To specify $a,b,k$ in $acos(kx)+b$












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 13:03









rtybase

10.5k21533




10.5k21533










asked Jan 7 at 13:00









Steve Cheng 鄭宗弘Steve Cheng 鄭宗弘

407




407












  • $begingroup$
    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






  • 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










1 Answer
1






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$begingroup$

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.






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    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 15:44









        Robert IsraelRobert Israel

        319k23209459




        319k23209459






























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