Proof that $ pi > 2 cdot sqrt{2}$ and $ pi > 3 $












2












$begingroup$


I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$



what I did



I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"






real-analysis summation pi







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
    $endgroup$
    – Winther
    Dec 18 '18 at 22:48












  • $begingroup$
    Ok, but as I have mentioned, I can't use it...
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:51






  • 1




    $begingroup$
    Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
    $endgroup$
    – Winther
    Dec 18 '18 at 22:52












  • $begingroup$
    "only in use of definition of ..."
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:52






  • 1




    $begingroup$
    The second inequality implies the first. Forget the first.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 23:01
















2












$begingroup$


I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$



what I did



I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"






real-analysis summation pi







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
    $endgroup$
    – Winther
    Dec 18 '18 at 22:48












  • $begingroup$
    Ok, but as I have mentioned, I can't use it...
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:51






  • 1




    $begingroup$
    Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
    $endgroup$
    – Winther
    Dec 18 '18 at 22:52












  • $begingroup$
    "only in use of definition of ..."
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:52






  • 1




    $begingroup$
    The second inequality implies the first. Forget the first.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 23:01














2












2








2





$begingroup$


I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$



what I did



I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"






real-analysis summation pi







share|cite|improve this question











$endgroup$




I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$



what I did



I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"






real-analysis summation pi




real-analysis summation pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 22:50









Bernard

121k740116




121k740116










asked Dec 18 '18 at 22:40









VirtualUserVirtualUser

899114




899114








  • 3




    $begingroup$
    Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
    $endgroup$
    – Winther
    Dec 18 '18 at 22:48












  • $begingroup$
    Ok, but as I have mentioned, I can't use it...
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:51






  • 1




    $begingroup$
    Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
    $endgroup$
    – Winther
    Dec 18 '18 at 22:52












  • $begingroup$
    "only in use of definition of ..."
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:52






  • 1




    $begingroup$
    The second inequality implies the first. Forget the first.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 23:01














  • 3




    $begingroup$
    Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
    $endgroup$
    – Winther
    Dec 18 '18 at 22:48












  • $begingroup$
    Ok, but as I have mentioned, I can't use it...
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:51






  • 1




    $begingroup$
    Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
    $endgroup$
    – Winther
    Dec 18 '18 at 22:52












  • $begingroup$
    "only in use of definition of ..."
    $endgroup$
    – VirtualUser
    Dec 18 '18 at 22:52






  • 1




    $begingroup$
    The second inequality implies the first. Forget the first.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 23:01








3




3




$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48






$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48














$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51




$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51




1




1




$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52






$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52














$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52




$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52




1




1




$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01




$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01










2 Answers
2






active

oldest

votes


















6












$begingroup$

The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.



Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.




Approach using the series definition of cosine and pi.



We can use the fact that $cos(x)$ is continuous, and that
$$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
$$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$frac{3245071}{45875200}approx 0.0707$$
with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
    $endgroup$
    – Frpzzd
    Dec 18 '18 at 23:04



















1












$begingroup$

You can show that in $[0,2]$n the cosine function is decreasing. Then



$$fracpi2>frac32iff0<cosfrac32.$$



You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).



The first partial sums are
$$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$



Below, the last partial sum above, and the exact value.
$$0.0707369341169leftrightarrow 0.0707372016677$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.



    Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.




    Approach using the series definition of cosine and pi.



    We can use the fact that $cos(x)$ is continuous, and that
    $$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
    We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
    $$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
    By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
    $$frac{3245071}{45875200}approx 0.0707$$
    with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
      $endgroup$
      – Frpzzd
      Dec 18 '18 at 23:04
















    6












    $begingroup$

    The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.



    Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.




    Approach using the series definition of cosine and pi.



    We can use the fact that $cos(x)$ is continuous, and that
    $$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
    We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
    $$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
    By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
    $$frac{3245071}{45875200}approx 0.0707$$
    with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
      $endgroup$
      – Frpzzd
      Dec 18 '18 at 23:04














    6












    6








    6





    $begingroup$

    The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.



    Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.




    Approach using the series definition of cosine and pi.



    We can use the fact that $cos(x)$ is continuous, and that
    $$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
    We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
    $$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
    By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
    $$frac{3245071}{45875200}approx 0.0707$$
    with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.






    share|cite|improve this answer











    $endgroup$



    The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.



    Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.




    Approach using the series definition of cosine and pi.



    We can use the fact that $cos(x)$ is continuous, and that
    $$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
    We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
    $$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
    By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
    $$frac{3245071}{45875200}approx 0.0707$$
    with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 19 '18 at 21:16

























    answered Dec 18 '18 at 22:52









    FrpzzdFrpzzd

    23k841109




    23k841109












    • $begingroup$
      @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
      $endgroup$
      – Frpzzd
      Dec 18 '18 at 23:04


















    • $begingroup$
      @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
      $endgroup$
      – Frpzzd
      Dec 18 '18 at 23:04
















    $begingroup$
    @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
    $endgroup$
    – Frpzzd
    Dec 18 '18 at 23:04




    $begingroup$
    @VirtualUser I have edited to show you how to do it using the series definition of $pi$.
    $endgroup$
    – Frpzzd
    Dec 18 '18 at 23:04











    1












    $begingroup$

    You can show that in $[0,2]$n the cosine function is decreasing. Then



    $$fracpi2>frac32iff0<cosfrac32.$$



    You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).



    The first partial sums are
    $$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$



    Below, the last partial sum above, and the exact value.
    $$0.0707369341169leftrightarrow 0.0707372016677$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You can show that in $[0,2]$n the cosine function is decreasing. Then



      $$fracpi2>frac32iff0<cosfrac32.$$



      You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).



      The first partial sums are
      $$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$



      Below, the last partial sum above, and the exact value.
      $$0.0707369341169leftrightarrow 0.0707372016677$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You can show that in $[0,2]$n the cosine function is decreasing. Then



        $$fracpi2>frac32iff0<cosfrac32.$$



        You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).



        The first partial sums are
        $$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$



        Below, the last partial sum above, and the exact value.
        $$0.0707369341169leftrightarrow 0.0707372016677$$






        share|cite|improve this answer









        $endgroup$



        You can show that in $[0,2]$n the cosine function is decreasing. Then



        $$fracpi2>frac32iff0<cosfrac32.$$



        You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).



        The first partial sums are
        $$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$



        Below, the last partial sum above, and the exact value.
        $$0.0707369341169leftrightarrow 0.0707372016677$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 23:10









        Yves DaoustYves Daoust

        129k675227




        129k675227






























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