Recent content by sana2476

It doesn't say anything about that. And as a matter of fact, I was getting (0,0) as one of my equilibrium points.

well let me state the entire problem: Consider the predator prey type of system that's given above with a>0. The population x is prey. By itself, its rate of growth increases for small populations and then decreases for x>1. The predator is given by y, and it dies out when no prey is present...

Homework Statement x'=x(1+2x-x2-y) y'=y(x-a) I need help finding the equilibrium points of this system.

Ok so (1,τ,τ2) is the basis for L(α+βτ+γτ²)

The basis for B is (1,τ)

It's in the basis: (1, τ, τ²)?

you would get: [2α + γ] [β + 3γ] Correct?

α,β,γ are the components in the basis (1, τ, τ²).

a,b,c would be the components.

The components would just be α=1, β=1, γ=1. Isn't that right?

Homework Statement Let L : R3[τ] → R2[τ] be a linear transformation, where the bases for the polynomial vector spaces R3[τ] and R2[τ] are (1,τ,τ2) and (1,τ) respectively. We also know the matrix representation for L is: A=[2 0 1] [0 1 3] What is the result of L(α+βτ+γτ2)...

So I guess we can achieve that by saying the following: -1*(u) + (-1)*v + 1*(u+v) = 0 Since none of the constants are actually zero. Therefore, (u,v,u+v) are infact linealy dependent.

Homework Statement Let V=<sinx,cosx> and T: V --> V be a transformation defined by T(f)=df/dx +f. Prove T is linear. The Attempt at a Solution T(f+g) = cosx-sinx+sinx+cosx T(f)+T(g) = (sinx+cosx)'+sinx+cosx = T(sinx)+T(cosx) T(αf)=αcosx +αsinx αT(f)= α(cosx+sinx)...

Homework Statement Prove (u,v,u+v) can not be a basis for <u,v,u+v>. Homework Equations The Attempt at a Solution Let αu+βv+γ(u+v)=0 αu+βv=-γ(u+v) α/γ(u)+β/γ(v)=-(u+v) α/γ(u)+β/γ(v)+1*(u+v)=0 Since α/γ,β/γ,1 are not all zeros, therefore, (u,v,u+v) are...

I really need help...can someone try and help me with this please!