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 $pi$?




Please Sign here and Sin here XD






sequences-and-series trigonometry pi







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    – lab bhattacharjee
    Dec 30 '18 at 15:53
















0












$begingroup$


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 $pi$?




Please Sign here and Sin here XD






sequences-and-series trigonometry pi







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/3056784/…
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 15:53














0












0








0





$begingroup$


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 $pi$?




Please Sign here and Sin here XD






sequences-and-series trigonometry pi







share|cite|improve this question











$endgroup$




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 $pi$?




Please Sign here and Sin here XD






sequences-and-series trigonometry pi




sequences-and-series trigonometry pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 15:18









Blue

48.6k870156




48.6k870156










asked Dec 30 '18 at 15:09







user630471



















  • $begingroup$
    math.stackexchange.com/questions/3056784/…
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 15:53


















  • $begingroup$
    math.stackexchange.com/questions/3056784/…
    $endgroup$
    – lab bhattacharjee
    Dec 30 '18 at 15:53
















$begingroup$
math.stackexchange.com/questions/3056784/…
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 15:53




$begingroup$
math.stackexchange.com/questions/3056784/…
$endgroup$
– lab bhattacharjee
Dec 30 '18 at 15:53










2 Answers
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active

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$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 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





















2












$begingroup$

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



$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks again Siong! :D
    $endgroup$
    – user630471
    Dec 30 '18 at 15:16










  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 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


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 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
















2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 15:14









Ross MillikanRoss Millikan

297k23198371




297k23198371








  • 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













2












$begingroup$

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



$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks again Siong! :D
    $endgroup$
    – user630471
    Dec 30 '18 at 15:16










  • $begingroup$
    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
















2












$begingroup$

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



$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks again Siong! :D
    $endgroup$
    – user630471
    Dec 30 '18 at 15:16










  • $begingroup$
    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














2












2








2





$begingroup$

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



$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$






share|cite|improve this answer











$endgroup$



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



$$lim_{n to infty}w cdot 2^n cdot sin left( frac{180^circ}{wcdot 2^n}right)=pi$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 15:18

























answered Dec 30 '18 at 15:16









Siong Thye GohSiong Thye Goh

102k1466118




102k1466118












  • $begingroup$
    Thanks again Siong! :D
    $endgroup$
    – user630471
    Dec 30 '18 at 15:16










  • $begingroup$
    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










  • $begingroup$
    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












$begingroup$
so does this work for other series as well?
$endgroup$
– user630471
Dec 30 '18 at 15:16






$begingroup$
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




$begingroup$
oops, thanks for pointing out the typo.
$endgroup$
– Siong Thye Goh
Dec 30 '18 at 15:19


















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