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Math Help - Invent a function

  1. #1
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    Invent a function

    I'm supposed to invent a continuous function f: R into R such that its improper integral is zero, but which is unbounded as x approaches negative infinity and x approaches infinity.
    I have no clue how to do this problem. Could somebody help me? Thanks
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    Quote Originally Posted by namelessguy View Post
    I'm supposed to invent a continuous function f: R into R such that its improper integral is zero, but which is unbounded as x approaches negative infinity and x approaches infinity.
    I have no clue how to do this problem. Could somebody help me? Thanks
    How about taking f(x) = x if x is an integer and f(x)=0 otherwise?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Opalg View Post
    How about taking f(x) = x if x is an integer and f(x)=0 otherwise?
    i was thinking about something like that. but is that function continuous?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hi

    How about f:\mathbb{R}^+\mapsto\mathbb{R} such that for all n\in\mathbb{N}^*, f is defined on every [n,\,n+1] by f(x)=\left\{\begin{array}{ll}<br />
n^4(x-n)&\mbox{ if } x\in\left[n,\,n+\frac{1}{n^3}\right]\\<br />
-n^4\left(x-n-\frac{2}{n^3}\right)& \mbox{ if } x\in\left[n+\frac{1}{n^3},\,n+\frac{2}{n^3}\right]\\<br />
0 & x\in\left[n+\frac{2}{n^3},\,n+1\right]<br />
\end{array}\right. and f=0 on [0,\,1] ?

    On each interval [n,\,n+1], the curve of the function has the shape of a triangle which has one vertex at (n,0), one at \left(n+\frac{1}{n^3},\,n\right) and one at \left(n+\frac{2}{n^3},\,0\right). As n approaches \infty, the height of the triangle does the same (see the second vertex) so the function is not bounded. It can be checked that this function is continuous on \mathbb{R}^+. As, for n>0, \int_n^{n+1}f =\frac{1}{2}\frac{2}{n^3}n (it's the area of a triangle which height is n and which base is \frac{2}{n^3}), we get that \int_0^{\infty}f=\sum_{k=1}^{\infty}\frac{1}{k^2}=  \frac{\pi^2}{6} and the function is integrable.

    As we want \int_{\mathbb{R}}f=0, we define for x<0, f(x)=-f(-x) and that's it.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    How about f:\mathbb{R}^+\mapsto\mathbb{R} such that for all n\in\mathbb{N}^*, f is defined on every [n,\,n+1] by f(x)=\left\{\begin{array}{ll}<br />
n^4(x-n)&\mbox{ if } x\in\left[n,\,n+\frac{1}{n^3}\right]\\<br />
-n^4\left(x-n-\frac{2}{n^3}\right)& \mbox{ if } x\in\left[n+\frac{1}{n^3},\,n+\frac{2}{n^3}\right]\\<br />
0 & x\in\left[n+\frac{2}{n^3},\,n+1\right]<br />
\end{array}\right. and f=0 on [0,\,1] ?

    On each interval [n,\,n+1], the curve of the function has the shape of a triangle which has one vertex at (n,0), one at \left(n+\frac{1}{n^3},\,n\right) and one at \left(n+\frac{2}{n^3},\,0\right). As n approaches \infty, the height of the triangle does the same (see the second vertex) so the function is not bounded. It can be checked that this function is continuous on \mathbb{R}^+. As, for n>0, \int_n^{n+1}f =\frac{1}{2}\frac{2}{n^3}n (it's the area of a triangle which height is n and which base is \frac{2}{n^3}), we get that \int_0^{\infty}f=\sum_{k=1}^{\infty}\frac{1}{k^2}=  \frac{\pi^2}{6} and the function is integrable.

    As we want \int_{\mathbb{R}}f=0, we define for x<0, f(x)=-f(-x) and that's it.
    Oooh
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    Quote Originally Posted by Jhevon View Post
    Oooh
    Look at my example in this tread.
    http://www.mathhelpforum.com/math-he...tml#post124737
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  7. #7
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    Quote Originally Posted by Opalg View Post
    How about taking f(x) = x if x is an integer and f(x)=0 otherwise?
    Quote Originally Posted by Jhevon View Post
    i was thinking about something like that. but is that function continuous?
    Sorry, I overlooked the continuity requirement. To get round that, you can to replace the isolated values f(n)=n at integer points by narrow triangular spikes. For example, for every n (apart from n=0 or 1) the function could go linearly from 0 to n in the interval (n-\frac1{n^3},n), and then linearly from n to 0 in the interval (n,n+\frac1{n^3}). The area of this triangle is 1/n^2, which means that the sum of the areas converges, so the function is integrable. Also, the function is odd, so its integral is zero.

    If you write out the formulas for this construction, you get exactly what flyingsquirrel wrote.
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