Using mean value theorem to show that $cos (x)>1-x^2/2$












3












$begingroup$


I have a question, by applying the mean value theorem to $f(x)=frac{x^2}{2}+cos (x)$, on the interval $[0,x]$, show that $cos (x)>1-frac{x^2}{2}$.



We know that $frac{text{df}(x)}{text{dx}}=x-sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.



This is where I get confused:



so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)cdot x=1$ or what is going on?






real-analysis inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    The sign of $f'(c)$ gives the result (and the approach works for $xlt0$ as well).
    $endgroup$
    – Did
    Aug 5 '14 at 9:47










  • $begingroup$
    Meaning that f′(c) is positive?
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:50










  • $begingroup$
    Tell me. $ $ $ $
    $endgroup$
    – Did
    Aug 5 '14 at 9:53










  • $begingroup$
    @Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:57






  • 1




    $begingroup$
    see it as $f(x)>[f(0)=1]$
    $endgroup$
    – RE60K
    Aug 5 '14 at 11:17
















3












$begingroup$


I have a question, by applying the mean value theorem to $f(x)=frac{x^2}{2}+cos (x)$, on the interval $[0,x]$, show that $cos (x)>1-frac{x^2}{2}$.



We know that $frac{text{df}(x)}{text{dx}}=x-sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.



This is where I get confused:



so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)cdot x=1$ or what is going on?






real-analysis inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    The sign of $f'(c)$ gives the result (and the approach works for $xlt0$ as well).
    $endgroup$
    – Did
    Aug 5 '14 at 9:47










  • $begingroup$
    Meaning that f′(c) is positive?
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:50










  • $begingroup$
    Tell me. $ $ $ $
    $endgroup$
    – Did
    Aug 5 '14 at 9:53










  • $begingroup$
    @Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:57






  • 1




    $begingroup$
    see it as $f(x)>[f(0)=1]$
    $endgroup$
    – RE60K
    Aug 5 '14 at 11:17














3












3








3


1



$begingroup$


I have a question, by applying the mean value theorem to $f(x)=frac{x^2}{2}+cos (x)$, on the interval $[0,x]$, show that $cos (x)>1-frac{x^2}{2}$.



We know that $frac{text{df}(x)}{text{dx}}=x-sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.



This is where I get confused:



so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)cdot x=1$ or what is going on?






real-analysis inequality







share|cite|improve this question











$endgroup$




I have a question, by applying the mean value theorem to $f(x)=frac{x^2}{2}+cos (x)$, on the interval $[0,x]$, show that $cos (x)>1-frac{x^2}{2}$.



We know that $frac{text{df}(x)}{text{dx}}=x-sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.



This is where I get confused:



so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)cdot x=1$ or what is going on?






real-analysis inequality




real-analysis inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 17:57









amWhy

1




1










asked Aug 5 '14 at 9:45









ALEXANDERALEXANDER

8951921




8951921












  • $begingroup$
    The sign of $f'(c)$ gives the result (and the approach works for $xlt0$ as well).
    $endgroup$
    – Did
    Aug 5 '14 at 9:47










  • $begingroup$
    Meaning that f′(c) is positive?
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:50










  • $begingroup$
    Tell me. $ $ $ $
    $endgroup$
    – Did
    Aug 5 '14 at 9:53










  • $begingroup$
    @Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:57






  • 1




    $begingroup$
    see it as $f(x)>[f(0)=1]$
    $endgroup$
    – RE60K
    Aug 5 '14 at 11:17


















  • $begingroup$
    The sign of $f'(c)$ gives the result (and the approach works for $xlt0$ as well).
    $endgroup$
    – Did
    Aug 5 '14 at 9:47










  • $begingroup$
    Meaning that f′(c) is positive?
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:50










  • $begingroup$
    Tell me. $ $ $ $
    $endgroup$
    – Did
    Aug 5 '14 at 9:53










  • $begingroup$
    @Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __
    $endgroup$
    – ALEXANDER
    Aug 5 '14 at 9:57






  • 1




    $begingroup$
    see it as $f(x)>[f(0)=1]$
    $endgroup$
    – RE60K
    Aug 5 '14 at 11:17
















$begingroup$
The sign of $f'(c)$ gives the result (and the approach works for $xlt0$ as well).
$endgroup$
– Did
Aug 5 '14 at 9:47




$begingroup$
The sign of $f'(c)$ gives the result (and the approach works for $xlt0$ as well).
$endgroup$
– Did
Aug 5 '14 at 9:47












$begingroup$
Meaning that f′(c) is positive?
$endgroup$
– ALEXANDER
Aug 5 '14 at 9:50




$begingroup$
Meaning that f′(c) is positive?
$endgroup$
– ALEXANDER
Aug 5 '14 at 9:50












$begingroup$
Tell me. $ $ $ $
$endgroup$
– Did
Aug 5 '14 at 9:53




$begingroup$
Tell me. $ $ $ $
$endgroup$
– Did
Aug 5 '14 at 9:53












$begingroup$
@Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __
$endgroup$
– ALEXANDER
Aug 5 '14 at 9:57




$begingroup$
@Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __
$endgroup$
– ALEXANDER
Aug 5 '14 at 9:57




1




1




$begingroup$
see it as $f(x)>[f(0)=1]$
$endgroup$
– RE60K
Aug 5 '14 at 11:17




$begingroup$
see it as $f(x)>[f(0)=1]$
$endgroup$
– RE60K
Aug 5 '14 at 11:17










2 Answers
2






active

oldest

votes


















3












$begingroup$

$ f(x)=x^2/2+cos(x)$



Note that $f(0)=0^2/2+1=1$
From your equation:
$$f(x)-f(0)=(x)f'(c)=x(c-sin c)$$




Let $g(x)=x-sin x$
Again you can show that $g'(x)=1-cos x$ which is always greater than
$0$ due to bounded nature of $cos x$.As $g(0)=0$ and it is an
increasing function ${g'(x)>0;forall x>0$}, thus $g(x)>0 ;forall x>0$.




So $f(x)-f(0)>0;forall x>0$ as $x>0$ and $c-sin c >0;forall c>0${as $0<c<x$}.



So $f(x)>f(0)=1$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    You started off well.



    Notice that, by MVT:



    $$f'(c) = frac{f(x) - f(0)}{x - 0}$$
    S0



    $$xf'(c) = f(x) - f(0)$$



    Notice that x is positive, and since $$f'(x) = x - sin(x)$$



    Also, note that $x > sin(x)$, so $f'(x) > 0$



    Therefore,



    We can conclude that



    $$f(x) > f(0)$$



    And



    $$cos(x) > 1- frac{x^2}{2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why not factor out the constant in the derivative
      $endgroup$
      – ALEXANDER
      Aug 5 '14 at 10:16










    • $begingroup$
      @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
      $endgroup$
      – Varun Iyer
      Aug 5 '14 at 10:17










    • $begingroup$
      @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
      $endgroup$
      – Varun Iyer
      Aug 5 '14 at 10:19










    • $begingroup$
      You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
      $endgroup$
      – ALEXANDER
      Aug 5 '14 at 10:19










    • $begingroup$
      That I can see, but I do not get where the number 1 is coming from.
      $endgroup$
      – ALEXANDER
      Aug 5 '14 at 10:22











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $ f(x)=x^2/2+cos(x)$



    Note that $f(0)=0^2/2+1=1$
    From your equation:
    $$f(x)-f(0)=(x)f'(c)=x(c-sin c)$$




    Let $g(x)=x-sin x$
    Again you can show that $g'(x)=1-cos x$ which is always greater than
    $0$ due to bounded nature of $cos x$.As $g(0)=0$ and it is an
    increasing function ${g'(x)>0;forall x>0$}, thus $g(x)>0 ;forall x>0$.




    So $f(x)-f(0)>0;forall x>0$ as $x>0$ and $c-sin c >0;forall c>0${as $0<c<x$}.



    So $f(x)>f(0)=1$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      $ f(x)=x^2/2+cos(x)$



      Note that $f(0)=0^2/2+1=1$
      From your equation:
      $$f(x)-f(0)=(x)f'(c)=x(c-sin c)$$




      Let $g(x)=x-sin x$
      Again you can show that $g'(x)=1-cos x$ which is always greater than
      $0$ due to bounded nature of $cos x$.As $g(0)=0$ and it is an
      increasing function ${g'(x)>0;forall x>0$}, thus $g(x)>0 ;forall x>0$.




      So $f(x)-f(0)>0;forall x>0$ as $x>0$ and $c-sin c >0;forall c>0${as $0<c<x$}.



      So $f(x)>f(0)=1$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        $ f(x)=x^2/2+cos(x)$



        Note that $f(0)=0^2/2+1=1$
        From your equation:
        $$f(x)-f(0)=(x)f'(c)=x(c-sin c)$$




        Let $g(x)=x-sin x$
        Again you can show that $g'(x)=1-cos x$ which is always greater than
        $0$ due to bounded nature of $cos x$.As $g(0)=0$ and it is an
        increasing function ${g'(x)>0;forall x>0$}, thus $g(x)>0 ;forall x>0$.




        So $f(x)-f(0)>0;forall x>0$ as $x>0$ and $c-sin c >0;forall c>0${as $0<c<x$}.



        So $f(x)>f(0)=1$






        share|cite|improve this answer











        $endgroup$



        $ f(x)=x^2/2+cos(x)$



        Note that $f(0)=0^2/2+1=1$
        From your equation:
        $$f(x)-f(0)=(x)f'(c)=x(c-sin c)$$




        Let $g(x)=x-sin x$
        Again you can show that $g'(x)=1-cos x$ which is always greater than
        $0$ due to bounded nature of $cos x$.As $g(0)=0$ and it is an
        increasing function ${g'(x)>0;forall x>0$}, thus $g(x)>0 ;forall x>0$.




        So $f(x)-f(0)>0;forall x>0$ as $x>0$ and $c-sin c >0;forall c>0${as $0<c<x$}.



        So $f(x)>f(0)=1$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 5 '14 at 11:09









        Martin Sleziak

        44.8k10119272




        44.8k10119272










        answered Aug 5 '14 at 10:58









        RE60KRE60K

        14k22155




        14k22155























            2












            $begingroup$

            You started off well.



            Notice that, by MVT:



            $$f'(c) = frac{f(x) - f(0)}{x - 0}$$
            S0



            $$xf'(c) = f(x) - f(0)$$



            Notice that x is positive, and since $$f'(x) = x - sin(x)$$



            Also, note that $x > sin(x)$, so $f'(x) > 0$



            Therefore,



            We can conclude that



            $$f(x) > f(0)$$



            And



            $$cos(x) > 1- frac{x^2}{2}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why not factor out the constant in the derivative
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:16










            • $begingroup$
              @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:17










            • $begingroup$
              @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:19










            • $begingroup$
              You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:19










            • $begingroup$
              That I can see, but I do not get where the number 1 is coming from.
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:22
















            2












            $begingroup$

            You started off well.



            Notice that, by MVT:



            $$f'(c) = frac{f(x) - f(0)}{x - 0}$$
            S0



            $$xf'(c) = f(x) - f(0)$$



            Notice that x is positive, and since $$f'(x) = x - sin(x)$$



            Also, note that $x > sin(x)$, so $f'(x) > 0$



            Therefore,



            We can conclude that



            $$f(x) > f(0)$$



            And



            $$cos(x) > 1- frac{x^2}{2}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why not factor out the constant in the derivative
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:16










            • $begingroup$
              @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:17










            • $begingroup$
              @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:19










            • $begingroup$
              You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:19










            • $begingroup$
              That I can see, but I do not get where the number 1 is coming from.
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:22














            2












            2








            2





            $begingroup$

            You started off well.



            Notice that, by MVT:



            $$f'(c) = frac{f(x) - f(0)}{x - 0}$$
            S0



            $$xf'(c) = f(x) - f(0)$$



            Notice that x is positive, and since $$f'(x) = x - sin(x)$$



            Also, note that $x > sin(x)$, so $f'(x) > 0$



            Therefore,



            We can conclude that



            $$f(x) > f(0)$$



            And



            $$cos(x) > 1- frac{x^2}{2}$$






            share|cite|improve this answer











            $endgroup$



            You started off well.



            Notice that, by MVT:



            $$f'(c) = frac{f(x) - f(0)}{x - 0}$$
            S0



            $$xf'(c) = f(x) - f(0)$$



            Notice that x is positive, and since $$f'(x) = x - sin(x)$$



            Also, note that $x > sin(x)$, so $f'(x) > 0$



            Therefore,



            We can conclude that



            $$f(x) > f(0)$$



            And



            $$cos(x) > 1- frac{x^2}{2}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 5 '14 at 10:32

























            answered Aug 5 '14 at 10:15









            Varun IyerVarun Iyer

            5,317826




            5,317826












            • $begingroup$
              Why not factor out the constant in the derivative
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:16










            • $begingroup$
              @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:17










            • $begingroup$
              @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:19










            • $begingroup$
              You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:19










            • $begingroup$
              That I can see, but I do not get where the number 1 is coming from.
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:22


















            • $begingroup$
              Why not factor out the constant in the derivative
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:16










            • $begingroup$
              @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:17










            • $begingroup$
              @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
              $endgroup$
              – Varun Iyer
              Aug 5 '14 at 10:19










            • $begingroup$
              You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:19










            • $begingroup$
              That I can see, but I do not get where the number 1 is coming from.
              $endgroup$
              – ALEXANDER
              Aug 5 '14 at 10:22
















            $begingroup$
            Why not factor out the constant in the derivative
            $endgroup$
            – ALEXANDER
            Aug 5 '14 at 10:16




            $begingroup$
            Why not factor out the constant in the derivative
            $endgroup$
            – ALEXANDER
            Aug 5 '14 at 10:16












            $begingroup$
            @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
            $endgroup$
            – Varun Iyer
            Aug 5 '14 at 10:17




            $begingroup$
            @ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out
            $endgroup$
            – Varun Iyer
            Aug 5 '14 at 10:17












            $begingroup$
            @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
            $endgroup$
            – Varun Iyer
            Aug 5 '14 at 10:19




            $begingroup$
            @ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$
            $endgroup$
            – Varun Iyer
            Aug 5 '14 at 10:19












            $begingroup$
            You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
            $endgroup$
            – ALEXANDER
            Aug 5 '14 at 10:19




            $begingroup$
            You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant
            $endgroup$
            – ALEXANDER
            Aug 5 '14 at 10:19












            $begingroup$
            That I can see, but I do not get where the number 1 is coming from.
            $endgroup$
            – ALEXANDER
            Aug 5 '14 at 10:22




            $begingroup$
            That I can see, but I do not get where the number 1 is coming from.
            $endgroup$
            – ALEXANDER
            Aug 5 '14 at 10:22


















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