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Math Help - Check whether function is invertible...?

  1. #1
    Super Member fardeen_gen's Avatar
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    Check whether function is invertible...?

    Check whether f(x) = [x] + \sqrt{\{x\}}, f:\mathbb{R}\rightarrow\mathbb{R}, (where [.] & {.} represent greatest integral and fractional part function respectively) is invertible or not, if yes, then (i)find its inverse. (ii)solve the equation f(x) = f^{-1}(x).

    Spoiler:
    \bold{Answer:}\ \mbox{(i)}f^{-1} = [x] + \{x\}^2;\mbox{(ii)}\ x\in \mathbb{Z}


    I was unable to do it. Any ideas?
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  2. #2
    Senior Member pankaj's Avatar
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    Let n\leq x<n+1. \therefore[x]=n

    f(x)=n+\sqrt{x-n}

    f'(x)=\frac{1}{2\sqrt{x-n}}>0,when x is not an integer

    Incase f(x) is an integer then f(x)=x which is obviously invertible.

    Thus f(x) is clearly invertible.

    Let g(x) be the inverse of f(x).

    f(g(x))=x

    n+\sqrt{g(x)-n}=x

    \sqrt{g(x)-n}=x-n

    g(x)-n=x^2-2nx+n^2

    g(x)=n+(x-n)^2

    g(x)=[x]+(x-[x])^2

    f^{-1}(x)=[x]+\{x\}^2

    Since f^{-1}(x) is not identically equal to f(x).Therefore,solving f(x)=f^{-1}(x) is same as solving for f(x)=x.

    [x]+\sqrt{\{x\}}=x

    \sqrt{\{x\}}=\{x\}

    On squaring,
    \{x\}=0,1

    \{x\}=1 is impossible

    \{x\}=0

    i.e. x is an integer
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  3. #3
    MHF Contributor chisigma's Avatar
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    According to...

    Nearest Integer Function -- from Wolfram MathWorld

    the notation f(x)=[x] indicates the 'nearest integer function' , so that ...

    [x]= n , n-\frac{1}{2} \le x < n + \frac{1}{2}

    The function defined as...

    \lfloor x \rfloor = n , n \le x < n+1

    is the so called 'floor function' and it is indicated as f(x) = \lfloor x \rfloor

    Kind regards

    \chi \sigma
    Last edited by chisigma; June 22nd 2009 at 06:30 AM.
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  4. #4
    Senior Member pankaj's Avatar
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    I think it is up to you to define a particular notation for any function.fardeen_gen has chosen to define floor function by [x],so let it be.

    In textbooks of our country(India) this is how the floor function is defined
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  5. #5
    MHF Contributor chisigma's Avatar
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    Fardeen_gen in his post first used the notation f(x)=[x] and after 'verbally' described the function as 'greatest integral'... not 'floor function'... at this point some sort of doubt is legitimate...

    Kind regards

    \chi \sigma
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  6. #6
    Senior Member pankaj's Avatar
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    I agree.Actually, I was famaliar with this question and went on to write the solution right away.
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