Fourier transform of $mathrme^-x^2 + i x^4$ The Next CEO of Stack OverflowWhich function's Fourier transform is the function itself?The graph of Fourier TransformTwo Fourier Transform QuestionsRange of Fourier transform: criterion to belong to the range of the Fourier transform on L1What is difference between Fourier Transform and Fast Fourier Transform?Formulation of Fourier transformRelation between Laplace and Fourier transformSimplifying an expression with Fourier transformLaplace and Fourier transformComputing Fourier Transform
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Fourier transform of $mathrme^-x^2 + i x^4$
The Next CEO of Stack OverflowWhich function's Fourier transform is the function itself?The graph of Fourier TransformTwo Fourier Transform QuestionsRange of Fourier transform: criterion to belong to the range of the Fourier transform on L1What is difference between Fourier Transform and Fast Fourier Transform?Formulation of Fourier transformRelation between Laplace and Fourier transformSimplifying an expression with Fourier transformLaplace and Fourier transformComputing Fourier Transform
$begingroup$
I would like to know if there is an analytical solution to the Fourier transform of
$$mathrme^-x^2 + i x^4,,$$
where $i$ is the imaginary unit.
More generally, is there an analytical solution to the Fourier transform of a function
$$mathrme^p_1(x) + i p_2(x),,$$
where $p_1(x)$ and $p_2(x)$ are polynomial functions of orders $n_1$ and $n_2$?
Thank you.
complex-analysis fourier-analysis linear-transformations
$endgroup$
add a comment |
$begingroup$
I would like to know if there is an analytical solution to the Fourier transform of
$$mathrme^-x^2 + i x^4,,$$
where $i$ is the imaginary unit.
More generally, is there an analytical solution to the Fourier transform of a function
$$mathrme^p_1(x) + i p_2(x),,$$
where $p_1(x)$ and $p_2(x)$ are polynomial functions of orders $n_1$ and $n_2$?
Thank you.
complex-analysis fourier-analysis linear-transformations
$endgroup$
1
$begingroup$
Do you have any reason to think that it can be computed explicitly?
$endgroup$
– Kavi Rama Murthy
Mar 19 at 10:07
1
$begingroup$
WolframAlpha can't find a solution to $mathscrF_xleft[mathrme^-x^2 + mathrmix^4right](omega)$, so I don't think there is a closed form answer to your question.
$endgroup$
– Peiffap
Mar 19 at 10:26
$begingroup$
Maybe you can complete square, square substitute and Fourier-transform a gaussian. The other idea would be to write $e^p_1(x)cdot e^ip_2(x)$ and use convolution theorem. If you can find Fourier transform of any exponential family function.
$endgroup$
– mathreadler
Mar 19 at 10:58
$begingroup$
$e^−x^2+i*x^2$ can be computed explicitly so I think there could be solution also for higher powers
$endgroup$
– Štěpán Venos
Mar 19 at 11:26
add a comment |
$begingroup$
I would like to know if there is an analytical solution to the Fourier transform of
$$mathrme^-x^2 + i x^4,,$$
where $i$ is the imaginary unit.
More generally, is there an analytical solution to the Fourier transform of a function
$$mathrme^p_1(x) + i p_2(x),,$$
where $p_1(x)$ and $p_2(x)$ are polynomial functions of orders $n_1$ and $n_2$?
Thank you.
complex-analysis fourier-analysis linear-transformations
$endgroup$
I would like to know if there is an analytical solution to the Fourier transform of
$$mathrme^-x^2 + i x^4,,$$
where $i$ is the imaginary unit.
More generally, is there an analytical solution to the Fourier transform of a function
$$mathrme^p_1(x) + i p_2(x),,$$
where $p_1(x)$ and $p_2(x)$ are polynomial functions of orders $n_1$ and $n_2$?
Thank you.
complex-analysis fourier-analysis linear-transformations
complex-analysis fourier-analysis linear-transformations
edited Mar 19 at 10:25
Peiffap
10312
10312
asked Mar 19 at 10:05
Štěpán VenosŠtěpán Venos
1
1
1
$begingroup$
Do you have any reason to think that it can be computed explicitly?
$endgroup$
– Kavi Rama Murthy
Mar 19 at 10:07
1
$begingroup$
WolframAlpha can't find a solution to $mathscrF_xleft[mathrme^-x^2 + mathrmix^4right](omega)$, so I don't think there is a closed form answer to your question.
$endgroup$
– Peiffap
Mar 19 at 10:26
$begingroup$
Maybe you can complete square, square substitute and Fourier-transform a gaussian. The other idea would be to write $e^p_1(x)cdot e^ip_2(x)$ and use convolution theorem. If you can find Fourier transform of any exponential family function.
$endgroup$
– mathreadler
Mar 19 at 10:58
$begingroup$
$e^−x^2+i*x^2$ can be computed explicitly so I think there could be solution also for higher powers
$endgroup$
– Štěpán Venos
Mar 19 at 11:26
add a comment |
1
$begingroup$
Do you have any reason to think that it can be computed explicitly?
$endgroup$
– Kavi Rama Murthy
Mar 19 at 10:07
1
$begingroup$
WolframAlpha can't find a solution to $mathscrF_xleft[mathrme^-x^2 + mathrmix^4right](omega)$, so I don't think there is a closed form answer to your question.
$endgroup$
– Peiffap
Mar 19 at 10:26
$begingroup$
Maybe you can complete square, square substitute and Fourier-transform a gaussian. The other idea would be to write $e^p_1(x)cdot e^ip_2(x)$ and use convolution theorem. If you can find Fourier transform of any exponential family function.
$endgroup$
– mathreadler
Mar 19 at 10:58
$begingroup$
$e^−x^2+i*x^2$ can be computed explicitly so I think there could be solution also for higher powers
$endgroup$
– Štěpán Venos
Mar 19 at 11:26
1
1
$begingroup$
Do you have any reason to think that it can be computed explicitly?
$endgroup$
– Kavi Rama Murthy
Mar 19 at 10:07
$begingroup$
Do you have any reason to think that it can be computed explicitly?
$endgroup$
– Kavi Rama Murthy
Mar 19 at 10:07
1
1
$begingroup$
WolframAlpha can't find a solution to $mathscrF_xleft[mathrme^-x^2 + mathrmix^4right](omega)$, so I don't think there is a closed form answer to your question.
$endgroup$
– Peiffap
Mar 19 at 10:26
$begingroup$
WolframAlpha can't find a solution to $mathscrF_xleft[mathrme^-x^2 + mathrmix^4right](omega)$, so I don't think there is a closed form answer to your question.
$endgroup$
– Peiffap
Mar 19 at 10:26
$begingroup$
Maybe you can complete square, square substitute and Fourier-transform a gaussian. The other idea would be to write $e^p_1(x)cdot e^ip_2(x)$ and use convolution theorem. If you can find Fourier transform of any exponential family function.
$endgroup$
– mathreadler
Mar 19 at 10:58
$begingroup$
Maybe you can complete square, square substitute and Fourier-transform a gaussian. The other idea would be to write $e^p_1(x)cdot e^ip_2(x)$ and use convolution theorem. If you can find Fourier transform of any exponential family function.
$endgroup$
– mathreadler
Mar 19 at 10:58
$begingroup$
$e^−x^2+i*x^2$ can be computed explicitly so I think there could be solution also for higher powers
$endgroup$
– Štěpán Venos
Mar 19 at 11:26
$begingroup$
$e^−x^2+i*x^2$ can be computed explicitly so I think there could be solution also for higher powers
$endgroup$
– Štěpán Venos
Mar 19 at 11:26
add a comment |
0
active
oldest
votes
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1
$begingroup$
Do you have any reason to think that it can be computed explicitly?
$endgroup$
– Kavi Rama Murthy
Mar 19 at 10:07
1
$begingroup$
WolframAlpha can't find a solution to $mathscrF_xleft[mathrme^-x^2 + mathrmix^4right](omega)$, so I don't think there is a closed form answer to your question.
$endgroup$
– Peiffap
Mar 19 at 10:26
$begingroup$
Maybe you can complete square, square substitute and Fourier-transform a gaussian. The other idea would be to write $e^p_1(x)cdot e^ip_2(x)$ and use convolution theorem. If you can find Fourier transform of any exponential family function.
$endgroup$
– mathreadler
Mar 19 at 10:58
$begingroup$
$e^−x^2+i*x^2$ can be computed explicitly so I think there could be solution also for higher powers
$endgroup$
– Štěpán Venos
Mar 19 at 11:26