Results 1 to 5 of 5

Thread: Analysis simple problem

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Analysis simple problem

    How can I prove (or show) that f has a unique positive root, if $\displaystyle f(x)=4sin(x)+1-x$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by arbolis View Post
    How can I prove (or show) that f has a unique positive root, if $\displaystyle f(x)=4sin(x)+1-x$?
    Intermediate value theorem. Find a $\displaystyle a$ so that $\displaystyle f(a) < 0$ and a $\displaystyle b$ so that $\displaystyle f(b)>0$ then the function must take a zero in that interval.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Find a so that and a so that then the function must take a zero in that interval.
    Yeah I thought about that, but I think that the function could take more than 1 zero, say in a very little interval close the another zero. I know it won't happen in this example, but it would be possible in another one. In other words it wouldn't be a rigorous proof of the uniqueness of a positive zero. During the night I thought about finding more or less where is the zero and then chose a close interval to it and derivative the function to check its sign. For showing that the function $\displaystyle f$ is strictly increasing or decreasing in the little interval, then it cross only once the x axis.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by arbolis View Post
    Yeah I thought about that, but I think that the function could take more than 1 zero, say in a very little interval close the another zero. I know it won't happen in this example, but it would be possible in another one. In other words it wouldn't be a rigorous proof of the uniqueness of a positive zero. During the night I thought about finding more or less where is the zero and then chose a close interval to it and derivative the function to check its sign. For showing that the function $\displaystyle f$ is strictly increasing or decreasing in the little interval, then it cross only once the x axis.
    To establish uniqueness of the positive root I think you could consider the coordinates of the turning points of the function f(x) = 4 sin(x) + 1 - x for x > a, where a > 0 and f(a) = 0 for the first time .......
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    I could spend some time today on the problem. Finally it wasn't that easy, but ain't hard neither. As I didn't have a calculator, it was too long, because without the calculator it's more complicated.
    First, we notice that $\displaystyle -4\leq 4sin(x)\leq 4 \Rightarrow -3-x \leq 4sin(x)+1-x \leq 5-x$. From this, we know that our function $\displaystyle f$ is bounded by the 2 lines. Equaling them to 0, we find that $\displaystyle x=-3$ for the first and $\displaystyle x=5$ for the second. This implies that $\displaystyle f$ has at least a zero between $\displaystyle x=-3$ and $\displaystyle x=5$, but as we only consider positive roots in the exercise, we can say that f has at least a zero on $\displaystyle [0,5]$.
    To be rigorous, I don't see anything else than finding the derivative of $\displaystyle f$ and to study its sign over $\displaystyle [0,5]$. But before this, I will make a verification. As I assumed $\displaystyle f$ had a zero on $\displaystyle [0,5]$, I can verify it multiplying $\displaystyle f(0)$ by $\displaystyle f(5)$. If the result is negative, then we're right assuming it has at least a zero on $\displaystyle [0,5]$. So $\displaystyle f(0)*f(5)=1*(4sin(5)-4)$ which is negative.
    Now $\displaystyle f'(x)=4cos(x)-1$.
    $\displaystyle f'(x)=0$ when $\displaystyle x=arcos(\frac{1}{4})$.
    When $\displaystyle x<arcos(\frac{1}{4}) \approx 1,32$, $\displaystyle f'(x)>0$ and when $\displaystyle x>arcos(\frac{1}{4})$, $\displaystyle f'(x)<0$. Therefore we conclude that $\displaystyle f$ is increasing on $\displaystyle [0,arcos(\frac{1}{4})]$ and decreasing on $\displaystyle [arcos(\frac{1}{4}),5]$, which implies that $\displaystyle f$ cross only once the x axis somewhere on $\displaystyle [0,5]$.

    Edit : I didn't understand well
    you could consider the coordinates of the turning points of the function f(x) = 4 sin(x) + 1 - x for x > a, where a > 0 and f(a) = 0 for the first time
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple circuit analysis Q
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: Jun 22nd 2011, 03:10 AM
  2. Having trouble with simple analysis problem
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 17th 2010, 11:33 AM
  3. Simple Complex Analysis Question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 14th 2010, 11:03 PM
  4. Simple complex analysis question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 31st 2009, 04:22 PM
  5. analysis: rather simple question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 12th 2008, 08:59 PM

Search Tags


/mathhelpforum @mathhelpforum