_k) \in [0,1]^\mathbb N$](http://latex.artofproblemsolving.com/a/d/8/ad8a80d847c60ee8e916a1f9b08b7b03ec17c814.png)
with 
converge.
converge?
Y by Adventure10, Mango247, kiyoras_2001
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 
is a polynomial with complex coefficients. The claim that at least one of 
and 
has degree at least 
is precisely the claim that the degree of 
for some fixed 
(the center of the polygon), some 
(the radius of the circle that circumscribes the polygon) and some 
(a measure of how far the polygon is rotated from standard position.) Subtracting a constant such as 
or 
does not change the degree of a polynomial. Hence WLOG we may assume that 
for 
By the binomial theorem, 
equals 
plus terms of degree lower than 
It follows that if 
for 
then 
is a polynomial of degree exactly 
)
at any point, then the degree of 
![\[ P(k) = \zeta^{k - 1}\text{ for }0\le k\le n - 1\]](http://latex.artofproblemsolving.com/c/1/9/c19fb238b952aa9aab82a0690b0319201dfb1356.png)
![\[ \Delta P(k) = \zeta^{k - 1}(\zeta - 1)\text{ for }0\le k\le n - 2
\]](http://latex.artofproblemsolving.com/e/6/f/e6fc4e828d43dc863198d85c36a6def0fec962a3.png)
![\[ \Delta^2 P(k) = \zeta^{k - 1}(\zeta - 1)^2\text{ for }0\le k\le n - 3\]](http://latex.artofproblemsolving.com/1/3/6/136c173795cfcfce38205e1ea5f632634932497b.png)
![\[ \vdots\]](http://latex.artofproblemsolving.com/6/5/9/659828d91cea2fe7fb8c7d07b9a501c7b808e818.png)
![\[ \Delta^{n - 1}P(0) = (\zeta - 1)^{n - 1}\]](http://latex.artofproblemsolving.com/8/c/e/8ce1e67c41f485ea684ccb156721b8a9d5652538.png)
this shows that 
which means that the degree of 
.
difference of any sequence with 
without changing its degree so that the polygon is centered at the origin. Call the new polynomial 
.
,
,
.
.
,
,
,
.
.![\[ \sum_{k = 1}^{n - 1} ( - 1)^{k - n + 1} \left(\!\!\begin{array}{c}n - 1 \\
k - 1\end{array}\!\!\right) \cos{\left(\theta + \frac {2\pi (k - 1)}{n}\right)} = \cos{\left(\theta - \frac {2\pi}{n}\right)}
\]](http://latex.artofproblemsolving.com/8/7/0/870b6daf5e6d64332a0219d76c15a904d060090a.png)
![\[ \sum_{k = 1}^{n - 1} ( - 1)^{k - n + 1} \left(\!\!\begin{array}{c}n - 1 \\
k - 1\end{array}\!\!\right) \sin{\left(\theta + \frac {2\pi (k - 1)}{n}\right)} = \sin{\left(\theta - \frac {2\pi}{n}\right)}
\]](http://latex.artofproblemsolving.com/f/b/f/fbf6c6a547fcc90fcea275e7d2788c8e41ed04f3.png)
times the second equation is 
on the LHS and 
on the RHS. The factor on the LHS has absolute value 
if and only if 
.
is not identically zero, but nice.
times the leading coefficient of 
![\[ P(k) = \zeta^{k - 1}\text{ for }0\le k\le n - 1
\]](http://latex.artofproblemsolving.com/c/6/0/c60d80eaebefb6ee46e428ee822efd20f6705ea9.png)
![\[ \Delta^2 P(k) = \zeta^{k - 1}(\zeta - 1)^2\text{ for }0\le k\le n - 3
\]](http://latex.artofproblemsolving.com/7/e/b/7ebef2df348f147fe751c709c0130c6f68185d13.png)
![\[ \vdots
\]](http://latex.artofproblemsolving.com/5/0/6/5064015d4d81a8fc828f9bbc4fba771bbe94505a.png)
![\[ \Delta^{n - 1}P(0) = (\zeta - 1)^{n - 1}
\]](http://latex.artofproblemsolving.com/3/a/2/3a2047a3c2b1c88cf1ccd6115eb09e431150b63e.png)
instead of 
and 
instead of 
?
instead, although that's a cosmetic distinction.![\[ L(z) = \sum_{k=1}^n e^{2 \pi i (k-1)/n}
\frac{
(z-1)
(z-2)
\cdots
(z-(k-1))
(z-(k+1))
\cdots
(z-n)}
{(k-1)
(k - 2)
\cdots(k - (k-1) )
(k - (k+1))
\cdots (k - n)
}\]](http://latex.artofproblemsolving.com/c/0/2/c02d0aebe1de766f07c76438a5fca90af7eebf66.png)
![\[ \sum_{k=1}^n e^{2 \pi i (k-1)/n}
\frac{1}
{(k-1)
(k - 2)
\cdots(k - (k-1) )
(k - (k+1))
\cdots (k - n)
}\]](http://latex.artofproblemsolving.com/c/f/8/cf879ef679a30a03e7d07775df50b326314c821d.png)
to make it a binomial coefficient times -1 to the something, then apply the binomial theorem.
coordinates are well-spaced, the interpolation is much easier using Newton polynomials than using Lagrange interpolation. See 
doesn't affect its degree.
be the angle of rotation. Then we have the identities:
and 
for 
.
and 
.
has at least 
roots, it is either of degree 
is the same as the degree of 
,
have degree at least 
,
.
,
for all 
for all integer 
,
,
has degree at least 
or 
have degree at least 
.
.
,
and 
are some complex numbers. for k = 1,2,...,n. Now lagrange interpolation tells us that![\[ P(x) = \sum_{k=1}^n P(k) \cdot \prod_{j = 1, j \neq k}^n \frac{x - j}{k - j}\]](http://latex.artofproblemsolving.com/4/e/9/4e926c9bea59fa22d22a811bd3c15c7d7f623660.png)
in this polynomial is non-zero, which will imply the result. It is not hard to see that coefficient of ![\[ \sum_{k = 1}^n P(k) \cdot \prod_{j = 1, j \neq k}^n \frac{1}{k-j}\]](http://latex.artofproblemsolving.com/1/a/2/1a2d9ceb3a1af8e951d025414e5599aa4888aa6e.png)
![\[ \prod_{j = 1, j \neq k}^n \frac{1}{k-j} = \frac{(-1)^{n-k}}{(k-1)!(n - k)!}\]](http://latex.artofproblemsolving.com/5/b/4/5b4c60035324a978db71bf134a42355502463d7c.png)
,![\[ \sum_{k = 1}^n P(k) \cdot (-1)^{n-k} \cdot \binom{n-1}{k-1}\]](http://latex.artofproblemsolving.com/9/d/4/9d45c1f2ffaa46b0c6fb4d86ec9059259543056f.png)
and multiplying it by some non-zero complex number clearly does not change the degree of the polynomial, so we may just assume that 
for k = 1,2,...,n. Now, all we have to show is that![\[\sum_{k=1}^n \omega^{k-1} \cdot (-1)^{n-k} \cdot \binom{n-1}{k-1}\]](http://latex.artofproblemsolving.com/a/8/9/a89addce8c157962072b806d439996bcfb94906a.png)
is nonzero. But by the binomial theorem, this is just 
,
,
.
and the vertices are 
,
for 
where 
for some 
and for all 
then 
.
;
,
by inductive hypothesis.
we have 
must have degree at least 
that satisfies 
for 
would have 
consider 
.
is a polynomial in 
.
,
.
,
,
.
,
,
,![\[\zeta g(X)-g(X+1)=0\qquad\text{for }X=0, 1, 2,\dots,n-2.\]](http://latex.artofproblemsolving.com/f/b/a/fbaaea9d66c46a1ae5db615aded0ac7cf61a0863.png)
Therefore 
divides 
,
implies that 
,
and 
we obtain the following:
.
by 
and equating real and imaginary parts with the complex number 
,
.
.
.
be a center of regular 
and let 
to 
,
be a real number such that 
.
,
.
.
,
.
of 
.
.
be coefficient of 
,
.
is not equal to 0 means 
f
g
,
.
forms a regular 
,![\[m(x+1) = \operatorname{cis } \left(\frac{2\pi}{n}\right) \cdot m(x)\]](http://latex.artofproblemsolving.com/1/5/3/1532b0c1fcf8d889f2f646a39e80352add40c262.png)
![\[n(x) = m(x+1) - m(x) \cdot \operatorname{cis } \left(\frac{2\pi}{n}\right)\]](http://latex.artofproblemsolving.com/8/3/0/830f83aa4b52fc5ad10077e3e8229ab7ac0bddd2.png)
.![\[n-1 \leq \deg(n) \leq \deg(m) \leq \max(\deg (f), \deg (g)).~\blacksquare\]](http://latex.artofproblemsolving.com/0/1/0/01098e2427e370532a34c6f975c80b5191d0dfe1.png)
.![\[h(x) = e^{\frac{2i\pi}{n}} h(x-1),\]](http://latex.artofproblemsolving.com/a/2/2/a22d11e281c46d82b939773161584bbf60600247.png)
.![\[h(x)-e^{\frac{2i\pi}{n}} h(x-1)\]](http://latex.artofproblemsolving.com/9/2/4/9247e34642bfac40c4e5c92bf4c8913e76ba5e53.png)
is the center of the polygon. Redefine 
.
where 
.
.
implies 
.
with the base case 
being trivial. Let 
.
.
,
.
and as 
,
-
.![\[H(x) = f(m) - a + ig(m) - bi\]](http://latex.artofproblemsolving.com/5/6/c/56c70dab1a29fb09c554ef6fa4a2276a4ae47613.png)
fulfills the property that 
where 
is a primitive root of unity of 
with 
,
denotes some random integer. We shall show that this polynomial has degree at least 
,
.
.![\[H(x) = \sum_{j=0}^{n-1} \omega^{j+k} \cdot \prod_{0 \leq r \neq j \leq n-1} \frac{x-r}{j - r}\]](http://latex.artofproblemsolving.com/3/0/f/30f31f0ef6da609ed10b5013e38ab3e5039568dc.png)
if this polynomial doesn't have degree ![\[0 = \sum_{j=0}^{n-1} \omega^{j+k} \cdot \prod_{0 \leq r \neq j \leq n-1} \frac{1}{j - r}\]](http://latex.artofproblemsolving.com/4/f/c/4fcf5f25b3fb40284929b0f98d954d307a171e81.png)
however we can just divide by 
and simplify to obtain![\[0 = \sum_{j=0}^{n-1} \omega^{j} \cdot \frac{1}{(j!){(-1)}^{n-1-j}(n-1-j)!} = \frac{1}{(n-1)!} \cdot \sum_{j=0}^{n-1} \binom{n-1}{j} \cdot {(-1)}^{n-1-j} \cdot \omega^{j}\]](http://latex.artofproblemsolving.com/a/5/c/a5c38e60ed2a72942c320f4453dbccb2da4fab22.png)
Getting rid of the 
term and using the binomial theorem gives us![\[0 = {(w-1)}^{n-1}\]](http://latex.artofproblemsolving.com/5/6/7/567422d62a1de4407fab481b0a3dfe2d0413d981.png)
which is plainly a contradiction.
where 
is a 
By finite differences, since
Hence, since 
we must have that 
