Mixing
Limit of $frac {e^{in θ}} {1+ e^{i2nθ}}$
$begingroup$
I see that when $θ$ is a multiple of $2π$, $frac {e^{in θ}} {1+ e^{i2nθ}}$ is constant and thus has limit $frac 1 2$.
When $θ$ isn't a multiple of $2π$, however, it seems that the value $frac {e^{in θ}} {1+ e^{i2nθ}}$ of oscillates in a way that the limit doesn't exist.
For certain special cases (i.e. when $e^{iθ}=-1$ or $i$ or $i^{0.3}$), this can be demonstrated by brute force. But I am struggling to prove it for the general case, especially since $θ$ isn't necessarily rational.
complex-analysis limits
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
$endgroup$
add a comment |
$begingroup$
I see that when $θ$ is a multiple of $2π$, $frac {e^{in θ}} {1+ e^{i2nθ}}$ is constant and thus has limit $frac 1 2$.
When $θ$ isn't a multiple of $2π$, however, it seems that the value $frac {e^{in θ}} {1+ e^{i2nθ}}$ of oscillates in a way that the limit doesn't exist.
For certain special cases (i.e. when $e^{iθ}=-1$ or $i$ or $i^{0.3}$), this can be demonstrated by brute force. But I am struggling to prove it for the general case, especially since $θ$ isn't necessarily rational.
complex-analysis limits
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
$endgroup$
$begingroup$
You are taking the limit as $nto infty$?
$endgroup$
– Xander Henderson
Jan 17 at 1:33
$begingroup$
@XanderHenderson yes
$endgroup$
– Pascal's Wager
Jan 17 at 1:34
$begingroup$
The expression is the same as $frac{1}{2cos(ntheta)}$. Does that help?
$endgroup$
– herb steinberg
Jan 17 at 1:36
$begingroup$
@herb This expression doesn’t seem to have a limit, but it does have a range. The range will be $(-infinity , -{1over2}] U [{1over2} , infinity )$
$endgroup$
– Math Lover
Jan 17 at 1:39
$begingroup$
I mean, $infty$
$endgroup$
– Math Lover
Jan 17 at 1:46
add a comment |
$begingroup$
I see that when $θ$ is a multiple of $2π$, $frac {e^{in θ}} {1+ e^{i2nθ}}$ is constant and thus has limit $frac 1 2$.
When $θ$ isn't a multiple of $2π$, however, it seems that the value $frac {e^{in θ}} {1+ e^{i2nθ}}$ of oscillates in a way that the limit doesn't exist.
For certain special cases (i.e. when $e^{iθ}=-1$ or $i$ or $i^{0.3}$), this can be demonstrated by brute force. But I am struggling to prove it for the general case, especially since $θ$ isn't necessarily rational.
complex-analysis limits
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
$endgroup$
I see that when $θ$ is a multiple of $2π$, $frac {e^{in θ}} {1+ e^{i2nθ}}$ is constant and thus has limit $frac 1 2$.
When $θ$ isn't a multiple of $2π$, however, it seems that the value $frac {e^{in θ}} {1+ e^{i2nθ}}$ of oscillates in a way that the limit doesn't exist.
For certain special cases (i.e. when $e^{iθ}=-1$ or $i$ or $i^{0.3}$), this can be demonstrated by brute force. But I am struggling to prove it for the general case, especially since $θ$ isn't necessarily rational.
complex-analysis limits
complex-analysis limits
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
asked Jan 17 at 1:28
Pascal's WagerPascal's Wager
362315
362315
$begingroup$
You are taking the limit as $nto infty$?
$endgroup$
– Xander Henderson
Jan 17 at 1:33
$begingroup$
@XanderHenderson yes
$endgroup$
– Pascal's Wager
Jan 17 at 1:34
$begingroup$
The expression is the same as $frac{1}{2cos(ntheta)}$. Does that help?
$endgroup$
– herb steinberg
Jan 17 at 1:36
$begingroup$
@herb This expression doesn’t seem to have a limit, but it does have a range. The range will be $(-infinity , -{1over2}] U [{1over2} , infinity )$
$endgroup$
– Math Lover
Jan 17 at 1:39
$begingroup$
I mean, $infty$
$endgroup$
– Math Lover
Jan 17 at 1:46
add a comment |
$begingroup$
You are taking the limit as $nto infty$?
$endgroup$
– Xander Henderson
Jan 17 at 1:33
$begingroup$
@XanderHenderson yes
$endgroup$
– Pascal's Wager
Jan 17 at 1:34
$begingroup$
The expression is the same as $frac{1}{2cos(ntheta)}$. Does that help?
$endgroup$
– herb steinberg
Jan 17 at 1:36
$begingroup$
@herb This expression doesn’t seem to have a limit, but it does have a range. The range will be $(-infinity , -{1over2}] U [{1over2} , infinity )$
$endgroup$
– Math Lover
Jan 17 at 1:39
$begingroup$
I mean, $infty$
$endgroup$
– Math Lover
Jan 17 at 1:46
$begingroup$
You are taking the limit as $nto infty$?
$endgroup$
– Xander Henderson
Jan 17 at 1:33
$begingroup$
You are taking the limit as $nto infty$?
$endgroup$
– Xander Henderson
Jan 17 at 1:33
$begingroup$
@XanderHenderson yes
$endgroup$
– Pascal's Wager
Jan 17 at 1:34
$begingroup$
@XanderHenderson yes
$endgroup$
– Pascal's Wager
Jan 17 at 1:34
$begingroup$
The expression is the same as $frac{1}{2cos(ntheta)}$. Does that help?
$endgroup$
– herb steinberg
Jan 17 at 1:36
$begingroup$
The expression is the same as $frac{1}{2cos(ntheta)}$. Does that help?
$endgroup$
– herb steinberg
Jan 17 at 1:36
$begingroup$
@herb This expression doesn’t seem to have a limit, but it does have a range. The range will be $(-infinity , -{1over2}] U [{1over2} , infinity )$
$endgroup$
– Math Lover
Jan 17 at 1:39
$begingroup$
@herb This expression doesn’t seem to have a limit, but it does have a range. The range will be $(-infinity , -{1over2}] U [{1over2} , infinity )$
$endgroup$
– Math Lover
Jan 17 at 1:39
$begingroup$
I mean, $infty$
$endgroup$
– Math Lover
Jan 17 at 1:46
$begingroup$
I mean, $infty$
$endgroup$
– Math Lover
Jan 17 at 1:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $e^{intheta}ne 0$. Therefore,
$$
f_n(theta)=frac{e^{intheta}}{1+e^{2intheta}}=frac{1}{e^{-intheta}+e^{intheta}}=frac{1}{2}frac{2}{e^{intheta}+e^{-intheta}}=frac{1}{2}frac{1}{cos ntheta}.
$$
Just a reminder that this result would be invalid if $1+e^{2intheta}=0$, or equivalently, if $cos ntheta=0$.
To explore the convergence of $f_n(theta)$ for a fixed $theta$, it suffices to explore the convergence of $g_n(theta)=cos ntheta$. In this case, we may generalize the answers in this thread.
Suppose
$$
lim_{ntoinfty}g_n(theta)=g(theta)
$$
for some function $g$ of $theta$. Then it is a must that
$$
lim_{ntoinfty}g_{n+1}(theta)=g(theta).
$$
Now, note that
$$
g_{n+1}(theta)=cosleft(n+1right)theta=cos nthetacostheta-sin nthetasintheta,
$$
or equivalently,
$$
sin ntheta=frac{g_n(theta)costheta-g_{n+1}(theta)}{sintheta}
$$
for $sinthetane 0$. In this case, the convergence of $g_n(theta)$ implies the convergence of $h_n(theta)=sin ntheta$, i.e.,
$$
lim_{ntoinfty}h_n(theta)=h(theta)
$$
for some function $h$ of $theta$.
Besides, note that
$$
g_n^2(theta)+h_n^2(theta)=cos^2ntheta+sin^2ntheta=1.
$$
It follows that
begin{equation}
g^2(theta)+h^2(theta)=1.tag{1}
end{equation}
Finally, if $sintheta=0$, we have $theta=kpi$ with $kinmathbb{Z}$. In this case, $g_n(theta)=pm 1$, whose convergence depends on the choice of $k$, which is obvious to figure out. If $sinthetane 0$, recall that
$$
g_{n+1}(theta)=g_n(theta)costheta-h_n(theta)sintheta.
$$
Taking the limit on both sides yields
$$
g(theta)=g(theta)costheta-h(theta)sintheta,
$$
or equivalently,
begin{equation}
h(theta)=-g(theta)frac{1-costheta}{sin{theta}}.tag{2}
end{equation}
Besides, apply the same trick to
begin{align}
h_{n+1}(theta)=sinleft(n+1right)theta&=sin nthetacostheta+cos nthetasintheta\
&=h_n(theta)costheta+g_n(theta)sintheta,
end{align}
and we obtain
$$
h(theta)=h(theta)costheta+g(theta)sintheta,
$$
or equivalently,
begin{equation}
g(theta)=h(theta)frac{1-costheta}{sintheta}.tag{3}
end{equation}
You may find an immediate contradiction between (1), (2), and (3).
In conclusion, the convergence of $g_n(theta)$ could be discussed for $theta=kpi$ with $kinmathbb{Z}$. Otherwise, $g_n(theta)$ always diverges.
answered Jan 17 at 1:35
hypernovahypernova
4,834414
$endgroup$
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
$begingroup$
@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $e^{intheta}ne 0$. Therefore,
$$
f_n(theta)=frac{e^{intheta}}{1+e^{2intheta}}=frac{1}{e^{-intheta}+e^{intheta}}=frac{1}{2}frac{2}{e^{intheta}+e^{-intheta}}=frac{1}{2}frac{1}{cos ntheta}.
$$
Just a reminder that this result would be invalid if $1+e^{2intheta}=0$, or equivalently, if $cos ntheta=0$.
To explore the convergence of $f_n(theta)$ for a fixed $theta$, it suffices to explore the convergence of $g_n(theta)=cos ntheta$. In this case, we may generalize the answers in this thread.
Suppose
$$
lim_{ntoinfty}g_n(theta)=g(theta)
$$
for some function $g$ of $theta$. Then it is a must that
$$
lim_{ntoinfty}g_{n+1}(theta)=g(theta).
$$
Now, note that
$$
g_{n+1}(theta)=cosleft(n+1right)theta=cos nthetacostheta-sin nthetasintheta,
$$
or equivalently,
$$
sin ntheta=frac{g_n(theta)costheta-g_{n+1}(theta)}{sintheta}
$$
for $sinthetane 0$. In this case, the convergence of $g_n(theta)$ implies the convergence of $h_n(theta)=sin ntheta$, i.e.,
$$
lim_{ntoinfty}h_n(theta)=h(theta)
$$
for some function $h$ of $theta$.
Besides, note that
$$
g_n^2(theta)+h_n^2(theta)=cos^2ntheta+sin^2ntheta=1.
$$
It follows that
begin{equation}
g^2(theta)+h^2(theta)=1.tag{1}
end{equation}
Finally, if $sintheta=0$, we have $theta=kpi$ with $kinmathbb{Z}$. In this case, $g_n(theta)=pm 1$, whose convergence depends on the choice of $k$, which is obvious to figure out. If $sinthetane 0$, recall that
$$
g_{n+1}(theta)=g_n(theta)costheta-h_n(theta)sintheta.
$$
Taking the limit on both sides yields
$$
g(theta)=g(theta)costheta-h(theta)sintheta,
$$
or equivalently,
begin{equation}
h(theta)=-g(theta)frac{1-costheta}{sin{theta}}.tag{2}
end{equation}
Besides, apply the same trick to
begin{align}
h_{n+1}(theta)=sinleft(n+1right)theta&=sin nthetacostheta+cos nthetasintheta\
&=h_n(theta)costheta+g_n(theta)sintheta,
end{align}
and we obtain
$$
h(theta)=h(theta)costheta+g(theta)sintheta,
$$
or equivalently,
begin{equation}
g(theta)=h(theta)frac{1-costheta}{sintheta}.tag{3}
end{equation}
You may find an immediate contradiction between (1), (2), and (3).
In conclusion, the convergence of $g_n(theta)$ could be discussed for $theta=kpi$ with $kinmathbb{Z}$. Otherwise, $g_n(theta)$ always diverges.
answered Jan 17 at 1:35
hypernovahypernova
4,834414
$endgroup$
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
$begingroup$
@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
add a comment |
$begingroup$
Note that $e^{intheta}ne 0$. Therefore,
$$
f_n(theta)=frac{e^{intheta}}{1+e^{2intheta}}=frac{1}{e^{-intheta}+e^{intheta}}=frac{1}{2}frac{2}{e^{intheta}+e^{-intheta}}=frac{1}{2}frac{1}{cos ntheta}.
$$
Just a reminder that this result would be invalid if $1+e^{2intheta}=0$, or equivalently, if $cos ntheta=0$.
To explore the convergence of $f_n(theta)$ for a fixed $theta$, it suffices to explore the convergence of $g_n(theta)=cos ntheta$. In this case, we may generalize the answers in this thread.
Suppose
$$
lim_{ntoinfty}g_n(theta)=g(theta)
$$
for some function $g$ of $theta$. Then it is a must that
$$
lim_{ntoinfty}g_{n+1}(theta)=g(theta).
$$
Now, note that
$$
g_{n+1}(theta)=cosleft(n+1right)theta=cos nthetacostheta-sin nthetasintheta,
$$
or equivalently,
$$
sin ntheta=frac{g_n(theta)costheta-g_{n+1}(theta)}{sintheta}
$$
for $sinthetane 0$. In this case, the convergence of $g_n(theta)$ implies the convergence of $h_n(theta)=sin ntheta$, i.e.,
$$
lim_{ntoinfty}h_n(theta)=h(theta)
$$
for some function $h$ of $theta$.
Besides, note that
$$
g_n^2(theta)+h_n^2(theta)=cos^2ntheta+sin^2ntheta=1.
$$
It follows that
begin{equation}
g^2(theta)+h^2(theta)=1.tag{1}
end{equation}
Finally, if $sintheta=0$, we have $theta=kpi$ with $kinmathbb{Z}$. In this case, $g_n(theta)=pm 1$, whose convergence depends on the choice of $k$, which is obvious to figure out. If $sinthetane 0$, recall that
$$
g_{n+1}(theta)=g_n(theta)costheta-h_n(theta)sintheta.
$$
Taking the limit on both sides yields
$$
g(theta)=g(theta)costheta-h(theta)sintheta,
$$
or equivalently,
begin{equation}
h(theta)=-g(theta)frac{1-costheta}{sin{theta}}.tag{2}
end{equation}
Besides, apply the same trick to
begin{align}
h_{n+1}(theta)=sinleft(n+1right)theta&=sin nthetacostheta+cos nthetasintheta\
&=h_n(theta)costheta+g_n(theta)sintheta,
end{align}
and we obtain
$$
h(theta)=h(theta)costheta+g(theta)sintheta,
$$
or equivalently,
begin{equation}
g(theta)=h(theta)frac{1-costheta}{sintheta}.tag{3}
end{equation}
You may find an immediate contradiction between (1), (2), and (3).
In conclusion, the convergence of $g_n(theta)$ could be discussed for $theta=kpi$ with $kinmathbb{Z}$. Otherwise, $g_n(theta)$ always diverges.
answered Jan 17 at 1:35
hypernovahypernova
4,834414
$endgroup$
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
$begingroup$
@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
add a comment |
$begingroup$
Note that $e^{intheta}ne 0$. Therefore,
$$
f_n(theta)=frac{e^{intheta}}{1+e^{2intheta}}=frac{1}{e^{-intheta}+e^{intheta}}=frac{1}{2}frac{2}{e^{intheta}+e^{-intheta}}=frac{1}{2}frac{1}{cos ntheta}.
$$
Just a reminder that this result would be invalid if $1+e^{2intheta}=0$, or equivalently, if $cos ntheta=0$.
To explore the convergence of $f_n(theta)$ for a fixed $theta$, it suffices to explore the convergence of $g_n(theta)=cos ntheta$. In this case, we may generalize the answers in this thread.
Suppose
$$
lim_{ntoinfty}g_n(theta)=g(theta)
$$
for some function $g$ of $theta$. Then it is a must that
$$
lim_{ntoinfty}g_{n+1}(theta)=g(theta).
$$
Now, note that
$$
g_{n+1}(theta)=cosleft(n+1right)theta=cos nthetacostheta-sin nthetasintheta,
$$
or equivalently,
$$
sin ntheta=frac{g_n(theta)costheta-g_{n+1}(theta)}{sintheta}
$$
for $sinthetane 0$. In this case, the convergence of $g_n(theta)$ implies the convergence of $h_n(theta)=sin ntheta$, i.e.,
$$
lim_{ntoinfty}h_n(theta)=h(theta)
$$
for some function $h$ of $theta$.
Besides, note that
$$
g_n^2(theta)+h_n^2(theta)=cos^2ntheta+sin^2ntheta=1.
$$
It follows that
begin{equation}
g^2(theta)+h^2(theta)=1.tag{1}
end{equation}
Finally, if $sintheta=0$, we have $theta=kpi$ with $kinmathbb{Z}$. In this case, $g_n(theta)=pm 1$, whose convergence depends on the choice of $k$, which is obvious to figure out. If $sinthetane 0$, recall that
$$
g_{n+1}(theta)=g_n(theta)costheta-h_n(theta)sintheta.
$$
Taking the limit on both sides yields
$$
g(theta)=g(theta)costheta-h(theta)sintheta,
$$
or equivalently,
begin{equation}
h(theta)=-g(theta)frac{1-costheta}{sin{theta}}.tag{2}
end{equation}
Besides, apply the same trick to
begin{align}
h_{n+1}(theta)=sinleft(n+1right)theta&=sin nthetacostheta+cos nthetasintheta\
&=h_n(theta)costheta+g_n(theta)sintheta,
end{align}
and we obtain
$$
h(theta)=h(theta)costheta+g(theta)sintheta,
$$
or equivalently,
begin{equation}
g(theta)=h(theta)frac{1-costheta}{sintheta}.tag{3}
end{equation}
You may find an immediate contradiction between (1), (2), and (3).
In conclusion, the convergence of $g_n(theta)$ could be discussed for $theta=kpi$ with $kinmathbb{Z}$. Otherwise, $g_n(theta)$ always diverges.
answered Jan 17 at 1:35
hypernovahypernova
4,834414
$endgroup$
Note that $e^{intheta}ne 0$. Therefore,
$$
f_n(theta)=frac{e^{intheta}}{1+e^{2intheta}}=frac{1}{e^{-intheta}+e^{intheta}}=frac{1}{2}frac{2}{e^{intheta}+e^{-intheta}}=frac{1}{2}frac{1}{cos ntheta}.
$$
Just a reminder that this result would be invalid if $1+e^{2intheta}=0$, or equivalently, if $cos ntheta=0$.
To explore the convergence of $f_n(theta)$ for a fixed $theta$, it suffices to explore the convergence of $g_n(theta)=cos ntheta$. In this case, we may generalize the answers in this thread.
Suppose
$$
lim_{ntoinfty}g_n(theta)=g(theta)
$$
for some function $g$ of $theta$. Then it is a must that
$$
lim_{ntoinfty}g_{n+1}(theta)=g(theta).
$$
Now, note that
$$
g_{n+1}(theta)=cosleft(n+1right)theta=cos nthetacostheta-sin nthetasintheta,
$$
or equivalently,
$$
sin ntheta=frac{g_n(theta)costheta-g_{n+1}(theta)}{sintheta}
$$
for $sinthetane 0$. In this case, the convergence of $g_n(theta)$ implies the convergence of $h_n(theta)=sin ntheta$, i.e.,
$$
lim_{ntoinfty}h_n(theta)=h(theta)
$$
for some function $h$ of $theta$.
Besides, note that
$$
g_n^2(theta)+h_n^2(theta)=cos^2ntheta+sin^2ntheta=1.
$$
It follows that
begin{equation}
g^2(theta)+h^2(theta)=1.tag{1}
end{equation}
Finally, if $sintheta=0$, we have $theta=kpi$ with $kinmathbb{Z}$. In this case, $g_n(theta)=pm 1$, whose convergence depends on the choice of $k$, which is obvious to figure out. If $sinthetane 0$, recall that
$$
g_{n+1}(theta)=g_n(theta)costheta-h_n(theta)sintheta.
$$
Taking the limit on both sides yields
$$
g(theta)=g(theta)costheta-h(theta)sintheta,
$$
or equivalently,
begin{equation}
h(theta)=-g(theta)frac{1-costheta}{sin{theta}}.tag{2}
end{equation}
Besides, apply the same trick to
begin{align}
h_{n+1}(theta)=sinleft(n+1right)theta&=sin nthetacostheta+cos nthetasintheta\
&=h_n(theta)costheta+g_n(theta)sintheta,
end{align}
and we obtain
$$
h(theta)=h(theta)costheta+g(theta)sintheta,
$$
or equivalently,
begin{equation}
g(theta)=h(theta)frac{1-costheta}{sintheta}.tag{3}
end{equation}
You may find an immediate contradiction between (1), (2), and (3).
In conclusion, the convergence of $g_n(theta)$ could be discussed for $theta=kpi$ with $kinmathbb{Z}$. Otherwise, $g_n(theta)$ always diverges.
answered Jan 17 at 1:35
hypernovahypernova
4,834414
edited Jan 17 at 2:52
answered Jan 17 at 1:35
hypernovahypernova
4,834414
answered Jan 17 at 1:35
hypernovahypernova
4,834414
answered Jan 17 at 1:35
hypernovahypernova
4,834414
4,834414
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
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@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
add a comment |
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
$begingroup$
@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
+1 I think this is close to what I need! But how can I show that $cos nθ$ oscillates?
$endgroup$
– Pascal's Wager
Jan 17 at 1:50
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
It would be easy to show if our limit $n to infty$ were over the real numbers, but I'm taking the limit over the natural numbers.
$endgroup$
– Pascal's Wager
Jan 17 at 1:57
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
$begingroup$
So if $θ=1$, for instance, $cos nθ$ will never be equal to $-1$. So I can't say that it oscillates between $-1$ and $1$.
$endgroup$
– Pascal's Wager
Jan 17 at 2:04
$begingroup$
@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
$begingroup$
@Pascal'sWager: I edited my answer. Perhaps you may want to have a look at it. Thanks :-)
$endgroup$
– hypernova
Jan 17 at 2:52
add a comment |
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$begingroup$
You are taking the limit as $nto infty$?
$endgroup$
– Xander Henderson
Jan 17 at 1:33
$begingroup$
@XanderHenderson yes
$endgroup$
– Pascal's Wager
Jan 17 at 1:34
$begingroup$
The expression is the same as $frac{1}{2cos(ntheta)}$. Does that help?
$endgroup$
– herb steinberg
Jan 17 at 1:36
$begingroup$
@herb This expression doesn’t seem to have a limit, but it does have a range. The range will be $(-infinity , -{1over2}] U [{1over2} , infinity )$
$endgroup$
– Math Lover
Jan 17 at 1:39
$begingroup$
I mean, $infty$
$endgroup$
– Math Lover
Jan 17 at 1:46