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Serius problem

  1. Apr 30, 2008 #1
    Serius problem!!

    Can you help me in solving this??

    Is there a positive and twice differentiable function f defned on [0,∞) such
    that f(x)f’’(x) ≤ -1 on [0,∞) ? Why or why not???
     
  2. jcsd
  3. May 2, 2008 #2
    Guys,no one can solve this problem??!!
     
  4. May 2, 2008 #3

    HallsofIvy

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    What does that tell you about f"(x)? Where is f concave up or down?
     
  5. May 2, 2008 #4

    Is anybody supposed to solve it.........besides you????
     
  6. May 2, 2008 #5
    Any positive constant funtion works.
     
  7. May 2, 2008 #6

    What do you mean with constant?

    say

    f(x)=b, where b is a constant, this defenitely doesn't work, since f'=f''=0 so also

    f(x)f''(x)=0, so this defenitely is not smaller than -1.
     
  8. May 2, 2008 #7
    picking up on Halls advice.

    SInce f(x)>0 on[0,infty) this means that the sign of f(x)f''(x) is determined only by the sign of f''. So, since the problem requires that f(x)f''(x) be smaller or equal to -1 on the interval [0,infty) it means that for all x's on this interval f''(x)<0. What does this tell us about the concavity of f? this means that f is concave down on the whole interval. But if f is concave down on the whole interval it means that at some point it should defenitely cross the x-axis, and therefore at some point be also negative, but this contradicts the fact that f>0, so i would say that such a function does not exist at all.
     
  9. May 2, 2008 #8
    Sorry I reversed the inequality. :-)
     
  10. May 9, 2008 #9
    that is wt I did,pls send me ur help
     

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  11. May 9, 2008 #10
    This is not what you did, but exactly what sutupidmath wrote in his post.... you should at least credit him if you literally copy his words....

    Anyways, could you prove that a function on [0,inf) which is always concave down has to be negative somewhere..? This is the key argument in your reasoning and you did not justify it.
     
  12. Feb 8, 2010 #11
    Re: Serius problem!!

    The answer is NO:

    Suppose the statement is true. Then [tex]f(x)>0, f''(x)<0 [/tex] for all [tex] x>0 [/tex]. Moreover, [tex]\lim_{x\rightarrow +\infty} f''(x)<0 [/tex] if it exists.

    Let [tex] f(0)>0, f'(x) [/tex] is monotonically decreasing, then [tex]\lim_{x\rightarrow +\infty} f'(x) [/tex] is either [tex] -\infty [/tex] or a constant.

    If [tex]\lim_{x\rightarrow +\infty} f'(x)=-\infty [/tex], then [tex]f [/tex] is monotonically decreasing on [tex] [T,\,+\infty)[/tex] for some [tex]T [/tex] large enough and [tex]\lim_{x\rightarrow+\infty} f(x) [/tex] must be negative. Otherwise, [tex]\lim_{x\rightarrow+\infty} f(x)=c_0\geq 0 [/tex] and there exists a unbounded sequence [tex]\{x_n\}[/tex] so that [tex]\lim_{n\rightarrow +\infty} f'(x_n)=0 [/tex]. Contradiction;

    if [tex]\lim_{x\rightarrow +\infty} f'(x)=c [/tex], where [tex]c[/tex] is a constant, then there exists a unbounded sequence [tex]\{x_n\}[/tex] so that [tex]\lim_{n\rightarrow +\infty} f''(x_n)=0 [/tex], but we already have [tex]\lim_{n\rightarrow +\infty} f''(x_n)<0 [/tex]. Contradiction.

    P. S.

    Note that the assumption is [tex]f(x)\cdot f''(x)\leq -1 [/tex] on [tex][0,\,+\infty)[/tex]. If we change it into [tex]f(x)\cdot f''(x)<0 [/tex], then we cannot have [tex]\lim_{x\rightarrow +\infty} f''(x)<0 [/tex] and in this case the statement is true:
    e.g.: Let [tex]f(x)=\ln (x+2)[/tex]. Then we have
    [tex]\begin{align*}
    f'(x)&=1/(x+2),\\
    f''(x)&=-1/(x+2)^2.
    \end{align*}[/tex] ​
    Therefore, for all [tex]x\geq 0[/tex], we have
    [tex]\begin{align*}f(x)\cdot f''(x)&=-\frac{\ln (x+2)}{(2+x)^2}<0 \\
    \intertext{and}
    f(x)&>0.
    \end{align*}[/tex]​
     
    Last edited: Feb 8, 2010
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