Real Analysis – A Fourier Transform Question: When do we have $ int f (x) e_n = int f (x) overline {e_n} $?

To let $ f (x) $ be a true 1-periodic function, define $ e_n: = e ^ {2 pi i n x} $ and $ overline {e_n} $ is his conjugate. The $ n $Fourier coefficient is
$$ has f_n = int_0 ^ 1f (x) overline {e_n} dx = int_0 ^ 1f (x) e ^ {- 2 pi i nx} dx. $$
My question is, for which function $ f $the following applies:
$$ int_0 ^ 1f (x) e ^ {- 2 pi i nx} dx = int_0 ^ 1f (x) e ^ {2 pi i nx} dx tag {$ * $} $$
trivially $ f (x) = 1 $ is working. Another, less trivial example is:
$$ int_0 ^ 1f (x) e ^ {- 2 pi n x} dx = int_0 ^ 1f (x) e ^ {2 pi n x} dx = frac {1} {n ^ 2}, $$
to the $ f (x) = frac { pi ^ 2} {3} (1-6x + 6x ^ 2) $,

I do not see any obvious reason.
Are these features also satisfactory ($ * $) in any way special?

Fourier analysis – The first zero crossing of a combination of sinewave

To let $ {c_i } _ {i = 1} ^ n $ be a sequence of real numbers, so that $ c_i geq 0 $ for each $ i $ and $ sum_ {i = 1} ^ n c_i = 1 $, To let $ omega_i in [delta, Delta]$ for each $ i $, from where $ delta $ and $ Delta $ are absolutely positive reals. Define $$ f (t) = sum_ {i = 1} ^ c_i sin ( omega_i t) $$ to the $ t in[0infty)$[0infty)$[0infty)$[0infty)$, and let $ t & # 39; $ the smallest value is not equal to zero $ t $ for what $ f (t) = 0 $,

  • How can we show that there is a constant $ T> 0 $ so that $ t & # 39; leq T $ for all possible choices of $ c_i $ and $ omega_i $?
  • Can we specify upper / lower bounds? $ T $ in terms of one of the other variables?
  • In particular, how can we determine which dependency is on $ T $ on $ n $?

For the first question, someone suggested the compactness (from WhatI'm not sure), along with the claim $$ lim _ { overline T rightarrow infty} frac {1} { overline T} int_0 ^ { overline T} f (t) = 0, $$ I am not sure yet what to do with these ideas. It is possible that I misunderstood the person, but I am unable to reach that person for further comments.

Fourier Transform Exponent – MathOverflow

How would I complete the Fourier transform in the following image?

I understand this so that in this situation the 1 in the red circle in the image must be negative, so that e ^ -Infinity approaches 0 and thus equals 0.

That's not the case. Since the 1 is positive. I found online that the Fourier transform of this function is X (W) = (1) / (1 + jW). although I can not get this result due to infinity.

What is the effect of e ^ -jw (infinite)? this will be cosine and imaginary sin function with the same w, but at present infinitely valued what is undefined?

Image processing – convolution in the Fourier domain

I think that's because of the easy assymetry that comes when your image dimensions are straight. With an even number $ n = 2k $ For the degrees of freedom per dimension, an even number of calculated Fourier coefficients is also calculated for the basis functions $ mathrm {e} ^ {- mathrm {i} (k-1) , t}, dotsc, mathrm {e} ^ {- mathrm {i} , t}, 1, mathrm { e} ^ { mathrm {i} , t}, dotsc, mathrm {e} ^ {- mathrm {i} k , t} $, That is, the reflection of $ mathrm {e} ^ {- mathrm {i} k , t} $ can not be displayed.

For example, the use of odd image dimensions seems to work well:

ImFourier[Img_] : = Module[{KerFun, Ker, FourierMul, Convolved, m, n},
   {m, n} = Dimensions[Img];
Ker = appearance[{x, y} [Function] (E ^ (- 500 x ^ 2 - 700 y ^ 2)),
Divide[-0.5, 0.5, m - 1]Divide[-0.5, 0.5, n - 1]];
FourierMul = Inverse Fourier[Fourier[Img]* Fourier[Ker]];
Folded = rescale[Chop[ifftshift[FourierMul]]];
{Return[Convolved], Return[FourierMul]}];

n = 33;
{Convolved, FourierMul} = ImFourier[DiskMatrix[13, n]];
ArrayPlot[Convolved, ColorFunction -> GrayLevel]

Enter the image description here

Fourier ship's expansion V / S Fourier series?

I am currently working on a project that includes the extension of Fourier Bessel. When I try to solve the coefficients using substitutes of Bessel functions as in the attached picture. I observe that the coefficients are scaled imaginary parts of Fourier series coefficients.

Bessel's function form that I used

How do they differ fundamentally? (I know that we have changed the basic function of Sinus to Bessel)

Simplify Expressions – Evaluate Fourier coefficients when exponential expressions exist

I'm trying an analytical solution for a 3D Laplace Equation and just wanted to know if I can use the symbolic computational functions of Mathematica to calculate the Fourier coefficients

context
(Please wear with the length)


The last step of solving a three-dimensional Laplace equation ($ nabla ^ 2T = 0 $, from where $ nabla ^ 2 = frac { partial ^ 2} { partial x ^ 2} + frac { partial ^ 2} { partial y ^ 2} + frac { partial ^ 2} { partial z ^ 2} $) with BC (s) as

$ T (0, y, z) = T (L, y, z) = T_a $,

$ T (x, 0, z) = T (x, l, z) = T_a $,

$ frac { partial T (x, y, 0)} { partial z} = p_c bigg (e ^ {- b_cy / l} left[T_{ci} + frac{b_c}{l}int_0^y e^{b_cs/l}T(x,s,z)dsright] – T (x, y, 0) bigg) $

$ frac { partial T (x, y, w)} { partial z} = p_h bigg (e ^ {- b_hx / L} left[T_{hi} + frac{b_h}{l}int_0^x e^{b_hs/l}T(s,y,z)dsright] – T (x, y, w) bigg) $


When using a solution form

$ T (x, y, z) = sum_ {m, n = 1} ^ { infty} (A_ {nm} e ^ { gamma z} + B_ {nm} e ^ {- gamma z}) sin ( frac {n pi x} {L}) sin ( frac {m pi y} {l}) + T_a. $

I have to rate the Fourier coefficients $ A_ {nm} $ and $ B_ {nm} $ using the following equation obtained after applying the $ z = 0 $ BC

$$
frac {1} {p_c} sum_ {n, m = 1} ^ infty sin ( frac {n pi x} {L}) sin ( frac {m pi y} {l}) gamma (A_ {nm} -B_ {nm}) = delta e ^ { frac {-b_cy} {l}} + PQ + RS rightarrow ( mathrm {1})
$$

$ P = sum_ {n, m = 1} ^ infty frac {b_c ^ 2} {b_c ^ 2 + (m pi) ^ 2} (A_ {nm} + B_ {nm}) sin ( frac {n pi x} {L}) sin ( frac {m pi y} {l}) $

$ Q = sum_ {n, m = 1} ^ infty frac {b_c m pi} {b_c ^ 2 + (m pi) ^ 2} (A_ {nm} + B_ {nm}) sin ( frac {n pi x} {L}) cos ( frac {m pi y} {l}) $

$ R = sum_ {n, m = 1} ^ infty frac {b_c m pi} {b_c ^ 2 + (m pi) ^ 2} (A_ {nm} + B_ {nm}) e ^ { frac {-b_c y} {l}} $

$ S = sum_ {n, m = 1} ^ infty (A_ {nm} + B_ {nm}) sin ( frac {n pi x} {L}) sin ( frac {m pi .) y} {l}) $

Here ,$ gamma ^ 2 = ( frac {n pi} {L}) ^ 2 + ( frac {m pi} {l}) ^ 2 $ and $ c, delta = (T_ {ci} -T_a) $ are constants.

Of course, there would be another relationship that would be a linear equation in $ A $ and $ B $ Finally, the values ​​of $ A_ {nm} $ and $ B_ {nm} $ (of the $ z = w $ BC) separately. My question relates to the way I should handle it $ e ^ { frac {-b_c y} {l}} $ Here.

attempt

I can multiply both sides of the equation $ mathrm {(1)} $ With $ int_0 ^ L sin ( frac {k pi x} {L}) $ and $ int_0 ^ l sin ( frac {j pi y} {l}) $ and then use the principle of orthogonality to finally arrive

For some $ n = k $ and $ m = j $, $ mathrm {(1)} $ becomes

$$
frac {1} {p_c} frac {L} {2} frac {l} {2} gamma (A_ {kj} -B_ {kj}) = delta e ^ { frac {-b_c y} {l}} int_0 ^ L sin ( frac {k pi x} {L}) mathrm {d} x int_0 ^ l sin ( frac {j pi y} {l}) mathrm {d} y + underbrace { frac {b_c ^ 2} {b_c ^ 2 + (j pi) ^ 2} (A_ {kj} + B_ {kj}) frac {L} {2} frac { l} {2}} _ {P} – overbrace {0} ^ {Q} + underbrace { sum_ {n, m = 1} ^ infty frac {b_c m pi} {b_c ^ 2 + ( m pi) ^ 2} (A_ {nm} + B_ {nm}) e ^ { frac {-b_cy} {l}} int_0 ^ L sin ( frac {k pi x} {L}) mathrm {d} x int_0 ^ l sin ( frac {j pi y} {l}) mathrm {d} y} _ {R} – overbrace {(A_ {kj} + B_ {kj} ) frac {L} {2} frac {l} {2}} ^ {S}
$$


Fourier extension of $ e ^ { frac {-b_c y} {l}} $ in the break $ y in [0,l]$

I expanded $ e ^ { frac {-b_c y} {l}} $

$$ e ^ { frac {-b_c y} {l}} = frac {(1-e ^ {- b_c})} {b_c} + sum_ {r = 1} ^ { infty} bigg[frac{2b_c(1-e^{-b_c})}{(b_c)^2 + (2rpi)^2}cosbigg(frac{2rpi y}{l}bigg) + frac{4rpi(1-e^{-b_c})}{(b_c)^2 + (2rpi)^2}sinbigg(frac{2rpi y}{l}bigg)bigg]$$

I think I have to replace that (1) $ to the $ e ^ { frac {-b_c y} {l}} $


Suggestions or comments would be very helpful and would be very happy.

When you see (1) $, you can see how, when the extension of $ e ^ { frac {-b_c y} {l}} $ this will be replaced $ R $ The term would result in a triple summation, which complicates the way. Can there be an alternative way to perform these calculations in? Mathematica, I know that the PDE solvers can use discretization schemes to get a result. I hoped that I could use it Mathematica to help me in finding these Fourier coefficients in the last step.