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  1. #1
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    Limit

    Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets.

    would someone give me a brief explanation on how to solve such equations?
    (note: these are greatest integer brackets)
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Calc1 View Post
    Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets.

    would someone give me a brief explanation on how to solve such equations?
    (note: these are greatest integer brackets)




    g(x)=\left\{<br />
\begin{array}{cc}<br />
0 & \text{if } -3 \leq x \leq 3 \end{array} \right.<br />

    I can't get the latex right but it looks to me like the function is zero for all x

    1-x is the additive invers of x-1
    Last edited by mr fantastic; June 11th 2009 at 02:28 PM. Reason: I only fixed the latex ....
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    g(x)=\left\{\begin{array}{cc}0,&\mbox{if}{-3\leq{x}\leq{3}}\end{array}\right

    I can't get the latex right but it looks to me like the function is zero for all x

    1-x is the additive invers of x-1
    Counterexample: g(2.5)=[1.5]+[1-2.5]
    =1+[-1.5]
    =1+(-2)
    =-1
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  4. #4
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    i understand the greatest integer theory but i still don't get how it would be solved in this question

    as you stated g(2.5) = -1

    ok i know how you got that but how is it exactly related to the question?
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Calc1 View Post
    Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets.

    would someone give me a brief explanation on how to solve such equations?
    (note: these are greatest integer brackets)
    let

    f(x)=[x-1]
    h(x)=[1-x] x\in [-3,3]

    f(x)=\begin{Bmatrix}<br /> <br />
-4&...if...-3\leq x<-2\\<br /> <br />
-3&...if...-2\leq x<-1\\<br /> <br />
-2&...if...-1\leq x<0\\<br /> <br />
-1&...if...0\leq x<1\\<br /> <br />
0&...if...1\leq x<2\\<br /> <br />
1&...if...2\leq x<3\\<br /> <br />
2&...if...x=3<br /> <br />
\end{Bmatrix}<br /> <br /> <br />


    h(x)=\begin{Bmatrix}<br /> <br />
4&...if...x=-3\\<br /> <br />
3&...if...-3< x\leq -2\\<br /> <br />
2&...if...-2< x\leq -1\\<br /> <br />
1&...if...-1< x\leq 0\\<br /> <br />
0&...if...0< x\leq 1\\<br /> <br />
-1&...if...1< x\leq 2\\<br /> <br />
-2&...if...2< x\leq 3 \\ <br /> <br />
\end{Bmatrix}


    g(x)=\begin{Bmatrix}<br /> <br />
0&...if...x=-3\\<br /> <br />
-1&...if...-3< x< -2\\<br /> <br />
0&...if...x=-2\\<br /> <br />
 -1&...if...-2< x< -1\\<br /> <br />
0&...if...x=-1\\<br /> <br />
 -1&...if...-1< x< 0\\<br /> <br />
0&...if...x=0\\<br /> <br />
\end{Bmatrix}
    <br />
............\begin{Bmatrix}<br /> <br />
 -1&...if...0< x< 1\\<br /> <br />
0&...if...x=1\\<br /> <br />
 -1&...if...1< x< 2\\<br /> <br />
0&...if...x=2\\<br /> <br />
 -1&...if...2< x<3\\<br /> <br />
0&...if...x=3\\<br /> <br /> <br />
\end{Bmatrix} since we can write f(x) and h(x) like below then find the sum of f(x) and h(x) because g(x)=f(x)+h(x)


    f(x)=\begin{Bmatrix}<br /> <br />
-4&...if... x=-3\\<br /> <br />
-4&...if...-3< x<-2\\<br /> <br />
-3&...if...x=-2\\<br /> <br />
-3&...if...-2< x<-1\\<br /> <br />
-2&...if...x=-1\\<br /> <br />
-2&...if...-1< x<0\\<br /> <br />
-1&...if...x=0\\<br /> <br />
\end{Bmatrix}
    <br />
...........\begin{Bmatrix}<br /> <br />
-1&...if...0< x<1\\<br /> <br />
0&...if...x=1\\<br /> <br />
0&...if...1< x<2\\<br /> <br />
1&...if...x=2\\<br /> <br />
1&...if...2< x<3\\<br /> <br />
2&...if...x=3<br /> <br />
\end{Bmatrix}<br /> <br />


    h(x)=\begin{Bmatrix}<br /> <br />
 4&...if... x=-3\\<br /> <br />
3&...if...-3< x<-2\\<br /> <br />
3&...if...x=-2\\<br /> <br />
 2&...if...-2< x<-1\\<br /> <br />
2&...if...x=-1\\<br /> <br />
 1&...if...-1< x<0\\<br /> <br />
1&...if...x=0\\<br /> <br />
\end{Bmatrix}
    <br />
...........\begin{Bmatrix}<br /> <br /> <br />
 0&...if...0< x<1\\<br /> <br />
0&...if...x=1\\<br /> <br />
 -1&...if...1< x<2\\<br /> <br />
-1&...if...x=2\\<br /> <br />
 -2&...if...2< x<3\\<br /> <br />
 -2&...if...x=3<br /> <br />
 \end{Bmatrix}


    it is clear or not






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  6. #6
    MHF Contributor Amer's Avatar
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    you can take 1 from the greatest integer function since the factor of x is one

    so you can write g(x) like this


    g(x)=[x-1]+[1-x]=[x]-1+1+[-x]=[x]+[-x]

    but the result will be the same as I solve before
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by alexmahone View Post
    Counterexample: g(2.5)=[1.5]+[1-2.5]
    =1+[-1.5]
    =1+(-2)
    =-1
    I've made three mistakes within the last 24 hours. I must be losing it.
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  8. #8
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    Thanks Amer, it makes sense to me now
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