When I learned about matrices, that sum property was part of the defenition of the determanent. We also never learned about the geometric meaning of the determanent, which is weird
2:37 There is a problem with the proof by induction. There is no need to a strong induction (for all n < k) but there is a need for both n=k-1 and n=k-2. Because of that, the base case should include both n=2 and n=3! Otherwise, we can't begin the induction properly because we can't infer case n=3 from n=2 only.
There is a simple example where we need more than the simplest base case: proof by induction that all hats are the same color. Base case: n=1. If there is only one hat, it has the same color as itself. Induction. Suppose there are n hats, numbered from 1 to n. The first (n-1) hats are all the same color, but the last (n-1) hats are also the same color. Thus all n hats are the same color.
We do need two base cases, but we also need strong induction. One could avoid the strong induction by rewording the proposition and making the claim for two consecutive numbers at once. With that rewording a single base case would suffice, but it would be the same as the two base cases for the standard wording.
Determinant sum property is special case of determinant being the unique alternating multilinear form on row vectors that evaluates to one at the identity matrix
When I learned about matrices, that sum property was part of the defenition of the determanent. We also never learned about the geometric meaning of the determanent, which is weird
2:37 There is a problem with the proof by induction. There is no need to a strong induction (for all n < k) but there is a need for both n=k-1 and n=k-2. Because of that, the base case should include both n=2 and n=3! Otherwise, we can't begin the induction properly because we can't infer case n=3 from n=2 only.
There is a simple example where we need more than the simplest base case: proof by induction that all hats are the same color.
Base case: n=1. If there is only one hat, it has the same color as itself.
Induction. Suppose there are n hats, numbered from 1 to n. The first (n-1) hats are all the same color, but the last (n-1) hats are also the same color. Thus all n hats are the same color.
We do need two base cases, but we also need strong induction. One could avoid the strong induction by rewording the proposition and making the claim for two consecutive numbers at once. With that rewording a single base case would suffice, but it would be the same as the two base cases for the standard wording.
Determinant sum property is special case of determinant being the unique alternating multilinear form on row vectors that evaluates to one at the identity matrix
The third identity on the board related to Fibonacci numbers is known in mathematical literature as d'Ocagne's Identity.
Sorry if I sound picky ,Do we have to prove that not only when n equals 2 but also when n equals 3? 4:10
I usually take the 0th fibonacci number f_0=0 so we have two steps before n=2.
THANKS
I misunderstood and thought that
g_n=g_0*f_(n-1)+g_1*f_n holds when n is 2 or more
Hello Michael, unrelated question here: it is possible to submit to you a problem?
And of course thank you and keep it up!
I don’t understand the significance of the variable “r”. Can you please define it and give some examples of how it is used within the “f” equation.
Sir can you explain me Lambert w function.
Please
Please😫🙏🙏💓