Hint 1:[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function
Actually i am not able to get how to solve it
I tried but I got stuck
I see you have unstuck your [CAPS] key.[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function
Actually i am not able to get how to solve it
I tried but I got stuck
Yes which is 2=1 which is not possibleCan you answer the simpler question 2x=x?
Instead of dividing both sides by x, try subtraction instead. You'll see why you can't divide x.Yes which is 2=1 which is not possible
I'd start by trying to graph each side on the same pair of axes. That would either give me the solution (which I could confirm by algebraic methods), or at least a hint as to what to consider. At the least, it would give you a better sense of how these functions work.[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function
Actually i am not able to get how to solve it
I tried but I got stuck
Before you can even consider solving your problem, you need to be able to solve the one I gave you.Yes which is 2=1 which is not possible
Ah, that is what the OP did.Instead of dividing both sides by x, try subtraction instead. You'll see why you can't divide x.
I got it, I made a mistake, I am sorryBefore you can even consider solving your problem, you need to be able to solve the one I gave you.
I can only hope that what you wrote was based on the fact that you thought x=1 and found out that was wrong.
2x=x is a very simple equation that you should be able to solve. And yes, it does have a solution
i did it in cases , just realised what I was doing wrong, rectified it and got the answerI'd start by trying to graph each side on the same pair of axes. That would either give me the solution (which I could confirm by algebraic methods), or at least a hint as to what to consider. At the least, it would give you a better sense of how these functions work.
Algebraically, I'd probably break it into cases.
Possibly you've done all this, In that case, we'll want to see where you got stuck!
Well, let's be very careful here. Check that endpoint, x=1/2:well thanks, everybody I got the answer it
x belongs to[-1/2,1/2]
[2x] = [2*1/2] = [1] = 1[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denotes greatest integer function
Okay okay then it will be x belongs to [-1/2,1/2)Well, let's be very careful here. Check that endpoint, x=1/2:
[2x] = [2*1/2] = [1] = 1
[x] = [1/2] = 0
They aren't equal!
I assume you've checked over your work to make sure there aren't any other errors, and that you won't make the same little mistake next time ...Okay okay then it will be x belongs to [-1/2,1/2)
Here's my solution.Hint 1:
For positive [imath]m[/imath],
\(\displaystyle \lfloor{mx\rfloor}=\lfloor{x\rfloor}+ \left\lfloor x + \frac{1}{m} \right\rfloor + \dots + \left\lfloor x + \frac{m-1}{m} \right\rfloor \\\)
Dr. PetersonI assume you've checked over your work to make sure there aren't any other errors, and that you won't make the same little mistake next time ...
If you'd like, you can show us your work, so we can see if there are any subtle errors in the method. But you probably did fine.
And next time you have a question, of course, you'll want to show your work from the start, so that if you just made some little slip, we can save time by just pointing it out.
When did you learn this identity ?Here's my solution.
[math] \lfloor2x\rfloor = \lfloor x \rfloor\\ \underbrace{\lfloor{x\rfloor}+ \left\lfloor x + \frac{1}{2} \right\rfloor}_{\text{Hermite's Identity}} = \lfloor x \rfloor\\ \left\lfloor x + \frac{1}{2} \right\rfloor=0 \iff 0 \le x + \frac{1}{2}<1\\ \therefore x\in \left[-\frac{1}{2},\frac{1}{2}\right) [/math]
I don't recall learning about the floor and ceiling function in grade school. I probably stumbled upon it outside of academic settings.When did you learn this identity ?
I mean in which grade