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

MATH437 December 2011

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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.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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