+483 A polynomial product of the form \((1 - z)^{b_1} (1 - z^2)^{b_2} (1 - z^3)^{b_3} (1 - z^4)^{b_4} (1 - z^5)^{b_5} \dotsm (1 - z^{32})^{b_{32}},\)
where the \(b_k\) are positive integers, has the surprising property that if we multiply it out and discard all terms involving \(z\) to a power larger than 32, what is left is just \(1-2z.\) Determine \(b_{32}.\)
You can enter your answer using exponential notation.
+483 (reposting because it has been a long time since I last posted here: https://web2.0calc.com/questions/polynomials-question-pls-help-really-hard )
+20 I don't have a definative anwswer, but this was how I attempted this problem.
The first thing to notice is that it is impossible to derive 2z from any term other than \((1-z)^{b_1}\) due to how the binomials expand.
Moreover, for any \(b_1\) other than 2, it is impossible to derive -2z (getting coefficients from binomial theorem).
In particular, when \(b_1\)= 2, then the (1-z) term expands to \(1-2z+z^2\).
Moreover, when we expand \((1-z^{32})^{b_{32}}\), the only terms that matter are the 1, and \(b_{32}C1-z^{32}\), or \(b_{32}z^{32}\). (we again expand using the binomial theorem)
All the other terms contain z to a power greater than 32, so they are irrelevant once we expand and discard the terms. Moreover, the 1 simply "maintains" all the terms in the expansion of things before the term containing \(b_{32}\), so it keeps the \(1-2z\) preserved.
The \(b_{32}z^{32}\) only matters when we multiply by the term "1" in the expansion of \((1-z)^{b_1}(1-z^2)^{b_2}(1-z^3)^{b_3}(1-z^4)^{b_4}...(1-z^{31})^{b_{31}}\)
Because of this fact, we know that the exponent of (1-z^2) must be 1 in order to make the coefficient of the z^2 term 0 (the z^2 term is not altered in any way after we multiply past (1-z^2)).
Afterward, using a similar strategy (and admittedly, wolfram alpha to hopefully find a pattern within the exponents), I found that the coefficient of the term with z^3 is 2
z^1 is 2
z^2 is 1
z^3 is 2
z^4 is 3
z^5 is 6
z^6 is 9
z^7 is 18
z^8 is 36
and this is where the calculator could not compute the expression any further.
I can't find a pattern for the numbers 2, 1, 2, 3, 6, 9, 18, 36, and so on, but hopefully you can continue this approach. \(b_{32}\) is the coefficient of the \(z^{32}\) term in the expansion
\((1-z)^{b_1}(1-z^2)^{b_2}(1-z^3)^{b_3}(1-z^4)^{b_4}...(1-z^{31})^{b_{31}}\)
, but I have no clue on how to get there. The reason why this is the case is because we have to cancel out the z^32 term, in accordance with the situation descibed in the problem. If we substitute the values for the b's we found earlier into the expression, it becomes
\((1-z)^{2}(1-z^2)^1(1-z^3)^{2}(1-z^4)^{3}(1-z^5)^{6}(1-z^6)^{9}(1-z^7)^{18}...(1-z^{31})^{b_{31}}\)
Im sure there are some types of combinotorical ways to use the binomial theorem in order to calculate the rest of the terms, but this is my best shot at this question.
Sorry I couldn't find a more specific answer, but tell me what the answer is if you manage to get it, this problem is pretty interesting, especially for a concise, efficient solution.
