Is it cubic?

Algebra Level 4

Let $f(x)$ be a polynomial with integer coefficients such that

$\begin{aligned} f(1) &= 1 \\ f(2) &= 8 \\ f(3) &= 27 \\ f(5) &= 125 \\ f(6) &= 216 \\ f(7) &= 343. \end{aligned}$

What is the minimum possible value of $|f(4)|?$

Inspiration.

The answer is 8.

2 solutions

Steven Yuan
Jul 26, 2017

Notice that $f(x) = x^3$ holds for $x = 1, 2, 3, 5, 6, 7.$ That means we can factor $f$ as

$f(x) = x^3 + g(x)(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7),$

for some polynomial $g$ with integer coefficients. Substituting in $x = 4$ yields $f(4) = 64 - 36g(4).$

Since $g$ consists of integer coefficients, $g(4)$ is an integer. If we set $g(4) = 2,$ we get $|f(4)| = |64 - 36(2)| = \boxed{8},$ which is the minimum value of $|f(4)|.$ We can attain this value if, for instance, $g(x) = 2,$ yielding the polynomial

$f(x) = x^3 + 2(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7).$

I follow the logic above, but it is based on the assumption that $f\left( x \right)$ can ONLY take the form ${ x }^{ 3 }+g(x)(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ . Couldn't we find a different polynomial without an ${ x }^{ 3}$ term that, while less obvious, nevertheless satisfies all six stated conditions from $f\left( 1 \right) =1$ to $f\left( 7 \right) =343$ ?

For example, if the only stated conditions were $f\left( 2 \right) =8$ and $f\left( 7 \right) =343$ , your logic would lead you to factor $f$ as

$f\left( x \right) ={ x }^{ 3 }+g(x)(x-2)(x-7)$

making $f\left( 4 \right) ={ 4 }^{ 3 }+g\left( 4 \right) (4-2)(4-7)=64-6g\left( 4 \right)$ . You would conclude that the minimum value of $\left| f\left( x \right) \right|$ is $\boxed {2}$ (which is achievable for any $g\left( x \right)$ such that $g\left( 4 \right) =11$ ). However, you could also factor $f$ as

$f\left( x \right) =6{ x }^{ 2 }+13x-42+g(x)(x-2)(x-7)$

making $f\left( 4 \right) =6{ 4 }^{ 2 }+13(4)-42+g(x)(4-2)(4-7)=106 - 6g\left( 4 \right)$ . This turns out to give the same minimum of $\boxed {2}$ for $\left| f\left( x \right) \right|$ (this occurs so long as $g\left( 4 \right) =18$ ), and I have a feeling that any such polynomial would, but I am not sure how to prove that. Until that is proved, the assumption upon which this proof is based seems unjustified.

Aric Floyd - 3 years, 10 months ago

(Misinterpreted the issue at hand. Sorry!)

Steven Yuan - 3 years, 10 months ago

Thanks for your reply Steven. The polynomials $f\left( x \right)={ x }^{ 3 }+g(x)(x-2)(x-7)$ and $f\left( x \right)=6{ x }^{ 2 }+13x-42+g(x)(x-2)(x-7)$ I mentioned in the second paragraph of my comment were not solutions to the original problem. Rather, I was trying to illustrate through a simpler problem - in which the only stated conditions were $f\left( 2 \right) =8$ and $f\left( 7 \right) =343$ - that ${ x }^{ 3 }$ is not the only possible remainder of $\frac { f(x) }{ g(x) }$ . It would of course take a polynomial of higher degree to fit all six conditions stated in the original problem, but there are infinitely many polynomials $q\left( x \right)$ of minimum fifth-degree that will pass through all six given points. So, how can we prove that factoring $f\left( x \right)$ as $q\left( x \right) +g(x)(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ (where $q\left( x \right)$ is any of these possible polynomials) rather than ${ x }^{ 3 }+g(x)(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ will not result in a different solution? Again, I have a feeling that we will get the same answer no matter what, but I do not know how to show that rigorously.

Aric Floyd - 3 years, 10 months ago

@Aric Floyd – Ah I see your concern now. Sorry for misinterpreting you at first!

Any such polynomial $q(x)$ that you describe must be of the form $x^3 + r(x)(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7),$ where $r$ is a polynomial with integer coefficients. Suppose not; then $q(x) = x^3 + r(x)(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7) + s(x)$ for some polynomial $s$ with integer coefficients such that at least one of $1, 2, 3, 5, 6, 7$ is not a root of the polynomial.

Let's let $k$ be one of those excluded roots. Then, $f(k) = q(k) = k^3 + s(k) \neq k^3.$ However, since $k \in \{1, 2, 3, 5, 6, 7\},$ we must have $f(k) = k^3,$ which is a contradiction. Therefore, $s$ has roots at $x = 1, 2, 3, 5, 6, 7,$ and $q(x) = x^3 + r(x)(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7).$

If $r(x) = 0,$ then $q(x) = x^3.$ Otherwise, the non- $x^3$ part of $q$ can be "absorbed" into the $g(x)$ part to get $f(x) = x^3 + (g(x) + r(x))(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7),$ which is essentially the same form as we had before since $g(x) + r(x)$ can be any polynomial with integer coefficients.

Steven Yuan - 3 years, 10 months ago

@Steven Yuan – Thank you so much for taking the time to write this out! I totally follow your logic, but for some reason when I got to this point myself I had the following hang-up: Shouldn't it be possible to create polynomial $q\left( x\right)$ of minimum degree six that satisfies the six conditions but lacks an ${ x }^{ 3}$ term? That is, a polynomial of the form

$q\left( x\right) ={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }\dots { a }_{ 1 }x+{ a }_{ 0 }$ where $n\ge 6$ and ${ a }_{ 3 }=0$

I realize now, though, that such a polynomial could still be written in the form

$q\left( x\right) ={ x }^{ 3 }+r(x)(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)+s(x)$

so long as $s\left( x\right) =-{ x }^{ 3 } + t\left( x\right)$ . That would, however, not change the fact that $s\left( x\right)$ would still need to have roots of $1,2,3,5,6,7$ . Thank you for helping me see that!

Aric Floyd - 3 years, 10 months ago

@Aric Floyd – No problem! I'm always glad to help.

Steven Yuan - 3 years, 10 months ago

@Aric Floyd – Phrased a different way, what I am asking is: how do we prove that $q\left( 4 \right) (mod\quad 36)$ is the same for every polynomial $q\left( x \right)$ that fits the six given conditions (not just $q\left( x \right) ={ x }^{ 3 }$ ), and that therefore the minimum of $\left| q\left( 4 \right) -36g\left( 4 \right) \right|$ is the same?

Aric Floyd - 3 years, 10 months ago

@Aric Floyd – Of course you can have a different polynomial to $x^3$ . Instead of $x^3$ and $g(x)$ , you could have $x^3 + a(x)(x-1)(x-2)(x-3)(x-5)(x-6)(x-7) \hspace{1cm} \mathrm{and} \hspace{1cm} g(x)-a(x)$ for any polynomial $a(x)$ . It does not matter what value of $a(x)$ you choose, you will still get $64-36g(x)$ as the value at $4$ . The advantage of $x^3$ is that it is easy to find.

Just because there is more than one way to prove a result does not mean that you have to write them all down.

Mark Hennings - 3 years, 10 months ago

The feeling when you forget the absolute value sign and write 64-36=28 as the answer :facepalms: Question though: how does that mean we can factor it like that? I just jumped to an assumption.

Josiah Kiok - 3 years, 10 months ago

Use the Remainder Theorem: $x-a$ divides a polynomial $f(x)$ if and only if $f(a) = 0$ . Then use the fact that if you divide a polynomial in $\mathbb{Z}[x]$ by a monic polynomial in $\mathbb{Z}[x]$ , then the quotient and remainder both belong to $\mathbb{Z}[x]$ .

Mark Hennings - 3 years, 10 months ago

Can i know what's the rule that you have used to find the polynomial I mean the the part of g(x)(x-1)(x-2)()()()() till the end Where did that come from sir?

Tetbirt Hassen - 3 years, 10 months ago

See my comment to Josiah just below...

Mark Hennings - 3 years, 10 months ago

But g(x) can be 1,2,3,.....

Manan Agrawal - 3 years, 10 months ago

Yes, we can choose the polynomial $g(x)$ so that $g(4)$ can take any integer value. However $f(4) = 64 - 36g(4)$ will have smallest modulus when we have chosen the polynomial so that $g(4) = 2$ .

Mark Hennings - 3 years, 10 months ago

Ayon Ghosh
Aug 7, 2017

observe that,

$f(x)$ $=$ $x^3$ for $x$ belonging to $(1,2,3,5,6,7)$

means $f(x)$ $-$ $x^3$ $=$ $0$ for $x$ belonging to $(1,2,3,5,6,7)$ .

means $x$ $=$ $1/2/3/5/6/7$ are some zeroes of $g(x)$ $=$ $f(x)$ $-$ $x^3$ .

we can write,

$g(x)$ $=$ $f(x)$ $-$ $x^3$ $=$ $K(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some integer $K$ (since co-efficients are integers so this forces K to be an integer.)

now we have our true $f(x)$ that is,

$f(x)$ $=$ $K(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ $+$ $x^3$

$f(4)$ $=$ $K(4-1)(4-2)(4-3)(4-5)(4-6)(4-7)$ $+$ $4^3$ $=$ $-36K$ $+$ $64$ .

this has a minima for integer $K$ at $K$ $=$ $2$ that is $f(4)$ $=$ $-8$ so $|f(4)|$ $=$ $8$ .

Edit: Richard Desper 's argument is fine.The constant K does not have to really be a constant it could be any function,say, h(x) of any degree satisfying $|h(x)|_{min}$ $= 2$ .In my case h(x) = K and it works fine but is a special case.

Nitpick: K doesn't have to be an integer. It can be any polynomial with integer coefficients. But we don't care since we only care about the value at x=4.

Richard Desper - 3 years, 10 months ago

@Richard Desper how ? we must have K as an integer...as an example suppose we have K = 2/3 then the polynomial will be of the form ax^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx +g,and we can be certain that a = 2/3.so this would make a polynomial with non-integer coefficients,and that is not possible.(see the first sentence of the question.)

Ayon Ghosh - 3 years, 10 months ago

All we are told about f(x) is that it has integral coefficients and what its values are at six points. What you call g(x) is a polynomial of degree at least six. But there is no reason to think that g(x) is of degree exactly 6. One would have to prove that this must be the case. The proper amount of generality would say g(x) = f(x) - x^3 = h(x)*(x-1)(x-2)(x-3)(x-5)(x-6)(x-7) for some polynomial h(x), not some constant K. All we see is that the optimal result is achieved is |h(x)| = 2, or more specifically if h(x) = 2. You can get this with a constant, or you can have h(x) be any polynomial with integral coefficients satisfying h(4) = 2. For example, h(x) = x-2 works fine. We can have the degree of h be arbitrarily high and still have h(4) = 2. For example, h(x)= (x-4)^n + 2 will work for any value of n.

Richard Desper - 3 years, 10 months ago

1cubed=1, 2cubed=8, 3cubed=27, how is 4cubed=8?

Henry Grabarz - 3 years, 10 months ago

The function doesn't satisfy x^3 for all values, only for the values listed on the right hand side. That's why it doesn't have a factor of (x-4) (because we don't know what the value). However for the rest of the values given, we know they follow the graph of x^3. So we know that the function f(x) - x^3 has zeros at all those values, which is what is said above.

Aneesh Saripalli - 3 years, 10 months ago

@Henry Grabarz it doesn't work that way.see the solutions for the actual way to do it.

Ayon Ghosh - 3 years, 10 months ago

I don't understand the concept u used

I mean particularly and only this part

" we can write, g(x)=f(x)- x^3=

K(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)"

How did u come up with that idea of "K()()()()()" please, what's the concept used!?

Tetbirt Hassen - 3 years, 10 months ago

How did you get the minima for integer K at K=2?

Abhinav Jain - 3 years, 10 months ago

@Abhinav Jain no i just put K = 2 and saw it is local minima.(f(x 0) > f(x) for all x sufficiently close to x 0.)I don't have a formal rigorous proof for that.

Ayon Ghosh - 3 years, 10 months ago

Very well stated answer!...Thanks

Ashutosh Kapre - 3 years, 8 months ago

Is it cubic?

The answer is 8.

2 solutions

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