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Math Help - find a function from the limit

  1. #1
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    find a function from the limit

    Suppose

    lim_{x -> infty} { log f(x) / log x } = c ,

    where c is a constant.

    Can we prove that:

    f(x) = b x^c + lower order terms,

    where b is something independent of x ?

    can anyone show me how to prove this? or maybe it is incorrect?

    tks
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by jill View Post
    Suppose

    lim_{x -> infty} { log f(x) / log x } = c ,

    where c is a constant.

    Can we prove that:

    f(x) = b x^c + lower order terms,

    where b is something independent of x ?
    It doesn't come out like this for me, because if I use Landau's little-oh notation the given limit means that

    \frac{\log f(x)}{\log x}=c+o(1) for x\rightarrow\infty

    in other words

    \log f(x)=\big(c+o(1)\big)\cdot \log x

    and thus

     f(x)=x^{c+o(1)}=x^{o(1)}\cdot x^c for x\rightarrow \infty

    Now, if o(1) goes to 0 veeeery slooowly, x^{o(1)} will actually get arbitrarily large, so: no, b cannot generally be independent of x.
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  3. #3
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    Quote Originally Posted by Failure View Post
    It doesn't come out like this for me, because if I use Landau's little-oh notation the given limit means that

    \frac{\log f(x)}{\log x}=c+o(1) for x\rightarrow\infty

    in other words

    \log f(x)=\big(c+o(1)\big)\cdot \log x

    and thus

     f(x)=x^{c+o(1)}=x^{o(1)}\cdot x^c for x\rightarrow \infty

    Now, if o(1) goes to 0 veeeery slooowly, x^{o(1)} will actually get arbitrarily large, so: no, b cannot generally be independent of x.

    Thanks a lot!

    So that if we know

    \lim_{x \rightarrow \infty}  \frac{log f_1(x)}{ log x } = c_1 , and

    \lim_{x \rightarrow \infty}  \frac{log f_2(x)}{ log x } = c_2  ,

    then we can prove (if c_2>c_1):

    \lim_{x \rightarrow \infty} \frac{ log [ a1 f_1(x) + a2 f_2(x)  ] }{ [ log x ] } = c_2 .

    Since

    f_1(x) = x^{ c_1 + o(1) }, and

    f_2(x) = x^{ c_2 + o(1) } .

    so

    \lim_{x \rightarrow \infty} \frac{ log [ a_1 f_1(x) + a_2 f_2(x)  ] }{ [ log x ] } =
    \lim_{x \rightarrow \infty} \frac{ log [ a_1 x^{ c_1 + o(1) } + a_2 x^{ c_2 + o(1) }  ] }{ [ log x ] } =
    \lim_{x \rightarrow \infty} \frac{ log [ a_2 x^{ c_2 + o(1) } ( \frac{a_1}{a_2} x^{ c_1 -c_2 + o(1) } + 1 )  ] }{ [ log x ] } =
    \lim_{x \rightarrow \infty} \frac{ log a_2 + log [x^{ c_2 + o(1) }] + log 1  }{ [ log x ] } =
    c_2

    please correct me if anything is wrong
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by jill View Post
    Thanks a lot!

    So that if we know

    \lim_{x \rightarrow \infty}  \frac{log f_1(x)}{ log x } = c_1 , and

    \lim_{x \rightarrow \infty}  \frac{log f_2(x)}{ log x } = c_2  ,

    then we can prove (if c_2>c_1):

    \lim_{x \rightarrow \infty} \frac{ log [ a1 f_1(x) + a2 f_2(x)  ] }{ [ log x ] } = c_2 .

    Since

    f_1(x) = x^{ c_1 + o(1) }, and

    f_2(x) = x^{ c_2 + o(1) } .

    so

    \lim_{x \rightarrow \infty} \frac{ log [ a_1 f_1(x) + a_2 f_2(x)  ] }{ [ log x ] } =
    \lim_{x \rightarrow \infty} \frac{ log [ a_1 x^{ c_1 + o(1) } + a_2 x^{ c_2 + o(1) }  ] }{ [ log x ] } =
    \lim_{x \rightarrow \infty} \frac{ log [ a_2 x^{ c_2 + o(1) } ( \frac{a_1}{a_2} x^{ c_1 -c_2 + o(1) } + 1 )  ] }{ [ log x ] } =
    \lim_{x \rightarrow \infty} \frac{ log a_2 + log [x^{ c_2 + o(1) }] + log 1  }{ [ log x ] } =
    c_2

    please correct me if anything is wrong
    Looks ok to me. - But I didn't quite figure that this is what you really wanted to prove.
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  5. #5
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    Quote Originally Posted by Failure View Post
    Looks ok to me. - But I didn't quite figure that this is what you really wanted to prove.
    This really is what i want to prove.
    It looks must-be-true, but i did get stuck when i tried to write down a rigorous proof,
    since i didn't know the small oh notation is available here and i was worried about the
    fact that log x does not exist for x->infty, so...

    Your answer helped me a lot. tks again
    Last edited by jill; July 20th 2010 at 12:04 PM.
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