Linear Independence
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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\(\LaTeX\)
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Exercise:
Prove that the functions phi_a :mathbbR_ rightarrow mathbbRquad x mapsto fraca+x for all a in mathbbR_geq are linear indepent. H: Use that a non-zero polynomial only has finitely many zeros.
Solution:
Take a finite number of u_i in mathbbR and pairwise distinct a_i in mathbbR_geq where limits_i^m u_iphi_a_i . For all x in mathbbR_ it must hold _i^m u_i fraca_i+x If we multiply this term with prodlimits_i ^ma_i+x for the first two terms we get: leftu_ fraca_+x+u_ fraca_+xrighta_+xa_+x u_a_+x+u_a_+x In general this results in: _i^m u_i prodlimits_j neq ia_j+x The left side of the is a polynomial so we can now use the h from above and deduce that it is valid for all x in mathbbR. The is valid if x-a_k quad textfor quad leq k leq m. If i neq k it follows prodlimits_j neq ia_j-a_k so we can reformulate the to u_k prodlimits_j neq ka_j-a_k Since we first set that a_i should be pairwise distinct it follows that prodlimits_j neq ka_j-a_kneq so u_k . Furthermore since k is randomly chosen it is clear that u_u_...u_m So phi_ phi_... are linear indepent.
Prove that the functions phi_a :mathbbR_ rightarrow mathbbRquad x mapsto fraca+x for all a in mathbbR_geq are linear indepent. H: Use that a non-zero polynomial only has finitely many zeros.
Solution:
Take a finite number of u_i in mathbbR and pairwise distinct a_i in mathbbR_geq where limits_i^m u_iphi_a_i . For all x in mathbbR_ it must hold _i^m u_i fraca_i+x If we multiply this term with prodlimits_i ^ma_i+x for the first two terms we get: leftu_ fraca_+x+u_ fraca_+xrighta_+xa_+x u_a_+x+u_a_+x In general this results in: _i^m u_i prodlimits_j neq ia_j+x The left side of the is a polynomial so we can now use the h from above and deduce that it is valid for all x in mathbbR. The is valid if x-a_k quad textfor quad leq k leq m. If i neq k it follows prodlimits_j neq ia_j-a_k so we can reformulate the to u_k prodlimits_j neq ka_j-a_k Since we first set that a_i should be pairwise distinct it follows that prodlimits_j neq ka_j-a_kneq so u_k . Furthermore since k is randomly chosen it is clear that u_u_...u_m So phi_ phi_... are linear indepent.
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Exercise:
Prove that the functions phi_a :mathbbR_ rightarrow mathbbRquad x mapsto fraca+x for all a in mathbbR_geq are linear indepent. H: Use that a non-zero polynomial only has finitely many zeros.
Solution:
Take a finite number of u_i in mathbbR and pairwise distinct a_i in mathbbR_geq where limits_i^m u_iphi_a_i . For all x in mathbbR_ it must hold _i^m u_i fraca_i+x If we multiply this term with prodlimits_i ^ma_i+x for the first two terms we get: leftu_ fraca_+x+u_ fraca_+xrighta_+xa_+x u_a_+x+u_a_+x In general this results in: _i^m u_i prodlimits_j neq ia_j+x The left side of the is a polynomial so we can now use the h from above and deduce that it is valid for all x in mathbbR. The is valid if x-a_k quad textfor quad leq k leq m. If i neq k it follows prodlimits_j neq ia_j-a_k so we can reformulate the to u_k prodlimits_j neq ka_j-a_k Since we first set that a_i should be pairwise distinct it follows that prodlimits_j neq ka_j-a_kneq so u_k . Furthermore since k is randomly chosen it is clear that u_u_...u_m So phi_ phi_... are linear indepent.
Prove that the functions phi_a :mathbbR_ rightarrow mathbbRquad x mapsto fraca+x for all a in mathbbR_geq are linear indepent. H: Use that a non-zero polynomial only has finitely many zeros.
Solution:
Take a finite number of u_i in mathbbR and pairwise distinct a_i in mathbbR_geq where limits_i^m u_iphi_a_i . For all x in mathbbR_ it must hold _i^m u_i fraca_i+x If we multiply this term with prodlimits_i ^ma_i+x for the first two terms we get: leftu_ fraca_+x+u_ fraca_+xrighta_+xa_+x u_a_+x+u_a_+x In general this results in: _i^m u_i prodlimits_j neq ia_j+x The left side of the is a polynomial so we can now use the h from above and deduce that it is valid for all x in mathbbR. The is valid if x-a_k quad textfor quad leq k leq m. If i neq k it follows prodlimits_j neq ia_j-a_k so we can reformulate the to u_k prodlimits_j neq ka_j-a_k Since we first set that a_i should be pairwise distinct it follows that prodlimits_j neq ka_j-a_kneq so u_k . Furthermore since k is randomly chosen it is clear that u_u_...u_m So phi_ phi_... are linear indepent.
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