Reelles Integral mit Residuensatz
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Exercise:
Seien mn inN mit le m n. Zeige dass _^infty fract^m+t^n rm dt fracpinsinleftfracm+pinright.
Solution:
usetikzlibrarydecorations.markings newcommandResrm Resleft # right renewcommandemathrme renewcommandimathrmi renewcommanddmathrmd newcommandftdfract^m+t^n newcommandfzdfracz^m+z^n newcommandzke^ipileftfrack+nright newcommandzme^ipileftfracm+nright newcommandnzme^-ipileftfracm+nright newcommandzmie^ipileftfracm+nright tikzset--/.styledecoration markings markat position . with arrowpostactiondecorate Der Integrand ist gerade deswegen lässt sich schreiben: _^infty fract^m+t^n d t frac _-infty^infty fract^m+t^n d t : frac I. Betrachte den Integrationsweg über den Rand der oberen Halbkreisscheibe mit Radius R. center tikzpicturelatex draw- -. -- . noderightrm Rez; draw- -. -- . nodeaboverm Imz; drawthin --. nodebelowfootnotesize -R -- - .; drawthin -. nodebelowfootnotesize R -- .; draw--stealthredthick - -- ; draw--stealthredthick nodeyshiftcmGamma_R arc ::cm; tikzpicture center Dann lässt sich das Integral über den Halbkreisbogen wie folgt abschätzen: lim_Rtoinfty left|_Gamma_Rfz d zright| le lim_Rtoinfty pi R fracR^m+R^n . Die Singularitäten von fz erfüllen die Gleichung +z^n quad Leftrightarrow quad z^n - e^ipi. Die Lösungen sind nach der Formel von De Moivre z_k e^ileftfracpin + fracpi knright zk qquad kdotsn-. Diejenigen die in der oberen Halbebene liegen müssen ein Argument pi haben also folges erfüllen: frack+n quad Leftrightarrow quad k n -frac quad Leftrightarrow quad k n. Nach dem Residuensatz gilt deshalb I piilimits_k^n- Resfz zk. Da z^m analytisch ist und +z^n gerade einfache Nullstellen in den Singularitäten hat gilt: * Resfz zk leftfracz^mn z^n-right_zzk frac n leftzkright^m-n+ -frac n leftzkright^m+ -frac n zm leftzmiright^k. * Aufmieren unter Verwung der Partialme einer geometrischen Reihe s_n limits_k^n- q^k frac-q^n-q führt schlusslich zu * I -piifrac n zmlimits_k^n- leftzmiright^k -fracpiin zm frac-e^ipim+-zmi fracpiinfraczm-nzm fracpiinfracisinleftfracm+pinright fracpinsinleftfracm+pinright * und damit ist unser Integral _^infty ft d t frac I fracpinsinleftfracm+pinright.
Seien mn inN mit le m n. Zeige dass _^infty fract^m+t^n rm dt fracpinsinleftfracm+pinright.
Solution:
usetikzlibrarydecorations.markings newcommandResrm Resleft # right renewcommandemathrme renewcommandimathrmi renewcommanddmathrmd newcommandftdfract^m+t^n newcommandfzdfracz^m+z^n newcommandzke^ipileftfrack+nright newcommandzme^ipileftfracm+nright newcommandnzme^-ipileftfracm+nright newcommandzmie^ipileftfracm+nright tikzset--/.styledecoration markings markat position . with arrowpostactiondecorate Der Integrand ist gerade deswegen lässt sich schreiben: _^infty fract^m+t^n d t frac _-infty^infty fract^m+t^n d t : frac I. Betrachte den Integrationsweg über den Rand der oberen Halbkreisscheibe mit Radius R. center tikzpicturelatex draw- -. -- . noderightrm Rez; draw- -. -- . nodeaboverm Imz; drawthin --. nodebelowfootnotesize -R -- - .; drawthin -. nodebelowfootnotesize R -- .; draw--stealthredthick - -- ; draw--stealthredthick nodeyshiftcmGamma_R arc ::cm; tikzpicture center Dann lässt sich das Integral über den Halbkreisbogen wie folgt abschätzen: lim_Rtoinfty left|_Gamma_Rfz d zright| le lim_Rtoinfty pi R fracR^m+R^n . Die Singularitäten von fz erfüllen die Gleichung +z^n quad Leftrightarrow quad z^n - e^ipi. Die Lösungen sind nach der Formel von De Moivre z_k e^ileftfracpin + fracpi knright zk qquad kdotsn-. Diejenigen die in der oberen Halbebene liegen müssen ein Argument pi haben also folges erfüllen: frack+n quad Leftrightarrow quad k n -frac quad Leftrightarrow quad k n. Nach dem Residuensatz gilt deshalb I piilimits_k^n- Resfz zk. Da z^m analytisch ist und +z^n gerade einfache Nullstellen in den Singularitäten hat gilt: * Resfz zk leftfracz^mn z^n-right_zzk frac n leftzkright^m-n+ -frac n leftzkright^m+ -frac n zm leftzmiright^k. * Aufmieren unter Verwung der Partialme einer geometrischen Reihe s_n limits_k^n- q^k frac-q^n-q führt schlusslich zu * I -piifrac n zmlimits_k^n- leftzmiright^k -fracpiin zm frac-e^ipim+-zmi fracpiinfraczm-nzm fracpiinfracisinleftfracm+pinright fracpinsinleftfracm+pinright * und damit ist unser Integral _^infty ft d t frac I fracpinsinleftfracm+pinright.
Meta Information
Exercise:
Seien mn inN mit le m n. Zeige dass _^infty fract^m+t^n rm dt fracpinsinleftfracm+pinright.
Solution:
usetikzlibrarydecorations.markings newcommandResrm Resleft # right renewcommandemathrme renewcommandimathrmi renewcommanddmathrmd newcommandftdfract^m+t^n newcommandfzdfracz^m+z^n newcommandzke^ipileftfrack+nright newcommandzme^ipileftfracm+nright newcommandnzme^-ipileftfracm+nright newcommandzmie^ipileftfracm+nright tikzset--/.styledecoration markings markat position . with arrowpostactiondecorate Der Integrand ist gerade deswegen lässt sich schreiben: _^infty fract^m+t^n d t frac _-infty^infty fract^m+t^n d t : frac I. Betrachte den Integrationsweg über den Rand der oberen Halbkreisscheibe mit Radius R. center tikzpicturelatex draw- -. -- . noderightrm Rez; draw- -. -- . nodeaboverm Imz; drawthin --. nodebelowfootnotesize -R -- - .; drawthin -. nodebelowfootnotesize R -- .; draw--stealthredthick - -- ; draw--stealthredthick nodeyshiftcmGamma_R arc ::cm; tikzpicture center Dann lässt sich das Integral über den Halbkreisbogen wie folgt abschätzen: lim_Rtoinfty left|_Gamma_Rfz d zright| le lim_Rtoinfty pi R fracR^m+R^n . Die Singularitäten von fz erfüllen die Gleichung +z^n quad Leftrightarrow quad z^n - e^ipi. Die Lösungen sind nach der Formel von De Moivre z_k e^ileftfracpin + fracpi knright zk qquad kdotsn-. Diejenigen die in der oberen Halbebene liegen müssen ein Argument pi haben also folges erfüllen: frack+n quad Leftrightarrow quad k n -frac quad Leftrightarrow quad k n. Nach dem Residuensatz gilt deshalb I piilimits_k^n- Resfz zk. Da z^m analytisch ist und +z^n gerade einfache Nullstellen in den Singularitäten hat gilt: * Resfz zk leftfracz^mn z^n-right_zzk frac n leftzkright^m-n+ -frac n leftzkright^m+ -frac n zm leftzmiright^k. * Aufmieren unter Verwung der Partialme einer geometrischen Reihe s_n limits_k^n- q^k frac-q^n-q führt schlusslich zu * I -piifrac n zmlimits_k^n- leftzmiright^k -fracpiin zm frac-e^ipim+-zmi fracpiinfraczm-nzm fracpiinfracisinleftfracm+pinright fracpinsinleftfracm+pinright * und damit ist unser Integral _^infty ft d t frac I fracpinsinleftfracm+pinright.
Seien mn inN mit le m n. Zeige dass _^infty fract^m+t^n rm dt fracpinsinleftfracm+pinright.
Solution:
usetikzlibrarydecorations.markings newcommandResrm Resleft # right renewcommandemathrme renewcommandimathrmi renewcommanddmathrmd newcommandftdfract^m+t^n newcommandfzdfracz^m+z^n newcommandzke^ipileftfrack+nright newcommandzme^ipileftfracm+nright newcommandnzme^-ipileftfracm+nright newcommandzmie^ipileftfracm+nright tikzset--/.styledecoration markings markat position . with arrowpostactiondecorate Der Integrand ist gerade deswegen lässt sich schreiben: _^infty fract^m+t^n d t frac _-infty^infty fract^m+t^n d t : frac I. Betrachte den Integrationsweg über den Rand der oberen Halbkreisscheibe mit Radius R. center tikzpicturelatex draw- -. -- . noderightrm Rez; draw- -. -- . nodeaboverm Imz; drawthin --. nodebelowfootnotesize -R -- - .; drawthin -. nodebelowfootnotesize R -- .; draw--stealthredthick - -- ; draw--stealthredthick nodeyshiftcmGamma_R arc ::cm; tikzpicture center Dann lässt sich das Integral über den Halbkreisbogen wie folgt abschätzen: lim_Rtoinfty left|_Gamma_Rfz d zright| le lim_Rtoinfty pi R fracR^m+R^n . Die Singularitäten von fz erfüllen die Gleichung +z^n quad Leftrightarrow quad z^n - e^ipi. Die Lösungen sind nach der Formel von De Moivre z_k e^ileftfracpin + fracpi knright zk qquad kdotsn-. Diejenigen die in der oberen Halbebene liegen müssen ein Argument pi haben also folges erfüllen: frack+n quad Leftrightarrow quad k n -frac quad Leftrightarrow quad k n. Nach dem Residuensatz gilt deshalb I piilimits_k^n- Resfz zk. Da z^m analytisch ist und +z^n gerade einfache Nullstellen in den Singularitäten hat gilt: * Resfz zk leftfracz^mn z^n-right_zzk frac n leftzkright^m-n+ -frac n leftzkright^m+ -frac n zm leftzmiright^k. * Aufmieren unter Verwung der Partialme einer geometrischen Reihe s_n limits_k^n- q^k frac-q^n-q führt schlusslich zu * I -piifrac n zmlimits_k^n- leftzmiright^k -fracpiin zm frac-e^ipim+-zmi fracpiinfraczm-nzm fracpiinfracisinleftfracm+pinright fracpinsinleftfracm+pinright * und damit ist unser Integral _^infty ft d t frac I fracpinsinleftfracm+pinright.
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