Why do $4cdot 2^nsinfrac{45}{2^n}$, $2cdot 2^nsinfrac{90}{2^n}$, and $1cdot 2^nsinfrac{180}{2^n}$ all tend to...
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I am not sure what question or inquiries to ask actually, but I just think this is really awesome
Can someone explain to me why the graphs of
$$4cdot 2^nsinfrac{45}{2^n}, qquad 2cdot 2^nsinfrac{90}{2^n}, qquadtext{and}qquad 1cdot 2^nsinfrac{180}{2^n}$$
all tend to $pi$?
sequences-and-series trigonometry pi
edited Dec 30 '18 at 15:18
Blue
49.8k870158
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I am not sure what question or inquiries to ask actually, but I just think this is really awesome
Can someone explain to me why the graphs of
$$4cdot 2^nsinfrac{45}{2^n}, qquad 2cdot 2^nsinfrac{90}{2^n}, qquadtext{and}qquad 1cdot 2^nsinfrac{180}{2^n}$$
all tend to $pi$?
sequences-and-series trigonometry pi
edited Dec 30 '18 at 15:18
Blue
49.8k870158
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math.stackexchange.com/questions/3056784/…
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– lab bhattacharjee
Dec 30 '18 at 15:53
|
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I am not sure what question or inquiries to ask actually, but I just think this is really awesome
Can someone explain to me why the graphs of
$$4cdot 2^nsinfrac{45}{2^n}, qquad 2cdot 2^nsinfrac{90}{2^n}, qquadtext{and}qquad 1cdot 2^nsinfrac{180}{2^n}$$
all tend to $pi$?
sequences-and-series trigonometry pi
edited Dec 30 '18 at 15:18
Blue
49.8k870158
$endgroup$
I am not sure what question or inquiries to ask actually, but I just think this is really awesome
Can someone explain to me why the graphs of
$$4cdot 2^nsinfrac{45}{2^n}, qquad 2cdot 2^nsinfrac{90}{2^n}, qquadtext{and}qquad 1cdot 2^nsinfrac{180}{2^n}$$
all tend to $pi$?
sequences-and-series trigonometry pi
sequences-and-series trigonometry pi
edited Dec 30 '18 at 15:18
Blue
49.8k870158
edited Dec 30 '18 at 15:18
Blue
49.8k870158
edited Dec 30 '18 at 15:18
Blue
49.8k870158
edited Dec 30 '18 at 15:18
Blue
49.8k870158
edited Dec 30 '18 at 15:18
Blue
49.8k870158
49.8k870158
asked Dec 30 '18 at 15:09
user630471
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– lab bhattacharjee
Dec 30 '18 at 15:53
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– lab bhattacharjee
Dec 30 '18 at 15:53
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– lab bhattacharjee
Dec 30 '18 at 15:53
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– lab bhattacharjee
Dec 30 '18 at 15:53
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2 Answers
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Because $sin x approx x$ when $x$ is small and measured in radians. When $x$ is measured in degrees $sin x approx frac {pi x}{180}.$ When $n$ gets large the argument of $sin$ becomes small.
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
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(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
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– Yanko
Dec 30 '18 at 15:18
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This is due to $lim_{h to 0}frac{sin h}h = 1$.
Hence begin{align}lim_{n to infty}4 cdot 2^n cdot sinleft(frac{45^circ}{2^n}right)&=lim_{n to infty}4 cdot 2^n cdot sinleft(frac{pi}{2^{n+2}}right) \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot lim_{n to infty}frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&= pi cdot 1\
&= piend{align}
Similarly for the other sequences.
That is we have
$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
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Thanks again Siong! :D
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– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
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– user630471
Dec 30 '18 at 15:16
1
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nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
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– The_Sympathizer
Dec 30 '18 at 15:17
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oops, thanks for pointing out the typo.
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– Siong Thye Goh
Dec 30 '18 at 15:19
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because $sin x approx x$ when $x$ is small and measured in radians. When $x$ is measured in degrees $sin x approx frac {pi x}{180}.$ When $n$ gets large the argument of $sin$ becomes small.
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
$endgroup$
1
$begingroup$
(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
$endgroup$
– Yanko
Dec 30 '18 at 15:18
|
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Because $sin x approx x$ when $x$ is small and measured in radians. When $x$ is measured in degrees $sin x approx frac {pi x}{180}.$ When $n$ gets large the argument of $sin$ becomes small.
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
$endgroup$
1
$begingroup$
(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
$endgroup$
– Yanko
Dec 30 '18 at 15:18
|
$begingroup$
Because $sin x approx x$ when $x$ is small and measured in radians. When $x$ is measured in degrees $sin x approx frac {pi x}{180}.$ When $n$ gets large the argument of $sin$ becomes small.
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
$endgroup$
Because $sin x approx x$ when $x$ is small and measured in radians. When $x$ is measured in degrees $sin x approx frac {pi x}{180}.$ When $n$ gets large the argument of $sin$ becomes small.
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
answered Dec 30 '18 at 15:14
Ross MillikanRoss Millikan
302k24201375
302k24201375
1
$begingroup$
(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
$endgroup$
– Yanko
Dec 30 '18 at 15:18
|
1
$begingroup$
(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
$endgroup$
– Yanko
Dec 30 '18 at 15:18
1
1
$begingroup$
(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
$endgroup$
– Yanko
Dec 30 '18 at 15:18
$begingroup$
(+1) Just to clarify, saying that $sin x approx x$ when $x$ is small is an informal way for saying that $lim_{xrightarrow 0} frac{sin x}{x} = 1$.
$endgroup$
– Yanko
Dec 30 '18 at 15:18
|
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This is due to $lim_{h to 0}frac{sin h}h = 1$.
Hence begin{align}lim_{n to infty}4 cdot 2^n cdot sinleft(frac{45^circ}{2^n}right)&=lim_{n to infty}4 cdot 2^n cdot sinleft(frac{pi}{2^{n+2}}right) \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot lim_{n to infty}frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&= pi cdot 1\
&= piend{align}
Similarly for the other sequences.
That is we have
$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
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Thanks again Siong! :D
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– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
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– user630471
Dec 30 '18 at 15:16
1
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nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
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– The_Sympathizer
Dec 30 '18 at 15:17
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oops, thanks for pointing out the typo.
$endgroup$
– Siong Thye Goh
Dec 30 '18 at 15:19
|
$begingroup$
This is due to $lim_{h to 0}frac{sin h}h = 1$.
Hence begin{align}lim_{n to infty}4 cdot 2^n cdot sinleft(frac{45^circ}{2^n}right)&=lim_{n to infty}4 cdot 2^n cdot sinleft(frac{pi}{2^{n+2}}right) \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot lim_{n to infty}frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&= pi cdot 1\
&= piend{align}
Similarly for the other sequences.
That is we have
$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Thanks again Siong! :D
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– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
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– user630471
Dec 30 '18 at 15:16
1
$begingroup$
nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
$endgroup$
– The_Sympathizer
Dec 30 '18 at 15:17
$begingroup$
oops, thanks for pointing out the typo.
$endgroup$
– Siong Thye Goh
Dec 30 '18 at 15:19
|
$begingroup$
This is due to $lim_{h to 0}frac{sin h}h = 1$.
Hence begin{align}lim_{n to infty}4 cdot 2^n cdot sinleft(frac{45^circ}{2^n}right)&=lim_{n to infty}4 cdot 2^n cdot sinleft(frac{pi}{2^{n+2}}right) \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot lim_{n to infty}frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&= pi cdot 1\
&= piend{align}
Similarly for the other sequences.
That is we have
$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
This is due to $lim_{h to 0}frac{sin h}h = 1$.
Hence begin{align}lim_{n to infty}4 cdot 2^n cdot sinleft(frac{45^circ}{2^n}right)&=lim_{n to infty}4 cdot 2^n cdot sinleft(frac{pi}{2^{n+2}}right) \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&=lim_{n to infty}4 cdot 2^n cdot frac{pi}{2^{n+2}}cdot lim_{n to infty}frac{sinleft(frac{pi}{2^{n+2}}right)}{frac{pi}{2^{n+2}}} \
&= pi cdot 1\
&= piend{align}
Similarly for the other sequences.
That is we have
$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
edited Dec 30 '18 at 15:18
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 30 '18 at 15:16
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
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Thanks again Siong! :D
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– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
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– user630471
Dec 30 '18 at 15:16
1
$begingroup$
nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
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– The_Sympathizer
Dec 30 '18 at 15:17
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oops, thanks for pointing out the typo.
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– Siong Thye Goh
Dec 30 '18 at 15:19
|
$begingroup$
Thanks again Siong! :D
$endgroup$
– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
$endgroup$
– user630471
Dec 30 '18 at 15:16
1
$begingroup$
nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
$endgroup$
– The_Sympathizer
Dec 30 '18 at 15:17
$begingroup$
oops, thanks for pointing out the typo.
$endgroup$
– Siong Thye Goh
Dec 30 '18 at 15:19
$begingroup$
Thanks again Siong! :D
$endgroup$
– user630471
Dec 30 '18 at 15:16
$begingroup$
Thanks again Siong! :D
$endgroup$
– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
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– user630471
Dec 30 '18 at 15:16
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so does this work for other series as well?
$endgroup$
– user630471
Dec 30 '18 at 15:16
1
1
$begingroup$
nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
$endgroup$
– The_Sympathizer
Dec 30 '18 at 15:17
$begingroup$
nb: $lim_{h rightarrow 0} frac{sin(h)}{h} = 1$, not 0.
$endgroup$
– The_Sympathizer
Dec 30 '18 at 15:17
$begingroup$
oops, thanks for pointing out the typo.
$endgroup$
– Siong Thye Goh
Dec 30 '18 at 15:19
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oops, thanks for pointing out the typo.
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– Siong Thye Goh
Dec 30 '18 at 15:19
|
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math.stackexchange.com/questions/3056784/…
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– lab bhattacharjee
Dec 30 '18 at 15:53