Question

show that the following function is an inner product

‹f,g›= ʃf(x)v(x)g(x) dx

(there is a line on top of g(x), and from x=a to x=b

where f(x)and g(x) are continous functions on [a,b]

Answer 1

o gahd these

Answer 2

you must love linear algebra eh? lol let me see if i can remember this lol

Answer 3

i typed it incorrectly

Answer 4

o hmm retype it then?

Answer 5

it should be f(x)g(x) under the integral sign not v(x)

Answer 6

This?\[
<f,g>=\int_a^bf(x)g(x)dx
\]

Answer 7

yes thanks

Answer 8

So does it mean show that: \[
\int_a^b f(x)g(x)dx = ||f(x)||\cdot ||g(x)||\cdot \cos\theta
\]Or something?

Answer 9

How are we defining inner product here?

Answer 10

there is a bar over g(x) and the question is show that the function is an inner product.

where f(x) and g(x)are continous functions on [a,b]

Answer 11

explanation of inner product https://ccrma.stanford.edu/~jos/st/Inner_Product.html

Answer 12

Well I can tell you that: \[
\int f(x) = \sum f(x)\Delta x
\]

Answer 13

To show something is in inner product it has to satisfy 3 properties:

1) It has to be bilinear. That is, if f,g and h are functions, and c is a scalar, then:\[\langle cf+g,h\rangle=c\langle f,h\rangle + \langle g,h\rangle\]The reason its bilinear (and not just linear) is because this must be true for the second slot as well.

2)\[\langle f,g\rangle = \overline{\langle g,f\rangle}\]The bar denotes complex conjugation. In the event we are dealing with a vector space where the field is the real numbers, then the bar doesnt matter.

3) for any function f that isnt the zero function, we need:\[\langle f, f\rangle \ne 0\]

Answer 14

So here we go. Let f,g, and h be continuous functions on [a,b] and c be a scalar. Then:\[\langle cf+g,h\rangle=\int\limits_a^b(cf(x)+g(x))\overline{h(x)}dx\]\[=c\int\limits_a^{b}f(x)\overline{h(x)}dx+\int\limits_a^{b}g(x)\overline{h(x)}dx\]\[c\langle f, g\rangle+\langle g,h\rangle\] So property one is done. The other properties are just as formal.

Property 2 you can do using the fact that:\[\overline {f(x)g(x)}=\overline {f(x)}\cdot \overline{g(x)}\]and \[\overline{(\overline{f(x)})}=f(x)\]

Property 3 can be shown by noting that:\[f(x)\cdot \overline{f(x)}=|f(x)|^2\]where the absolute values bars denote the modulus of a complex number.