Polynomials with many variables over R
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
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_rquad c_underlineiin mathbbR underlineii_...i_r textmulti-index mathcalIsubseteq mathbbZ_geq ^r textfinite set mathbbRx_...x_r set of all polynomials in x_...x_r and coeffs in mathbbR. bf Lemma . Suppose that fx_...x_rin mathbbRx_...x_r satisfies that fa_...a_r forall a_...a_rin mathbbR^r. Then fx_...x_r as a polynomial i.e. all the coeffs of fx_...x_r are .
Solution:
Proof. Write fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_r. Let underlineii_...i_rin mathcalI. We'll show that c_underlinei. Since fa_...a_r forall a_...a_rin mathbbR^r then all the partial derivatives of f of any order are all over mathbbR^r Longrightarrow fracpartial^m fpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r mi_+...+i_r. * But fracpartial^m fpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_r cases quad textif exists k:j_k i_k j_j_-...j_-i_+x_^j_-i_... j_rj_r-... j_r-i_r+x_r^j_r-i_r textif forall k i_jgeq i_kcases &Longrightarrow leftfracpartial^kpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_rrightBiggr|_x_...x_r cases quad underlinejneq i i_!... i_r!quad underlinejunderlineicases &Longrightarrow fracpartial^kfpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r textfirst term is because of * c_underlinei &Longrightarrow c_underlinei.
fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_rquad c_underlineiin mathbbR underlineii_...i_r textmulti-index mathcalIsubseteq mathbbZ_geq ^r textfinite set mathbbRx_...x_r set of all polynomials in x_...x_r and coeffs in mathbbR. bf Lemma . Suppose that fx_...x_rin mathbbRx_...x_r satisfies that fa_...a_r forall a_...a_rin mathbbR^r. Then fx_...x_r as a polynomial i.e. all the coeffs of fx_...x_r are .
Solution:
Proof. Write fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_r. Let underlineii_...i_rin mathcalI. We'll show that c_underlinei. Since fa_...a_r forall a_...a_rin mathbbR^r then all the partial derivatives of f of any order are all over mathbbR^r Longrightarrow fracpartial^m fpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r mi_+...+i_r. * But fracpartial^m fpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_r cases quad textif exists k:j_k i_k j_j_-...j_-i_+x_^j_-i_... j_rj_r-... j_r-i_r+x_r^j_r-i_r textif forall k i_jgeq i_kcases &Longrightarrow leftfracpartial^kpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_rrightBiggr|_x_...x_r cases quad underlinejneq i i_!... i_r!quad underlinejunderlineicases &Longrightarrow fracpartial^kfpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r textfirst term is because of * c_underlinei &Longrightarrow c_underlinei.
Meta Information
Exercise:
fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_rquad c_underlineiin mathbbR underlineii_...i_r textmulti-index mathcalIsubseteq mathbbZ_geq ^r textfinite set mathbbRx_...x_r set of all polynomials in x_...x_r and coeffs in mathbbR. bf Lemma . Suppose that fx_...x_rin mathbbRx_...x_r satisfies that fa_...a_r forall a_...a_rin mathbbR^r. Then fx_...x_r as a polynomial i.e. all the coeffs of fx_...x_r are .
Solution:
Proof. Write fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_r. Let underlineii_...i_rin mathcalI. We'll show that c_underlinei. Since fa_...a_r forall a_...a_rin mathbbR^r then all the partial derivatives of f of any order are all over mathbbR^r Longrightarrow fracpartial^m fpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r mi_+...+i_r. * But fracpartial^m fpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_r cases quad textif exists k:j_k i_k j_j_-...j_-i_+x_^j_-i_... j_rj_r-... j_r-i_r+x_r^j_r-i_r textif forall k i_jgeq i_kcases &Longrightarrow leftfracpartial^kpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_rrightBiggr|_x_...x_r cases quad underlinejneq i i_!... i_r!quad underlinejunderlineicases &Longrightarrow fracpartial^kfpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r textfirst term is because of * c_underlinei &Longrightarrow c_underlinei.
fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_rquad c_underlineiin mathbbR underlineii_...i_r textmulti-index mathcalIsubseteq mathbbZ_geq ^r textfinite set mathbbRx_...x_r set of all polynomials in x_...x_r and coeffs in mathbbR. bf Lemma . Suppose that fx_...x_rin mathbbRx_...x_r satisfies that fa_...a_r forall a_...a_rin mathbbR^r. Then fx_...x_r as a polynomial i.e. all the coeffs of fx_...x_r are .
Solution:
Proof. Write fx_...x_r_underlineiin mathcalIc_underlinei x_^i_... x_r^i_r. Let underlineii_...i_rin mathcalI. We'll show that c_underlinei. Since fa_...a_r forall a_...a_rin mathbbR^r then all the partial derivatives of f of any order are all over mathbbR^r Longrightarrow fracpartial^m fpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r mi_+...+i_r. * But fracpartial^m fpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_r cases quad textif exists k:j_k i_k j_j_-...j_-i_+x_^j_-i_... j_rj_r-... j_r-i_r+x_r^j_r-i_r textif forall k i_jgeq i_kcases &Longrightarrow leftfracpartial^kpartial x_^i_...partial x_r^i_rx_^j_... x_r^j_rrightBiggr|_x_...x_r cases quad underlinejneq i i_!... i_r!quad underlinejunderlineicases &Longrightarrow fracpartial^kfpartial x_^i_...partial x_r^i_rBiggr|_x_...x_r textfirst term is because of * c_underlinei &Longrightarrow c_underlinei.
Contained in these collections
| Title | Matched on |
|---|---|
| Eigenvalues and direct sum | tags |
| Geometric and algebraic multiplicities of eigenvalues | tags |
| Eigenvalues and linear independence | tags |
| Trigonalization over C | tags |
| Diagonalizable endomorphisms and eigenvalues | tags |