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Why does the Maclaurin series of sin(x) converge to sin(x)? [duplicate]
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Sine taylor series
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When looking up how the extremely famous series $$sin(x)=sum_k=0^infty(-1)^kfracx^2k+1(2k+1)!$$ is derived, I found this great explanation by Proof Wiki.
My question is this: the explanation shows clearly how to derive the Maclaurin series for $sin(x)$ and how it converges for all real arguments, however - as someone new to the intricacies of Maclaurin series - it does not prove that whatever the series converges to at the real number $a$ is $sin(a)$. Why is this true?
functions trigonometry power-series taylor-expansion
edited Jul 22 at 19:08
Bernard
110k635103
asked Jul 22 at 18:52
Isky Mathews
777214
marked as duplicate by José Carlos Santos, Community♦ Jul 22 at 19:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Sine taylor series
1 answer
When looking up how the extremely famous series $$sin(x)=sum_k=0^infty(-1)^kfracx^2k+1(2k+1)!$$ is derived, I found this great explanation by Proof Wiki.
My question is this: the explanation shows clearly how to derive the Maclaurin series for $sin(x)$ and how it converges for all real arguments, however - as someone new to the intricacies of Maclaurin series - it does not prove that whatever the series converges to at the real number $a$ is $sin(a)$. Why is this true?
functions trigonometry power-series taylor-expansion
edited Jul 22 at 19:08
Bernard
110k635103
asked Jul 22 at 18:52
Isky Mathews
777214
marked as duplicate by José Carlos Santos, Community♦ Jul 22 at 19:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
To this and similar questions I always respond with: What is your definition of $sin>$?
– Christian Blatter
Jul 22 at 18:57
Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles.
– Isky Mathews
Jul 22 at 19:04
One approach is to look at to show that for any complex $z$, $left(1+frac znright)^n$ converges to $sum_k=0^inftyfracz^kk!.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $cos x + isin x = 1+x+ o(x^2),$ which is enough to show that $left(1+frac ix nright)^n$ converges to the same value as $left(cos x/n + isin x/nright)^n=cos x + isin x.$
– Thomas Andrews
Jul 22 at 19:19
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Sine taylor series
1 answer
When looking up how the extremely famous series $$sin(x)=sum_k=0^infty(-1)^kfracx^2k+1(2k+1)!$$ is derived, I found this great explanation by Proof Wiki.
My question is this: the explanation shows clearly how to derive the Maclaurin series for $sin(x)$ and how it converges for all real arguments, however - as someone new to the intricacies of Maclaurin series - it does not prove that whatever the series converges to at the real number $a$ is $sin(a)$. Why is this true?
functions trigonometry power-series taylor-expansion
edited Jul 22 at 19:08
Bernard
110k635103
asked Jul 22 at 18:52
Isky Mathews
777214
This question already has an answer here:
Sine taylor series
1 answer
When looking up how the extremely famous series $$sin(x)=sum_k=0^infty(-1)^kfracx^2k+1(2k+1)!$$ is derived, I found this great explanation by Proof Wiki.
My question is this: the explanation shows clearly how to derive the Maclaurin series for $sin(x)$ and how it converges for all real arguments, however - as someone new to the intricacies of Maclaurin series - it does not prove that whatever the series converges to at the real number $a$ is $sin(a)$. Why is this true?
This question already has an answer here:
Sine taylor series
1 answer
functions trigonometry power-series taylor-expansion
edited Jul 22 at 19:08
Bernard
110k635103
asked Jul 22 at 18:52
Isky Mathews
777214
edited Jul 22 at 19:08
Bernard
110k635103
edited Jul 22 at 19:08
Bernard
110k635103
edited Jul 22 at 19:08
Bernard
110k635103
110k635103
asked Jul 22 at 18:52
Isky Mathews
777214
asked Jul 22 at 18:52
Isky Mathews
777214
asked Jul 22 at 18:52
Isky Mathews
777214
777214
marked as duplicate by José Carlos Santos, Community♦ Jul 22 at 19:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, Community♦ Jul 22 at 19:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
To this and similar questions I always respond with: What is your definition of $sin>$?
– Christian Blatter
Jul 22 at 18:57
Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles.
– Isky Mathews
Jul 22 at 19:04
One approach is to look at to show that for any complex $z$, $left(1+frac znright)^n$ converges to $sum_k=0^inftyfracz^kk!.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $cos x + isin x = 1+x+ o(x^2),$ which is enough to show that $left(1+frac ix nright)^n$ converges to the same value as $left(cos x/n + isin x/nright)^n=cos x + isin x.$
– Thomas Andrews
Jul 22 at 19:19
add a comment |Â
3
To this and similar questions I always respond with: What is your definition of $sin>$?
– Christian Blatter
Jul 22 at 18:57
Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles.
– Isky Mathews
Jul 22 at 19:04
One approach is to look at to show that for any complex $z$, $left(1+frac znright)^n$ converges to $sum_k=0^inftyfracz^kk!.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $cos x + isin x = 1+x+ o(x^2),$ which is enough to show that $left(1+frac ix nright)^n$ converges to the same value as $left(cos x/n + isin x/nright)^n=cos x + isin x.$
– Thomas Andrews
Jul 22 at 19:19
3
3
To this and similar questions I always respond with: What is your definition of $sin>$?
– Christian Blatter
Jul 22 at 18:57
To this and similar questions I always respond with: What is your definition of $sin>$?
– Christian Blatter
Jul 22 at 18:57
Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles.
– Isky Mathews
Jul 22 at 19:04
Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles.
– Isky Mathews
Jul 22 at 19:04
One approach is to look at to show that for any complex $z$, $left(1+frac znright)^n$ converges to $sum_k=0^inftyfracz^kk!.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $cos x + isin x = 1+x+ o(x^2),$ which is enough to show that $left(1+frac ix nright)^n$ converges to the same value as $left(cos x/n + isin x/nright)^n=cos x + isin x.$
– Thomas Andrews
Jul 22 at 19:19
One approach is to look at to show that for any complex $z$, $left(1+frac znright)^n$ converges to $sum_k=0^inftyfracz^kk!.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $cos x + isin x = 1+x+ o(x^2),$ which is enough to show that $left(1+frac ix nright)^n$ converges to the same value as $left(cos x/n + isin x/nright)^n=cos x + isin x.$
– Thomas Andrews
Jul 22 at 19:19
add a comment |Â
1 Answer
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For acute $x$ a geometric proof that $0<sin x<x$ is easy, and we can alternately use $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ to bound both trigonometric functions with increasingly high-degree polynomials. In each case the even- and odd-$n$ sums of the first $n$ terms provide subsequences that bound the functions on either side, so a ratio-test proof of convergence and the squeeze theorem proves each series converges to exactly what you think it does
Extending the above to non-acute $x$ is an exercise left for the reader. (Hint: prove compound-angle formulae for the Taylor series, then verify the effects of adding multiples of $pi/2$ are the same as for the trigonometric functions.)
answered Jul 22 at 19:07
J.G.
13.2k11424
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For acute $x$ a geometric proof that $0<sin x<x$ is easy, and we can alternately use $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ to bound both trigonometric functions with increasingly high-degree polynomials. In each case the even- and odd-$n$ sums of the first $n$ terms provide subsequences that bound the functions on either side, so a ratio-test proof of convergence and the squeeze theorem proves each series converges to exactly what you think it does
Extending the above to non-acute $x$ is an exercise left for the reader. (Hint: prove compound-angle formulae for the Taylor series, then verify the effects of adding multiples of $pi/2$ are the same as for the trigonometric functions.)
answered Jul 22 at 19:07
J.G.
13.2k11424
add a comment |Â
up vote
0
down vote
For acute $x$ a geometric proof that $0<sin x<x$ is easy, and we can alternately use $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ to bound both trigonometric functions with increasingly high-degree polynomials. In each case the even- and odd-$n$ sums of the first $n$ terms provide subsequences that bound the functions on either side, so a ratio-test proof of convergence and the squeeze theorem proves each series converges to exactly what you think it does
Extending the above to non-acute $x$ is an exercise left for the reader. (Hint: prove compound-angle formulae for the Taylor series, then verify the effects of adding multiples of $pi/2$ are the same as for the trigonometric functions.)
answered Jul 22 at 19:07
J.G.
13.2k11424
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For acute $x$ a geometric proof that $0<sin x<x$ is easy, and we can alternately use $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ to bound both trigonometric functions with increasingly high-degree polynomials. In each case the even- and odd-$n$ sums of the first $n$ terms provide subsequences that bound the functions on either side, so a ratio-test proof of convergence and the squeeze theorem proves each series converges to exactly what you think it does
Extending the above to non-acute $x$ is an exercise left for the reader. (Hint: prove compound-angle formulae for the Taylor series, then verify the effects of adding multiples of $pi/2$ are the same as for the trigonometric functions.)
answered Jul 22 at 19:07
J.G.
13.2k11424
For acute $x$ a geometric proof that $0<sin x<x$ is easy, and we can alternately use $cos x=1-int_0^xsin t dt,,sin x=int_0^xcos t dt$ to bound both trigonometric functions with increasingly high-degree polynomials. In each case the even- and odd-$n$ sums of the first $n$ terms provide subsequences that bound the functions on either side, so a ratio-test proof of convergence and the squeeze theorem proves each series converges to exactly what you think it does
Extending the above to non-acute $x$ is an exercise left for the reader. (Hint: prove compound-angle formulae for the Taylor series, then verify the effects of adding multiples of $pi/2$ are the same as for the trigonometric functions.)
answered Jul 22 at 19:07
J.G.
13.2k11424
answered Jul 22 at 19:07
J.G.
13.2k11424
answered Jul 22 at 19:07
J.G.
13.2k11424
answered Jul 22 at 19:07
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
3
To this and similar questions I always respond with: What is your definition of $sin>$?
– Christian Blatter
Jul 22 at 18:57
Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles.
– Isky Mathews
Jul 22 at 19:04
One approach is to look at to show that for any complex $z$, $left(1+frac znright)^n$ converges to $sum_k=0^inftyfracz^kk!.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $cos x + isin x = 1+x+ o(x^2),$ which is enough to show that $left(1+frac ix nright)^n$ converges to the same value as $left(cos x/n + isin x/nright)^n=cos x + isin x.$
– Thomas Andrews
Jul 22 at 19:19