3
$\begingroup$

This equations comes from my other question, and I thought it was ok to create another question about the same exercise. So I have to solve the equation: $$\int_0^{\lfloor x\rfloor}\lfloor t\rfloor^2\mathrm dt+\lfloor x\rfloor^2(x-\lfloor x\rfloor)=2(x-1),$$which is the same as $$\frac{(\lfloor x\rfloor^2-\lfloor x\rfloor)(2\lfloor x\rfloor-1)}{6}+\lfloor x\rfloor^2(x-\lfloor x\rfloor)=2(x-1),$$ here's the graph of $f(x)=\int_0^{\lfloor x\rfloor}\lfloor t\rfloor^2\mathrm dt+\lfloor x\rfloor^2(x-\lfloor x\rfloor)$:

The graph of f(x)

One solution is obviously $1$, which is easy, but I don't know how to find other solutions.

Thanks for @Ross Millikan's answer, here's the graph of both functions: graphs of both functions

They cross each other at points $(1,0)$ and $(\frac{5}{2},3)$, so solutions are $x=1, x=\frac{5}{2}$.

$\endgroup$
3
$\begingroup$

To use the graph, you can just overplot $2(x-1)$ on it, which passes through $(1,0)$ with a slope of $2$. It also passes through $(5/2,3)$, which is another solution, then stays below the curve forever. You can plug in $x=5/2$ to verify the solution.

$\endgroup$
2
  • $\begingroup$ They cross each other on $1$ and $5/2$ which are indeed the solutions, Thanks! I am wondering how would one solve the equation without plotting 2 graphs, I guess I mean algebraically. $\endgroup$ – George Apriashvili Dec 13 '14 at 15:16
  • $\begingroup$ It is hard to read the graph sometimes, but would be easier at a different scale. Another approach is to recognize that all solutions will be small as the $x^3$ term on the left will swamp the right side quickly. Then you can assume $x$ is in a particular interval, for example $[2,3)$. Now you know $\lfloor x \rfloor=2$, plug that in and solve. Reject any solution outside your interval. It is the same idea as breaking absolute value signs into cases. $\endgroup$ – Ross Millikan Dec 13 '14 at 15:29
1
$\begingroup$

Here is an algebraically solution. Put $n=\lfloor x\rfloor$; hence we get $\displaystyle x=\frac{4n^3+3n^2-n-12}{6(n^2-2)}$. Now we have $x=n+a$ with $a\in [0,1[$. We have: $$a=-\frac{2n^3-3n^2-11n+12}{6(n^2-2)}=-c_n$$ We have $c_n\geq n(2n^2-3n-11)$, it is easy to see that $2n^2-3n-11>0$, hence $c_n>0$ if $n\geq 4$, and that $c_3>0$ by direct computation. Hence for $n\geq 3$, we get $a<0$, this show that we must have $n\leq 2$. It is easy to finish.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.