Existenz der totalen Ableitung
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:
Sei U subseteq mathbbR^n offen und f:U rightarrow mathbbR^m eine Funktion. Falls für jedes j in ...n die partielle Ableitung partial_j f auf ganz U existiert und eine stetige Funktion definiert so ist f auf ganz U diffbar. siehe Skript wegen Skizze!
Solution:
Beweis. Wegen Lemma . kann man m annehmen. Für xx_...x_n^t in U und hinreich kleine hh_...h_n^t in mathbbR^n gilt dann fx+h-fxfx_+h_x_+h_...x_n+h_n-fx_x_+h_...x_n+h_n &+fx_x_+h_x_+h_...x_n+h_n-fx_x_x_+h_...x_n+h_n &+...+fx_...x_n-x_n+h_n-fx partial_ fx_+xi_hx_+h_...x_n+h_nh_ &+partial_ fx_+x_+xi_hx_+h_...x_n+h_nh_ &+...+partial_n fx_...x_n-x_n+xi_nhh_n wobei für jedes j in ...n nach dem MWS ein Zwischenpunkt xi_j h zwischen und h_j gewählt wurde. Wegen Stetigkeit der partiellen Ableitungen kann man nun in obigen Ausdrücken stattdessen die partiellen Ableitungen bei x betrachten. Tatsächlich gilt für alphaxhpartial_ fx_+ xi_hx_+h_...x_n+h_n-partial_fxh_ &+partial_ fx_x_+xi_hx_+h_...x_n+h_n-partial_fxh_ &+...+partial_n fx_...x_n-x_n+ xi_nh-partial_nfxh_n nach den Annahmen des Satzes und wegen frac|h_k|||h||leq für alle h in mathbbR^n und k in ...n die Asymptotik lim limits_^h rightarrow fracalphaxh||h|| Daher ist schlusslich fx+h-fx partial_fxh_+...+partial_nfxh_n+alphaxhLh+alphaxh wobei Lpartial_fx...partial_nfx in textMat_nmathbbR. Also ist f bei x diffbar und da x in U beliebig war ist f also diffbar.
Sei U subseteq mathbbR^n offen und f:U rightarrow mathbbR^m eine Funktion. Falls für jedes j in ...n die partielle Ableitung partial_j f auf ganz U existiert und eine stetige Funktion definiert so ist f auf ganz U diffbar. siehe Skript wegen Skizze!
Solution:
Beweis. Wegen Lemma . kann man m annehmen. Für xx_...x_n^t in U und hinreich kleine hh_...h_n^t in mathbbR^n gilt dann fx+h-fxfx_+h_x_+h_...x_n+h_n-fx_x_+h_...x_n+h_n &+fx_x_+h_x_+h_...x_n+h_n-fx_x_x_+h_...x_n+h_n &+...+fx_...x_n-x_n+h_n-fx partial_ fx_+xi_hx_+h_...x_n+h_nh_ &+partial_ fx_+x_+xi_hx_+h_...x_n+h_nh_ &+...+partial_n fx_...x_n-x_n+xi_nhh_n wobei für jedes j in ...n nach dem MWS ein Zwischenpunkt xi_j h zwischen und h_j gewählt wurde. Wegen Stetigkeit der partiellen Ableitungen kann man nun in obigen Ausdrücken stattdessen die partiellen Ableitungen bei x betrachten. Tatsächlich gilt für alphaxhpartial_ fx_+ xi_hx_+h_...x_n+h_n-partial_fxh_ &+partial_ fx_x_+xi_hx_+h_...x_n+h_n-partial_fxh_ &+...+partial_n fx_...x_n-x_n+ xi_nh-partial_nfxh_n nach den Annahmen des Satzes und wegen frac|h_k|||h||leq für alle h in mathbbR^n und k in ...n die Asymptotik lim limits_^h rightarrow fracalphaxh||h|| Daher ist schlusslich fx+h-fx partial_fxh_+...+partial_nfxh_n+alphaxhLh+alphaxh wobei Lpartial_fx...partial_nfx in textMat_nmathbbR. Also ist f bei x diffbar und da x in U beliebig war ist f also diffbar.
Meta Information
Exercise:
Sei U subseteq mathbbR^n offen und f:U rightarrow mathbbR^m eine Funktion. Falls für jedes j in ...n die partielle Ableitung partial_j f auf ganz U existiert und eine stetige Funktion definiert so ist f auf ganz U diffbar. siehe Skript wegen Skizze!
Solution:
Beweis. Wegen Lemma . kann man m annehmen. Für xx_...x_n^t in U und hinreich kleine hh_...h_n^t in mathbbR^n gilt dann fx+h-fxfx_+h_x_+h_...x_n+h_n-fx_x_+h_...x_n+h_n &+fx_x_+h_x_+h_...x_n+h_n-fx_x_x_+h_...x_n+h_n &+...+fx_...x_n-x_n+h_n-fx partial_ fx_+xi_hx_+h_...x_n+h_nh_ &+partial_ fx_+x_+xi_hx_+h_...x_n+h_nh_ &+...+partial_n fx_...x_n-x_n+xi_nhh_n wobei für jedes j in ...n nach dem MWS ein Zwischenpunkt xi_j h zwischen und h_j gewählt wurde. Wegen Stetigkeit der partiellen Ableitungen kann man nun in obigen Ausdrücken stattdessen die partiellen Ableitungen bei x betrachten. Tatsächlich gilt für alphaxhpartial_ fx_+ xi_hx_+h_...x_n+h_n-partial_fxh_ &+partial_ fx_x_+xi_hx_+h_...x_n+h_n-partial_fxh_ &+...+partial_n fx_...x_n-x_n+ xi_nh-partial_nfxh_n nach den Annahmen des Satzes und wegen frac|h_k|||h||leq für alle h in mathbbR^n und k in ...n die Asymptotik lim limits_^h rightarrow fracalphaxh||h|| Daher ist schlusslich fx+h-fx partial_fxh_+...+partial_nfxh_n+alphaxhLh+alphaxh wobei Lpartial_fx...partial_nfx in textMat_nmathbbR. Also ist f bei x diffbar und da x in U beliebig war ist f also diffbar.
Sei U subseteq mathbbR^n offen und f:U rightarrow mathbbR^m eine Funktion. Falls für jedes j in ...n die partielle Ableitung partial_j f auf ganz U existiert und eine stetige Funktion definiert so ist f auf ganz U diffbar. siehe Skript wegen Skizze!
Solution:
Beweis. Wegen Lemma . kann man m annehmen. Für xx_...x_n^t in U und hinreich kleine hh_...h_n^t in mathbbR^n gilt dann fx+h-fxfx_+h_x_+h_...x_n+h_n-fx_x_+h_...x_n+h_n &+fx_x_+h_x_+h_...x_n+h_n-fx_x_x_+h_...x_n+h_n &+...+fx_...x_n-x_n+h_n-fx partial_ fx_+xi_hx_+h_...x_n+h_nh_ &+partial_ fx_+x_+xi_hx_+h_...x_n+h_nh_ &+...+partial_n fx_...x_n-x_n+xi_nhh_n wobei für jedes j in ...n nach dem MWS ein Zwischenpunkt xi_j h zwischen und h_j gewählt wurde. Wegen Stetigkeit der partiellen Ableitungen kann man nun in obigen Ausdrücken stattdessen die partiellen Ableitungen bei x betrachten. Tatsächlich gilt für alphaxhpartial_ fx_+ xi_hx_+h_...x_n+h_n-partial_fxh_ &+partial_ fx_x_+xi_hx_+h_...x_n+h_n-partial_fxh_ &+...+partial_n fx_...x_n-x_n+ xi_nh-partial_nfxh_n nach den Annahmen des Satzes und wegen frac|h_k|||h||leq für alle h in mathbbR^n und k in ...n die Asymptotik lim limits_^h rightarrow fracalphaxh||h|| Daher ist schlusslich fx+h-fx partial_fxh_+...+partial_nfxh_n+alphaxhLh+alphaxh wobei Lpartial_fx...partial_nfx in textMat_nmathbbR. Also ist f bei x diffbar und da x in U beliebig war ist f also diffbar.
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