Proof that $sin {x}$ is infinitely continuously differentiable over $[m,n]$
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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
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
$endgroup$
add a comment |
$begingroup$
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
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
$endgroup$
3
$begingroup$
Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
$endgroup$
– user29418
Jan 7 at 21:06
add a comment |
$begingroup$
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
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
$endgroup$
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
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
asked Jan 7 at 21:02
GnumbertesterGnumbertester
565113
565113
3
$begingroup$
Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
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– user29418
Jan 7 at 21:06
add a comment |
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
add a comment |
1 Answer
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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.
answered Jan 7 at 21:19
D.B.D.B.
1,25018
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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.
answered Jan 7 at 21:19
D.B.D.B.
1,25018
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 7 at 21:19
D.B.D.B.
1,25018
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 7 at 21:19
D.B.D.B.
1,25018
$endgroup$
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.
answered Jan 7 at 21:19
D.B.D.B.
1,25018
answered Jan 7 at 21:19
D.B.D.B.
1,25018
answered Jan 7 at 21:19
D.B.D.B.
1,25018
answered Jan 7 at 21:19
D.B.D.B.
1,25018
1,25018
add a comment |
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Yup just differentiate it infinitely many times math.stackexchange.com/questions/13815/…
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– user29418
Jan 7 at 21:06