Determinant

[weak robin]

Help pls 😭

[frosty anchor]

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[lime cedar]

@weak robin induct

[weak robin]

What is that 🤔

[lime cedar]

Prove for n=1

Then assume true for n=k

Then prove for n=k+1

Got it?

[weak robin]

But that's the difficult part

[lime cedar]

Actually no lol

[weak robin]

How do we know the n=k+1 part

[lime cedar]

It’s not that difficult

[lime cedar]

You know for n=k

Wait @weak robin have you ever used induction before?

[weak robin]

Actually can you elaborate a bit more on this specifically?

[weak robin]

Not really

[lime cedar]

Ah um, you’ll want to see a fee examples first

[weak robin]

What about this question as an example

[lime cedar]

I meant simple example lol

Wait I’ll solve this as well

[weak robin]

Omg thanks 😩

[lime cedar]

Wait

@weak robin on the right side

What would happen if you actually did the long division?

[weak robin]

Idk

I never did long division in two variables

[lime cedar]

a^n + ba^(n-1) + b^2a^(n-2) + … + b^n

[weak robin]

Okay?

Yeah nvm fck this

I give up math

You suck

[lime cedar]

Ouch

[lime cedar]

I solved it

It was simple lol

If a=b…

[lime cedar]

If a=b then using L’Hopital you get (n+1)b^n

[merry wraith]

Let $\Delta_n(a,b)$ denote the déterminant of the nxn matrice as shown here

[vague arrow]

Rotor

[merry wraith]

Developing the déterminant with the first columns (using cofactors) you can actually prove that for $n\geq 2$ $\Delta_n(a,b)=(a+b)\Delta_{n-1}(a,b) -ab\Delta_{n-2}(a,b)$

[vague arrow]

Rotor

[merry wraith]

You can now explicitly find $\Delta_n(a,b)$ solving the second order recurrent relation (finding the characteristic equation which has roots a and b)

[vague arrow]

Rotor

[merry wraith]

And with initial conditions $\delta_0(a,b)=1$ and $\delta_1(a,b)=a+b$ you will have the scalars ( the solutions of the récurent relation is a vectorial space so do get an explicit solutions we need the scalars)

[vague arrow]

Rotor

[merry wraith]

First column *

And like @lime cedar said if you don’t know reccurent sequences like I presented you can prove it via induction ( double induction with the relation I gave)

Sorry for the ping

[lime cedar]

Yeah I assumed true for n=k and n=k-1 and got it

(Proved for n=1,2)

[merry wraith]

And for the case a=b you can litterally simplify the expression by using Bernoulli factorisation or using the reccurent relation we see that the characteristic equation has only 1 double root a so vectorial space of solutions is different but solving it once again gives the same solutions as the Bernoulli factorization

Normally it should be (n+1)a^n

[merry wraith]

Naaah bro really said « you suck » and left the server that’s quite rude imo

[lime cedar]

wait he LEFT-

[merry wraith]

Yes I think

[lime cedar]

that's why I couldn't ping him

Lmao

I bet he wanted the answer

[merry wraith]

And I gave the whole process like a dumbass but I don’t think séquences defined by a recurrent linear relation are asked here because they answer is already given so induction is more adapted

[weak robin]

@lime cedar I am back

I accidentally left the server

@merry wraith I am back

Thanks for the help ☺️

[lime cedar]

Lmao alr