The value of k such that two vectors are orthogonal in integral product spaces












1












$begingroup$


For $f,g in C[0,1]$ we define inner product space $langle f,g rangle =int_0^1 f(x)g(x) dx$. Find the value of $k$ such that $f(x)=sin(kx)$ and $g(x)=cos(7x)$ are orthogonal in this inner product space.



I have tried to solve the integral for the given $f(x)$ and $g(x)$ function, and got this result:
$$k-kcos k cos 7-28sin k sin 7 = 0$$



I don't know how to find the value of $k$. I hope you can help me.






linear-algebra inner-product-space







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  • $begingroup$
    Would you give more details of how you found the integral of $f(x)g(x)$ on $[0,1]$? I'd expect the product of the two trig functions to be written as a sum.
    $endgroup$
    – hardmath
    Jan 2 at 4:16






  • 1




    $begingroup$
    I write $sin(kx) cos(7x)$ as $frac{1}{2} sin(k+7)x + frac{1}{2} sin(k-7)x$ first.
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 4:28
















1












$begingroup$


For $f,g in C[0,1]$ we define inner product space $langle f,g rangle =int_0^1 f(x)g(x) dx$. Find the value of $k$ such that $f(x)=sin(kx)$ and $g(x)=cos(7x)$ are orthogonal in this inner product space.



I have tried to solve the integral for the given $f(x)$ and $g(x)$ function, and got this result:
$$k-kcos k cos 7-28sin k sin 7 = 0$$



I don't know how to find the value of $k$. I hope you can help me.






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$












  • $begingroup$
    Would you give more details of how you found the integral of $f(x)g(x)$ on $[0,1]$? I'd expect the product of the two trig functions to be written as a sum.
    $endgroup$
    – hardmath
    Jan 2 at 4:16






  • 1




    $begingroup$
    I write $sin(kx) cos(7x)$ as $frac{1}{2} sin(k+7)x + frac{1}{2} sin(k-7)x$ first.
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 4:28














1












1








1





$begingroup$


For $f,g in C[0,1]$ we define inner product space $langle f,g rangle =int_0^1 f(x)g(x) dx$. Find the value of $k$ such that $f(x)=sin(kx)$ and $g(x)=cos(7x)$ are orthogonal in this inner product space.



I have tried to solve the integral for the given $f(x)$ and $g(x)$ function, and got this result:
$$k-kcos k cos 7-28sin k sin 7 = 0$$



I don't know how to find the value of $k$. I hope you can help me.






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$




For $f,g in C[0,1]$ we define inner product space $langle f,g rangle =int_0^1 f(x)g(x) dx$. Find the value of $k$ such that $f(x)=sin(kx)$ and $g(x)=cos(7x)$ are orthogonal in this inner product space.



I have tried to solve the integral for the given $f(x)$ and $g(x)$ function, and got this result:
$$k-kcos k cos 7-28sin k sin 7 = 0$$



I don't know how to find the value of $k$. I hope you can help me.






linear-algebra inner-product-space




linear-algebra inner-product-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 5:39









zipirovich

11.4k11731




11.4k11731










asked Jan 2 at 3:55









Agung Izzul HaqAgung Izzul Haq

563




563












  • $begingroup$
    Would you give more details of how you found the integral of $f(x)g(x)$ on $[0,1]$? I'd expect the product of the two trig functions to be written as a sum.
    $endgroup$
    – hardmath
    Jan 2 at 4:16






  • 1




    $begingroup$
    I write $sin(kx) cos(7x)$ as $frac{1}{2} sin(k+7)x + frac{1}{2} sin(k-7)x$ first.
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 4:28


















  • $begingroup$
    Would you give more details of how you found the integral of $f(x)g(x)$ on $[0,1]$? I'd expect the product of the two trig functions to be written as a sum.
    $endgroup$
    – hardmath
    Jan 2 at 4:16






  • 1




    $begingroup$
    I write $sin(kx) cos(7x)$ as $frac{1}{2} sin(k+7)x + frac{1}{2} sin(k-7)x$ first.
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 4:28
















$begingroup$
Would you give more details of how you found the integral of $f(x)g(x)$ on $[0,1]$? I'd expect the product of the two trig functions to be written as a sum.
$endgroup$
– hardmath
Jan 2 at 4:16




$begingroup$
Would you give more details of how you found the integral of $f(x)g(x)$ on $[0,1]$? I'd expect the product of the two trig functions to be written as a sum.
$endgroup$
– hardmath
Jan 2 at 4:16




1




1




$begingroup$
I write $sin(kx) cos(7x)$ as $frac{1}{2} sin(k+7)x + frac{1}{2} sin(k-7)x$ first.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:28




$begingroup$
I write $sin(kx) cos(7x)$ as $frac{1}{2} sin(k+7)x + frac{1}{2} sin(k-7)x$ first.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

First of all, while integrating at one point you had to divide by both $kpm7$ when finding antiderivatives of $sin(kpm7)x$. Such division is undefined when $k=pm7$, so these two cases have to be considered separately. Integrating
$$int_0^1sin(pm7x)cos(7x),dx=pmint_0^1sin(7x)cos(7x),dx,$$
you will see that these integrals are not equal to zero.



Second, you made a slight mistake somewhere in your calculations, because the coefficient of $color{red}{28}$ is incorrect. The correct equation should be
$$k-kcos kcos7-color{magenta}{7}sin ksin7=0.$$
This is a transcendental equation, and I seriously doubt it can be solved for exact answers, where by "solving" I mean performing a sequence of steps to get to the roots of the equation. One root is easy to guess, though: $k=0$ clearly works. Are there any more?



Graphing the left-hand side or asking a numerical solver to find the roots, we'll see that there are no more integer roots, but four more real roots. For example, Mathematica input



NSolve[k - k Cos[k] Cos[7] - 7 Sin[k] Sin[7] == 0, k, Reals]



produces (remember that $k=pm7$ have been discarded for our purposes)



{{k -> -7.}, {k -> -6.891}, {k -> -2.29558}, {k -> 0.}, {k -> 2.29558}, {k -> 6.891}, {k -> 7.}}



but I'm honestly not sure how to prove manually that these are the only roots.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 8:55










  • $begingroup$
    @AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
    $endgroup$
    – zipirovich
    Jan 2 at 15:15












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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

First of all, while integrating at one point you had to divide by both $kpm7$ when finding antiderivatives of $sin(kpm7)x$. Such division is undefined when $k=pm7$, so these two cases have to be considered separately. Integrating
$$int_0^1sin(pm7x)cos(7x),dx=pmint_0^1sin(7x)cos(7x),dx,$$
you will see that these integrals are not equal to zero.



Second, you made a slight mistake somewhere in your calculations, because the coefficient of $color{red}{28}$ is incorrect. The correct equation should be
$$k-kcos kcos7-color{magenta}{7}sin ksin7=0.$$
This is a transcendental equation, and I seriously doubt it can be solved for exact answers, where by "solving" I mean performing a sequence of steps to get to the roots of the equation. One root is easy to guess, though: $k=0$ clearly works. Are there any more?



Graphing the left-hand side or asking a numerical solver to find the roots, we'll see that there are no more integer roots, but four more real roots. For example, Mathematica input



NSolve[k - k Cos[k] Cos[7] - 7 Sin[k] Sin[7] == 0, k, Reals]



produces (remember that $k=pm7$ have been discarded for our purposes)



{{k -> -7.}, {k -> -6.891}, {k -> -2.29558}, {k -> 0.}, {k -> 2.29558}, {k -> 6.891}, {k -> 7.}}



but I'm honestly not sure how to prove manually that these are the only roots.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 8:55










  • $begingroup$
    @AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
    $endgroup$
    – zipirovich
    Jan 2 at 15:15
















1












$begingroup$

First of all, while integrating at one point you had to divide by both $kpm7$ when finding antiderivatives of $sin(kpm7)x$. Such division is undefined when $k=pm7$, so these two cases have to be considered separately. Integrating
$$int_0^1sin(pm7x)cos(7x),dx=pmint_0^1sin(7x)cos(7x),dx,$$
you will see that these integrals are not equal to zero.



Second, you made a slight mistake somewhere in your calculations, because the coefficient of $color{red}{28}$ is incorrect. The correct equation should be
$$k-kcos kcos7-color{magenta}{7}sin ksin7=0.$$
This is a transcendental equation, and I seriously doubt it can be solved for exact answers, where by "solving" I mean performing a sequence of steps to get to the roots of the equation. One root is easy to guess, though: $k=0$ clearly works. Are there any more?



Graphing the left-hand side or asking a numerical solver to find the roots, we'll see that there are no more integer roots, but four more real roots. For example, Mathematica input



NSolve[k - k Cos[k] Cos[7] - 7 Sin[k] Sin[7] == 0, k, Reals]



produces (remember that $k=pm7$ have been discarded for our purposes)



{{k -> -7.}, {k -> -6.891}, {k -> -2.29558}, {k -> 0.}, {k -> 2.29558}, {k -> 6.891}, {k -> 7.}}



but I'm honestly not sure how to prove manually that these are the only roots.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 8:55










  • $begingroup$
    @AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
    $endgroup$
    – zipirovich
    Jan 2 at 15:15














1












1








1





$begingroup$

First of all, while integrating at one point you had to divide by both $kpm7$ when finding antiderivatives of $sin(kpm7)x$. Such division is undefined when $k=pm7$, so these two cases have to be considered separately. Integrating
$$int_0^1sin(pm7x)cos(7x),dx=pmint_0^1sin(7x)cos(7x),dx,$$
you will see that these integrals are not equal to zero.



Second, you made a slight mistake somewhere in your calculations, because the coefficient of $color{red}{28}$ is incorrect. The correct equation should be
$$k-kcos kcos7-color{magenta}{7}sin ksin7=0.$$
This is a transcendental equation, and I seriously doubt it can be solved for exact answers, where by "solving" I mean performing a sequence of steps to get to the roots of the equation. One root is easy to guess, though: $k=0$ clearly works. Are there any more?



Graphing the left-hand side or asking a numerical solver to find the roots, we'll see that there are no more integer roots, but four more real roots. For example, Mathematica input



NSolve[k - k Cos[k] Cos[7] - 7 Sin[k] Sin[7] == 0, k, Reals]



produces (remember that $k=pm7$ have been discarded for our purposes)



{{k -> -7.}, {k -> -6.891}, {k -> -2.29558}, {k -> 0.}, {k -> 2.29558}, {k -> 6.891}, {k -> 7.}}



but I'm honestly not sure how to prove manually that these are the only roots.






share|cite|improve this answer









$endgroup$



First of all, while integrating at one point you had to divide by both $kpm7$ when finding antiderivatives of $sin(kpm7)x$. Such division is undefined when $k=pm7$, so these two cases have to be considered separately. Integrating
$$int_0^1sin(pm7x)cos(7x),dx=pmint_0^1sin(7x)cos(7x),dx,$$
you will see that these integrals are not equal to zero.



Second, you made a slight mistake somewhere in your calculations, because the coefficient of $color{red}{28}$ is incorrect. The correct equation should be
$$k-kcos kcos7-color{magenta}{7}sin ksin7=0.$$
This is a transcendental equation, and I seriously doubt it can be solved for exact answers, where by "solving" I mean performing a sequence of steps to get to the roots of the equation. One root is easy to guess, though: $k=0$ clearly works. Are there any more?



Graphing the left-hand side or asking a numerical solver to find the roots, we'll see that there are no more integer roots, but four more real roots. For example, Mathematica input



NSolve[k - k Cos[k] Cos[7] - 7 Sin[k] Sin[7] == 0, k, Reals]



produces (remember that $k=pm7$ have been discarded for our purposes)



{{k -> -7.}, {k -> -6.891}, {k -> -2.29558}, {k -> 0.}, {k -> 2.29558}, {k -> 6.891}, {k -> 7.}}



but I'm honestly not sure how to prove manually that these are the only roots.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 6:01









zipirovichzipirovich

11.4k11731




11.4k11731








  • 1




    $begingroup$
    My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 8:55










  • $begingroup$
    @AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
    $endgroup$
    – zipirovich
    Jan 2 at 15:15














  • 1




    $begingroup$
    My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
    $endgroup$
    – Agung Izzul Haq
    Jan 2 at 8:55










  • $begingroup$
    @AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
    $endgroup$
    – zipirovich
    Jan 2 at 15:15








1




1




$begingroup$
My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
$endgroup$
– Agung Izzul Haq
Jan 2 at 8:55




$begingroup$
My mistake. Thanks for answering my question. But, is there another way to solve it, without using a software?
$endgroup$
– Agung Izzul Haq
Jan 2 at 8:55












$begingroup$
@AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
$endgroup$
– zipirovich
Jan 2 at 15:15




$begingroup$
@AgungIzzulHaq: As I said in the end of my answer, I don't know for sure, but I seriously doubt that there's a way to solve it analytically.
$endgroup$
– zipirovich
Jan 2 at 15:15


















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