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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    Hi,

    I would like to know how to proof the following:

    Show that sin x > x if x < 0.

    I think I have to start by
    let x<0.
    let f(t) = Sin t
    and f(t) is continuous on [- infinity,0) and differentiable on (-infinity,0)
    and ...
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  2. #2
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    Quote Originally Posted by zxcv View Post
    Hi,

    I would like to know how to proof the following:

    Show that sin x > x if x < 0.

    I think I have to start by
    let x<0.
    let f(t) = Sin t
    and f(t) is continuous on (- infinity,0] and differentiable on (-infinity,0] (note I changed the intervals)
    and ...
    ... you want to show, for t<0, f(t)<t, which is equivalent to f(t)-t<0. Let g(t)=f(t)-t. You have g(0)=0. If you can prove that g is strictly increasing, this will imply what you need. Can you carry on?
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  3. #3
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    So what I want is f(t)>t and f(t)-t>0.
    Let g(t) = f(t)-t. Let g(0) =0.

    By the Mean Value Theorem, since f(t) is cont. and differentiable, then there is a c in (- infinity, 0) such that

    f(0) - f(-infinity) = f '(c) (0 - -infinity)

    And for any x1 and x2 in [- infinity, 0] with x1<x2, there exists a c in (x1,x2) such that g(x2)-g(x1) = g '(x)(x2-x1).
    g ' (x) = cos x-1 <0 for x in (- infinity, 0).
    if g '(x) <0 for all x in (- infinity, 0), then f is strictly increasing.

    Can you guide me more, I find myself a bit confusing at times ...
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  4. #4
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    Quote Originally Posted by zxcv View Post
    So what I want is f(t)>t and f(t)-t>0.
    Let g(t) = f(t)-t. Let g(0) =0.

    By the Mean Value Theorem, since f(t) is cont. and differentiable, then there is a c in (- infinity, 0) such that

    f(0) - f(-infinity) = f '(c) (0 - -infinity)

    And for any x1 and x2 in [- infinity, 0] with x1<x2, there exists a c in (x1,x2) such that g(x2)-g(x1) = g '(x)(x2-x1).
    g ' (x) = cos x-1 <0 for x in (- infinity, 0).
    if g '(x) <0 for all x in (- infinity, 0), then f is strictly increasing.

    Can you guide me more, I find myself a bit confusing at times ...
    First you shouldn't write "f(-infinity)" or take x1=-infinity... this makes no sense indeed (especially since f has no limit at -infinity).

    I didn't take the title "Mean value theorem" into account when I wrote my previous post; I thought you could use the result "If g is differentiable on (-\infty,0] and g'(x)<0 for every x<0 then g is strictly decreasing on (-\infty,0]". This would give you g(x)>g(0) for x<0, which is what you want.

    Now, using the mean value theorem... If you apply it to f=\sin on [x,0] where x<0, you get \sin x-\sin 0=(\cos c)(x-0) for some c\in(x,0), hence \sin x= x \cos c\geq x (since x<0 and \cos c\leq 1). It doesn't give you the strict inequality though (you may a priori have \cos c=1). I don't know how you could get it easily from the mean value theorem.
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  5. #5
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    If using your method, when you mention using "If g is differentiable on (-infinity,0] and g '(x)<0 for every x<0 then g is strictly decreasing on (-infinity,0]".

    I will then start by saying that let g be differentiable on (-infinity,0], that is, g ' (x) = cos x-1 <0 for x in (- infinity, 0).
    Then do you know how I should continue on so to really show that g '(x)<0 from your way?
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  6. #6
    Super Member Showcase_22's Avatar
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    Show sin(x)>x for x<0.
    Define f(t):=sin(t)-t

    By algebra of continuity if g(t)=sin(t) and h(t)=-t then since f(t)=g(t)+h(t) then f(t) is continuous on \mathbb{R}.

    Say f(t) is continuous on [a,b] where a<b<0 and a,b \in \mathbb{R}.

    f(t) is differentiable on (a,b). In fact f'(t)=cos(t)-1

    Hence, by the Mean Value Theorem \exists \ t_0 \in (a,b) \ s.t \ f'(t_0)=\frac{f(b)-f(a)}{b-a}.

    f'(t)=cos(t)-1<1-1=0.

    Hence f'(t_0)=\frac{f(b)-f(a)}{b-a}<0 \Rightarrow \ f(b)<f(a) (ie. strictly decreasing)

    f(b)=sin(b)-b and f(a)=sin(a)-a.

    Now we have sin(b)-b<sin(a)-a \Rightarrow \ a-b<sin(a)-sin(b)

    a-b<2cos(\frac{a+b}{2})sin(\frac{a-b}{2})<2sin(\frac{a-b}{2})

    Hence 2sin(\frac{a-b}{2})>a-b \Rightarrow \ sin(\frac{a-b}{2})>\frac{a-b}{2} .

    Therefore define x:=\frac{a-b}{2} so we get sin(x)>x as required.
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