Proving that the power series for the cosine function is greater than zero, for $x$ in $[0, pi/2)$.












1












$begingroup$



I'm trying to prove the cosine power series
$$sum_{k=0}^infty (-1)^k frac{x^{2k}}{(2k)!} ;>;0$$
for all $x in [0, pi/2)$. Here, $pi$ is defined as the smallest positive real such that $cosfrac{pi}{2} = 0$.




My initial thoughts for an attempt at a solution:



Showing that the series achieves an absolute maximum and minimum at its endpoints, respectfully, and that both of these are positive. Thus by the intermediate value theorem, all points between must also be greater than zero.



I'm at a bit of a loss as to how to formally write this as a proof, however.










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  • $begingroup$
    To solve this question, the way you define $pi$ is important.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 20:14










  • $begingroup$
    Right. Forgot to metion that we have defined $pi$ as the smallest positive real number such that cos($pi$/2) = 0.
    $endgroup$
    – user624612
    Dec 9 '18 at 20:15










  • $begingroup$
    Did you prove that the $cos$ function is continuous?
    $endgroup$
    – Botond
    Dec 9 '18 at 20:21










  • $begingroup$
    Well since series converges by the absolute series convergence test, the series would be continuous, correct?
    $endgroup$
    – user624612
    Dec 9 '18 at 20:32






  • 2




    $begingroup$
    If $pi/2$ is defined as the smallest positive zero of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=f(x)$ there is nothing to prove: such function is continuous and (positive where positive).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:16


















1












$begingroup$



I'm trying to prove the cosine power series
$$sum_{k=0}^infty (-1)^k frac{x^{2k}}{(2k)!} ;>;0$$
for all $x in [0, pi/2)$. Here, $pi$ is defined as the smallest positive real such that $cosfrac{pi}{2} = 0$.




My initial thoughts for an attempt at a solution:



Showing that the series achieves an absolute maximum and minimum at its endpoints, respectfully, and that both of these are positive. Thus by the intermediate value theorem, all points between must also be greater than zero.



I'm at a bit of a loss as to how to formally write this as a proof, however.










share|cite|improve this question











$endgroup$












  • $begingroup$
    To solve this question, the way you define $pi$ is important.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 20:14










  • $begingroup$
    Right. Forgot to metion that we have defined $pi$ as the smallest positive real number such that cos($pi$/2) = 0.
    $endgroup$
    – user624612
    Dec 9 '18 at 20:15










  • $begingroup$
    Did you prove that the $cos$ function is continuous?
    $endgroup$
    – Botond
    Dec 9 '18 at 20:21










  • $begingroup$
    Well since series converges by the absolute series convergence test, the series would be continuous, correct?
    $endgroup$
    – user624612
    Dec 9 '18 at 20:32






  • 2




    $begingroup$
    If $pi/2$ is defined as the smallest positive zero of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=f(x)$ there is nothing to prove: such function is continuous and (positive where positive).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:16
















1












1








1





$begingroup$



I'm trying to prove the cosine power series
$$sum_{k=0}^infty (-1)^k frac{x^{2k}}{(2k)!} ;>;0$$
for all $x in [0, pi/2)$. Here, $pi$ is defined as the smallest positive real such that $cosfrac{pi}{2} = 0$.




My initial thoughts for an attempt at a solution:



Showing that the series achieves an absolute maximum and minimum at its endpoints, respectfully, and that both of these are positive. Thus by the intermediate value theorem, all points between must also be greater than zero.



I'm at a bit of a loss as to how to formally write this as a proof, however.










share|cite|improve this question











$endgroup$





I'm trying to prove the cosine power series
$$sum_{k=0}^infty (-1)^k frac{x^{2k}}{(2k)!} ;>;0$$
for all $x in [0, pi/2)$. Here, $pi$ is defined as the smallest positive real such that $cosfrac{pi}{2} = 0$.




My initial thoughts for an attempt at a solution:



Showing that the series achieves an absolute maximum and minimum at its endpoints, respectfully, and that both of these are positive. Thus by the intermediate value theorem, all points between must also be greater than zero.



I'm at a bit of a loss as to how to formally write this as a proof, however.







real-analysis sequences-and-series trigonometry proof-writing taylor-expansion






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share|cite|improve this question













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edited Dec 9 '18 at 20:24









Blue

48k870153




48k870153










asked Dec 9 '18 at 20:09







user624612



















  • $begingroup$
    To solve this question, the way you define $pi$ is important.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 20:14










  • $begingroup$
    Right. Forgot to metion that we have defined $pi$ as the smallest positive real number such that cos($pi$/2) = 0.
    $endgroup$
    – user624612
    Dec 9 '18 at 20:15










  • $begingroup$
    Did you prove that the $cos$ function is continuous?
    $endgroup$
    – Botond
    Dec 9 '18 at 20:21










  • $begingroup$
    Well since series converges by the absolute series convergence test, the series would be continuous, correct?
    $endgroup$
    – user624612
    Dec 9 '18 at 20:32






  • 2




    $begingroup$
    If $pi/2$ is defined as the smallest positive zero of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=f(x)$ there is nothing to prove: such function is continuous and (positive where positive).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:16




















  • $begingroup$
    To solve this question, the way you define $pi$ is important.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 20:14










  • $begingroup$
    Right. Forgot to metion that we have defined $pi$ as the smallest positive real number such that cos($pi$/2) = 0.
    $endgroup$
    – user624612
    Dec 9 '18 at 20:15










  • $begingroup$
    Did you prove that the $cos$ function is continuous?
    $endgroup$
    – Botond
    Dec 9 '18 at 20:21










  • $begingroup$
    Well since series converges by the absolute series convergence test, the series would be continuous, correct?
    $endgroup$
    – user624612
    Dec 9 '18 at 20:32






  • 2




    $begingroup$
    If $pi/2$ is defined as the smallest positive zero of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=f(x)$ there is nothing to prove: such function is continuous and (positive where positive).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:16


















$begingroup$
To solve this question, the way you define $pi$ is important.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:14




$begingroup$
To solve this question, the way you define $pi$ is important.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:14












$begingroup$
Right. Forgot to metion that we have defined $pi$ as the smallest positive real number such that cos($pi$/2) = 0.
$endgroup$
– user624612
Dec 9 '18 at 20:15




$begingroup$
Right. Forgot to metion that we have defined $pi$ as the smallest positive real number such that cos($pi$/2) = 0.
$endgroup$
– user624612
Dec 9 '18 at 20:15












$begingroup$
Did you prove that the $cos$ function is continuous?
$endgroup$
– Botond
Dec 9 '18 at 20:21




$begingroup$
Did you prove that the $cos$ function is continuous?
$endgroup$
– Botond
Dec 9 '18 at 20:21












$begingroup$
Well since series converges by the absolute series convergence test, the series would be continuous, correct?
$endgroup$
– user624612
Dec 9 '18 at 20:32




$begingroup$
Well since series converges by the absolute series convergence test, the series would be continuous, correct?
$endgroup$
– user624612
Dec 9 '18 at 20:32




2




2




$begingroup$
If $pi/2$ is defined as the smallest positive zero of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=f(x)$ there is nothing to prove: such function is continuous and (positive where positive).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 21:16






$begingroup$
If $pi/2$ is defined as the smallest positive zero of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=f(x)$ there is nothing to prove: such function is continuous and (positive where positive).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 21:16












4 Answers
4






active

oldest

votes


















2












$begingroup$

Denote $f(x)= cos x$. You have $f(0)=1$. As $f$ is continuous (converging power series are continuous), $f$ is strictly positive around $0$.



You can then conclude with $pi$ definition.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:05










  • $begingroup$
    @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 21:07










  • $begingroup$
    Positive real and not positive integer!
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 21:14



















0












$begingroup$

We can use geometry to show $cos(x)$ is $C^{(infty)}.$ Then we can use Lagrange Remainder Theorem to show the series equals the cosine function on all $mathbb{R}$. Then we automatically get that the series is positive whenever $cos(x)$ is positive.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:06










  • $begingroup$
    Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
    $endgroup$
    – Melody
    Dec 9 '18 at 21:08










  • $begingroup$
    That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:12










  • $begingroup$
    Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
    $endgroup$
    – Melody
    Dec 9 '18 at 21:13



















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$begingroup$

A famous diagram-based argument shows any acute $x$ satisfies $0lesin xle x$. Alternate use of $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ allows us to prove by induction that $c_{2n+1}lecos xle c_{2n}$ with $c_n:=sum_{k=0}^nfrac{(-x^2)^k}{(2k)!}$, and a similar result for $sin x$. Since the ratio test implies $lim_{ntoinfty} c_n$ is finite for any acute $x$, by the squeeze theorem this limit is $cos xge 0$.






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$endgroup$













  • $begingroup$
    You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:09










  • $begingroup$
    @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
    $endgroup$
    – J.G.
    Dec 9 '18 at 21:11










  • $begingroup$
    Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:13










  • $begingroup$
    @JackD'Aurizio I think I spelled it out pretty well.
    $endgroup$
    – J.G.
    Dec 9 '18 at 21:19



















0












$begingroup$

$cos x$, defined as its Maclaurin series, is an entire solution of the differential equation $f''(x)+f(x)=0$, $f(0)=1$, $f'(0)=1$. $cos x$ is even and concave where positive, and we may denote with $alpha$ the smallest positive zero of $cos x$. The problem boils down to showing that $alpha=frac{pi}{2}$. Now, by denoting through $sin x$ the opposite of the derivative of $cos x$, we have that $f''+f=0$ implies that $f^2+f'^2$ is constant, in particular equal to one (Pythagorean theorem, $sin^2+cos^2=1$). Since $cos x$ is decreasing on $(0,alpha)$ and the Pythagorean theorem plus the chain rule imply that the derivative of the inverse function of $cos x$ is $-frac{1}{sqrt{1-x^2}}$, we have
$$ alpha = int_{0}^{1}frac{dx}{sqrt{1-x^2}}.$$
Integration by parts relates $alpha$ with the area of the unit circle:
$$ alpha = 2int_{0}^{1}sqrt{1-x^2},dx = frac{pi}{2} $$
and we are done: the series defining $cos x$ is positive on $[0,alpha)=left[0,frac{pi}{2}right)$.

As a side-effect, we have the hypergeometric identities
$$ frac{pi}{2}=sum_{ngeq 0}frac{binom{2n}{n}}{(2n+1)4^n},qquad frac{pi}{2}=2-sum_{ngeq 0}frac{binom{2n}{n}}{4^n(n+1)(2n+3)}$$
provided by the Maclaurin series of $frac{1}{sqrt{1-x^2}}$ and $sqrt{1-x^2}$.
The latter is faster-convergent, of course.






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    4 Answers
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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Denote $f(x)= cos x$. You have $f(0)=1$. As $f$ is continuous (converging power series are continuous), $f$ is strictly positive around $0$.



    You can then conclude with $pi$ definition.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:05










    • $begingroup$
      @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:07










    • $begingroup$
      Positive real and not positive integer!
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:14
















    2












    $begingroup$

    Denote $f(x)= cos x$. You have $f(0)=1$. As $f$ is continuous (converging power series are continuous), $f$ is strictly positive around $0$.



    You can then conclude with $pi$ definition.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:05










    • $begingroup$
      @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:07










    • $begingroup$
      Positive real and not positive integer!
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:14














    2












    2








    2





    $begingroup$

    Denote $f(x)= cos x$. You have $f(0)=1$. As $f$ is continuous (converging power series are continuous), $f$ is strictly positive around $0$.



    You can then conclude with $pi$ definition.






    share|cite|improve this answer









    $endgroup$



    Denote $f(x)= cos x$. You have $f(0)=1$. As $f$ is continuous (converging power series are continuous), $f$ is strictly positive around $0$.



    You can then conclude with $pi$ definition.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 20:24









    mathcounterexamples.netmathcounterexamples.net

    26.2k21955




    26.2k21955












    • $begingroup$
      Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:05










    • $begingroup$
      @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:07










    • $begingroup$
      Positive real and not positive integer!
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:14


















    • $begingroup$
      Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:05










    • $begingroup$
      @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:07










    • $begingroup$
      Positive real and not positive integer!
      $endgroup$
      – mathcounterexamples.net
      Dec 9 '18 at 21:14
















    $begingroup$
    Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:05




    $begingroup$
    Which "$pi$ definition"? The area of the unit circle is not immediately related to the first positive root of $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:05












    $begingroup$
    @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 21:07




    $begingroup$
    @JackD'Aurizio See the question I ask to the OP and his answer. He defines $pi$ as the smallest positive integer such that $cos pi/2 =0$.
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 21:07












    $begingroup$
    Positive real and not positive integer!
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 21:14




    $begingroup$
    Positive real and not positive integer!
    $endgroup$
    – mathcounterexamples.net
    Dec 9 '18 at 21:14











    0












    $begingroup$

    We can use geometry to show $cos(x)$ is $C^{(infty)}.$ Then we can use Lagrange Remainder Theorem to show the series equals the cosine function on all $mathbb{R}$. Then we automatically get that the series is positive whenever $cos(x)$ is positive.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:06










    • $begingroup$
      Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:08










    • $begingroup$
      That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:12










    • $begingroup$
      Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:13
















    0












    $begingroup$

    We can use geometry to show $cos(x)$ is $C^{(infty)}.$ Then we can use Lagrange Remainder Theorem to show the series equals the cosine function on all $mathbb{R}$. Then we automatically get that the series is positive whenever $cos(x)$ is positive.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:06










    • $begingroup$
      Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:08










    • $begingroup$
      That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:12










    • $begingroup$
      Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:13














    0












    0








    0





    $begingroup$

    We can use geometry to show $cos(x)$ is $C^{(infty)}.$ Then we can use Lagrange Remainder Theorem to show the series equals the cosine function on all $mathbb{R}$. Then we automatically get that the series is positive whenever $cos(x)$ is positive.






    share|cite|improve this answer









    $endgroup$



    We can use geometry to show $cos(x)$ is $C^{(infty)}.$ Then we can use Lagrange Remainder Theorem to show the series equals the cosine function on all $mathbb{R}$. Then we automatically get that the series is positive whenever $cos(x)$ is positive.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 20:18









    MelodyMelody

    78812




    78812












    • $begingroup$
      If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:06










    • $begingroup$
      Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:08










    • $begingroup$
      That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:12










    • $begingroup$
      Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:13


















    • $begingroup$
      If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:06










    • $begingroup$
      Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:08










    • $begingroup$
      That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:12










    • $begingroup$
      Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
      $endgroup$
      – Melody
      Dec 9 '18 at 21:13
















    $begingroup$
    If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:06




    $begingroup$
    If $cos x$ is defined as its Maclaurin series, there is no geometry we may invoke unless we prove it.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:06












    $begingroup$
    Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
    $endgroup$
    – Melody
    Dec 9 '18 at 21:08




    $begingroup$
    Then call the geometric version of $cos(x)$ banana and use the same argument to show that $text{banana}(x)=cos(x).$ The statement I gave was how to show the geometric definition equals the series. That's the crux, it doesn't matter what you call them.
    $endgroup$
    – Melody
    Dec 9 '18 at 21:08












    $begingroup$
    That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:12




    $begingroup$
    That's the crux - I do not agree, since you do not need the geometric version of $cos x$ to relate its first positive zero with the area of the circle. Yours (properly worded) is just one way.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:12












    $begingroup$
    Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
    $endgroup$
    – Melody
    Dec 9 '18 at 21:13




    $begingroup$
    Well, I meant the crux of my argument. Sorry if it seemed I meant every argument. Also, I do think my statement is properly worded, it's just missing details.
    $endgroup$
    – Melody
    Dec 9 '18 at 21:13











    0












    $begingroup$

    A famous diagram-based argument shows any acute $x$ satisfies $0lesin xle x$. Alternate use of $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ allows us to prove by induction that $c_{2n+1}lecos xle c_{2n}$ with $c_n:=sum_{k=0}^nfrac{(-x^2)^k}{(2k)!}$, and a similar result for $sin x$. Since the ratio test implies $lim_{ntoinfty} c_n$ is finite for any acute $x$, by the squeeze theorem this limit is $cos xge 0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:09










    • $begingroup$
      @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:11










    • $begingroup$
      Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:13










    • $begingroup$
      @JackD'Aurizio I think I spelled it out pretty well.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:19
















    0












    $begingroup$

    A famous diagram-based argument shows any acute $x$ satisfies $0lesin xle x$. Alternate use of $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ allows us to prove by induction that $c_{2n+1}lecos xle c_{2n}$ with $c_n:=sum_{k=0}^nfrac{(-x^2)^k}{(2k)!}$, and a similar result for $sin x$. Since the ratio test implies $lim_{ntoinfty} c_n$ is finite for any acute $x$, by the squeeze theorem this limit is $cos xge 0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:09










    • $begingroup$
      @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:11










    • $begingroup$
      Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:13










    • $begingroup$
      @JackD'Aurizio I think I spelled it out pretty well.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:19














    0












    0








    0





    $begingroup$

    A famous diagram-based argument shows any acute $x$ satisfies $0lesin xle x$. Alternate use of $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ allows us to prove by induction that $c_{2n+1}lecos xle c_{2n}$ with $c_n:=sum_{k=0}^nfrac{(-x^2)^k}{(2k)!}$, and a similar result for $sin x$. Since the ratio test implies $lim_{ntoinfty} c_n$ is finite for any acute $x$, by the squeeze theorem this limit is $cos xge 0$.






    share|cite|improve this answer









    $endgroup$



    A famous diagram-based argument shows any acute $x$ satisfies $0lesin xle x$. Alternate use of $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ allows us to prove by induction that $c_{2n+1}lecos xle c_{2n}$ with $c_n:=sum_{k=0}^nfrac{(-x^2)^k}{(2k)!}$, and a similar result for $sin x$. Since the ratio test implies $lim_{ntoinfty} c_n$ is finite for any acute $x$, by the squeeze theorem this limit is $cos xge 0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 20:24









    J.G.J.G.

    25.3k22539




    25.3k22539












    • $begingroup$
      You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:09










    • $begingroup$
      @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:11










    • $begingroup$
      Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:13










    • $begingroup$
      @JackD'Aurizio I think I spelled it out pretty well.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:19


















    • $begingroup$
      You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:09










    • $begingroup$
      @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:11










    • $begingroup$
      Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
      $endgroup$
      – Jack D'Aurizio
      Dec 9 '18 at 21:13










    • $begingroup$
      @JackD'Aurizio I think I spelled it out pretty well.
      $endgroup$
      – J.G.
      Dec 9 '18 at 21:19
















    $begingroup$
    You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:09




    $begingroup$
    You are (more or less) starting by assuming the thesis, i.e. that $$ 0 leq sum_{ngeq 0}frac{(-1)^n x^{2n+1}}{(2n+1)!}leq x$$ holds for any $xin[0,pi/2]$. We need to prove the geometric properties of $cos x$ or $sin x$ if they are just defined through their Maclaurin series.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:09












    $begingroup$
    @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
    $endgroup$
    – J.G.
    Dec 9 '18 at 21:11




    $begingroup$
    @JackD'Aurizio No, I'm defining the trigonometric functions from geometry, and deducing their range on acute angles from geometry, and proving what the series converges to from geometry.
    $endgroup$
    – J.G.
    Dec 9 '18 at 21:11












    $begingroup$
    Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:13




    $begingroup$
    Then how do you relate $sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}$ with the geometric definition of $cos x$? I agree that if we show $$sum_{ngeq 0}frac{(-1)^n x^{2n}}{(2n)!}=cos xtext{ defined by geometry}$$ we have nothing to do more.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 21:13












    $begingroup$
    @JackD'Aurizio I think I spelled it out pretty well.
    $endgroup$
    – J.G.
    Dec 9 '18 at 21:19




    $begingroup$
    @JackD'Aurizio I think I spelled it out pretty well.
    $endgroup$
    – J.G.
    Dec 9 '18 at 21:19











    0












    $begingroup$

    $cos x$, defined as its Maclaurin series, is an entire solution of the differential equation $f''(x)+f(x)=0$, $f(0)=1$, $f'(0)=1$. $cos x$ is even and concave where positive, and we may denote with $alpha$ the smallest positive zero of $cos x$. The problem boils down to showing that $alpha=frac{pi}{2}$. Now, by denoting through $sin x$ the opposite of the derivative of $cos x$, we have that $f''+f=0$ implies that $f^2+f'^2$ is constant, in particular equal to one (Pythagorean theorem, $sin^2+cos^2=1$). Since $cos x$ is decreasing on $(0,alpha)$ and the Pythagorean theorem plus the chain rule imply that the derivative of the inverse function of $cos x$ is $-frac{1}{sqrt{1-x^2}}$, we have
    $$ alpha = int_{0}^{1}frac{dx}{sqrt{1-x^2}}.$$
    Integration by parts relates $alpha$ with the area of the unit circle:
    $$ alpha = 2int_{0}^{1}sqrt{1-x^2},dx = frac{pi}{2} $$
    and we are done: the series defining $cos x$ is positive on $[0,alpha)=left[0,frac{pi}{2}right)$.

    As a side-effect, we have the hypergeometric identities
    $$ frac{pi}{2}=sum_{ngeq 0}frac{binom{2n}{n}}{(2n+1)4^n},qquad frac{pi}{2}=2-sum_{ngeq 0}frac{binom{2n}{n}}{4^n(n+1)(2n+3)}$$
    provided by the Maclaurin series of $frac{1}{sqrt{1-x^2}}$ and $sqrt{1-x^2}$.
    The latter is faster-convergent, of course.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $cos x$, defined as its Maclaurin series, is an entire solution of the differential equation $f''(x)+f(x)=0$, $f(0)=1$, $f'(0)=1$. $cos x$ is even and concave where positive, and we may denote with $alpha$ the smallest positive zero of $cos x$. The problem boils down to showing that $alpha=frac{pi}{2}$. Now, by denoting through $sin x$ the opposite of the derivative of $cos x$, we have that $f''+f=0$ implies that $f^2+f'^2$ is constant, in particular equal to one (Pythagorean theorem, $sin^2+cos^2=1$). Since $cos x$ is decreasing on $(0,alpha)$ and the Pythagorean theorem plus the chain rule imply that the derivative of the inverse function of $cos x$ is $-frac{1}{sqrt{1-x^2}}$, we have
      $$ alpha = int_{0}^{1}frac{dx}{sqrt{1-x^2}}.$$
      Integration by parts relates $alpha$ with the area of the unit circle:
      $$ alpha = 2int_{0}^{1}sqrt{1-x^2},dx = frac{pi}{2} $$
      and we are done: the series defining $cos x$ is positive on $[0,alpha)=left[0,frac{pi}{2}right)$.

      As a side-effect, we have the hypergeometric identities
      $$ frac{pi}{2}=sum_{ngeq 0}frac{binom{2n}{n}}{(2n+1)4^n},qquad frac{pi}{2}=2-sum_{ngeq 0}frac{binom{2n}{n}}{4^n(n+1)(2n+3)}$$
      provided by the Maclaurin series of $frac{1}{sqrt{1-x^2}}$ and $sqrt{1-x^2}$.
      The latter is faster-convergent, of course.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $cos x$, defined as its Maclaurin series, is an entire solution of the differential equation $f''(x)+f(x)=0$, $f(0)=1$, $f'(0)=1$. $cos x$ is even and concave where positive, and we may denote with $alpha$ the smallest positive zero of $cos x$. The problem boils down to showing that $alpha=frac{pi}{2}$. Now, by denoting through $sin x$ the opposite of the derivative of $cos x$, we have that $f''+f=0$ implies that $f^2+f'^2$ is constant, in particular equal to one (Pythagorean theorem, $sin^2+cos^2=1$). Since $cos x$ is decreasing on $(0,alpha)$ and the Pythagorean theorem plus the chain rule imply that the derivative of the inverse function of $cos x$ is $-frac{1}{sqrt{1-x^2}}$, we have
        $$ alpha = int_{0}^{1}frac{dx}{sqrt{1-x^2}}.$$
        Integration by parts relates $alpha$ with the area of the unit circle:
        $$ alpha = 2int_{0}^{1}sqrt{1-x^2},dx = frac{pi}{2} $$
        and we are done: the series defining $cos x$ is positive on $[0,alpha)=left[0,frac{pi}{2}right)$.

        As a side-effect, we have the hypergeometric identities
        $$ frac{pi}{2}=sum_{ngeq 0}frac{binom{2n}{n}}{(2n+1)4^n},qquad frac{pi}{2}=2-sum_{ngeq 0}frac{binom{2n}{n}}{4^n(n+1)(2n+3)}$$
        provided by the Maclaurin series of $frac{1}{sqrt{1-x^2}}$ and $sqrt{1-x^2}$.
        The latter is faster-convergent, of course.






        share|cite|improve this answer











        $endgroup$



        $cos x$, defined as its Maclaurin series, is an entire solution of the differential equation $f''(x)+f(x)=0$, $f(0)=1$, $f'(0)=1$. $cos x$ is even and concave where positive, and we may denote with $alpha$ the smallest positive zero of $cos x$. The problem boils down to showing that $alpha=frac{pi}{2}$. Now, by denoting through $sin x$ the opposite of the derivative of $cos x$, we have that $f''+f=0$ implies that $f^2+f'^2$ is constant, in particular equal to one (Pythagorean theorem, $sin^2+cos^2=1$). Since $cos x$ is decreasing on $(0,alpha)$ and the Pythagorean theorem plus the chain rule imply that the derivative of the inverse function of $cos x$ is $-frac{1}{sqrt{1-x^2}}$, we have
        $$ alpha = int_{0}^{1}frac{dx}{sqrt{1-x^2}}.$$
        Integration by parts relates $alpha$ with the area of the unit circle:
        $$ alpha = 2int_{0}^{1}sqrt{1-x^2},dx = frac{pi}{2} $$
        and we are done: the series defining $cos x$ is positive on $[0,alpha)=left[0,frac{pi}{2}right)$.

        As a side-effect, we have the hypergeometric identities
        $$ frac{pi}{2}=sum_{ngeq 0}frac{binom{2n}{n}}{(2n+1)4^n},qquad frac{pi}{2}=2-sum_{ngeq 0}frac{binom{2n}{n}}{4^n(n+1)(2n+3)}$$
        provided by the Maclaurin series of $frac{1}{sqrt{1-x^2}}$ and $sqrt{1-x^2}$.
        The latter is faster-convergent, of course.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 21:02

























        answered Dec 9 '18 at 20:46









        Jack D'AurizioJack D'Aurizio

        1




        1






























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