Thread: Composing Linear Transformations

Hi, this is not an assignment question (nothing I have to hand in anyway), but I am a bit confused about how to compose this:

S: p(x) -> p(x + 1)
R: p(x) -> (x-1)p(x)

Goal is to get the rule for RS: p(x)

R(S(p(x))) = R(p(x+1)) but, is it equal (x + 1 -1)p(x + 1) = xp(x+1) or (x - 1)p(x+1). i.e. is the x in (x-1) whatever is in the brackets, or always x?

I am seeing reasoning for both answers, and am confused.

Originally Posted by rhfish

Hi, this is not an assignment question (nothing I have to hand in anyway), but I am a bit confused about how to compose this:

S: p(x) -> p(x + 1)
R: p(x) -> (x-1)p(x)

Goal is to get the rule for RS: p(x)

R(S(p(x))) = R(p(x+1)) but, is it equal (x + 1 -1)p(x + 1) = xp(x+1) or (x - 1)p(x+1). i.e. is the x in (x-1) whatever is in the brackets, or always x?

I am seeing reasoning for both answers, and am confused.

R(whatever)= (x-1)*whatever. If "whatever" is p(x+1), then R(p(x+1))= (x-1)p(x+1).

Originally Posted by rhfish

Hi, this is not an assignment question (nothing I have to hand in anyway), but I am a bit confused about how to compose this:

S: p(x) -> p(x + 1)
R: p(x) -> (x-1)p(x)

Goal is to get the rule for RS: p(x)

R(S(p(x))) = R(p(x+1)) but, is it equal (x + 1 -1)p(x + 1) = xp(x+1) or (x - 1)p(x+1). i.e. is the x in (x-1) whatever is in the brackets, or always x?

I am seeing reasoning for both answers, and am confused.

Yes, (x-1) will do. The composition of R with S is , therefore you will have .

Thank you, I guess what confused me is that p(x) -> (x-1)p(x) would suggest that the x in (x-1) should be whatever is in the brackets of p.

Since p has "x + 1" in the brackets after the previous transformation, I wanted it to carry over.

I can see what you guys are saying, but I am confused why it doesn't carry over. I know your answer is correct, but I don't know why it's making my brain feel weird.

Maybe to phrase it better...

S: p(x) -> p(x + 1)
R: p(y) -> (y - 1)p(y)

So after S y actually = x + 1... sigh. I feel very stupid.