USAMO 2002 Problem 3

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MithsApprentice

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MithsApprentice

#1 h Sep 30, 2005, 7:49 PM • 5 Y

Y by mijail, samrocksnature, Adventure10, hurdler, megarnie

Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $n$ with real coefficients is the average of two monic polynomials of degree $n$ with $n$ real roots.

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Erken

#2 h Jul 23, 2009, 6:39 AM • 22 Y

Y by PRO2000, ali.agh, Delray, froont, Supercali, OlympusHero, mijail, Pluto1708, Kobayashi, samrocksnature, SPHS1234, hakN, Lamboreghini, everythingpi3141592, Adventure10, Mango247, signifance, and 5 other users

Let $P$ be the given polynomial, and we're about to prove that there exist $Q(x),R(x)\in\mathbb{R}[x]$ , both having $n$ real roots, and
$R(x) = 2\cdot P(x) - Q(x)$
Let $x_1< x_2 < \dots < x_n$ be arbitrary real numbers, and suppose that $a_i = P(x_i)$ , and let's take some real numbers $b_i$ , such that for all odd $i$ :
$2\cdot a_i < b_i, b_i > 0$
and for all even $i$ we have:
$2\cdot a_i > b_i, b_i < 0$
According to the Interpolational Lagrange Formula, there does exist the polynomial, such that $Q(x_i) = b_i$ . As we know,

$Q(x_1), Q(x_3), \dots > 0$ and $Q(x_2),Q(x_4),\dots < 0$ , so $Q$ has at least $n - 1$ real roots, consequently it has precisely $n$ real roots. On the other hand, $R(x_1) = 2a_1 - b_1 < 0$ , $R(x_2) = 2a_2 - b_2 > 0$ and so on. Therefore, $R(x)$ has exactly $n$ real roots either.

So the only problem left is to both of the polynomials to have leading coefficients equal $1$ . Note that when using Interpolational Formula, we used only $n$ points, $x_1,x_2,\dots,x_n$ so we may choose another point $x$ , such that $Q$ is monic. Hence, $R$ is monic either, seeing as $R(x) = 2P(x) - Q(x)$ .

P.S I managed to come up with this solution after thinkin' for 30 minutes, so it may contain mistakes.

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dysfunctionalequations

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dysfunctionalequations

#3 h Jan 15, 2011, 8:45 PM • 6 Y

Y by TheStrangeCharm, vsathiam, samrocksnature, Adventure10, Mango247, and 1 other user

Let the polynomial be $P(x)$ . Let $M$ be any number greater than the maximum value of $|P(x)|$ as $-1 \leq x \leq 1$ . Then, let $Q(x) = P(x) + M T_{n-1}(x)$ , where $T_{n-1}(x)$ is the $n-1$ st Chebyshev polynomial. Then define $x_k = \cos\left(\frac{\pi (n-1-k)}{n-1}\right)$ . We get $Q(x_k) = P(x_k) + M T_{n-1}(x_k) = P(x_k) + M (-1)^{n-1 - k},$ which has the same sign as $(-1)^{n - 1 - k}$ .

Additionally, because $Q$ is monic of degree $n$ , there is some (large negative) $x < x_0$ such that $Q(x)$ has the same sign as $(-1)^n$ , and some (large positive) $X > x_{n-1}$ such that $Q(X)$ has the same sign as $1$ (i.e. is positive).

Since $x < x_0 < x_1 < \cdots < x_{n-1} < X$ , by the Intermediate Value Theorem, the number of real roots $N$ of $Q$ is at least the number of pairs of consecutive members in the sequence $Q(x), Q(x_0), Q(x_1), \cdots, Q(x_{n-1}), Q(X)$ with opposite signs. By our previous results, the members of this sequence have the same signs as the members of the sequence $(-1)^n, (-1)^{n-1}, (-1)^{n-2}, \cdots, -1, 1, 1.$ Since this sequence has $n$ pairs of consecutive members with opposite sign, $Q$ has at least $n$ real roots. Since it has degree $n$ , it has exactly $n$ real roots. By the same argument, $R(x) = P(x) - M T_{n-1}(x)$ has exactly $n$ real roots.

Therefore, $P(x)$ is the average of two polynomials ( $Q(x)$ and $R(x)$ ) of degree $n$ with $n$ real roots.

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#4 h Sep 28, 2015, 12:18 AM • 10 Y

Y by Delray, mijail, Pluto1708, samrocksnature, nguyennam_2020, HamstPan38825, GeoMetrix, veirab, Ru83n05, Adventure10

Let $f(n)$ be the monic polynomial and let $M > 1000\max_{t=1, \dots, n} |f(t)|+1000$ .
Then we may select reals $a_1, \dots, a_n$ and $b_1, \dots, b_n$ such that for each $k = 1, \dots, n$ , we have
$\begin{align*} a_k + b_k &= 2f(k) \\ (-1)^k a_k & > M \\ (-1)^{k+1} b_k & > M. \end{align*}$ We may interpolate monic polynomials $g$ and $h$ through the $a_k$ and $b_k$ (if the $a_k$ , $b_k$ are selected "generically" from each other).
Then one can easily check $f = \tfrac12(g+h)$ works.

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#5 h Feb 14, 2017, 2:34 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Does this work for a solution?

We use induction on $n$ . $n=1$ is trivial.
Now suppose it is true for $n-1$ . If $n$ is odd or $P(x)$ has a real root, then $P(x)$ has a factor $x-r$ ( $r$ real), in which we are done by factoring out $x-r$ and applying the induction hypothesis on $n-1$ . Now assume that $n=2m$ is even and $P(x)>0$ for all real $x$ .
Let $Q(x)=(x-q_1)(x-q_2)...(x-q_{2m})$ where the $q_i$ 's are strictly increasing and positive reals and $q_{2m}$ is arbitrarily large and $R(x)=2P(x)-Q(x)$ . Note that $R(q_{2i-1})=2P(q_{2i-1})>0$ , since $P$ does not have any real factors.
Also, since $q_{2m}$ is arbitrarily large, $Q((q_{2i-2}+q_{2i-1})/2)>2P(q_{2i-2}+q_{2i-1})/2)$ i.e. $R(q_{2i-2}+q_{2i-1})/2)<0$ . Also, $Q(0) < 0$ . Thus, there are two roots of $R$ in each of the intervals $(-\infty, q_1), (q_2, q_3), ..., (q_{2m-2}, q_{2m-1})$ , so we are done.
I don't know if this is right since $q_1, q_2, ..., q_{2m-1}$ are virtually random. Could someone tell me if this is right? Thank you!

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#7 h Jun 7, 2019, 1:56 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

v_Enhance wrote:

Let $f(n)$ be the monic polynomial and let $M > 1000\max_{t=1, \dots, n} |f(t)|+1000$ .

Why was the $M > 1000 \ \text{"something"} + 1000$ necessary in the first place? Couldn't we just state that $M$ is arbitrarily large?

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#8 h Nov 10, 2020, 3:21 PM • 1 Y

Y by samrocksnature

MithsApprentice wrote:

Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $n$ with real coefficients is the average of two monic polynomials of degree $n$ with $n$ real roots.

Let's do some thorough analysis and motivation on this problem.

First, the statement of this problem is nice

Pretty straightforward if you know two general things about polynomial: the main problem is to $\textbf{construct}$ polynomials.
Now, if we were to find 2 polynomials of degree $n$ with $n$ real roots, the first thing that comes to mind is indeed, Intermediate Value Theorem. Why do we think of this?
Basically you wanted to construct a polynomial with $n$ real roots, but to do this you just need $n+1$ distinct points such that $f(a_i) < 0$ if $i \le n + 1$ is odd and $f(a_i) > 0$ if $i \le n + 1$ is even. Then, by Intermediate Value Theorem, there would be a root in the interval $(a_i , a_{i + 1})$ for every $1 \le i \le n$ , satisfying the hypothesis that we wanted.

Keeping that in mind, suppose that we have pick $a_i$ such numbers that $g$ passes through, by $\textbf{Lagrange Interpolation Theorem}$ , we are pretty much done, since there must exists polynomial of degree at most $n$ that pass these $n + 1$ points.
So, we are pretty much done since we can construct two polynomials $g$ and $h$ that has $n$ roots, and we can indeed choose $2f = g + h$ . Formalizing this, choose $g_1, g_2, \dots, g_{n + 1}$ being roots of $g$ , and $h_1, h_2, \dots, h_{n + 1}$ being roots of $h$ , such that
$g_i + h_i = 2f(x_i)$ for some $x_i$ and $|g_i|, |h_i| > M$ for a large value of $M > 100 \cdot | \max_{x_i, 1 \le i \le n } f(x_i) |^{100} + 200$ and $g_i$ switch signs depend on parity ( $h_i$ as well).

Now, there's just one problem left: the polynomial $g$ is not necessarily monic. But this is easy to fixed. Suppose we have pick $n$ points: $g_1, g_2, \dots, g_n$ . This could determine a polynomial of degree $n$ and we could somehow control $g_{n + 1}$ such that this is monic.

Now, if $g$ is monic, we could simply check the leading coefficient of $h = 2f - g$ , is monic as well. So we are done.

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Eyed

#12 h Jan 24, 2021, 12:55 AM • 2 Y

Y by centslordm, samrocksnature

Cute problem

We will prove this using inducktion on $n$ . Our base case is $n = 1$ , where the polynomial $P(x) = x-k$ , we can just take $x-k, x-k$ as our two monic polynomials.

Now, onto the inducktive step. Assume it is true for $n-1$ , we will prove it for polynomials of degree $n$ . Let the polynomial be $P(x)$ . If $P$ had a real root $r$ , then let $P = (x-r)Q$ . By our inducktive hypothesis, there exists 2 polynomials $f, g$ with all real roots such that $f+g = 2Q$ . Then take $f(x-r), g(x-r)$ , which will sum to $2P$ , and are both monic and have all real roots, which proves the problem. Thus, assume $P$ has no real roots (which also means $n$ is even), and since $P$ is monic this means $P(x) > 0$ for all real $x$ . Now, consider points $(x_{1}, y_{1}), (x_{2}, y_{2}), \ldots (x_{n}, y_{n})$ that satisfy the following conditions:
$y_{i} > P(x_{i}), x_{i} < x_{i+1}$ Let $R$ be the degree $n-1$ polynomial that passes through $(x_{i}, -(-1)^{i}y_{i})$ for each $i$ . This means
$R(x_{2k+1}) > P(x_{2k+1}), R(x_{2k+2}) < P(x_{2k+2})$ and by the intermediate value theorem, this means $P$ and $R$ must intersect at least $n-1$ times. Furthermore, since $R$ is an $n-1$ degree polynomial, which is odd degree, then there exists some $t < x_{1}$ such that $R(t) < P(t)$ , but since $R(x_{1}) > P(x_{1})$ , they must intersect before $x_{1}$ as well, so $P$ and $R$ intersect $n$ times, which means $P - R$ has $n$ real roots. We can similarly deduce that $P$ and $-R$ intersect $n$ times, so $P+R$ has $n$ real roots. Then, taking $P-R, P+R$ as our two polynomials, they both have $n$ real roots and average to $P$ , which completes our inducktive step.

Thus, by inducktion, there exists such polynomials.

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#13 h May 15, 2021, 3:53 PM • 5 Y

Y by aie8920, samrocksnature, Bradygho, Lucky0123, hellomath010118

Solution

Remarks

Attachments:

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#14 h Oct 22, 2021, 5:40 PM

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MithsApprentice wrote:

Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $n$ with real coefficients is the average of two monic polynomials of degree $n$ with $n$ real roots.

In other words,given any monic polynomial $f$ we need to show that $f \equiv \frac{1}{2}\left[g+h \right]$ i.e $2f=g+h$ By Lagrange interpolation choose points $\{a_1,b_1\},........,\{a_n,b_n\}$ such that $\begin{align*} a_k + b_k &= 2f(k) \\ (-1)^k a_k & > \text{some sufficiently large number} \\ (-1)^{k+1} b_k & > \text{some sufficiently large number} \end{align*}$ The first condition ensures $2f=g+h$ and the second condition by IVT ensures both $g$ and $h$ have an real root. $\blacksquare$

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#15 h Jun 30, 2022, 1:55 AM

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bruh what is this problem :skull:
too lazy to write-up details lel, just outline

Let $P(x)$ be the polynomial. Pick $-1 < x_1 < x_2 \cdots < x_n < 1$ ; and let $a_1 \cdots a_n$ , $b_1 \cdots b_n$ be reals such that $a_i < -1, b_i > 1$ and $a_i + b_i = 2(P(x_i)-x^n)$ for all $1 \le i \le n$ .

Use Lagrange Interpolation to let $Q_1(x)$ be the unique polynomial with degree at most n-1 that passes through points $(x_1, a_1), (x_2, b_2), (x_3, a_3), (x_4, b_4), \cdots$ in alternating order. Similarly define $Q_2(x)$ to be the polynomial that passes through points $(x_1, b_1), (x_2, a_2), (x_3, b_3), (x_4, a_4), \cdots$ . Then, we let $P_1(x) = Q_1(x) + x^n$ and $P_2(x) = Q_2(x) + x^n$ .

It is easy to see that $Q_1$ and $Q_2$ each have $n-1$ roots by Intermediate Value Theorem, and since $-1 < x^n < 1$ for $-1 < x < 1$ , the roots are preserved in $P_1$ and $P_2$ . Casework on the parity of $n$ yields that in either case, the end-behavior of $P_1$ and $P_2$ yields another root, giving $n$ roots.

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#16 h Oct 26, 2022, 3:45 PM • 2 Y

Y by kamatadu, Mango247

First note that if we have monic $p(x),q(x),r(x)$ such that $2p(x)=q(x)+r(x)$ —i.e. the statement holds for $p(x)$ —then the statement holds for $p(x)(x-r)$ for any real $r$ as well, as $2p(x)(x-r)=q(x)(x-r)+r(x)(x-r)$ , and the two polynomials on the RHS are still monic and have all real roots. Hence it suffices to consider the case where our monic polynomial $P(x)$ (that we want to express as the average) has no real roots. In particular, its degree must be even, so it's always positive (by looking at end behavior).
Define $y_i=P(i)$ for $1 \leq i \leq n$ , so $y_i>0$ for all $i$ . By Lagrange Interpolation, there exist (not necessarily monic) polynomials $q(x),r(x)$ of degree at most $n-1$ such that
$q(1)=3y_1,q(2)=-y_2,\ldots,q(n-1)=3y_{n-1},q(n)=-y_n,$ $r(1)=-y_1,r(2)=3y_2,\ldots,r(n-1)=-y_{n-1},r(n)=3y_n.$ Then let $Q(x)=q(x)+(x-1)\ldots(x-n)$ and $R(x)=r(x)+(x-1)\ldots(x-n)$ , so $Q,R$ are monic polynomials of degree $n$ which agree with $q,r$ respectively over $\{1,\ldots,n\}$ . Then,
$Q(1)>0,Q(2)<0,\ldots,Q(n-1)>0,Q(n)<0,\lim_{x \to \infty} Q(x)=\infty,$ hence $Q$ has a root in each of the $n$ intervals $(1,2),\ldots,(n-1,n),(n,\infty)$ by IVT. LIkewise,
$\lim_{x \to -\infty} R(x)=\infty,R(1)<0,R(2)>0,\ldots,R(n-1)<0,R(n)>0,$ hence $R$ has a root in each of the $n$ intervals $(1,2),\ldots,(n-1,n),(-\infty,1)$ .
Further, since $P,Q,R$ are all monic, $2P(x)-Q(x)-R(x)$ is a polynomial of degree $n-1$ which is evidently zero at $x=1,\ldots,n$ , hence it is identically zero and we have $2P(x)=Q(x)+R(x)$ for all $x$ as desired. $\blacksquare$

Remark: graphing op

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yugrey

#17 h Dec 17, 2022, 9:54 AM

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I have an email from September 25, 2014 (wow, long time ago) where I describe the solution.

We induct on the degree, it's easy for $n=0,1$ . If $P$ has a real root, WLOG $P(0)=0$ . Then write $P(x)=xQ(x)$ , and let $Q(x)=\frac{R(x)+S(x)}{2}$ be such an average. So $P(x)=\frac{xR(x)+xS(x)}{2}$ is the average of two polynomials with $n-1$ real roots, since $R,S$ have $n-1$ real roots. Replace $(R,S)$ by $(R+\varepsilon,S-\varepsilon)$ if need be so that $R,S$ do not have a root at $0$ . Then $xR(x), xS(x)$ have $n$ real roots.

If $P$ does not have a real root, it is of even degree $n=2d \ge 2$ , and we can take $Q(x)=(x-2)...(x-2d)(x-r)$ for some sufficiently large huge $r$ . Now, $P$ is the average of $Q$ and $R=2P-Q$ . Clearly $Q$ has $n$ real roots. We claim that $2P-Q$ has two real roots on each of $(-\infty,2), (3,4), (5,6), \cdots, (2d-1,2d)$ . Note that $Q$ is positive on these intervals $(3,4), \cdots, (2d-1,2d)$ but $0$ at their endpoints. As $r \to \infty$ , it blows up on each of these intervals, and so intersects the graph of $2P$ at two points, when $r$ is big enough. We have $R(2)=2P(2)-Q(2)=2P(2)>0$ , and as $R$ is monic of even degree, $\lim_{x \to -\infty}R(x)=\infty$ . However, if $r$ is sufficiently large, $R(1)=2P(1)-Q(1)<0$ , since $\lim_{r \to \infty}Q(1)=\lim_{r \to \infty}r(2d-1)!=\infty$ .

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awesomeming327.

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awesomeming327.

#18 h Dec 26, 2022, 4:53 PM • 2 Y

Y by Mango247, Mango247

Let $P(x)$ be the given monic polynomial, and select some random $n$ pairwise distinct $x$ -values $x_1<x_2<\dots < x_n$ . At each value $x_i$ , select two other values $a_i,b_i$ for which $a_i+b_i = 2P(x_i).$ If $i$ is even, make $a_i <0,b_i>0$ and if $i$ is odd, make $a_i>0,b_i<0$ . Clearly, if we can find monic polynomials for which $Q(x_i)=a_i$ , $R(x_i)=b_i$ then we are done by the Intermediate Value Theorem. Fortunately, we can let $Q(x)=c_0+c_1x+c_2x^2+\dots + c_{n-1}x^{n-1}+x^n$ and plug in values $x_1, \dots, x_n$ into that, and we get a system of $n$ equations and $n$ variables ( $c_0,\dots c_{n-1}.$ ) Since the coefficients behind the variables have different ratios in each equation, we have a unique solution. We are done.

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#20 h Jul 5, 2023, 2:26 PM

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Feels a bit superficial, easy to do directly.

Pick some very large $N_1, N_2, \dots, N_{n+1} \gg 0$ and real numbers $x_1, x_2, \dots, x_{n+1}$ . Let $P(x_i) = y_i$ for each $i$ . Set two polynomials $Q, R$ such that $Q(x_i) = N_i$ and $R(x_i) = 2y_i - N_i$ if $i$ is odd, and vice versa if $i$ is even by Lagrange interpolation.

Now, observe that by intermediate value theorem, $Q$ and $R$ both have at least one root in $[x_i, x_{i+1}]$ , hence they both have at least $i$ real roots. Furthermore, $P(x_i) = \frac{Q(x_i) + R(x_i)}2$ for $n+1$ values of $i$ ; thus $P = \frac 12(Q+R)$ exactly, and we're done.

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#21 h Aug 24, 2023, 8:29 PM

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Really silly. The idea is to basically make the initial polynomial irrelevant.

I interpreted $n$ real roots as $n$ distinct real roots, but idk.

Let $P$ be the given polynomial with degree $n$ . Let $Q(x) = \prod_{i = 1}^{n - 1} (x-i)$ . Let $M$ be a gigantic real number which we will establish later.

Consider the two polynomials
$P(x) + MQ(x), P(x) - MQ(x).$ Note that these two polynomials are monic.

Let's consider $Q$ . Fix an arbitrarily small $\varepsilon > 0$ . For each $i \in \{ 1, 2, \dots, n-1\}$ , choose some $\frac{1}{1000} > \delta_i > 0$ such that $\forall x \in (i - \delta_i, i + \delta_i)$ , we have $Q(x) \in (-\varepsilon, \varepsilon )$ . This follows from the definition of continuity, and the fact that a subset of a set is in fact a subset of a set [that's craaazy]. Now take the set $U = \bigcup_{i = 1}^{n - 1} (i - \delta_i, i + \delta_i)$ , and define
$T := \{ t \in \mathbb R | \exists x \in U \text{ s.t. } |P(x)| = t \}.$ Observe that $T$ is nonempty as $|P(1)|$ is in this set. Thus it makes sense to take $T_M = \sup T$ , and choosing $M = \frac{2T_M}{\varepsilon}$ , we can guarantee that for $x \in (i - \delta_i, i + \delta_i)$ , $Q(x) - \varepsilon/2 \le \frac{P(x)}{M} + Q(x) \le Q(x) + \varepsilon/2$ .It follows then that $P(x) + MQ(x)$ has a root on each of these intervals, and hence has at least $n-1$ distinct roots. Likewise $P(x) - MQ(x)$ has a root on each of these intervals too.

Now we need to show that there's no double roots here. This is really tedious (we just basically remove the contribution of $P'(x)$ , by choosing $M$ such that the maximal absolute size of $P'(x)/M$ on any of the delta intervals is bounded by some $c$ such that $0 \not \in [Q'(x) - c, Q'(x) + c]$ , also tightening $\varepsilon$ even more such that the derivative is never zero on the interval.) so I'll leave it as an exercise to the reader to work out all the details.

The remaining root can't be complex as $P$ is a real polynomial. Thus the remaining root is real. We're done.

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HumanCalculator9

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HumanCalculator9

#22 h Sep 4, 2023, 5:32 PM

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I have a really goofy solution

First, note that every polynomial with real coefficients can be factored into linear and quadratic polynomials with real coefficients, and if the original was monic, all of these linear and quadratic polynomials are monic as well. Thus, it is sufficient to prove that all linear and quadratic polynomials work. Linears are trivial as all have 1 real root. For quadratics, let $x^2+ax+b$ be our quadratic, and say it is the average of $(x+i)(x+j)$ and $(x+k)(x+l)$ for some reals i, j, k, l. It is sufficient to prove that i, j, k, l exist for all a, b. Thus, $2a=i+j+k+l$ , $2b=ij+kl$ . Say that $i+j=a+n$ , $k+l=a-n$ for some arbitrary $n$ . By AM-GM, $ij\leq \frac{(a+n)^2}{4}$ , $kl\leq \frac{(a-n)^2}{4}$ , thus $ij+kl\leq \frac{a^2+n^2}{2}$ , or $4b\leq a^2+n^2$ . If $a^2>4b$ , we are simply done. Otherwise, set $n^2=4b-a^2$ , with $i=j=\frac{a+n}{2}$ , $k, l=\frac{a-n}{2}$ . We can confirm that this works. We are thus done.

someone tell me if I am having a skill issue or not

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#23 h Dec 7, 2023, 2:47 AM

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AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

So this is like my first realpoly problem thanks ApraTrip for giving me hint "how to find poly with all real roots"

I guess Intermediate Value Theorem is like extremely overpowered. Also thinking graphically actually works I guess.

So basically we just choose $n$ values where it keeps having one polynomial be really big then really small next so it keeps crossing the $x$ -axis. Then the last degree of freedom for Lagrange Interpolation is used to make the polynomials monic, gg.

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#24 h Apr 18, 2024, 5:15 PM

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Call $P$ the desired monic polynomial that we want to express as an average.

Claim (monic polynomial interpolation): Given any points $(a_1,b_1), \ldots, (a_n,b_n) \in \mathbb{R}^2$ where each $a_i$ are pairwise distinct, there exists a unique monic polynomial $Q \in \mathbb{R}[x]$ of degree $n$ where $Q(a_i)=b_i$ for each $i$ .

By polynomial interpolation, there is a unique polynomial $R \in \mathbb{R}[x]$ of degree $n-1$ where $R(a_i)=b_i-a_i^n$ . Thus, setting
$Q(x) = x^n + R(x)$ gives the unique monic polynomial of degree $n$ satisfying
$Q(a_i) = a_i^n + R(a_i) = a_i^n + (b_i-a_i^n) = b_i$ $\square$

Now, choose a real number $M > \max(|P(1)|, |P(2)|, \ldots, |P(n)|)$ and let $f$ and $g$ be the unique monic polynomials where
$f(i) = P(i)+ M \cdot (-1)^i \text{ } \text{ and } \text{ } g(i) = P(i) - M \cdot (-1)^i$ for each $i \in \{1,2, \ldots, n\}$ . We claim $f$ and $g$ are the desired monic polynomials. Indeed, first notice that for all $x \in \{1,2, \ldots, n\}$ we have
$\frac{f(x)+g(x)}{2} = \frac{P(x)+P(x)+M-M}{2} = P(x)$ Since $\frac{f+g}{2}$ and $P$ are both monic polynomials of degree $n$ and agree on $n$ distinct inputs, by monic polynomial interpolation they must be the same polynomial.

Now, all that is left to show is that $f$ and $g$ have $n$ distinct real roots. Indeed, notice that if $1 \leq k \leq n-1$ is even then $f(k) = P(k)+M$ and since $|M|>|P(k)|$ and $M$ is positive, we have $f(k)>0$ . Moreover, $f(k+1) = P(k+1)-M$ and for the same reason, this implies that $f(k+1)<0$ . Thus, the sequence $f(1),f(2), \ldots, f(n)$ has alternating sign. Thus, by IVT, $f$ has a root in the interval $(k,k+1)$ for $k=1,2, \ldots,n-1$ . This implies $f$ has $n-1$ distinct real roots. To get the final root, we look at the parity of $n$ .

If $n$ is even, then $f(X)$ is positive for all $X \to \infty$ and $X \to -\infty$ . But, $f(1)$ and $f(n)$ have opposite sign, implying it will hit $0$ at some other point.

If $n$ is odd, then $f(X)>0$ as $X \to \infty$ and $f(X) < 0$ as $X \to -\infty$ . But, $f(1)$ and $f(n)$ have the same sign, and thus must hit $0$ at another point.

By symmetry, the same holds for $g$ .
$\blacksquare$

This post has been edited 2 times. Last edited by blackbluecar, Apr 30, 2024, 5:33 PM

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N3bula

#25 h Jan 4, 2025, 8:47 AM

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I will first prove that for any $n$ points in the plane we can find a monic polynomial that passes through them. Consider the $n$ points,
$(x_1, y_1)$ , \dots, $(x_n, y_n)$ . Let $f(x)=x^n$ . Now consider the $n$ points $(x_1, y_1-f(x_1))$ , \dots, $(x_n, y_n-f(x_n))$ . Clearly there exists
an $n-1$ th degree polynomial passing through these $n$ points because of legrange interpolation, now let this polynomial be $P$ , clearly $x^n+P(x)$ is a
monic $n$ the degree polynomial and clearly it passes through $(x_1, y_1)$ , \dots, $(x_n, y_n)$ . Thus for any $n$ points in the plane we can find a
monic polynomial thats passes through all of them. Now let $Q(x)$ be a monic $n$ th degree polynomial. Now suppose that for two polynomials $P$ and $R$ .
$\frac{P(x)+R(x)}{2}=Q(x)$ . Clearly for any value $x$ , we have that $P(x)-Q(x)=Q(x)-R(x)$ . Thus for any polynomial, $P$ , the corresponding polynomial $R$ , such
that $\frac{P(x)+R(x)}{2}=Q(x)$ , for any $x$ the $y$ values are reflections over the $y$ value of $Q(x)$ . Thus for $P(x)$ , choose $n$ $(x_1, y_1)$ , \dots, $(x_n, y_n)$
such that $x_1<x_2<\dots<x_n$ , such that $y_{2m+1}$ for any $m\leq \frac{n-1}{2}$ is positive and its reflection over $Q(x_{2m+1})$ is negative, and
such that $y_{2m}$ for any $m\leq \frac{n}{2}$ is negative and its reflection over $Q(x_{2m})$ is positive. Clearly there exists a monic polynomial passing through
$(x_1, y_1)$ , \dots, $(x_n, y_n)$ . Clearly because of $IVT$ and polynomial stuff, this polynomial has $n$ roots, let this polynomial be $P(x)$ . Clearly
if $\frac{P(x)+R(x)}{2}=Q(x)$ , we have that due to the reflection property that $R(x)$ must also have $n$ roots.

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#26 h Apr 25, 2025, 8:22 PM

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Feels like an anti-problem
Subjective Difficulty Rating

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