Proof that $sin {x}$ is infinitely continuously differentiable over $[m,n]$












4












$begingroup$


I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing







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  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06
















4












$begingroup$


I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06














4












4








4


0



$begingroup$


I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing







share|cite|improve this question









$endgroup$




I am trying to prove that $sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.



Since,



$frac{d}{dx}sin{x} = cos{x}$,



$frac{d^2}{dx^2}sin{x} = -sin{x}$,



$frac{d^3}{dx^3}sin{x} = -cos{x}$,



and



$frac{d^4}{dx^4}sin{x} = sin{x}$,



the derivatives of $sin{x}$, are periodic. Since the first four derivatives of $sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.






calculus trigonometry proof-verification proof-writing




calculus trigonometry proof-verification proof-writing






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asked Jan 7 at 21:02









GnumbertesterGnumbertester

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6821114








  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06














  • 3




    $begingroup$
    Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
    $endgroup$
    – user29418
    Jan 7 at 21:06








3




3




$begingroup$
Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
$endgroup$
– user29418
Jan 7 at 21:06




$begingroup$
Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
$endgroup$
– user29418
Jan 7 at 21:06










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A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






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    $begingroup$

    A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






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      $begingroup$

      A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






      share|cite|improve this answer









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        2





        $begingroup$

        A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.






        share|cite|improve this answer









        $endgroup$



        A more formal way to show this is by induction. We know that $f(x) = sin(x)$ is continuous. Also, $f'(x) = cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-cos(x)) = (-1)^{(n+1)+1}cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-sin(x)) = (-1)^{n+2}sin(x)$, which is also continuous. So, for all $n geq 0$, $f^{(n)}(x)$ is continuous.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Jan 7 at 21:19









        D.B.D.B.

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