Proving the product expansion of $sintheta$. Where did I go wrong?
up vote
6
down vote
favorite
You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$
However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$
Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?
trigonometry
|
show 2 more comments
up vote
6
down vote
favorite
You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$
However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$
Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?
trigonometry
Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16
As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19
@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29
@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31
Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04
|
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$
However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$
Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?
trigonometry
You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$
However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$
Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?
trigonometry
trigonometry
edited Aug 14 '17 at 0:29
Blue
47.2k870149
47.2k870149
asked Aug 13 '17 at 16:12
Frank
3,7551631
3,7551631
Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16
As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19
@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29
@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31
Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04
|
show 2 more comments
Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16
As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19
@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29
@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31
Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04
Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16
Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16
As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19
As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19
@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29
@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29
@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31
@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31
Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04
Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}
Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}
and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
1
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}
Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}
and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
1
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
add a comment |
up vote
1
down vote
We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}
Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}
and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
1
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
add a comment |
up vote
1
down vote
up vote
1
down vote
We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}
Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}
and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$
We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}
Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}
and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$
edited Nov 27 at 4:24
answered Jan 4 at 17:46
Anant
147
147
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
1
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
add a comment |
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
1
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30
1
1
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2392398%2fproving-the-product-expansion-of-sin-theta-where-did-i-go-wrong%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16
As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19
@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29
@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31
Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04