Results 1 to 4 of 4

Math Help - proving : tg(x)>x using Lagrange / Cauchy's Lemma Use

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    168

    [SOLVED] proving : tg(x)>x using Lagrange / Cauchy's Lemma Use

    I need to prove that tg(x)>x when x \in (0,\frac{\pi}{2}).

    I'll use Cauchy's version for Lagrange's lemma:

    If f,g are continuous in [a,b] and derivative in (a,b), then \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)} , (there exists a 'c' : a<c<b for which...)

    So I'll take the range [0,x], while 0<x<\frac{\pi}{2}, and f(x):=sin(x), g(x):=xcos(x) :

    \frac{tg(x)}{x}=\frac{sinx}{xcosx}=\frac{sin(x)-sin(0)}{xcos(x)-0*cos(0)} = \frac{f'(c)}{g'(c)} = \frac{c}{cos(c)-c*sin(c)}

    Now, how can I prove that the last phrase is larger than 1? (so that \frac{tg(x)}{x}>1 \Rightarrow tg(x)>x) ?

    Thank you very much!
    Last edited by adam63; March 12th 2010 at 10:32 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by adam63 View Post
    I need to prove that tg(x)>x when x \in (0,\frac{\pi}{2}).

    I'll use Cauchy's version for Lagrange's lemma:

    If f,g are continuous in [a,b] and derivative in (a,b), then \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)} , (there exists a 'c' : a<c<b for which...)

    So I'll take the range [0,x], while 0<x<\frac{\pi}{2}, and f(x):=sin(x), g(x):=xcos(x) :

    \frac{tg(x)}{x}=\frac{sinx}{xcosx}=\frac{sin(x)-sin(0)}{xcos(x)-0*cos(0)} = \frac{f'(c)}{g'(c)} = \frac{c}{cos(c)-c*sin(c)}

    Now, how can I prove that the last phrase is larger than 1? (so that \frac{tg(x)}{x}>1 \Rightarrow tg(x)>x) ?

    Thank you very much!
    Don't really think there is any need for Lagrange here...

    Let f(x) = tan(x) -x, ~ x \in (0, \frac{\pi}{2}), then f'(x) = \frac{1}{cos^2(x)} - 1

    Now, note that:

    1. f(0) = tan(0) - 0 = 0

    2. Since for any x \in (0, \frac{\pi}{2}), ~ 0 < cos(x) < 1 \Rightarrow 0 < cos^2(x) < 1 \Rightarrow \frac{1}{cos^2(x)} > 1, we get that f'(x) > 0 ~\forall x \in (0, \frac{\pi}{2}).

    These two facts, along with the continuity of f(x), give us that f(x) > 0 ~ \forall x \in (0, \frac{\pi}{2}) and therefore tan(x) > x ~ \forall x \in (0, \frac{\pi}{2})
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2009
    Posts
    168
    Quote Originally Posted by Defunkt View Post
    Don't really think there is any need for Lagrange here...

    Let f(x) = tan(x) -x, ~ x \in (0, \frac{\pi}{2}), then f'(x) = \frac{1}{cos^2(x)} - 1

    Now, note that:

    1. f(0) = tan(0) - 0 = 0

    2. Since for any x \in (0, \frac{\pi}{2}), ~ 0 < cos(x) < 1 \Rightarrow 0 < cos^2(x) < 1 \Rightarrow \frac{1}{cos^2(x)} > 1, we get that f'(x) > 0 ~\forall x \in (0, \frac{\pi}{2}).

    These two facts, along with the continuity of f(x), give us that f(x) > 0 ~ \forall x \in (0, \frac{\pi}{2}) and therefore tan(x) > x ~ \forall x \in (0, \frac{\pi}{2})
    That's a great way to solve it, only I haven't 'officially' learned it, and need to use Lagrange \ Roll \ Cauchy's lemmas for functions. Can you help me?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    168
    I got it!! I was really close, but I couldn't see it :

    I need to prove:

    \frac{c}{cos(c)-csin(c)} > 1

    \frac{cos(c)-csin(c)}{c} < 1

    1-c*tg(c) < 1

    c*tg(c) > 0

    ^ Which is right for every 0<c<\frac{\pi}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: December 12th 2010, 01:16 PM
  2. Proving a Sequence is Cauchy proof Help!
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 22nd 2010, 03:58 AM
  3. Proving convergent seq. is a cauchy seq.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 17th 2010, 07:24 PM
  4. Proving the IVP using the lemma...
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 21st 2009, 01:28 PM
  5. proving a sequence is Cauchy
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: September 16th 2009, 04:14 AM

Search Tags


/mathhelpforum @mathhelpforum