# Find number of values of x satisfying the given equation.

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• March 14th 2011, 12:05 AM
findmehere.genius
Find number of values of x satisfying the given equation.
limit 0 to 2[x+14] $\int [\frac{t}{2}].dt$=limit 0 to {x} $\int [t+14] dt$

where {.} is fractional part of x and [.] is integer number less than or equal to x.

actually speaking, I am not able to get anything from the language of question whether is it asking integral values or real number.
But one thing I hope am doing correct is that R.H.S. comes out to be 14 (Happy).

P.S.: i tried my best to solve this question and also use LaTex to type it here. (Nerd)
• March 14th 2011, 01:19 AM
NOX Andrew
Is the following equation the equation you are trying to solve for in terms of $x$?

$\displaystyle \int_0^{2(x+14)} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^x \lfloor t + 14 \rfloor \, dt$
• March 14th 2011, 01:52 AM
findmehere.genius
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Quote:

Originally Posted by NOX Andrew
Is the following equation the equation you are trying to solve for in terms of $x$?

$\displaystyle \int_0^{2(x+14)} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^x \lfloor t + 14 \rfloor \, dt$

No..the upper limit in LHS is enclosed by [.] (greatest integer function)...and that of RHS by { }(fractional part of x)...so it will look like...i am nt so good at latex so...just trying to modify yours
Quote:

$\displaystyle \int_0^{[2x+14]} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^{/{x/}} \lfloor t + 14 \rfloor \, dt$
and if it still doesnt seems clear...so a jpg i am attaching!
• March 14th 2011, 02:35 AM
NOX Andrew
Don't worry about the LaTeX. It took me a while to familiarize myself with the syntax, and I'm still not that good! Here is the equation for anyone else who may be able to help:

$\displaystyle \int_0^{2 \lfloor x+14 \rfloor} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^{\{x\}} \lfloor t + 14 \rfloor \, dt$

I have to leave for school now, but I will work on the problem during the day. I'll let you know how I fare.

Edit: $x = -14$ is the only solution I found.
• March 14th 2011, 07:01 AM
findmehere.genius
answer given is 14 i.e. 14 solutions of x!
anyway, would please you elaborate, or rather just hint out how did you proceed with to get to that answer at least!
• March 14th 2011, 02:02 PM
NOX Andrew
I made an educated guess. At $x = -14$, $2\lfloor x - 14 \rfloor = 0$ and the left-hand side of the equation is $0$. Furthermore, at $x = -14$, $\{x\} = 0$ and the right-hand side is also $0$. Therefore, $x = -14$ is a solution to the equation.

I don't know how to solve the equation using analytic methods, but I will explore the properties of the greatest integer and fractional part functions. In the meantime, someone else with more knowledge may come along and help.
• March 14th 2011, 03:03 PM
Plato
Quote:

Originally Posted by findmehere.genius
limit 0 to 2[x+14] $\int [\frac{t}{2}].dt$=limit 0 to {x} $\int [t+14] dt$
where {.} is fractional part of x and [.] is integer number less than or equal to x.

Simplify by using some properties of the floor function.
Note that $0\le \left\{ x \right\} = x - \left\lfloor x \right\rfloor <1,\quad \left\lfloor {x + n} \right\rfloor = \left\lfloor x \right\rfloor + n,\;\& \,\left\{ {x + n} \right\} = \left\{ x \right\}$

Notice that $\displaystyle \int_0^{\left\{ x \right\}} {\left\lfloor {t + 14} \right\rfloor dt}= \int_0^{\left\{ x \right\}} {\left\lfloor t \right\rfloor dt} + \int_0^{\left\{ x \right\}} {14dt} = 0 + 14\left\{ x \right\} = 14\left\{ x \right\}$
• March 15th 2011, 03:42 AM
findmehere.genius
Thanks to you both of you for trying this question, even I finally got somewhere through it and the integration part reduced to(Do confirm it) [x+14]^2 =14{x}

then plotting the graph of both the function would give us 7 solutions?!

I would be pleased with just this much if I get to confirm that my integration is correct, rest answer given in book doesn't matter!
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