Is function $f(x)=frac{text{cos}(x)}{x}$ significant?
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I am undertaking course in digital signal processing. I've learned during my undergraduate maths the sinc function i.e.
$$text{sinc}(x)=frac{text{sin}(x)}{x}$$
But only now I am actually doing stuff with this function and finding it relevant. Hence my question. Is the function
$$f(x)=frac{text{cos}(x)}{x}$$
also significant in some areas? Usually sines and cosines go hand to hand hence do we use an original name for it i.e. "cosc"? I looked for it but can't really find anything about it. No mention about it in Wiki site about sinc function either.
functions trigonometry
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add a comment |
$begingroup$
I am undertaking course in digital signal processing. I've learned during my undergraduate maths the sinc function i.e.
$$text{sinc}(x)=frac{text{sin}(x)}{x}$$
But only now I am actually doing stuff with this function and finding it relevant. Hence my question. Is the function
$$f(x)=frac{text{cos}(x)}{x}$$
also significant in some areas? Usually sines and cosines go hand to hand hence do we use an original name for it i.e. "cosc"? I looked for it but can't really find anything about it. No mention about it in Wiki site about sinc function either.
functions trigonometry
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4
$begingroup$
Unlike $text{sinc}$ this has a singularity.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 20:41
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One of the main reasons that the sinc function is so useful is because it's Fourier transform is the rectangular function $mathbb{1}_{big[-tfrac{1}{2},tfrac{1}{2}big]}$. It's a nice exercise to compute the Fourier transform of $frac{cos(x)}{x}$. You'll see that you obtain something different, and "less useful" in the digital signals environment.
$endgroup$
– Kolja
Jan 13 at 20:56
1
$begingroup$
I don't know of any special applications. The "mathematical" reason sinc is special is because sin(x)/x can be continuously extended to x=0, indeed differentiably extended. I guess it has applications in signal processing but I don't know what they are. The cosine "counterpart" to sin(x)/x is probably [1-cos(x)]/x which also can be continuously extended to x=0. However I don't know of any practical applications for that function either.
$endgroup$
– Ben W
Jan 13 at 20:57
add a comment |
$begingroup$
I am undertaking course in digital signal processing. I've learned during my undergraduate maths the sinc function i.e.
$$text{sinc}(x)=frac{text{sin}(x)}{x}$$
But only now I am actually doing stuff with this function and finding it relevant. Hence my question. Is the function
$$f(x)=frac{text{cos}(x)}{x}$$
also significant in some areas? Usually sines and cosines go hand to hand hence do we use an original name for it i.e. "cosc"? I looked for it but can't really find anything about it. No mention about it in Wiki site about sinc function either.
functions trigonometry
$endgroup$
I am undertaking course in digital signal processing. I've learned during my undergraduate maths the sinc function i.e.
$$text{sinc}(x)=frac{text{sin}(x)}{x}$$
But only now I am actually doing stuff with this function and finding it relevant. Hence my question. Is the function
$$f(x)=frac{text{cos}(x)}{x}$$
also significant in some areas? Usually sines and cosines go hand to hand hence do we use an original name for it i.e. "cosc"? I looked for it but can't really find anything about it. No mention about it in Wiki site about sinc function either.
functions trigonometry
functions trigonometry
asked Jan 13 at 20:39
Scavenger23Scavenger23
544319
544319
4
$begingroup$
Unlike $text{sinc}$ this has a singularity.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 20:41
$begingroup$
One of the main reasons that the sinc function is so useful is because it's Fourier transform is the rectangular function $mathbb{1}_{big[-tfrac{1}{2},tfrac{1}{2}big]}$. It's a nice exercise to compute the Fourier transform of $frac{cos(x)}{x}$. You'll see that you obtain something different, and "less useful" in the digital signals environment.
$endgroup$
– Kolja
Jan 13 at 20:56
1
$begingroup$
I don't know of any special applications. The "mathematical" reason sinc is special is because sin(x)/x can be continuously extended to x=0, indeed differentiably extended. I guess it has applications in signal processing but I don't know what they are. The cosine "counterpart" to sin(x)/x is probably [1-cos(x)]/x which also can be continuously extended to x=0. However I don't know of any practical applications for that function either.
$endgroup$
– Ben W
Jan 13 at 20:57
add a comment |
4
$begingroup$
Unlike $text{sinc}$ this has a singularity.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 20:41
$begingroup$
One of the main reasons that the sinc function is so useful is because it's Fourier transform is the rectangular function $mathbb{1}_{big[-tfrac{1}{2},tfrac{1}{2}big]}$. It's a nice exercise to compute the Fourier transform of $frac{cos(x)}{x}$. You'll see that you obtain something different, and "less useful" in the digital signals environment.
$endgroup$
– Kolja
Jan 13 at 20:56
1
$begingroup$
I don't know of any special applications. The "mathematical" reason sinc is special is because sin(x)/x can be continuously extended to x=0, indeed differentiably extended. I guess it has applications in signal processing but I don't know what they are. The cosine "counterpart" to sin(x)/x is probably [1-cos(x)]/x which also can be continuously extended to x=0. However I don't know of any practical applications for that function either.
$endgroup$
– Ben W
Jan 13 at 20:57
4
4
$begingroup$
Unlike $text{sinc}$ this has a singularity.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 20:41
$begingroup$
Unlike $text{sinc}$ this has a singularity.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 20:41
$begingroup$
One of the main reasons that the sinc function is so useful is because it's Fourier transform is the rectangular function $mathbb{1}_{big[-tfrac{1}{2},tfrac{1}{2}big]}$. It's a nice exercise to compute the Fourier transform of $frac{cos(x)}{x}$. You'll see that you obtain something different, and "less useful" in the digital signals environment.
$endgroup$
– Kolja
Jan 13 at 20:56
$begingroup$
One of the main reasons that the sinc function is so useful is because it's Fourier transform is the rectangular function $mathbb{1}_{big[-tfrac{1}{2},tfrac{1}{2}big]}$. It's a nice exercise to compute the Fourier transform of $frac{cos(x)}{x}$. You'll see that you obtain something different, and "less useful" in the digital signals environment.
$endgroup$
– Kolja
Jan 13 at 20:56
1
1
$begingroup$
I don't know of any special applications. The "mathematical" reason sinc is special is because sin(x)/x can be continuously extended to x=0, indeed differentiably extended. I guess it has applications in signal processing but I don't know what they are. The cosine "counterpart" to sin(x)/x is probably [1-cos(x)]/x which also can be continuously extended to x=0. However I don't know of any practical applications for that function either.
$endgroup$
– Ben W
Jan 13 at 20:57
$begingroup$
I don't know of any special applications. The "mathematical" reason sinc is special is because sin(x)/x can be continuously extended to x=0, indeed differentiably extended. I guess it has applications in signal processing but I don't know what they are. The cosine "counterpart" to sin(x)/x is probably [1-cos(x)]/x which also can be continuously extended to x=0. However I don't know of any practical applications for that function either.
$endgroup$
– Ben W
Jan 13 at 20:57
add a comment |
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4
$begingroup$
Unlike $text{sinc}$ this has a singularity.
$endgroup$
– Lord Shark the Unknown
Jan 13 at 20:41
$begingroup$
One of the main reasons that the sinc function is so useful is because it's Fourier transform is the rectangular function $mathbb{1}_{big[-tfrac{1}{2},tfrac{1}{2}big]}$. It's a nice exercise to compute the Fourier transform of $frac{cos(x)}{x}$. You'll see that you obtain something different, and "less useful" in the digital signals environment.
$endgroup$
– Kolja
Jan 13 at 20:56
1
$begingroup$
I don't know of any special applications. The "mathematical" reason sinc is special is because sin(x)/x can be continuously extended to x=0, indeed differentiably extended. I guess it has applications in signal processing but I don't know what they are. The cosine "counterpart" to sin(x)/x is probably [1-cos(x)]/x which also can be continuously extended to x=0. However I don't know of any practical applications for that function either.
$endgroup$
– Ben W
Jan 13 at 20:57