Science:Math Exam Resources/Courses/MATH437/December 2011/Question 02

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH437 December 2011

  •  Q1  •  Q2  •  Q3  •  Q4  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  

   Other MATH437 Exams

  •  December 2006  •  December 2011  •  

Question 02
Let $\displaystyle k\in \mathbb {N}$, $\displaystyle a_{0},a_{1},...,a_{k}\in \mathbb {R}$ with $\displaystyle a_{k}\neq 0$. If $\displaystyle f:\mathbb {N} \to \mathbb {R}$ given by $\displaystyle f(n)=a_{k}n^{k}+a_{k-1}n^{k-1}+...+a_{1}n+a_{0}$ is a multiplicative function, show that $\displaystyle a_{k}=1$ and that $\displaystyle a_{i}=0$ for $\displaystyle 0\leq i<k$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Argue using the fact that $\displaystyle f((mn)^{\ell })=f(m^{\ell })f(n^{\ell })$ where $\displaystyle \gcd(m,n)=1$. Your proof should also have a feel of analysis in it.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

First we show that $\displaystyle a_{k}=1$. Let $\displaystyle \ell$ be an integer. Since our function is multiplicative, we know that $\displaystyle f(10^{\ell })=f(2^{\ell })f(5^{\ell })$. Expanding gives $\displaystyle {\begin{aligned}a_{k}10^{\ell k}+a_{k-1}10^{\ell (k-1)}+...+a_{0}&=(a_{k}2^{\ell k}+a_{k-1}2^{\ell (k-1)}+...+a_{0})(a_{k}5^{\ell k}+a_{k-1}5^{\ell (k-1)}+...+a_{0})\\&=a_{k}^{2}10^{\ell k}+a_{k-1}10^{\ell (k-1)}((2+5)a_{k}+a_{k-1})+..+a_{0}^{2}\end{aligned}}$ Isolating for $\displaystyle 10^{\ell k}$ gives $\displaystyle a_{k}(1-a_{k})10^{\ell k}=a_{k-1}10^{\ell (k-1)}((2+5)a_{k}+a_{k-1}-1)+...+a_{0}^{2}-a_{0}$ Dividing both sides by $\displaystyle 10^{\ell k}$ gives $\displaystyle a_{k}(1-a_{k})=10^{-\ell }(a_{k-1}((2+5)a_{k}+a_{k-1}-1)+...+(a_{0}^{2}-a_{0})10^{-\ell (k-1)})$ Making $\displaystyle \ell$ sufficiently large, we can see that the right hand side tends to 0. Since the left hand side is an integer, we see that the right hand side must in fact be 0 and hence so must the left hand side. Thus, $\displaystyle a_{k}(1-a_{k})=0$ giving $\displaystyle a_{k}=1$ since $\displaystyle a_{k}\neq 0$ as stated in the question. Next, we show that all the other coefficients must be 0. Let j be the largest index less that k such that $\displaystyle a_{j}\neq 0$. We assume that j exists and find a contradiction. Thus, $\displaystyle f(n)=x^{k}+a_{j}x^{j}+...+a_{0}$. Let m and n be arbitrary nonzero, nonunital positive integers such that $\displaystyle \gcd(m,n)=1$. Then, we see again that $\displaystyle f((mn)^{\ell })=f(m^{\ell })f(n^{\ell })$ and hence $\displaystyle {\begin{aligned}(mn)^{\ell k}+a_{j}(mn)^{\ell j}+...+a_{0}&=(m^{\ell k}+a_{j}m^{\ell j}+...+a_{0})(n^{\ell k}+a_{j}n^{\ell j}+...+a_{0})\\&=(mn)^{\ell k}+a_{j}(mn)^{\ell j}(m^{\ell (k-j)}+n^{\ell (k-j)}+a_{j})+..+a_{0}^{2}\end{aligned}}$ Cancelling and isolating for $\displaystyle a_{j}$ yields $\displaystyle {\begin{aligned}a_{j}(mn)^{\ell j}(1-m^{\ell (k-j)}-n^{\ell (k-j)}-a_{j})={\text{terms with order less than }}(mn)^{\ell j}\end{aligned}}$ As before dividing by $\displaystyle (mn)^{\ell j}$ and letting $\displaystyle \ell$ tend to infinity shows that the modified right hand side is 0 and thus, we have $\displaystyle a_{j}(mn)^{\ell j}(1-m^{\ell (k-j)}-n^{\ell (k-j)}-a_{j})=0$ Hence, either $\displaystyle a_{j}=0$ or $\displaystyle a_{j}=1-m^{\ell (k-j)}n^{\ell (k-j)}$. However, $\displaystyle a_{j}$ is completely independent of $\displaystyle \ell$. Thus as j is distinct from k (so $\displaystyle k-j\neq 0$), we must have that $\displaystyle a_{j}=0$ which shows that no such j can exists. Hence each and every $\displaystyle a_{j}$ for $\displaystyle 0\leq j<k$ must in fact be 0. This completes the proof. Keep in mind that $\displaystyle f(1)=1$ and this holds with this polynomial. Also it is clear that $\displaystyle f(n)=n^{k}$ is multiplicative.

{{#incat:MER CT flag||

Click here for similar questions

MER QGQ flag, MER RH flag, MER RS flag, MER RT flag, MER Tag Multiplicative functions, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag, Pages with math render errors

}}