Fourier Series: Is $cos(n pi)$ the same thing as $cos(-npi)$?
$begingroup$
For a Fourier series calculation, if you have bounds from $-1$ to $+1$ and you are plugging it into the function $cos(n pi)$, would it result in $0$? Thank you for any clarification.
ordinary-differential-equations
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
asked Dec 14 '18 at 0:28
JamesJames
1
$endgroup$
add a comment |
$begingroup$
For a Fourier series calculation, if you have bounds from $-1$ to $+1$ and you are plugging it into the function $cos(n pi)$, would it result in $0$? Thank you for any clarification.
ordinary-differential-equations
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
asked Dec 14 '18 at 0:28
JamesJames
1
$endgroup$
6
$begingroup$
Cosine is an even function.
$endgroup$
– T. Bongers
Dec 14 '18 at 0:28
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As Bongers said, $cos$ is an even function. Even more, we know that $$forall ninmathbb Z,cos(npi)=(-1)^n.$$
$endgroup$
– Scientifica
Dec 14 '18 at 0:33
add a comment |
$begingroup$
For a Fourier series calculation, if you have bounds from $-1$ to $+1$ and you are plugging it into the function $cos(n pi)$, would it result in $0$? Thank you for any clarification.
ordinary-differential-equations
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
asked Dec 14 '18 at 0:28
JamesJames
1
$endgroup$
For a Fourier series calculation, if you have bounds from $-1$ to $+1$ and you are plugging it into the function $cos(n pi)$, would it result in $0$? Thank you for any clarification.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
asked Dec 14 '18 at 0:28
JamesJames
1
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
asked Dec 14 '18 at 0:28
JamesJames
1
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
edited Dec 14 '18 at 0:36
Eevee Trainer
5,9471936
5,9471936
asked Dec 14 '18 at 0:28
JamesJames
1
asked Dec 14 '18 at 0:28
JamesJames
1
asked Dec 14 '18 at 0:28
JamesJames
1
1
6
$begingroup$
Cosine is an even function.
$endgroup$
– T. Bongers
Dec 14 '18 at 0:28
$begingroup$
As Bongers said, $cos$ is an even function. Even more, we know that $$forall ninmathbb Z,cos(npi)=(-1)^n.$$
$endgroup$
– Scientifica
Dec 14 '18 at 0:33
add a comment |
6
$begingroup$
Cosine is an even function.
$endgroup$
– T. Bongers
Dec 14 '18 at 0:28
$begingroup$
As Bongers said, $cos$ is an even function. Even more, we know that $$forall ninmathbb Z,cos(npi)=(-1)^n.$$
$endgroup$
– Scientifica
Dec 14 '18 at 0:33
6
6
$begingroup$
Cosine is an even function.
$endgroup$
– T. Bongers
Dec 14 '18 at 0:28
$begingroup$
Cosine is an even function.
$endgroup$
– T. Bongers
Dec 14 '18 at 0:28
$begingroup$
As Bongers said, $cos$ is an even function. Even more, we know that $$forall ninmathbb Z,cos(npi)=(-1)^n.$$
$endgroup$
– Scientifica
Dec 14 '18 at 0:33
$begingroup$
As Bongers said, $cos$ is an even function. Even more, we know that $$forall ninmathbb Z,cos(npi)=(-1)^n.$$
$endgroup$
– Scientifica
Dec 14 '18 at 0:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Cosine is an even function, i.e. $cos(-x) = cos(x)$.
Thus, as a result,
$$cos(-n pi) = cos(n pi)$$
The integrals of even functions $f$, further, can be given by
$$int_{-a}^a f(x)dx = 2 cdot int_0^a f(x)dx$$
If $f$ was odd (and thus $f(-x)=-f(x)$), as with the $sin(x)$ function, however, we'd have
$$int_{-a}^a f(x)dx = 0$$
Since $cos(x)$ is an even function, though, the first integral applies. (Not the second as you posit in your question.)
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
$endgroup$
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
add a comment |
Your Answer
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1 Answer
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$begingroup$
Cosine is an even function, i.e. $cos(-x) = cos(x)$.
Thus, as a result,
$$cos(-n pi) = cos(n pi)$$
The integrals of even functions $f$, further, can be given by
$$int_{-a}^a f(x)dx = 2 cdot int_0^a f(x)dx$$
If $f$ was odd (and thus $f(-x)=-f(x)$), as with the $sin(x)$ function, however, we'd have
$$int_{-a}^a f(x)dx = 0$$
Since $cos(x)$ is an even function, though, the first integral applies. (Not the second as you posit in your question.)
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
$endgroup$
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
add a comment |
$begingroup$
Cosine is an even function, i.e. $cos(-x) = cos(x)$.
Thus, as a result,
$$cos(-n pi) = cos(n pi)$$
The integrals of even functions $f$, further, can be given by
$$int_{-a}^a f(x)dx = 2 cdot int_0^a f(x)dx$$
If $f$ was odd (and thus $f(-x)=-f(x)$), as with the $sin(x)$ function, however, we'd have
$$int_{-a}^a f(x)dx = 0$$
Since $cos(x)$ is an even function, though, the first integral applies. (Not the second as you posit in your question.)
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
$endgroup$
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
add a comment |
$begingroup$
Cosine is an even function, i.e. $cos(-x) = cos(x)$.
Thus, as a result,
$$cos(-n pi) = cos(n pi)$$
The integrals of even functions $f$, further, can be given by
$$int_{-a}^a f(x)dx = 2 cdot int_0^a f(x)dx$$
If $f$ was odd (and thus $f(-x)=-f(x)$), as with the $sin(x)$ function, however, we'd have
$$int_{-a}^a f(x)dx = 0$$
Since $cos(x)$ is an even function, though, the first integral applies. (Not the second as you posit in your question.)
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
$endgroup$
Cosine is an even function, i.e. $cos(-x) = cos(x)$.
Thus, as a result,
$$cos(-n pi) = cos(n pi)$$
The integrals of even functions $f$, further, can be given by
$$int_{-a}^a f(x)dx = 2 cdot int_0^a f(x)dx$$
If $f$ was odd (and thus $f(-x)=-f(x)$), as with the $sin(x)$ function, however, we'd have
$$int_{-a}^a f(x)dx = 0$$
Since $cos(x)$ is an even function, though, the first integral applies. (Not the second as you posit in your question.)
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
answered Dec 14 '18 at 0:33
Eevee TrainerEevee Trainer
5,9471936
5,9471936
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
add a comment |
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
Thank you for the feedback. I have another followup question: when you are solving for a_(sub)n and b_(sub)n, is there a simple way to tell if one of them is equal to zero based on whether the function is even or odd? Or do you have to perform all the calculations? Thank you again
$endgroup$
– James
Dec 14 '18 at 1:23
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
I always perform all of the calculations just for its own sake. I guess if you're on a symmetric interval $(-a,a)$ and the function is odd, the cosine terms (the $a_n$) might disappear, and if the function is even then the sine terms (the $b_n$) would, but this is just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:12
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
There might be more conditions to it, though, so it's just a guess.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 2:13
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
$begingroup$
Okay thank you. I will just calculate it all out to be safe.
$endgroup$
– James
Dec 14 '18 at 2:16
add a comment |
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6
$begingroup$
Cosine is an even function.
$endgroup$
– T. Bongers
Dec 14 '18 at 0:28
$begingroup$
As Bongers said, $cos$ is an even function. Even more, we know that $$forall ninmathbb Z,cos(npi)=(-1)^n.$$
$endgroup$
– Scientifica
Dec 14 '18 at 0:33