Mixing
$|e^{-ina}-e^{-inb}|=2sin(ntheta_0)$
$begingroup$
How can I show that $|e^{-ina}-e^{-inb}|=2sin(ntheta_0)$ for some $theta_0 in[a,b]$?
I am trying this post Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally. and was very quickly lost by this fact among others.
Is this some application of the mean value theorem that I am not spotting? How can I see this?
calculus complex-analysis fourier-analysis
edited Jan 31 at 3:07
the_fox
2,90231538
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
$endgroup$
add a comment |
$begingroup$
How can I show that $|e^{-ina}-e^{-inb}|=2sin(ntheta_0)$ for some $theta_0 in[a,b]$?
I am trying this post Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally. and was very quickly lost by this fact among others.
Is this some application of the mean value theorem that I am not spotting? How can I see this?
calculus complex-analysis fourier-analysis
edited Jan 31 at 3:07
the_fox
2,90231538
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
$endgroup$
1
$begingroup$
The range of $theta_0$ doesn't seem correct. Just think about the case $n=1$ and $a,b$ very close to $pi/2$. In that case LHS is close to zero but right hand side would be very close to $1$ if $theta_0in[a,b]$. With few computations you get that $|e^{-ina}-e^{-inb}|=2|sin(n(a-b)/2)|$. So, depending on the sign of sine you can choose $theta_0=pm (a-b)/2$.
$endgroup$
– Julian Mejia
Jan 31 at 2:45
$begingroup$
@JulianMejia As I state in my answer, consider something like $a = 2$ and $b = 3$. In this case, $theta_0 = pm frac{1}{2}$ would not meet the requirement of $theta_0 in left[a,bright]$.
$endgroup$
– John Omielan
Jan 31 at 3:22
$begingroup$
@JulianMejia As I show in my answer, I believe that you need to change something, like $e^{-inb}$ to $e^{inb}$, plus you need to use absolute value signs around $sinleft(ntheta_0right)$.
$endgroup$
– John Omielan
Jan 31 at 3:46
$begingroup$
I made a mistake with my comment above. It should have gone to the OP, i.e., Jungleshrimp, instead.
$endgroup$
– John Omielan
Jan 31 at 5:53
$begingroup$
@JulianMejia Please ignore my earlier comment as I meant it to go to the OP instead.
$endgroup$
– John Omielan
Jan 31 at 5:54
add a comment |
$begingroup$
How can I show that $|e^{-ina}-e^{-inb}|=2sin(ntheta_0)$ for some $theta_0 in[a,b]$?
I am trying this post Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally. and was very quickly lost by this fact among others.
Is this some application of the mean value theorem that I am not spotting? How can I see this?
calculus complex-analysis fourier-analysis
edited Jan 31 at 3:07
the_fox
2,90231538
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
$endgroup$
How can I show that $|e^{-ina}-e^{-inb}|=2sin(ntheta_0)$ for some $theta_0 in[a,b]$?
I am trying this post Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally. and was very quickly lost by this fact among others.
Is this some application of the mean value theorem that I am not spotting? How can I see this?
calculus complex-analysis fourier-analysis
calculus complex-analysis fourier-analysis
edited Jan 31 at 3:07
the_fox
2,90231538
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
edited Jan 31 at 3:07
the_fox
2,90231538
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
edited Jan 31 at 3:07
the_fox
2,90231538
edited Jan 31 at 3:07
the_fox
2,90231538
edited Jan 31 at 3:07
the_fox
2,90231538
2,90231538
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
asked Jan 31 at 2:17
JungleshrimpJungleshrimp
366112
366112
1
$begingroup$
The range of $theta_0$ doesn't seem correct. Just think about the case $n=1$ and $a,b$ very close to $pi/2$. In that case LHS is close to zero but right hand side would be very close to $1$ if $theta_0in[a,b]$. With few computations you get that $|e^{-ina}-e^{-inb}|=2|sin(n(a-b)/2)|$. So, depending on the sign of sine you can choose $theta_0=pm (a-b)/2$.
$endgroup$
– Julian Mejia
Jan 31 at 2:45
$begingroup$
@JulianMejia As I state in my answer, consider something like $a = 2$ and $b = 3$. In this case, $theta_0 = pm frac{1}{2}$ would not meet the requirement of $theta_0 in left[a,bright]$.
$endgroup$
– John Omielan
Jan 31 at 3:22
$begingroup$
@JulianMejia As I show in my answer, I believe that you need to change something, like $e^{-inb}$ to $e^{inb}$, plus you need to use absolute value signs around $sinleft(ntheta_0right)$.
$endgroup$
– John Omielan
Jan 31 at 3:46
$begingroup$
I made a mistake with my comment above. It should have gone to the OP, i.e., Jungleshrimp, instead.
$endgroup$
– John Omielan
Jan 31 at 5:53
$begingroup$
@JulianMejia Please ignore my earlier comment as I meant it to go to the OP instead.
$endgroup$
– John Omielan
Jan 31 at 5:54
add a comment |
1
$begingroup$
The range of $theta_0$ doesn't seem correct. Just think about the case $n=1$ and $a,b$ very close to $pi/2$. In that case LHS is close to zero but right hand side would be very close to $1$ if $theta_0in[a,b]$. With few computations you get that $|e^{-ina}-e^{-inb}|=2|sin(n(a-b)/2)|$. So, depending on the sign of sine you can choose $theta_0=pm (a-b)/2$.
$endgroup$
– Julian Mejia
Jan 31 at 2:45
$begingroup$
@JulianMejia As I state in my answer, consider something like $a = 2$ and $b = 3$. In this case, $theta_0 = pm frac{1}{2}$ would not meet the requirement of $theta_0 in left[a,bright]$.
$endgroup$
– John Omielan
Jan 31 at 3:22
$begingroup$
@JulianMejia As I show in my answer, I believe that you need to change something, like $e^{-inb}$ to $e^{inb}$, plus you need to use absolute value signs around $sinleft(ntheta_0right)$.
$endgroup$
– John Omielan
Jan 31 at 3:46
$begingroup$
I made a mistake with my comment above. It should have gone to the OP, i.e., Jungleshrimp, instead.
$endgroup$
– John Omielan
Jan 31 at 5:53
$begingroup$
@JulianMejia Please ignore my earlier comment as I meant it to go to the OP instead.
$endgroup$
– John Omielan
Jan 31 at 5:54
1
1
$begingroup$
The range of $theta_0$ doesn't seem correct. Just think about the case $n=1$ and $a,b$ very close to $pi/2$. In that case LHS is close to zero but right hand side would be very close to $1$ if $theta_0in[a,b]$. With few computations you get that $|e^{-ina}-e^{-inb}|=2|sin(n(a-b)/2)|$. So, depending on the sign of sine you can choose $theta_0=pm (a-b)/2$.
$endgroup$
– Julian Mejia
Jan 31 at 2:45
$begingroup$
The range of $theta_0$ doesn't seem correct. Just think about the case $n=1$ and $a,b$ very close to $pi/2$. In that case LHS is close to zero but right hand side would be very close to $1$ if $theta_0in[a,b]$. With few computations you get that $|e^{-ina}-e^{-inb}|=2|sin(n(a-b)/2)|$. So, depending on the sign of sine you can choose $theta_0=pm (a-b)/2$.
$endgroup$
– Julian Mejia
Jan 31 at 2:45
$begingroup$
@JulianMejia As I state in my answer, consider something like $a = 2$ and $b = 3$. In this case, $theta_0 = pm frac{1}{2}$ would not meet the requirement of $theta_0 in left[a,bright]$.
$endgroup$
– John Omielan
Jan 31 at 3:22
$begingroup$
@JulianMejia As I state in my answer, consider something like $a = 2$ and $b = 3$. In this case, $theta_0 = pm frac{1}{2}$ would not meet the requirement of $theta_0 in left[a,bright]$.
$endgroup$
– John Omielan
Jan 31 at 3:22
$begingroup$
@JulianMejia As I show in my answer, I believe that you need to change something, like $e^{-inb}$ to $e^{inb}$, plus you need to use absolute value signs around $sinleft(ntheta_0right)$.
$endgroup$
– John Omielan
Jan 31 at 3:46
$begingroup$
@JulianMejia As I show in my answer, I believe that you need to change something, like $e^{-inb}$ to $e^{inb}$, plus you need to use absolute value signs around $sinleft(ntheta_0right)$.
$endgroup$
– John Omielan
Jan 31 at 3:46
$begingroup$
I made a mistake with my comment above. It should have gone to the OP, i.e., Jungleshrimp, instead.
$endgroup$
– John Omielan
Jan 31 at 5:53
$begingroup$
I made a mistake with my comment above. It should have gone to the OP, i.e., Jungleshrimp, instead.
$endgroup$
– John Omielan
Jan 31 at 5:53
$begingroup$
@JulianMejia Please ignore my earlier comment as I meant it to go to the OP instead.
$endgroup$
– John Omielan
Jan 31 at 5:54
$begingroup$
@JulianMejia Please ignore my earlier comment as I meant it to go to the OP instead.
$endgroup$
– John Omielan
Jan 31 at 5:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have
begin{align}
|e^{-ina}-e^{-inb}|^2
&=(cos (-na) - cos (-nb))^2+(sin (-nb)-sin (-na))^2\ \
&=2-2cos(na)cos(nb)-2sin(na)sin(nb)\ \
&=2-2cos n(a-b)\ \
&=2-2cos 2tfrac{n(a-b)}2\ \
&=2-2left(cos^2tfrac{n(a-b)}2-sin^2tfrac{n(a-b)}2right)\ \
&=2-2left(1-2sin^2tfrac{n(a-b)}2 right)\ \
&=4sin^2tfrac{n(a-b)}2.
end{align}
So
$$
|e^{-ina}-e^{-inb}|=2left|sintfrac{n(a-b)}2right|
$$
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
This is similar to the answer from Martin Argerami, but uses a different, possibly simpler, way to compute the result. However, at the end, I show the requirement of $theta_0 in left[a,bright]$ cannot be met in general, although I also show what I consider to be a likely typo and what the answer would be then.
Using Euler's formula, then gathering the real & imaginary parts together, and using $cosleft(-xright) = cosleft(xright)$ and $sinleft(-xright) = -sinleft(xright)$, we have
$$e^{-ina} - e^{-inb} = left(cosleft(naright) - cosleft(nbright)right) + ileft(sinleft(nbright) - sinleft(naright)right) tag{1}label{eq1}$$
Using the Sum and Difference Formulas for $sin$ and $cos$ to combine their differences into products, we have that
$$cosleft(naright) - cosleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)sinleft(frac{n}{2}left(b - aright)right) tag{2}label{eq2}$$
and
$$sinleft(nbright) - sinleft(naright) = 2sinleft(frac{n}{2}left(b - aright)right)cosleft(frac{n}{2}left(a + bright)right) tag{3}label{eq3}$$
Thus, for the magnitude, note there is a common factor of $2sinleft(frac{n}{2}left(b - aright)right)$ in both eqref{eq2} and eqref{eq3}, so it can be pulled out. Also, there is a $sin$ and $cos$ factor with the same argument of $frac{n}{2}left(a + bright)$, so the square root of the sum of their squares would be $1$. Thus,
$$leftlvert e^{-ina} - e^{-inb}rightrvert = 2leftlvert sinleft(nleft(frac{b - a}{2}right)right)rightrvert tag{4}label{eq4}$$
Note you need to use absolute value signs due to the magnitude being non-negative. However, there may not necessarily be any $theta_0 in left[a,bright]$ which works. For example, consider that $a = 2$ and $b = 3$. The result gives a $theta_0 = pm frac{b - a}{2} = pm frac{1}{2}$. As such, you cannot prove the question is true in general.
However, if you change $e^{-inb}$ to $e^{inb}$ instead, then it'll work, so it's likely a typo there. In this case,
$$e^{-ina} - e^{inb} = left(cosleft(naright) - cosleft(nbright)right) - ileft(sinleft(naright) + sinleft(nbright)right) tag{5}label{eq5}$$
This changes eqref{eq3} to
$$sinleft(naright) + sinleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)cosleft(frac{n}{2}left(b - aright)right) tag{6}label{eq6}$$
Using the similar arguments as before, along with using eqref{eq2}, we will now get
$$leftlvert e^{-ina} - e^{inb}rightrvert = 2leftlvert sinleft(nleft(frac{a + b}{2}right)right)rightrvert tag{7}label{eq7}$$
As such, in this case, $theta_0 = frac{a + b}{2}$ works as $theta_0 in left[a,bright]$.
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3094413%2fe-ina-e-inb-2-sinn-theta-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have
begin{align}
|e^{-ina}-e^{-inb}|^2
&=(cos (-na) - cos (-nb))^2+(sin (-nb)-sin (-na))^2\ \
&=2-2cos(na)cos(nb)-2sin(na)sin(nb)\ \
&=2-2cos n(a-b)\ \
&=2-2cos 2tfrac{n(a-b)}2\ \
&=2-2left(cos^2tfrac{n(a-b)}2-sin^2tfrac{n(a-b)}2right)\ \
&=2-2left(1-2sin^2tfrac{n(a-b)}2 right)\ \
&=4sin^2tfrac{n(a-b)}2.
end{align}
So
$$
|e^{-ina}-e^{-inb}|=2left|sintfrac{n(a-b)}2right|
$$
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
You have
begin{align}
|e^{-ina}-e^{-inb}|^2
&=(cos (-na) - cos (-nb))^2+(sin (-nb)-sin (-na))^2\ \
&=2-2cos(na)cos(nb)-2sin(na)sin(nb)\ \
&=2-2cos n(a-b)\ \
&=2-2cos 2tfrac{n(a-b)}2\ \
&=2-2left(cos^2tfrac{n(a-b)}2-sin^2tfrac{n(a-b)}2right)\ \
&=2-2left(1-2sin^2tfrac{n(a-b)}2 right)\ \
&=4sin^2tfrac{n(a-b)}2.
end{align}
So
$$
|e^{-ina}-e^{-inb}|=2left|sintfrac{n(a-b)}2right|
$$
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
You have
begin{align}
|e^{-ina}-e^{-inb}|^2
&=(cos (-na) - cos (-nb))^2+(sin (-nb)-sin (-na))^2\ \
&=2-2cos(na)cos(nb)-2sin(na)sin(nb)\ \
&=2-2cos n(a-b)\ \
&=2-2cos 2tfrac{n(a-b)}2\ \
&=2-2left(cos^2tfrac{n(a-b)}2-sin^2tfrac{n(a-b)}2right)\ \
&=2-2left(1-2sin^2tfrac{n(a-b)}2 right)\ \
&=4sin^2tfrac{n(a-b)}2.
end{align}
So
$$
|e^{-ina}-e^{-inb}|=2left|sintfrac{n(a-b)}2right|
$$
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
You have
begin{align}
|e^{-ina}-e^{-inb}|^2
&=(cos (-na) - cos (-nb))^2+(sin (-nb)-sin (-na))^2\ \
&=2-2cos(na)cos(nb)-2sin(na)sin(nb)\ \
&=2-2cos n(a-b)\ \
&=2-2cos 2tfrac{n(a-b)}2\ \
&=2-2left(cos^2tfrac{n(a-b)}2-sin^2tfrac{n(a-b)}2right)\ \
&=2-2left(1-2sin^2tfrac{n(a-b)}2 right)\ \
&=4sin^2tfrac{n(a-b)}2.
end{align}
So
$$
|e^{-ina}-e^{-inb}|=2left|sintfrac{n(a-b)}2right|
$$
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
edited Jan 31 at 2:40
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 2:30
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
$begingroup$
This is similar to the answer from Martin Argerami, but uses a different, possibly simpler, way to compute the result. However, at the end, I show the requirement of $theta_0 in left[a,bright]$ cannot be met in general, although I also show what I consider to be a likely typo and what the answer would be then.
Using Euler's formula, then gathering the real & imaginary parts together, and using $cosleft(-xright) = cosleft(xright)$ and $sinleft(-xright) = -sinleft(xright)$, we have
$$e^{-ina} - e^{-inb} = left(cosleft(naright) - cosleft(nbright)right) + ileft(sinleft(nbright) - sinleft(naright)right) tag{1}label{eq1}$$
Using the Sum and Difference Formulas for $sin$ and $cos$ to combine their differences into products, we have that
$$cosleft(naright) - cosleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)sinleft(frac{n}{2}left(b - aright)right) tag{2}label{eq2}$$
and
$$sinleft(nbright) - sinleft(naright) = 2sinleft(frac{n}{2}left(b - aright)right)cosleft(frac{n}{2}left(a + bright)right) tag{3}label{eq3}$$
Thus, for the magnitude, note there is a common factor of $2sinleft(frac{n}{2}left(b - aright)right)$ in both eqref{eq2} and eqref{eq3}, so it can be pulled out. Also, there is a $sin$ and $cos$ factor with the same argument of $frac{n}{2}left(a + bright)$, so the square root of the sum of their squares would be $1$. Thus,
$$leftlvert e^{-ina} - e^{-inb}rightrvert = 2leftlvert sinleft(nleft(frac{b - a}{2}right)right)rightrvert tag{4}label{eq4}$$
Note you need to use absolute value signs due to the magnitude being non-negative. However, there may not necessarily be any $theta_0 in left[a,bright]$ which works. For example, consider that $a = 2$ and $b = 3$. The result gives a $theta_0 = pm frac{b - a}{2} = pm frac{1}{2}$. As such, you cannot prove the question is true in general.
However, if you change $e^{-inb}$ to $e^{inb}$ instead, then it'll work, so it's likely a typo there. In this case,
$$e^{-ina} - e^{inb} = left(cosleft(naright) - cosleft(nbright)right) - ileft(sinleft(naright) + sinleft(nbright)right) tag{5}label{eq5}$$
This changes eqref{eq3} to
$$sinleft(naright) + sinleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)cosleft(frac{n}{2}left(b - aright)right) tag{6}label{eq6}$$
Using the similar arguments as before, along with using eqref{eq2}, we will now get
$$leftlvert e^{-ina} - e^{inb}rightrvert = 2leftlvert sinleft(nleft(frac{a + b}{2}right)right)rightrvert tag{7}label{eq7}$$
As such, in this case, $theta_0 = frac{a + b}{2}$ works as $theta_0 in left[a,bright]$.
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
This is similar to the answer from Martin Argerami, but uses a different, possibly simpler, way to compute the result. However, at the end, I show the requirement of $theta_0 in left[a,bright]$ cannot be met in general, although I also show what I consider to be a likely typo and what the answer would be then.
Using Euler's formula, then gathering the real & imaginary parts together, and using $cosleft(-xright) = cosleft(xright)$ and $sinleft(-xright) = -sinleft(xright)$, we have
$$e^{-ina} - e^{-inb} = left(cosleft(naright) - cosleft(nbright)right) + ileft(sinleft(nbright) - sinleft(naright)right) tag{1}label{eq1}$$
Using the Sum and Difference Formulas for $sin$ and $cos$ to combine their differences into products, we have that
$$cosleft(naright) - cosleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)sinleft(frac{n}{2}left(b - aright)right) tag{2}label{eq2}$$
and
$$sinleft(nbright) - sinleft(naright) = 2sinleft(frac{n}{2}left(b - aright)right)cosleft(frac{n}{2}left(a + bright)right) tag{3}label{eq3}$$
Thus, for the magnitude, note there is a common factor of $2sinleft(frac{n}{2}left(b - aright)right)$ in both eqref{eq2} and eqref{eq3}, so it can be pulled out. Also, there is a $sin$ and $cos$ factor with the same argument of $frac{n}{2}left(a + bright)$, so the square root of the sum of their squares would be $1$. Thus,
$$leftlvert e^{-ina} - e^{-inb}rightrvert = 2leftlvert sinleft(nleft(frac{b - a}{2}right)right)rightrvert tag{4}label{eq4}$$
Note you need to use absolute value signs due to the magnitude being non-negative. However, there may not necessarily be any $theta_0 in left[a,bright]$ which works. For example, consider that $a = 2$ and $b = 3$. The result gives a $theta_0 = pm frac{b - a}{2} = pm frac{1}{2}$. As such, you cannot prove the question is true in general.
However, if you change $e^{-inb}$ to $e^{inb}$ instead, then it'll work, so it's likely a typo there. In this case,
$$e^{-ina} - e^{inb} = left(cosleft(naright) - cosleft(nbright)right) - ileft(sinleft(naright) + sinleft(nbright)right) tag{5}label{eq5}$$
This changes eqref{eq3} to
$$sinleft(naright) + sinleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)cosleft(frac{n}{2}left(b - aright)right) tag{6}label{eq6}$$
Using the similar arguments as before, along with using eqref{eq2}, we will now get
$$leftlvert e^{-ina} - e^{inb}rightrvert = 2leftlvert sinleft(nleft(frac{a + b}{2}right)right)rightrvert tag{7}label{eq7}$$
As such, in this case, $theta_0 = frac{a + b}{2}$ works as $theta_0 in left[a,bright]$.
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
This is similar to the answer from Martin Argerami, but uses a different, possibly simpler, way to compute the result. However, at the end, I show the requirement of $theta_0 in left[a,bright]$ cannot be met in general, although I also show what I consider to be a likely typo and what the answer would be then.
Using Euler's formula, then gathering the real & imaginary parts together, and using $cosleft(-xright) = cosleft(xright)$ and $sinleft(-xright) = -sinleft(xright)$, we have
$$e^{-ina} - e^{-inb} = left(cosleft(naright) - cosleft(nbright)right) + ileft(sinleft(nbright) - sinleft(naright)right) tag{1}label{eq1}$$
Using the Sum and Difference Formulas for $sin$ and $cos$ to combine their differences into products, we have that
$$cosleft(naright) - cosleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)sinleft(frac{n}{2}left(b - aright)right) tag{2}label{eq2}$$
and
$$sinleft(nbright) - sinleft(naright) = 2sinleft(frac{n}{2}left(b - aright)right)cosleft(frac{n}{2}left(a + bright)right) tag{3}label{eq3}$$
Thus, for the magnitude, note there is a common factor of $2sinleft(frac{n}{2}left(b - aright)right)$ in both eqref{eq2} and eqref{eq3}, so it can be pulled out. Also, there is a $sin$ and $cos$ factor with the same argument of $frac{n}{2}left(a + bright)$, so the square root of the sum of their squares would be $1$. Thus,
$$leftlvert e^{-ina} - e^{-inb}rightrvert = 2leftlvert sinleft(nleft(frac{b - a}{2}right)right)rightrvert tag{4}label{eq4}$$
Note you need to use absolute value signs due to the magnitude being non-negative. However, there may not necessarily be any $theta_0 in left[a,bright]$ which works. For example, consider that $a = 2$ and $b = 3$. The result gives a $theta_0 = pm frac{b - a}{2} = pm frac{1}{2}$. As such, you cannot prove the question is true in general.
However, if you change $e^{-inb}$ to $e^{inb}$ instead, then it'll work, so it's likely a typo there. In this case,
$$e^{-ina} - e^{inb} = left(cosleft(naright) - cosleft(nbright)right) - ileft(sinleft(naright) + sinleft(nbright)right) tag{5}label{eq5}$$
This changes eqref{eq3} to
$$sinleft(naright) + sinleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)cosleft(frac{n}{2}left(b - aright)right) tag{6}label{eq6}$$
Using the similar arguments as before, along with using eqref{eq2}, we will now get
$$leftlvert e^{-ina} - e^{inb}rightrvert = 2leftlvert sinleft(nleft(frac{a + b}{2}right)right)rightrvert tag{7}label{eq7}$$
As such, in this case, $theta_0 = frac{a + b}{2}$ works as $theta_0 in left[a,bright]$.
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
$endgroup$
This is similar to the answer from Martin Argerami, but uses a different, possibly simpler, way to compute the result. However, at the end, I show the requirement of $theta_0 in left[a,bright]$ cannot be met in general, although I also show what I consider to be a likely typo and what the answer would be then.
Using Euler's formula, then gathering the real & imaginary parts together, and using $cosleft(-xright) = cosleft(xright)$ and $sinleft(-xright) = -sinleft(xright)$, we have
$$e^{-ina} - e^{-inb} = left(cosleft(naright) - cosleft(nbright)right) + ileft(sinleft(nbright) - sinleft(naright)right) tag{1}label{eq1}$$
Using the Sum and Difference Formulas for $sin$ and $cos$ to combine their differences into products, we have that
$$cosleft(naright) - cosleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)sinleft(frac{n}{2}left(b - aright)right) tag{2}label{eq2}$$
and
$$sinleft(nbright) - sinleft(naright) = 2sinleft(frac{n}{2}left(b - aright)right)cosleft(frac{n}{2}left(a + bright)right) tag{3}label{eq3}$$
Thus, for the magnitude, note there is a common factor of $2sinleft(frac{n}{2}left(b - aright)right)$ in both eqref{eq2} and eqref{eq3}, so it can be pulled out. Also, there is a $sin$ and $cos$ factor with the same argument of $frac{n}{2}left(a + bright)$, so the square root of the sum of their squares would be $1$. Thus,
$$leftlvert e^{-ina} - e^{-inb}rightrvert = 2leftlvert sinleft(nleft(frac{b - a}{2}right)right)rightrvert tag{4}label{eq4}$$
Note you need to use absolute value signs due to the magnitude being non-negative. However, there may not necessarily be any $theta_0 in left[a,bright]$ which works. For example, consider that $a = 2$ and $b = 3$. The result gives a $theta_0 = pm frac{b - a}{2} = pm frac{1}{2}$. As such, you cannot prove the question is true in general.
However, if you change $e^{-inb}$ to $e^{inb}$ instead, then it'll work, so it's likely a typo there. In this case,
$$e^{-ina} - e^{inb} = left(cosleft(naright) - cosleft(nbright)right) - ileft(sinleft(naright) + sinleft(nbright)right) tag{5}label{eq5}$$
This changes eqref{eq3} to
$$sinleft(naright) + sinleft(nbright) = 2sinleft(frac{n}{2}left(a + bright)right)cosleft(frac{n}{2}left(b - aright)right) tag{6}label{eq6}$$
Using the similar arguments as before, along with using eqref{eq2}, we will now get
$$leftlvert e^{-ina} - e^{inb}rightrvert = 2leftlvert sinleft(nleft(frac{a + b}{2}right)right)rightrvert tag{7}label{eq7}$$
As such, in this case, $theta_0 = frac{a + b}{2}$ works as $theta_0 in left[a,bright]$.
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
edited Jan 31 at 3:56
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
answered Jan 31 at 2:58
John OmielanJohn Omielan
4,6312215
4,6312215
add a comment |
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.
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%2f3094413%2fe-ina-e-inb-2-sinn-theta-0%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
1
$begingroup$
The range of $theta_0$ doesn't seem correct. Just think about the case $n=1$ and $a,b$ very close to $pi/2$. In that case LHS is close to zero but right hand side would be very close to $1$ if $theta_0in[a,b]$. With few computations you get that $|e^{-ina}-e^{-inb}|=2|sin(n(a-b)/2)|$. So, depending on the sign of sine you can choose $theta_0=pm (a-b)/2$.
$endgroup$
– Julian Mejia
Jan 31 at 2:45
$begingroup$
@JulianMejia As I state in my answer, consider something like $a = 2$ and $b = 3$. In this case, $theta_0 = pm frac{1}{2}$ would not meet the requirement of $theta_0 in left[a,bright]$.
$endgroup$
– John Omielan
Jan 31 at 3:22
$begingroup$
@JulianMejia As I show in my answer, I believe that you need to change something, like $e^{-inb}$ to $e^{inb}$, plus you need to use absolute value signs around $sinleft(ntheta_0right)$.
$endgroup$
– John Omielan
Jan 31 at 3:46
$begingroup$
I made a mistake with my comment above. It should have gone to the OP, i.e., Jungleshrimp, instead.
$endgroup$
– John Omielan
Jan 31 at 5:53
$begingroup$
@JulianMejia Please ignore my earlier comment as I meant it to go to the OP instead.
$endgroup$
– John Omielan
Jan 31 at 5:54