Applying sine orthogonality












0












$begingroup$


I have a confusion as to why this is a viable procedure in the following image:



$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$



To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.



Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.






fourier-analysis orthonormal







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:17










  • $begingroup$
    I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:19






  • 1




    $begingroup$
    Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:20










  • $begingroup$
    That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:26












  • $begingroup$
    Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:37
















0












$begingroup$


I have a confusion as to why this is a viable procedure in the following image:



$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$



To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.



Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.






fourier-analysis orthonormal







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:17










  • $begingroup$
    I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:19






  • 1




    $begingroup$
    Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:20










  • $begingroup$
    That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:26












  • $begingroup$
    Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:37














0












0








0





$begingroup$


I have a confusion as to why this is a viable procedure in the following image:



$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$



To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.



Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.






fourier-analysis orthonormal







share|cite|improve this question











$endgroup$




I have a confusion as to why this is a viable procedure in the following image:



$$
sum_{n} B_n int_{0}^{a} sin left( frac{n pi x}{a}right) sin left( frac{m pi x}{a}right)
= sum_{n} B_n left( frac{a}{2} right) delta_m
= frac{a B_m}{2}
$$



To my knowledge (which is likely the thing that needs to improve here), sine orthogonality can be applied when integrating $sin(m k x) sin(n kx)$ over a full period (where $m, n$ are integers, and $k$ is a wavenumber, where $k = frac{2pi}{lambda}$), not half of one.



Here the period is clearly $2a$, so why is this permissible as well? It seems I don't know when and when I can't apply sine orthogonality given certain limits of integration and wavenumbers.






fourier-analysis orthonormal




fourier-analysis orthonormal






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 21:49









Viktor Glombik

1,0931527




1,0931527










asked Dec 14 '18 at 21:14









sangstarsangstar

853215




853215








  • 3




    $begingroup$
    Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:17










  • $begingroup$
    I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:19






  • 1




    $begingroup$
    Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:20










  • $begingroup$
    That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:26












  • $begingroup$
    Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:37














  • 3




    $begingroup$
    Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:17










  • $begingroup$
    I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:19






  • 1




    $begingroup$
    Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 21:20










  • $begingroup$
    That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:26












  • $begingroup$
    Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
    $endgroup$
    – sangstar
    Dec 14 '18 at 21:37








3




3




$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17




$begingroup$
Why don't you just compute the integral explicitly? A sine addition formula will let you do this.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:17












$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19




$begingroup$
I certainly agree that the integral does turn out to be $a/2$, but I'm looking for a heuristic answer, basically. I don't want to have to check if my integral is equal to $delta_{nm}$ manually each time.
$endgroup$
– sangstar
Dec 14 '18 at 21:19




1




1




$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20




$begingroup$
Notice that the product of these two sine functions is even, so integrating over $[0, a]$ is the same as integrating over $[-a, 0]$. So you get half the value of integrating over a full period.
$endgroup$
– T. Bongers
Dec 14 '18 at 21:20












$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26






$begingroup$
That's a good point. The context for the image is that we're considering $0 le x le a$, so perhaps this is the motivation for this. So, here, the assumption that we must integrate over a full period for orthogonality is correct, but my lecturer is basically stating $int_0^a text{even function of x} dx = 1/2 int_0^{2a} text{even function of x} dx $? So I am true about having to integrate over a full period, I'm just forgetting the implicit step there?
$endgroup$
– sangstar
Dec 14 '18 at 21:26














$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37




$begingroup$
Am I right in assuming this applies to any fraction of $2a$, not just $frac{2a}{2}$?
$endgroup$
– sangstar
Dec 14 '18 at 21:37










2 Answers
2






active

oldest

votes


















0












$begingroup$

Your eigenfunctions are solutions of
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0) = 0,;; y(a)=0.
$$
That is,
$y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.



Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
$$
y''+lambda y= 0,;;; 0 le x le a \
y'(0)=0,;; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.



There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+lambda y = 0,;;; 0 le x le a \
y(0)=0,;;; Ay(a)+By'(a)=0.
$$
The functions $sin(alpha_n x)$ are solutions where
$$
Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
$$
The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.



    Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Your eigenfunctions are solutions of
      $$
      y''+lambda y = 0,;;; 0 le x le a \
      y(0) = 0,;; y(a)=0.
      $$
      That is,
      $y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.



      Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
      $$
      y''+lambda y= 0,;;; 0 le x le a \
      y'(0)=0,;; y'(a)=0
      $$
      So these functions also form an orthogonal basis of $L^2[0,a]$.



      There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
      $$
      y''+lambda y = 0,;;; 0 le x le a \
      y(0)=0,;;; Ay(a)+By'(a)=0.
      $$
      The functions $sin(alpha_n x)$ are solutions where
      $$
      Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
      $$
      The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Your eigenfunctions are solutions of
        $$
        y''+lambda y = 0,;;; 0 le x le a \
        y(0) = 0,;; y(a)=0.
        $$
        That is,
        $y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.



        Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
        $$
        y''+lambda y= 0,;;; 0 le x le a \
        y'(0)=0,;; y'(a)=0
        $$
        So these functions also form an orthogonal basis of $L^2[0,a]$.



        There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
        $$
        y''+lambda y = 0,;;; 0 le x le a \
        y(0)=0,;;; Ay(a)+By'(a)=0.
        $$
        The functions $sin(alpha_n x)$ are solutions where
        $$
        Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
        $$
        The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Your eigenfunctions are solutions of
          $$
          y''+lambda y = 0,;;; 0 le x le a \
          y(0) = 0,;; y(a)=0.
          $$
          That is,
          $y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.



          Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
          $$
          y''+lambda y= 0,;;; 0 le x le a \
          y'(0)=0,;; y'(a)=0
          $$
          So these functions also form an orthogonal basis of $L^2[0,a]$.



          There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
          $$
          y''+lambda y = 0,;;; 0 le x le a \
          y(0)=0,;;; Ay(a)+By'(a)=0.
          $$
          The functions $sin(alpha_n x)$ are solutions where
          $$
          Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
          $$
          The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.






          share|cite|improve this answer









          $endgroup$



          Your eigenfunctions are solutions of
          $$
          y''+lambda y = 0,;;; 0 le x le a \
          y(0) = 0,;; y(a)=0.
          $$
          That is,
          $y_n(x)=sin(npi x/a)$ for $n=1,2,3,cdots$ are the solutions where $lambda_n=n^2pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.



          Likewise, $y_n(x)=cos(npi x/a)$ for $n=0,1,2,3,cdots$ are solutions of
          $$
          y''+lambda y= 0,;;; 0 le x le a \
          y'(0)=0,;; y'(a)=0
          $$
          So these functions also form an orthogonal basis of $L^2[0,a]$.



          There are also orthogonal bases of $sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
          $$
          y''+lambda y = 0,;;; 0 le x le a \
          y(0)=0,;;; Ay(a)+By'(a)=0.
          $$
          The functions $sin(alpha_n x)$ are solutions where
          $$
          Asin(alpha_n a)+Balpha_ncos(alpha_n a)=0
          $$
          The solutions $alpha_n$ are not evenly spaced, but the corresponding $sin(alpha_a x)$ are mutually orthogonal.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 17:51









          DisintegratingByPartsDisintegratingByParts

          59.5k42580




          59.5k42580























              0












              $begingroup$

              If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.



              Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.



                Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.



                  Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.






                  share|cite|improve this answer









                  $endgroup$



                  If you start with a function $fin L^2[0,a]$ and you extend this to an odd function $f_o in L^2[-a,a]$, and then you expand this function in a Fourier series on $[-a,a]$, you get your sine series for $f$ on $[0,a]$.



                  Similarly, if you extend $f$ to an even function $f_e$, then you get an expansion of $f$ on $[0,a]$ in terms of cosines $cos(npi x/a)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 17:07









                  DisintegratingByPartsDisintegratingByParts

                  59.5k42580




                  59.5k42580






























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                      Bundesstraße 106

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                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
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                      Verónica Boquete

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                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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