Algebra Problem #2
by maxo, Jun 11, 2016, 10:51 PM
(Source: Mathematical Mayhem)
Solved with largely 'guesstimating' lol, probably not a valid solution, but just how I solved it.
Let![$\color[rgb]{0.35,0.35,0.35}f(x)$](//latex.artofproblemsolving.com/b/8/f/b8f24ff4b8a4ae7e76087e03fba11cc7c67a02fa.png)
be a polynomial in ![$\color[rgb]{0.35,0.35,0.35}x$](//latex.artofproblemsolving.com/b/d/0/bd0f2bb6c5661cf5ff93ce07099f910a458f8e79.png)
of degree greater than 1. Define ![$\color[rgb]{0.35,0.35,0.35}g_i(x)$](//latex.artofproblemsolving.com/f/3/d/f3d18a0e0b86378b304d8be5930b4403d21f3c6c.png)
by ![$\color[rgb]{0.35,0.35,0.35}g_1(x) = f(x)$](//latex.artofproblemsolving.com/4/8/1/48141a7d9da383028cdede6d8a84d5b294668763.png)
, and ![$\color[rgb]{0.35,0.35,0.35}g_{k + 1}(x) = f(g_k(x))$](//latex.artofproblemsolving.com/6/1/c/61cf758619c8bc1261fcf939e6170cdfecb2a5a0.png)
. Let ![$\color[rgb]{0.35,0.35,0.35}r_k$](//latex.artofproblemsolving.com/b/a/b/bab0b3167e07a9d4a7784160bf6708d3c01662a7.png)
be the average of the roots of ![$\color[rgb]{0.35,0.35,0.35}g_k$](//latex.artofproblemsolving.com/c/7/2/c72222930a9c1fbed785adce23e8e75da00e9f9f.png)
. Determine ![$\color[rgb]{0.35,0.35,0.35}r_{89}$](//latex.artofproblemsolving.com/e/8/6/e86c4918d7887213005f4d7b48739c7cebec443a.png)
if ![$\color[rgb]{0.35,0.35,0.35}r_{19} = 89$](//latex.artofproblemsolving.com/3/1/e/31e538d515f32c1050fd1c4b63de5f3cd118ebc1.png)
.
Solution:
Click to reveal hidden text
Solved with largely 'guesstimating' lol, probably not a valid solution, but just how I solved it.
Let
![$\color[rgb]{0.35,0.35,0.35}f(x)$](http://latex.artofproblemsolving.com/b/8/f/b8f24ff4b8a4ae7e76087e03fba11cc7c67a02fa.png)
be a polynomial in ![$\color[rgb]{0.35,0.35,0.35}x$](http://latex.artofproblemsolving.com/b/d/0/bd0f2bb6c5661cf5ff93ce07099f910a458f8e79.png)
of degree greater than 1. Define ![$\color[rgb]{0.35,0.35,0.35}g_i(x)$](http://latex.artofproblemsolving.com/f/3/d/f3d18a0e0b86378b304d8be5930b4403d21f3c6c.png)
by ![$\color[rgb]{0.35,0.35,0.35}g_1(x) = f(x)$](http://latex.artofproblemsolving.com/4/8/1/48141a7d9da383028cdede6d8a84d5b294668763.png)
, and ![$\color[rgb]{0.35,0.35,0.35}g_{k + 1}(x) = f(g_k(x))$](http://latex.artofproblemsolving.com/6/1/c/61cf758619c8bc1261fcf939e6170cdfecb2a5a0.png)
. Let ![$\color[rgb]{0.35,0.35,0.35}r_k$](http://latex.artofproblemsolving.com/b/a/b/bab0b3167e07a9d4a7784160bf6708d3c01662a7.png)
be the average of the roots of ![$\color[rgb]{0.35,0.35,0.35}g_k$](http://latex.artofproblemsolving.com/c/7/2/c72222930a9c1fbed785adce23e8e75da00e9f9f.png)
. Determine ![$\color[rgb]{0.35,0.35,0.35}r_{89}$](http://latex.artofproblemsolving.com/e/8/6/e86c4918d7887213005f4d7b48739c7cebec443a.png)
if ![$\color[rgb]{0.35,0.35,0.35}r_{19} = 89$](http://latex.artofproblemsolving.com/3/1/e/31e538d515f32c1050fd1c4b63de5f3cd118ebc1.png)
.Solution:
Click to reveal hidden text
First let's note our givens:
is the average of the roots of
....
So we know that and
Let's just write some random stuff down!
From this info we can conclude that:
![$\color[rgb]{0.35,0.35,0.35}g_{2}(x) = f(g_1(x))$](//latex.artofproblemsolving.com/6/c/4/6c49ebc1440afa5e849cc69c2612ee49d71ee414.png)
.. but wait, ..
that means![$\color[rgb]{0.35,0.35,0.35}g_{2}(x) = f(f(x))$](//latex.artofproblemsolving.com/6/3/b/63b75b66d45768b4cb78937b308a912c04c84954.png)
.. and looking at
![$\color[rgb]{0.35,0.35,0.35}g_{3}(x) = f(g_2(x))$](//latex.artofproblemsolving.com/b/f/7/bf72a4c8ac6e5e65e230996cfef523b4608c69eb.png)
, we can see that there is a pattern to
![$\color[rgb]{0.35,0.35,0.35}g_n(x)$](//latex.artofproblemsolving.com/2/3/e/23e5989d6d800eedaab94a9c93dd1f17cf54cc70.png)
= the repeated composition of f(x) n times.
So that means that![$\color[rgb]{0.35,0.35,0.35}g_{19}(x) = f(f(f....(x)))))))))))))))))))$](//latex.artofproblemsolving.com/5/7/7/577d10e893e0433e3ea569c74ca1636c95088fa6.png)
, where the ... represents the continuation of the 19 f's. This simplifies our problem some.
Let f(x) =![\[\color[rgb]{0.35,0.35,0.35} f(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0 \]](//latex.artofproblemsolving.com/e/7/7/e7764f9dce30626688355505bb7d1a1690da43c1.png)
If we perform![\[g_2\]](//latex.artofproblemsolving.com/d/0/6/d06075067e244737163dcb0d46e28f9575228ece.png)
, we get ![\[a_n (a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0)^n + a_{n - 1} (a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0)^{n - 1} + \dots + a_1 (a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0) + a_0\]](//latex.artofproblemsolving.com/8/f/9/8f9ac844a7d0d828551679243654143394821b24.png)
Whoa, that's messy. The problem talks about the average of roots, so we need to find the sum of roots of these big equations. We can use vietas to conclude that for a polynomial with in the form f(x) =![\[a_n x^n + a_{n - 1} a^{n - 1} + \dots + a_1 x + a_0\]](//latex.artofproblemsolving.com/0/a/f/0af52c4aded6ed896993719c612612af4a1f2c32.png)
, the sum is ![\[\frac{-(a_{n-1})}{a_n}\]](//latex.artofproblemsolving.com/0/9/0/090aac52cec79ea6ef7bf5cabb4e16d57b0c9459.png)
.. However, when the composition occurs, how will the sum change?
We look at a few smaller polynomials to observe this change. Click to reveal hidden text
So we see performing this composition, turn the original sum of the roots 2, to 4.
We see that this is true, since our original function could be factored into (x-1)(x-1) .. performing the composition, and disregarding all terms except the ones with the highest and second highest powers gives us![\[((x-1)(x-1))^2\]](//latex.artofproblemsolving.com/d/d/5/dd5af9f53cfb87ddd4e5d1e121831c1f23c28a48.png)
= (x-1)(x-1)(x-1)(x-1) .. so it adds on another sets. So for whatever sum is there originally, gets multiplied by n, where n is the degree of the polynomial.
That means the sum of the roots of![\[g_n(x)\]](//latex.artofproblemsolving.com/1/5/0/1509a2ea1f6874eb20e35a4a572d17ac12f1f10e.png)
= ![\[\frac{-(a_{n-1}(n))}{a_n}\]](//latex.artofproblemsolving.com/1/8/b/18ba211c6a1984f319c0122c6df3866b998b7888.png)
that means the average of the roots is simply![\[\frac{\frac{-(a_{n-1}(n))}{a_n}}{n^n}\]](//latex.artofproblemsolving.com/5/3/7/537cefab8bb997999e59a8aedd840d45247e8156.png)
, which simplifies to:
![\[r_n\]](//latex.artofproblemsolving.com/b/5/b/b5b9a2c03e4dc2a98f75a8ee544b53699d92bf77.png)
= ![\[\frac{\frac{-(a_{n-1})}{a_n}}{n^(n-1)}\]](//latex.artofproblemsolving.com/0/6/a/06a93b7682ed49c50b204eca930135b8012059b7.png)
..
Well what do we do now? Lets try the same method we did here to prove that= , to see how repeated composition of a single function affects the average of the roots.
Let's take the equation
![\[q(x)\]](//latex.artofproblemsolving.com/8/e/2/8e20b6e10b6bdbecdae6d59d6ed7cf3ab14d70d8.png)
= ![\[x^2 - 6x + 5\]](//latex.artofproblemsolving.com/a/8/1/a81b60e2c4bfa16d1a69fdc9c7f1020baa0a011d.png)
..
the sum of the polynomial is 6 by using vietas, so the average is 3. performing![\[q_2(x)\]](//latex.artofproblemsolving.com/b/3/0/b302c9743a07e34e7f5caa02d8dc727f85ffcc6f.png)
, and only taking in account the x^2, we get
![\[(x^2 - 6x + 5)^2\]](//latex.artofproblemsolving.com/8/b/f/8bf63e74e4c69e763868645332b2d0edc7890b05.png)
.. multiplying this out, and taking only in account the x^4 and x^3 terms, we get ![\[x^4 - 12x^3 ..\]](//latex.artofproblemsolving.com/2/a/6/2a60c98ebdd1a6eacaf2f3a130cf15f436612289.png)
.. we see that the sum of the roots of this is 12, but since there were originally 2 roots, and we squared this polynomial when performing composition, we got 4 roots instead since by the fundamental theorem of algebra a polynomial with nth degree has n roots.
That means 12/4 = 3. So the average of the roots of this new polynomial is still 3!
After plugging in the respetive coefficients into our equation:
=
we can verify that the average of the roots will stay the same despite repeated composition.
Thus, the average of the roots of is the same for all values of k.
Thus![$\color[rgb]{0.35,0.35,0.35}r_{19}$](//latex.artofproblemsolving.com/b/e/b/bebc74e9a0059877796c9487e33710c3cb55f6bc.png)
= = 89
is the average of the roots of ....
So we know that
and Let's just write some random stuff down!
From this info we can conclude that:
![$\color[rgb]{0.35,0.35,0.35}g_{2}(x) = f(g_1(x))$](http://latex.artofproblemsolving.com/6/c/4/6c49ebc1440afa5e849cc69c2612ee49d71ee414.png)
.. but wait, ..that means
![$\color[rgb]{0.35,0.35,0.35}g_{2}(x) = f(f(x))$](http://latex.artofproblemsolving.com/6/3/b/63b75b66d45768b4cb78937b308a912c04c84954.png)
.. and looking at![$\color[rgb]{0.35,0.35,0.35}g_{3}(x) = f(g_2(x))$](http://latex.artofproblemsolving.com/b/f/7/bf72a4c8ac6e5e65e230996cfef523b4608c69eb.png)
, we can see that there is a pattern to![$\color[rgb]{0.35,0.35,0.35}g_n(x)$](http://latex.artofproblemsolving.com/2/3/e/23e5989d6d800eedaab94a9c93dd1f17cf54cc70.png)
= the repeated composition of f(x) n times.So that means that
![$\color[rgb]{0.35,0.35,0.35}g_{19}(x) = f(f(f....(x)))))))))))))))))))$](http://latex.artofproblemsolving.com/5/7/7/577d10e893e0433e3ea569c74ca1636c95088fa6.png)
, where the ... represents the continuation of the 19 f's. This simplifies our problem some.Let f(x) =
![\[\color[rgb]{0.35,0.35,0.35} f(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0 \]](http://latex.artofproblemsolving.com/e/7/7/e7764f9dce30626688355505bb7d1a1690da43c1.png)
If we perform
![\[g_2\]](http://latex.artofproblemsolving.com/d/0/6/d06075067e244737163dcb0d46e28f9575228ece.png)
, we get ![\[a_n (a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0)^n + a_{n - 1} (a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0)^{n - 1} + \dots + a_1 (a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0) + a_0\]](http://latex.artofproblemsolving.com/8/f/9/8f9ac844a7d0d828551679243654143394821b24.png)
Whoa, that's messy. The problem talks about the average of roots, so we need to find the sum of roots of these big equations. We can use vietas to conclude that for a polynomial with in the form f(x) =
![\[a_n x^n + a_{n - 1} a^{n - 1} + \dots + a_1 x + a_0\]](http://latex.artofproblemsolving.com/0/a/f/0af52c4aded6ed896993719c612612af4a1f2c32.png)
, the sum is ![\[\frac{-(a_{n-1})}{a_n}\]](http://latex.artofproblemsolving.com/0/9/0/090aac52cec79ea6ef7bf5cabb4e16d57b0c9459.png)
.. However, when the composition occurs, how will the sum change?We look at a few smaller polynomials to observe this change. Click to reveal hidden text
Say we had the polynomial = ![\[x^2 - 2x + 1\]](//latex.artofproblemsolving.com/0/c/1/0c1054dadc57282e8e214879848c942739d75e39.png)
.. we know that the sum of the roots is 2 here from vietas. If we perform composition, , we get ![\[(x^2 - 2x + 1)^2 - 2(x^2 - 2x + 1) + 1\]](//latex.artofproblemsolving.com/1/9/6/1964bbf6852f533e0e33603d8a3c17310dc7278b.png)
.. we see here we only want the coefficient of the power of x one less than the maximum, so we disregard the rest of the polynomial, only focusing on
![\[(x^2 - 2x + 1)^2\]](//latex.artofproblemsolving.com/e/7/7/e77b59cf89d26d8c608125ba483f13ee7a0a9083.png)
, multiplying everything out gives us ![\[x^4 - 4x^3 + \dots\]](//latex.artofproblemsolving.com/c/0/3/c033a42ba5738a280232d016be65fe5843aead10.png)
![\[q(x)\]](http://latex.artofproblemsolving.com/8/e/2/8e20b6e10b6bdbecdae6d59d6ed7cf3ab14d70d8.png)
= ![\[x^2 - 2x + 1\]](http://latex.artofproblemsolving.com/0/c/1/0c1054dadc57282e8e214879848c942739d75e39.png)
.. we know that the sum of the roots is 2 here from vietas. If we perform composition, ![\[q_2(x)\]](http://latex.artofproblemsolving.com/b/3/0/b302c9743a07e34e7f5caa02d8dc727f85ffcc6f.png)
, we get ![\[(x^2 - 2x + 1)^2 - 2(x^2 - 2x + 1) + 1\]](http://latex.artofproblemsolving.com/1/9/6/1964bbf6852f533e0e33603d8a3c17310dc7278b.png)
.. we see here we only want the coefficient of the power of x one less than the maximum, so we disregard the rest of the polynomial, only focusing on![\[(x^2 - 2x + 1)^2\]](http://latex.artofproblemsolving.com/e/7/7/e77b59cf89d26d8c608125ba483f13ee7a0a9083.png)
, multiplying everything out gives us ![\[x^4 - 4x^3 + \dots\]](http://latex.artofproblemsolving.com/c/0/3/c033a42ba5738a280232d016be65fe5843aead10.png)
So we see performing this composition, turn the original sum of the roots 2, to 4.
We see that this is true, since our original function could be factored into (x-1)(x-1) .. performing the composition, and disregarding all terms except the ones with the highest and second highest powers gives us
![\[((x-1)(x-1))^2\]](http://latex.artofproblemsolving.com/d/d/5/dd5af9f53cfb87ddd4e5d1e121831c1f23c28a48.png)
= (x-1)(x-1)(x-1)(x-1) .. so it adds on another sets. So for whatever sum is there originally, gets multiplied by n, where n is the degree of the polynomial.That means the sum of the roots of
![\[g_n(x)\]](http://latex.artofproblemsolving.com/1/5/0/1509a2ea1f6874eb20e35a4a572d17ac12f1f10e.png)
= ![\[\frac{-(a_{n-1}(n))}{a_n}\]](http://latex.artofproblemsolving.com/1/8/b/18ba211c6a1984f319c0122c6df3866b998b7888.png)
that means the average of the roots is simply
![\[\frac{\frac{-(a_{n-1}(n))}{a_n}}{n^n}\]](http://latex.artofproblemsolving.com/5/3/7/537cefab8bb997999e59a8aedd840d45247e8156.png)
, which simplifies to:![\[r_n\]](http://latex.artofproblemsolving.com/b/5/b/b5b9a2c03e4dc2a98f75a8ee544b53699d92bf77.png)
= ![\[\frac{\frac{-(a_{n-1})}{a_n}}{n^(n-1)}\]](http://latex.artofproblemsolving.com/0/6/a/06a93b7682ed49c50b204eca930135b8012059b7.png)
..Well what do we do now? Lets try the same method we did here to prove that
= , to see how repeated composition of a single function affects the average of the roots.Let's take the equation
= ![\[x^2 - 6x + 5\]](http://latex.artofproblemsolving.com/a/8/1/a81b60e2c4bfa16d1a69fdc9c7f1020baa0a011d.png)
..the sum of the polynomial is 6 by using vietas, so the average is 3. performing
, and only taking in account the x^2, we get![\[(x^2 - 6x + 5)^2\]](http://latex.artofproblemsolving.com/8/b/f/8bf63e74e4c69e763868645332b2d0edc7890b05.png)
.. multiplying this out, and taking only in account the x^4 and x^3 terms, we get ![\[x^4 - 12x^3 ..\]](http://latex.artofproblemsolving.com/2/a/6/2a60c98ebdd1a6eacaf2f3a130cf15f436612289.png)
.. we see that the sum of the roots of this is 12, but since there were originally 2 roots, and we squared this polynomial when performing composition, we got 4 roots instead since by the fundamental theorem of algebra a polynomial with nth degree has n roots.That means 12/4 = 3. So the average of the roots of this new polynomial is still 3!
After plugging in the respetive coefficients into our equation:
= we can verify that the average of the roots will stay the same despite repeated composition.
Thus, the average of the roots of
is the same for all values of k.Thus
![$\color[rgb]{0.35,0.35,0.35}r_{19}$](http://latex.artofproblemsolving.com/b/e/b/bebc74e9a0059877796c9487e33710c3cb55f6bc.png)
= = 89This post has been edited 7 times. Last edited by maxo, Jun 11, 2016, 11:49 PM
June 2018
December 2017