Question

Vector space V and subspace W. Let C(1) be the set of all continuously differentiable functions.

V=C(1) and W consists of functions x(t) in C(1) satisfying the Integral w/ limits 4 to -2 of x(t)dt=0.

Answer 1

\[\vec v=\int_4^{-2}x(t)dt\]this?

Answer 2

\[\int\limits_{-2}^{4}x(t)dt=0\]

Answer 3

\[\vec v_1=\int_{-2}^4x_1(t)dt\]\[\vec v_2=\int_{-2}^4 x_2(t)dt\]what is \[\vec v_1+\vec v_2\]?

Answer 4

is it possible to even add those? I guess it has to be possible, but how do you do it since x(t) can be any function

Answer 5

well you could imagine some examples\[x_1(t)=t^2-t\]\[x_2(t)=t^3\]

Answer 6

what is\[\int_a^b t^2-tdt+\int_a^b t^3dt\]?

Answer 7

@zonazoo
what is your question?

Answer 8

\[=\int_a^bt^3+t^2-tdt\]right?

Answer 9

I have to show W is in the subspace of V

Answer 10

you mean that you have to show that W is a subspace of V, not "in" the subpace

Answer 11

subspace*

Answer 12

yes that W is a subspace of V, sorry, trying to feed the kid and type at the same time.

Answer 13

we can see examples, but i can't show it using examples can i

Answer 14

\[\int_a^bf(t)dt+\int_a^bg(t)dt=\int_a^bf(t)+g(t)dt\]this should ring a bell

Answer 15

clearly \(W\subseteq C(1)\) and \(W\) is nonempty (since \(\vec{0}\in W\))


now just show linearity

Answer 16

how do I know that it = 0 though? because in order to be a subspace under addition it has to = 0 right?

Answer 17

\[\vec u=\int_a^b x(t)dt=0\]\[\vec v=\int_a^b y(t)dt=0\]so\[\vec u+\vec v=?\]

Answer 18

\[x(t),y(t)\in C_1\]

Answer 19

let say for example instead of = 0 that it was = 1, if that it was the case then it wouldn't be a subspace right?

Answer 20

right

Answer 21

so really its more like anything except =0 it wouldn't be.

Answer 22

in this case yes, but you should be able to show why.

Answer 23

fantastic. not very comfortable with these as continuous functions. but it makes much more sense now. thank you. only a few more to go.

Answer 24

you got it? welcome then

Answer 25

yes I believe so.