Thursday, April 23, 2015

Floor Function Problem...

Solve the following equation:

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

Note: $\displaystyle \lfloor x \rfloor$ denotes the largest integer not greater than $x$. This function, referred to as the floor function, is also called the greatest integer function, and its value at $x$ is called the integral part or integer part of $x$.

 Method 1:

We may choose to set:

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

so that:

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

So now our equation is:

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

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

and this gives us:

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

$\displaystyle n^2-16\le x-\frac{9}{4}<n^2-15$

We may simplify the system as:

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

$\displaystyle n^2-\frac{55}{4}\le x<n^2-\frac{51}{4}$

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:

$\displaystyle n-\frac{7}{3}\le n^2-\frac{51}{4}\:\therefore\:0\le n^2-n-\frac{125}{12}$

and

$\displaystyle n^2-\frac{55}{4}\le n-\frac{4}{3}\:\therefore\:n^2-n-\frac{149}{12}\le0$

For $\displaystyle 0\le n^2-n-\frac{125}{12}$ we find, by equating the quadratic to zero to find its roots:

$\displaystyle n=\frac{3\pm8\sqrt{6}}{6}$

So, using decimal approximations, $n$ must be less than about $-2.766$ and greater than about $3.766$.

For $\displaystyle 0\le n^2-n-\frac{149}{12}$ we find, by equating the quadratic to zero to find its roots:

$\displaystyle n=\frac{3\pm2\sqrt{114}}{6}$

So, using decimal approximations, $n$ must be more than about $-3.059$ and less than about $4.059$.

So, we find the only integer solutions for $n$ are $-3$ and $4$.

We could also complete the square on the two quadratics, and combine the two inequalities as:

$\displaystyle \frac{32}{3}\le\left(n-\frac{1}{2} \right)^2\le\frac{38}{3}$

$\displaystyle 32\le3\left(n-\frac{1}{2} \right)^2\le38$

$128\le3(2n-1)^2\le152$

$\displaystyle 3\cdot6^2=108$

$\displaystyle 3\cdot7^2=147$

$\displaystyle 3\cdot8^2=192$

So, we find:

$\displaystyle 2n-1=\pm7$

$\displaystyle n=\frac{1\pm7}{2}=-3,\,4$

Case 1:

$\displaystyle n=-3$

$\displaystyle -\frac{16}{3}\le x<-\frac{13}{3}$

$\displaystyle -\frac{19}{4}\le x<-\frac{15}{4}$

Thus:

$\displaystyle -\frac{19}{4}\le x<-\frac{13}{3}$

Case 2:

$\displaystyle n=4$

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

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

Thus:

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

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

$\displaystyle \left[-\frac{19}{4},-\frac{13}{3} \right)\,\cup\,\left[\frac{9}{4},\frac{8}{3} \right)$

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:

$\displaystyle \left\lfloor x+\frac{1}{3}\right\rfloor^2+4\left\lfloor x+\frac{1}{3}\right\rfloor-\left\lfloor x-\frac{1}{4}\right\rfloor=10$

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:

$\displaystyle n^2+3n-10=0$

$\displaystyle (n+5)(n-2)=0$

Since this equation has integral roots, we may proceed.

First root: $n=-5$

$\displaystyle -5+\frac{1}{4}\le x<-5+\frac{2}{3}$

$\displaystyle -\frac{19}{4}\le x<-\frac{13}{3}$

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:

$\displaystyle \left[-\frac{19}{4},-\frac{13}{3} \right)\,\cup\,\left[\frac{9}{4},\frac{8}{3} \right)$

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:

$\displaystyle (n+1)^2+4(n+1)-n-10=0$

$\displaystyle n^2+2n+1+4n+4-n-10=0$

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

This equation has irrational roots, so there are no valid solutions to consider from this case.

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