Verifying dot product and finding h

[HallsofIvy]

The instructor has put a pdf file on the internet (which I didn't see), apparently the inner product is defined as above:

http://img221.imageshack.us/img221/2899/58958904pd7.png

So then the answer is $$||g|| = \sqrt{ \frac{1}{3} }$$

But for <f,g> the inner product can have two values depending which function you use for the overbar, right?

Yes, you can. The innerproduct, over the complex numbers, is NOT symmetric:
$$<f, g>= \overline{<g, f>}$$

 

[dirk_mec1]

Hmmm...I'm not too familiar with the notation here. What exactly do the []'s mean?

Here is how I would write the span:

$$\text{span} \{ f,g \} =\{ \alpha_n \left( \frac{-1}{2} \right) ^n+ \beta_n \left( \frac{i}{2} \right) ^n|\alpha_n,\beta_n \in \mathbb{C} \quad \forall \; n \in \mathbb{N} \}$$

I talked to the instructor and the expression above is incorrect.

It should be like this:

$$\text{span} \{ f,g \} =\{ \alpha \left( \frac{-1}{2} \right) ^n+ \beta \left( \frac{i}{2} \right) ^n|\alpha,\beta \in \mathbb{C} \quad \forall \; n \in \mathbb{N} \}$$

Notice that the constants are independent of n.

 

[dirk_mec1]

How many definitions of "inner product" are there?

The only one I know is
"An inner product on a vector space V is a mapping from VxV to C (or R is the vector space is over R) such that
1) <au+ bv,w>= a<u,v>+ b<v,w>

2) <u, v>= complex conjugate of <v, u> (or just <v,u> if the vector space is over R)

3) <u, u> $\ge$ 0

4) <u, u>= 0 if and only if u= 0."

If by "defining" the inner product, you mean defining such a function on a specific vector space, if it does not satisfy those, it is NOT an inner product.

Yes, this definition is used in my lecture notes.

 

[dirk_mec1]

Yes, you can. The innerproduct, over the complex numbers, is NOT symmetric:
$$<f, g>= \overline{<g, f>}$$

What do you mean by "yes, you can"? I presume you mean that only one value corresponds to <f,g> with f and g complex.