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Math Help - proof problem( I dont know what this belong to)

  1. #1
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    [SOLVED]proof problem( I dont know what this belong to)

    prove lnx < square root of x if x > 0
    Last edited by soleilion; April 5th 2008 at 02:42 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    ok

    ok you have to redefine them.... \ln(x)=\int_1^{x}\frac{1}{y}dy and \sqrt{x}=\int_0^{x}\frac{1}{2y}dy \forall{x}>0 from there you can see what to do..but then again \ln(x)\sim\sqrt{x}...so you could make a bid that this is untrue
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    ok you have to redefine them.... \ln(x)=\int_1^{x}\frac{1}{y}dy and \sqrt{x}=\int_0^{x}\frac{1}{2y}dy for every x>0 from there you can see what to do..but then again \ln(x)~ \sqrt{x}...so you could make a bid that this is untrue
    sorry, i misunderstood
    can u see this?
    http://img99.imageshack.us/img99/5271/13493377zy4.png

    this is how to prove by solution manual
    I dont understood how the minimum is 4

    can u tell me the reason?
    thank you
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  4. #4
    o_O
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    \frac{1}{2\sqrt{x}} - \frac{1}{x} = 0
    \frac{1}{2 x^{\frac{1}{2}}} = \frac{1}{x}
    2x^{\frac{1}{2}} = x
    2 = x^{\frac{1}{2}}
    x = 4

    Use your first derivative test and you'll see it's a local/absolute minimum.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Ok

    if f(x)=\sqrt{x}-\ln(x) then f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x} find the places where it is undefined and zero..this is at x=0(undefined) and x=4(zero) since 0 is not in the domain we can get rid of it..now we need to check where the original function is undefined...it is undefined nowhere on the domain of the question...so now all we have is the point x=4...we test that in the second derivative f''(x)=\frac{1}{x^2}-\frac{1}{4x^{\frac{3}{2}}} and we test the point 4 in there..since it gives \frac{1}{32} which is positive x=4 is a minimum point on (0,\infty)
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Someone good

    can you tell me if what I said was correct that since \ln(x)\sim\sqrt{x} that you could make an argument that \forall{x}>0,\ln(x)<\sqrt{x} is untrue?


    Edit:forget it I dont know what I was thinking they are not asymptotically equivalent...forget I asked
    Last edited by Mathstud28; April 5th 2008 at 07:26 PM.
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    \sqrt{x}=\int_0^{x}\frac{1}{2y}dy
    Careful when workin' with such parameters, this one is wrong.
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  8. #8
    Moo
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    Hi,

    Integrate from 1 to x, uh ?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    What?

    Integrate from 1 to x for what? and Krizalid is that wrong because the boundary restriction of that does not mesh well with this problem?
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  10. #10
    Moo
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    ..

    \int_0^x \frac{1}{2y} dy = \frac{1}{2} \ln(x) \big|^x_0

    Which is not defined at 0.

    Plus \ln(\sqrt{x}) is not equal to \sqrt{x}
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Oh

    I am sorry that is a typo it should read \sqrt{x}=\int_0^{x}\frac{1}{2\sqrt{y}}dy...sorry for the typo
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  12. #12
    Moo
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    I think that you should sometimes leave integrals to solve problems the way it's good to solve
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Haha

    I am not quite sure what you mean by that..but ok! haha =D
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