To specify $a,b,k$ in $acos(kx)+b$
$begingroup$
If we have known that $f(x)=acos(kx)+b$ for some $
f:(0,infty)to mathbb C$,
is there a method of choosing finite $(x,f(x))$ to know what $a,b,k$ are?
functions trigonometry
$endgroup$
|
show 2 more comments
$begingroup$
If we have known that $f(x)=acos(kx)+b$ for some $
f:(0,infty)to mathbb C$,
is there a method of choosing finite $(x,f(x))$ to know what $a,b,k$ are?
functions trigonometry
$endgroup$
$begingroup$
How about k in the formula?
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:04
1
$begingroup$
Not quite. You can always change $k$ to $-k$.
$endgroup$
– Robert Israel
Jan 7 at 13:04
$begingroup$
@RobertIsrael You're absolutely right.
$endgroup$
– Yanko
Jan 7 at 13:05
$begingroup$
what I mean is : if there is no k, because you can determine cosx, a and b are easy to be determind. But in this case, cos x cannot be determined because of the existence of k.
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:06
$begingroup$
@SteveCheng鄭宗弘 Looks like I was wrong. As Robert says, the function $f(x)=acos(kx)+b$ equals to the function $g(x)=acos ((-k)x)+b$ and so you can never determine $k$.
$endgroup$
– Yanko
Jan 7 at 13:06
|
show 2 more comments
$begingroup$
If we have known that $f(x)=acos(kx)+b$ for some $
f:(0,infty)to mathbb C$,
is there a method of choosing finite $(x,f(x))$ to know what $a,b,k$ are?
functions trigonometry
$endgroup$
If we have known that $f(x)=acos(kx)+b$ for some $
f:(0,infty)to mathbb C$,
is there a method of choosing finite $(x,f(x))$ to know what $a,b,k$ are?
functions trigonometry
functions trigonometry
edited Jan 7 at 13:03
rtybase
11.6k31534
11.6k31534
asked Jan 7 at 13:00
Steve Cheng 鄭宗弘Steve Cheng 鄭宗弘
628
628
$begingroup$
How about k in the formula?
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:04
1
$begingroup$
Not quite. You can always change $k$ to $-k$.
$endgroup$
– Robert Israel
Jan 7 at 13:04
$begingroup$
@RobertIsrael You're absolutely right.
$endgroup$
– Yanko
Jan 7 at 13:05
$begingroup$
what I mean is : if there is no k, because you can determine cosx, a and b are easy to be determind. But in this case, cos x cannot be determined because of the existence of k.
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:06
$begingroup$
@SteveCheng鄭宗弘 Looks like I was wrong. As Robert says, the function $f(x)=acos(kx)+b$ equals to the function $g(x)=acos ((-k)x)+b$ and so you can never determine $k$.
$endgroup$
– Yanko
Jan 7 at 13:06
|
show 2 more comments
$begingroup$
How about k in the formula?
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:04
1
$begingroup$
Not quite. You can always change $k$ to $-k$.
$endgroup$
– Robert Israel
Jan 7 at 13:04
$begingroup$
@RobertIsrael You're absolutely right.
$endgroup$
– Yanko
Jan 7 at 13:05
$begingroup$
what I mean is : if there is no k, because you can determine cosx, a and b are easy to be determind. But in this case, cos x cannot be determined because of the existence of k.
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:06
$begingroup$
@SteveCheng鄭宗弘 Looks like I was wrong. As Robert says, the function $f(x)=acos(kx)+b$ equals to the function $g(x)=acos ((-k)x)+b$ and so you can never determine $k$.
$endgroup$
– Yanko
Jan 7 at 13:06
$begingroup$
How about k in the formula?
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:04
$begingroup$
How about k in the formula?
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:04
1
1
$begingroup$
Not quite. You can always change $k$ to $-k$.
$endgroup$
– Robert Israel
Jan 7 at 13:04
$begingroup$
Not quite. You can always change $k$ to $-k$.
$endgroup$
– Robert Israel
Jan 7 at 13:04
$begingroup$
@RobertIsrael You're absolutely right.
$endgroup$
– Yanko
Jan 7 at 13:05
$begingroup$
@RobertIsrael You're absolutely right.
$endgroup$
– Yanko
Jan 7 at 13:05
$begingroup$
what I mean is : if there is no k, because you can determine cosx, a and b are easy to be determind. But in this case, cos x cannot be determined because of the existence of k.
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:06
$begingroup$
what I mean is : if there is no k, because you can determine cosx, a and b are easy to be determind. But in this case, cos x cannot be determined because of the existence of k.
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:06
$begingroup$
@SteveCheng鄭宗弘 Looks like I was wrong. As Robert says, the function $f(x)=acos(kx)+b$ equals to the function $g(x)=acos ((-k)x)+b$ and so you can never determine $k$.
$endgroup$
– Yanko
Jan 7 at 13:06
$begingroup$
@SteveCheng鄭宗弘 Looks like I was wrong. As Robert says, the function $f(x)=acos(kx)+b$ equals to the function $g(x)=acos ((-k)x)+b$ and so you can never determine $k$.
$endgroup$
– Yanko
Jan 7 at 13:06
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Even if you require $k>0$ to avoid the sign ambiguity, there will be trouble. After eliminating $a$ and $b$ from the equations $f(x_1) = y_1$, $f(x_2) = y_2$, $f(x_3) = y_3$ you get
$$ (y_2 - y_3) cos(k x_1) + (y_3 - y_1) cos(k x_2) + (y_1 - y_2) cos(k x_3) = 0 $$
from which you want to determine $k$. However, the left side will usually be $0$ for infinitely many $k$. The exceptions are when one of the $x_i$, let's say $x_1$, is $0$, and $y_3-y_1$ and $y_1-y_2$ have the same sign, in which case the only way to satisfy the equation is for $cos(k x_1)$ and $cos(k x_2)$ to both be $1$, and therefore (if $x_1/x_2$ is irrational) $k = 0$.
Thus if the answer happens to be $k = 0$ you can choose the $x_i$ and $y_i$ to prove that, but if the correct $k$ is nonzero you can't determine it.
$endgroup$
add a comment |
Your Answer
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%2f3064980%2fto-specify-a-b-k-in-a-coskxb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Even if you require $k>0$ to avoid the sign ambiguity, there will be trouble. After eliminating $a$ and $b$ from the equations $f(x_1) = y_1$, $f(x_2) = y_2$, $f(x_3) = y_3$ you get
$$ (y_2 - y_3) cos(k x_1) + (y_3 - y_1) cos(k x_2) + (y_1 - y_2) cos(k x_3) = 0 $$
from which you want to determine $k$. However, the left side will usually be $0$ for infinitely many $k$. The exceptions are when one of the $x_i$, let's say $x_1$, is $0$, and $y_3-y_1$ and $y_1-y_2$ have the same sign, in which case the only way to satisfy the equation is for $cos(k x_1)$ and $cos(k x_2)$ to both be $1$, and therefore (if $x_1/x_2$ is irrational) $k = 0$.
Thus if the answer happens to be $k = 0$ you can choose the $x_i$ and $y_i$ to prove that, but if the correct $k$ is nonzero you can't determine it.
$endgroup$
add a comment |
$begingroup$
Even if you require $k>0$ to avoid the sign ambiguity, there will be trouble. After eliminating $a$ and $b$ from the equations $f(x_1) = y_1$, $f(x_2) = y_2$, $f(x_3) = y_3$ you get
$$ (y_2 - y_3) cos(k x_1) + (y_3 - y_1) cos(k x_2) + (y_1 - y_2) cos(k x_3) = 0 $$
from which you want to determine $k$. However, the left side will usually be $0$ for infinitely many $k$. The exceptions are when one of the $x_i$, let's say $x_1$, is $0$, and $y_3-y_1$ and $y_1-y_2$ have the same sign, in which case the only way to satisfy the equation is for $cos(k x_1)$ and $cos(k x_2)$ to both be $1$, and therefore (if $x_1/x_2$ is irrational) $k = 0$.
Thus if the answer happens to be $k = 0$ you can choose the $x_i$ and $y_i$ to prove that, but if the correct $k$ is nonzero you can't determine it.
$endgroup$
add a comment |
$begingroup$
Even if you require $k>0$ to avoid the sign ambiguity, there will be trouble. After eliminating $a$ and $b$ from the equations $f(x_1) = y_1$, $f(x_2) = y_2$, $f(x_3) = y_3$ you get
$$ (y_2 - y_3) cos(k x_1) + (y_3 - y_1) cos(k x_2) + (y_1 - y_2) cos(k x_3) = 0 $$
from which you want to determine $k$. However, the left side will usually be $0$ for infinitely many $k$. The exceptions are when one of the $x_i$, let's say $x_1$, is $0$, and $y_3-y_1$ and $y_1-y_2$ have the same sign, in which case the only way to satisfy the equation is for $cos(k x_1)$ and $cos(k x_2)$ to both be $1$, and therefore (if $x_1/x_2$ is irrational) $k = 0$.
Thus if the answer happens to be $k = 0$ you can choose the $x_i$ and $y_i$ to prove that, but if the correct $k$ is nonzero you can't determine it.
$endgroup$
Even if you require $k>0$ to avoid the sign ambiguity, there will be trouble. After eliminating $a$ and $b$ from the equations $f(x_1) = y_1$, $f(x_2) = y_2$, $f(x_3) = y_3$ you get
$$ (y_2 - y_3) cos(k x_1) + (y_3 - y_1) cos(k x_2) + (y_1 - y_2) cos(k x_3) = 0 $$
from which you want to determine $k$. However, the left side will usually be $0$ for infinitely many $k$. The exceptions are when one of the $x_i$, let's say $x_1$, is $0$, and $y_3-y_1$ and $y_1-y_2$ have the same sign, in which case the only way to satisfy the equation is for $cos(k x_1)$ and $cos(k x_2)$ to both be $1$, and therefore (if $x_1/x_2$ is irrational) $k = 0$.
Thus if the answer happens to be $k = 0$ you can choose the $x_i$ and $y_i$ to prove that, but if the correct $k$ is nonzero you can't determine it.
answered Jan 7 at 15:44
Robert IsraelRobert Israel
331k23221477
331k23221477
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%2f3064980%2fto-specify-a-b-k-in-a-coskxb%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
$begingroup$
How about k in the formula?
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:04
1
$begingroup$
Not quite. You can always change $k$ to $-k$.
$endgroup$
– Robert Israel
Jan 7 at 13:04
$begingroup$
@RobertIsrael You're absolutely right.
$endgroup$
– Yanko
Jan 7 at 13:05
$begingroup$
what I mean is : if there is no k, because you can determine cosx, a and b are easy to be determind. But in this case, cos x cannot be determined because of the existence of k.
$endgroup$
– Steve Cheng 鄭宗弘
Jan 7 at 13:06
$begingroup$
@SteveCheng鄭宗弘 Looks like I was wrong. As Robert says, the function $f(x)=acos(kx)+b$ equals to the function $g(x)=acos ((-k)x)+b$ and so you can never determine $k$.
$endgroup$
– Yanko
Jan 7 at 13:06