help

[cinder compass]

I got a,d and e. Can someone check if they r right or not

[cinder compass]

ping me if anyone knows how to do this

[dry coral]

why do you think e is correct ? @cinder compass

also did you actually look into option b or were you too scared by the differential equation ?

[cinder compass]

i did but tbh i just made it into a quadratic equation then just followed the subspace test

/i said e because can u not have p(0)+1 as your solution

?

@dry coral

[cinder compass]

i wouldnt say i knew properly how to answer this ome tho

[dry coral]

ok but why do you think it's a subspace then

[clever mortar]

I got a,b,d

[dry coral]

yea

[clever mortar]

let's say p(t) is a polynomial such that p(t+1) = p(t) + 1.
I'll set P(t) = c p(t) for some constant c.
then P(t+1) = cp(t+1) = c(p(t) + 1 ) = cp(t) + c != cp(t) + 1

checking that things are vector spaces or vector subspaces are usually pretty boring, imo

it's always just checking that they're closed under addition and scalar multiplication

there's no trick, no nothing

it's always just that

[dry coral]

ah lol I thought you were OP that's why I just said yea

well i gtg anyway

[vast wave]

The easy way to see e) isn't a subspace is to see it doesn't contain the zero polynomial.

[cinder compass]

how do you know though?

do u sub in 0 for t?

but then wouldnt u just get p(0)+1

so it would just be the constant polynomial?

thats what i dont get

[vast wave]

P_n is a space of polynomials. The zero polynomial is the polynomial whose coefficients are all zero. If we call this polynomial z, then z(x) = 0 for EVERY real x. So, no, you don't have to sub anything in. You have z(t+1) = 0 and z(t) = 0 for the zero polynomial. So it doesn't satisfy the condition to be in S, so S isn't a subspace.

[cinder compass]

So space of polynomials

I was treating it like degree of polynomials

Up to n

Are the space polynomials and degree of poly that different and if so how?