Question

Alchemista can you please take a look at this?

Answer 1

http://assets.openstudy.com/updates/attachments/4e1ca72a0b8bc227574897e4-bahrom7893-1310500752834-image197.jpg
What do I do next?

Answer 2

Well since the square of the function is always positive, the integral will be positive as well.

Answer 3

I can't just integrate that, that's some function..

Answer 4

Also the zero function is always 0, the integral of the 0 function will be 0. So you can show that for the 0 vector the inner product will be 0

Answer 5

really? But if it's like x^2 right then integral is x^3/3 that's not necessarily positive..

Answer 6

maybe use the fact that (x(t))^2 is always an even function? even if x(t) is odd?

Answer 7

(x^3/3)^2 I don't see the problem.

Answer 8

No no if the function itself is x right, then (x)^2 = x^2

Answer 9

Integral of that x^3/3

Answer 10

\[\langle f(x) , f(x) \rangle = \int_{-1}^{1} f(x)^2\]

Answer 11

ohh i see from -1 to 1 it cannot be negative..

Answer 12

it can't be negative at all. f(x) is some number some number squared is positive.

Answer 13

okay thanks!

Answer 14

how do I prove the 2nd part? that it;s 0 only if the function itself is 0?

Answer 15

remember this is a vector space, the 0 vector is the 0 function

Answer 16

if you take the integral of a function thats always 0 its obviously 0

Answer 17

Ohh I see..

Answer 18

You guys won't see me for a while now, I know how to solve the next question..

Answer 19

ok

Answer 20

gogo :)