0
$\begingroup$

Say you have two sequences which are given by a polynomial of some degree. Both of these sequences share a term. Is there a way to find the missing term, and what is the minimum number of values (or information in general, such as degree of the polynomial) for each sequence for a unique solution to be found?

To make this a little bit clearer, I have an example of sequences which I know share a term but I do not know what that term is. The sequences are:

  • 0, 6, 30, __
  • 0, 12, 40, __

I also know that both of these sequences are increasing and that they may be third degree polynomials (they definitely are the same degree).
From the information given by these two sequences, is it possible to find out what the forth term (which is common to both of them) is, and how do you go about solving such a problem?

$\endgroup$
  • $\begingroup$ Are the two polynomials the same degree. Not sure we can do this if we don't know the degree. $\endgroup$ – fleablood Nov 27 '18 at 5:21
  • $\begingroup$ In the case of the example, yes. In general, maybe. If it's not possible without knowledge of them being the same, then assume the question requires them to be the same degree. I'll update the question to make it a bit clearer. $\endgroup$ – JolonB Nov 27 '18 at 5:27
  • $\begingroup$ I don't know but I don't think you can solve that. I might be wrong but I'm not sure. You know that $a_0=b_0=0$. You know $f(1)=6$ so the sum of the coefficients is $6$. and so on....If you replace $f(x)$ with $p(x)=f(x-1)$ we get that the first coeficient of $p$ is $64 and that thw sum of the coeficients of $p$ is $30$ but... I don't anticipate success. $\endgroup$ – fleablood Nov 27 '18 at 5:34
0
$\begingroup$

I assume you mean that $$a_n=v_3 n^3+v_2 n^2+v_1 n+v_0\\b_n=u_3 n^3+u_2 n^2+u_1 n+u_0$$which are also expressible as $$a_n=p_0(n-p_1)(n-p_2)(n-p_3)\\b_n=q_0(n-q_1)(n-q_2)(n-q_3)$$while $a_1=b_1=0$ we have $p_1=q_1=1$. Also for the rest of the terms we have $$p_0(2-p_1)(2-p_2)=6\\p_0(3-p_1)(3-p_2)=30$$and $$p_0(2-p_1)(2-p_2)=12\\q_0(3-q_1)(3-q_2)=40$$but we can not go further since we have 4 equations and 6 unknown variables. An extra constraint may be $p_0=q_0=1$.

$\endgroup$

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.