MHB My methods for solving the July 8, 2013 High School POTW

MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Hello MHB,

I wanted to post the 2 methods I provided for solving the High School POTW, which I submitted for use that week. :D

The problem was to solve:

$$\left\lfloor x+\frac{7}{3} \right\rfloor^2+\left\lfloor x-\frac{9}{4} \right\rfloor=16$$

Method 1:

We may choose to set (where $n\in\mathbb{Z}$):

$\displaystyle n\le x+\frac{7}{3}<n+1$

so that:

$\displaystyle \left \lfloor x+\frac{7}{3}\right \rfloor^2=n^2$

So now our equation is:

$\displaystyle n^2+\left \lfloor x-\frac{9}{4}\right \rfloor=16$

$\displaystyle 16-n^2=\left \lfloor x-\frac{9}{4}\right \rfloor$

and this gives us:

$\displaystyle 16-n^2\le x-\frac{9}{4}<16-n^2+1$

$\displaystyle 16-n^2\le x-\frac{9}{4}<17-n^2$

We may simplify the system as:

$\displaystyle n-\frac{7}{3}\le x<n-\frac{4}{3}$

$\displaystyle \frac{73}{4}-n^2\le x<\frac{77}{4}-n^2$

We now have two intervals where $x$ must be, and since $x$ must simultaneously satisfy both inequalities, we want to find the intersection of these two intervals, or where they overlap. In order for there to be any overlap, we require the lower bound of each interval to be less than or equal to the upper bound of the other, giving us:

$\displaystyle n-\frac{7}{3}\le\frac{77}{4}-n^2\:\therefore\:n^2+n-\frac{259}{12}\le0$

and

$\displaystyle \frac{73}{4}-n^2\le n-\frac{4}{3}\:\therefore\:0\le n^2+n-\frac{235}{12}$

For $\displaystyle n^2+n-\frac{259}{12}\le0$ we find, by equating the quadratic to zero to find its roots:

$\displaystyle n=\frac{-3\pm\sqrt{786}}{6}$

So, using decimal approximations, $n$ must be greater than about $-5.1726$ and less than about $4.1726$.

For $\displaystyle 0\le n^2+n-\frac{235}{12}$ we find, by equating the quadratic to zero to find its roots:

$\displaystyle n=\frac{-3\pm\sqrt{741}}{6}$

So, using decimal approximations, $n$ must be less than about $-4.9535$ and greater than about $3.9535$.

So, we find the only integer solutions for $n$ are $-5$ and $4$.

Case 1:

$\displaystyle n=-5$

$\displaystyle -\frac{22}{3}\le x<-\frac{19}{3}$

$\displaystyle -\frac{27}{4}\le x<-\frac{23}{4}$

Thus:

$\displaystyle -\frac{27}{4}\le x<-\frac{19}{3}$

Case 2:

$\displaystyle n=4$

$\displaystyle n-\frac{7}{3}\le x<n-\frac{4}{3}$

$\displaystyle \frac{73}{4}-n^2\le x<\frac{77}{4}-n^2$

$\displaystyle \frac{5}{3}\le x<\frac{8}{3}$

$\displaystyle \frac{9}{4}\le x<\frac{13}{4}$

Thus:

$\displaystyle \frac{9}{4}\le x<\frac{8}{3}$

So, the solution for $x$ in interval notation is:

$\displaystyle \left[-\frac{27}{4},-\frac{19}{3} \right)\,\cup\,\left[\frac{9}{4},\frac{8}{3} \right)$

Method 2:

The first step is to reduce the floor arguments using:

$\displaystyle \lfloor x\pm(n+k) \rfloor=\lfloor x\pm k \rfloor\pm n$ where $\displaystyle n\in\mathbb N$ and $\displaystyle k\in\mathbb R$, $0<k<1$.

So now we have:

$\displaystyle \left(\left\lfloor x+\frac{1}{3}\right\rfloor +2\right)^2+\left\lfloor x-\frac{1}{4}\right\rfloor-2=16$

$\displaystyle \left\lfloor x+\frac{1}{3}\right\rfloor^2+4\left\lfloor x+\frac{1}{3}\right\rfloor+
\left\lfloor x-\frac{1}{4}\right\rfloor=14$

Then we observe that:

$\displaystyle d\left(x-\frac{1}{4},x+\frac{1}{3} \right)=\left|\left(x+\frac{1}{3} \right)-\left(x-\frac{1}{4} \right) \right|=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$

Now since the difference between the two arguments is less than 1 we have to consider two cases. Either the two arguments are between successive integers or the two arguments lie on either side of an integer. In the first case, the smaller argument may be equal to the smaller of the two successive integers but the larger argument must be strictly less than the larger of the two successive integers.

Case 1:

$\displaystyle n\le x-\frac{1}{4}<n+\left(1-\frac{7}{12} \right)$

$\displaystyle n\le x-\frac{1}{4}<n+\frac{5}{12}$

$\displaystyle n+\frac{1}{4}\le x<n+\frac{2}{3}$

Since both arguments are between the same two successive integers, we must have:

$\displaystyle n=\left\lfloor x-\frac{1}{4}\right\rfloor=\left\lfloor x+\frac{1}{3}\right\rfloor$

Our equation becomes:

$\displaystyle n^2+5n-14=0$

$\displaystyle (n+7)(n-2)=0$

Since this equation has integral roots, we may proceed.

First root: $n=-7$

$\displaystyle -7+\frac{1}{4}\le x<-7+\frac{2}{3}$

$\displaystyle -\frac{27}{4}\le x<-\frac{19}{3}$

Second root: $n=2$

$\displaystyle 2+\frac{1}{4}\le x<2+\frac{2}{3}$

$\displaystyle \frac{9}{4}\le x<\frac{8}{3}$

Thus, from the first case, we find the solution for $x$ in interval notation as:

$\displaystyle \left[-\frac{27}{4},-\frac{19}{3} \right)\,\cup\,\left[\frac{9}{4},\frac{8}{3} \right)$

Case 2:

$\displaystyle n=\left\lfloor x-\frac{1}{4} \right\rfloor$

$\displaystyle n+1=\left\lfloor x+\frac{1}{3}\right\rfloor$

And our equation becomes:

$\displaystyle (n+1)^2+4(n+1)+n-14=0$

$\displaystyle n^2+2n+1+4n+4+n-14=0$

$\displaystyle n^2+7n-9=0$

This equation has irrational roots, so there are no valid solutions to consider from this case.
 
Mathematics news on Phys.org
I love it when the solutions are so elegant as these...mine looks like a bit childish!:o

Bravo, MarkFL!(Clapping)
 
anemone said:
I love it when the solutions are so elegant as these...mine looks like a bit childish!:o

Bravo, MarkFL!(Clapping)

(Sun) I love the way you incorporated the colors of my usergroups into my username.

I thought your solution was very well done too! Congrats on getting it correct! (Clapping)

I never formally studied the floor/ceiling functions and found this problem to be a way for me to try to become more familiar with the step functions. (Emo)
 
I've added these two solutions to the POTW as well. I was planning on doing this originally so quickly posted the solution from anemone. Sorry for not including your solutions at first Mark! :(

Anyway, I think that this problem is interesting enough as well as are the solutions that this thread is warranted still. :D
 
Jameson said:
I've added these two solutions to the POTW as well. I was planning on doing this originally so quickly posted the solution from anemone. Sorry for not including your solutions at first Mark! :(

Anyway, I think that this problem is interesting enough as well as are the solutions that this thread is warranted still. :D

Hey, no worries...I just figured you are in the habit of posting only the solutions of those who submitted a solution, but I did feel that MHB would benefit from having these methods publicly available too. :D

Also, since I had to redo the solutions because as you know I actually mistyped the problem, I just couldn't let all that work go unseen. (Tongueout)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top