Gram-Schmidt orthogonalisation process
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 -
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Exercise:
Let Vlangle rangle be an inner product space. abcliste abc If Ssubseteq V is an orthogonal system then S is linearly indepent. abc If v_...v_n is an orthogonal system and vin textSpv_...v_n and write va_v_+...+a_nv_n a_jin K KmathbbR or mathbbC. Then the coefficients a_j are given by a_j fraclangle vv_j ranglelangle v_j v_j rangle forall leq jleq n. abcliste
Solution:
Proof. abcliste abc Suppose _k^na_kv_k where ngeq v_...v_nin S are distinct and a_...a_nin K. By b a_jfraclangle v_j ranglelangle v_j v_j rangle forall leq jleq n. abc Write va_v_+...+a_nv_n. Let leq jleq n. We have langle vv_j rangle _k^n a_klangle v_kv_j rangle a_jlangle v_jv_j rangle Longrightarrow a_jfraclangle vv_j ranglelangle v_jv_j rangle abcliste
Let Vlangle rangle be an inner product space. abcliste abc If Ssubseteq V is an orthogonal system then S is linearly indepent. abc If v_...v_n is an orthogonal system and vin textSpv_...v_n and write va_v_+...+a_nv_n a_jin K KmathbbR or mathbbC. Then the coefficients a_j are given by a_j fraclangle vv_j ranglelangle v_j v_j rangle forall leq jleq n. abcliste
Solution:
Proof. abcliste abc Suppose _k^na_kv_k where ngeq v_...v_nin S are distinct and a_...a_nin K. By b a_jfraclangle v_j ranglelangle v_j v_j rangle forall leq jleq n. abc Write va_v_+...+a_nv_n. Let leq jleq n. We have langle vv_j rangle _k^n a_klangle v_kv_j rangle a_jlangle v_jv_j rangle Longrightarrow a_jfraclangle vv_j ranglelangle v_jv_j rangle abcliste
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Exercise:
Let Vlangle rangle be an inner product space. abcliste abc If Ssubseteq V is an orthogonal system then S is linearly indepent. abc If v_...v_n is an orthogonal system and vin textSpv_...v_n and write va_v_+...+a_nv_n a_jin K KmathbbR or mathbbC. Then the coefficients a_j are given by a_j fraclangle vv_j ranglelangle v_j v_j rangle forall leq jleq n. abcliste
Solution:
Proof. abcliste abc Suppose _k^na_kv_k where ngeq v_...v_nin S are distinct and a_...a_nin K. By b a_jfraclangle v_j ranglelangle v_j v_j rangle forall leq jleq n. abc Write va_v_+...+a_nv_n. Let leq jleq n. We have langle vv_j rangle _k^n a_klangle v_kv_j rangle a_jlangle v_jv_j rangle Longrightarrow a_jfraclangle vv_j ranglelangle v_jv_j rangle abcliste
Let Vlangle rangle be an inner product space. abcliste abc If Ssubseteq V is an orthogonal system then S is linearly indepent. abc If v_...v_n is an orthogonal system and vin textSpv_...v_n and write va_v_+...+a_nv_n a_jin K KmathbbR or mathbbC. Then the coefficients a_j are given by a_j fraclangle vv_j ranglelangle v_j v_j rangle forall leq jleq n. abcliste
Solution:
Proof. abcliste abc Suppose _k^na_kv_k where ngeq v_...v_nin S are distinct and a_...a_nin K. By b a_jfraclangle v_j ranglelangle v_j v_j rangle forall leq jleq n. abc Write va_v_+...+a_nv_n. Let leq jleq n. We have langle vv_j rangle _k^n a_klangle v_kv_j rangle a_jlangle v_jv_j rangle Longrightarrow a_jfraclangle vv_j ranglelangle v_jv_j rangle abcliste
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