7
$\begingroup$

Question: Is there an easier way to solve this problem?

Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.

My Attempt: I started off with the general cubic $ax^3+bx^2+cx+d=f(x)$ and manually plugged in each point to get the following system:$$\begin{align*} & 27a+9b+3c+d=2\\ & 64a+16b+4c+d=4\\ & 125a+25b+5c+d=-3\\ & 216a+36b+6c+d=8\end{align*}\tag1$$

Solving the system with the handy matrix gives the solutions as $a=\frac 92,b=-\frac {117}2,c=245,d=-328$. Thus, $f(0)=-328$.


Even though I (think) solved the problem correctly, this method seems a bit "bulky" especially when everything becomes a higher degree. So I'm wondering if there is a quicker way to evaluate this kind of problem.

$\endgroup$
16
$\begingroup$

Linear functions (degree $1$ polynomials) $p(x)$ have constant first differences $(\Delta^1f)(x)=f(x)-f(x-1)$, quadratic functions have constant second differences $(\Delta^2f)(x)=(\Delta^1f)(x)-(\Delta^1f)(x-1)$, and so on. You can use this fact to find $f(0)$ as shown below. First, from the known values, calculate the forward differences (red) from left to right, then copy the constant third difference (blue), and finally “back-calculate” to $f(0)$ (purple).

enter image description here

$\endgroup$
  • 2
    $\begingroup$ Best method indeed. $\endgroup$ – StubbornAtom Nov 27 '16 at 8:12
  • $\begingroup$ Do you know if the constant difference has a name? $\endgroup$ – Frank Nov 27 '16 at 21:06
  • $\begingroup$ No, I don’t know if a name other than the $k$-th (backward in this case, because it is the difference $f(x)-f(x-1)$ as opposed to the difference $f(x+1)-f(x)$) difference. $\endgroup$ – Steve Kass Nov 28 '16 at 3:47
  • $\begingroup$ @SteveKass Either way, this has got to be the best method possible. $\endgroup$ – Frank Dec 2 '16 at 19:00
7
$\begingroup$

There is a conceptually simpler way: do some linear algebra and polynomial arithmetic. Instead of a complex linear system, solve $4$ much simpler linear systems. Namely, solve the following problems: find polynomials $p(x), q(x), r(x), s(x)$ such that: \begin{align} (a)\enspace&\begin{cases}p(3)=1\\p(4)=0\\p(5)=0\\p(6)=0\end{cases} &(b)\enspace&\begin{cases}q(3)=0\\q(4)=1\\q(5)=0\\q(6)=0\end{cases} &(c)\enspace&\begin{cases}r(3)=0\\r(4)=0\\r(5)=1\\r(6)=0\end{cases} &(d)\enspace&\begin{cases}s(3)=0\\s(4)=0\\s(5)=0\\s(6)=1\end{cases} \end{align} Then the solution is $$f(x)=2p(x)+4q(x)-3r(x)+8s(x).$$

Reminder:

Let $K$ be a field, $\alpha\in K$, $f(x)$ a polynomial in $K[x]$. Then, $$f(\alpha)=0\iff x-\alpha\enspace\text{divides}\enspace f(x).$$

$\endgroup$
  • $\begingroup$ That's nice!${}{}$ $\endgroup$ – TonyK Nov 27 '16 at 2:06
  • 2
    $\begingroup$ From the linear algebra point of view, those systems aren't really simpler; they all have the same matrix. There is a fast way to solve them using the zeros of polynomials, but you should mention it in your answer, otherwise I am afraid it is not very helpful as it stands. $\endgroup$ – Federico Poloni Nov 27 '16 at 9:04
  • $\begingroup$ @Federico Poloni: You're right.. I supposed everyone has this result in mind, but it's not very realistic. Thanks for drawing my attention to this point! $\endgroup$ – Bernard Nov 27 '16 at 12:38
5
$\begingroup$

You can directly find out the polynomial $f$ by considering it according as the $x$-values available:

Let $f(x)=a_0+a_1(x-3)+a_2(x-3)(x-4)+a_3(x-3)(x-4)(x-5)$ for real constants $a_0,a_1,a_2,a_3$. Note that we don't need to take into account the value $x=6$ as this is already a cubic polynomial.

Then, $f(3)=2\Rightarrow a_0=2$

$f(4)=4\Rightarrow a_1=2$

$f(5)=-3\Rightarrow a_2=-\frac{9}{2}$

$f(6)=8\Rightarrow a_3=\frac{9}{2}$

Thus, $f(x)=2+2(x-3)-\frac{9}{2}(x-3)(x-4)+\frac{9}{2}(x-3)(x-4)(x-5)$.

I think the calculations are pretty simple this way as you have chosen $f$ to be such.

$\endgroup$
  • $\begingroup$ Basically this is Lagrange's interpolation formula. $\endgroup$ – StubbornAtom Jun 26 '18 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.