Inner product

1. Nov 3, 2008

dirk_mec1

1. The problem statement, all variables and given/known data

http://img152.imageshack.us/img152/3851/33495448dh9.png [Broken]

2. Relevant equations

http://img146.imageshack.us/img146/4655/37276835io7.png [Broken]

3. The attempt at a solution

Well I found:

$$||f|| = \frac{1}{ \sqrt{3}}$$

$$||g||=\frac{i}{ \sqrt{3}}$$

$$<f,g> = \frac{1-4i}{17}$$

Can someone verify the above?

The last question really bothers me. I know that the inproduct must be zero but what does the span explicitly mean? Is it the term below?

$$\alpha \cdot \left( \frac{-1}{2} \right) ^n + \beta \left( \frac{i}{2} \right) ^n$$

But if that's the case how can I find h?

Last edited by a moderator: May 3, 2017
2. Nov 3, 2008

gabbagabbahey

I'm not seeing any questions. I can't check your solutions if I don't know the question.

3. Nov 3, 2008

dirk_mec1

Oops, sorry! I've edited the post.

4. Nov 3, 2008

HallsofIvy

Staff Emeritus
Each of these is correct except ||g||. The norm is MUST be a real number. That is true even for a space over the complex numbers. Each of <f, f>, <f, g>, and <g, g> is a geometric series so you can use
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

5. Nov 3, 2008

gabbagabbahey

Your ||g|| and <f,g> are a little off, perhaps you should show me your work for those two.

6. Nov 3, 2008

dirk_mec1

$$||g|| = \sum_{k=0}^{\infty} \left( \frac{i}{2} \right) ^k \left( \frac{i}{2} \right)^k = \sum_{k=0}^{\infty} \left( \frac{-1}{2} \right) ^k = |-1/3| =1/3$$

7. Nov 3, 2008

gabbagabbahey

shouldn't you have (-1)^k/4 ? ;0) ....(and the summation should start at k=1, since n is a natural number)

8. Nov 3, 2008

dirk_mec1

Yes, you're right! I'll try again:

$$||g|| = | \sum_{k=1}^{\infty} \left( \frac{i}{2} \right) ^k \left( \frac{i}{2} \right)^k | =| \sum_{k=1}^{\infty} \left( \frac{-1}{4} \right) ^k |= |\frac{-1}{5}| = \frac{1}{5}$$

9. Nov 3, 2008

Office_Shredder

Staff Emeritus
That's a bold assertion to make in a math forum :tongue2:

For the last part, the span of f and g is all linear combinations of f and g (so what you put up in the OP describes the nth term of an arbitrary element in the span)

10. Nov 3, 2008

gabbagabbahey

||g|| looks good now, but I think you are still missing a neg sign in your <f,g>

11. Nov 3, 2008

dirk_mec1

$$<f,g> = | \sum_{m=1}^{\infty} \left( \frac{-1}{2}\ \right) ^m \left( \frac{i}{2} \right) ^m | =| \sum_{m=1}^{\infty} \left( \frac{-i}{4} \right) ^m |= | \frac{-i}{i+4} |= | \frac{-1-4i}{17} | =$$

12. Nov 3, 2008

gabbagabbahey

Why do you have modulus signs in that equation? ...And (4-i)(-i)= (-1)-4i not (+1)-4i ;0)

13. Nov 3, 2008

dirk_mec1

So the modulus sign is only inserted in the norm calculation?

$$= \sum_{m=1}^{\infty} \left( \frac{-1}{2}\ \right) ^m \left( \frac{i}{2} \right) ^m = \sum_{m=1}^{\infty} \left( \frac{-i}{4} \right) ^m = \frac{-i}{i+4} = \frac{-1-4i}{17} =$$

14. Nov 3, 2008

gabbagabbahey

Yes, if you were asked to find ||<f,g>|| then you would use them, but <f,g> doesn't even need to be real, let alone positive.

15. Nov 3, 2008

dirk_mec1

So it is:

$$\alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ \left( \frac{i}{2} \right) ^n \right] _{n \in \mathbb{N} }$$

But how are you then suppose to find h?

Last edited: Nov 3, 2008
16. Nov 3, 2008

gabbagabbahey

If you knew what h was, what would span{f,h} be?

17. Nov 3, 2008

dirk_mec1

I think this:

$$\alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ h(n)] _ {n \in \mathbb{N} }$$

18. Nov 3, 2008

gabbagabbahey

right, now use that along with the fact that <f,h>=0 (and span{f,g}=span{f,h}) to find h(n).

19. Nov 3, 2008

dirk_mec1

Wait something is wrong...

There holds: $$<g,g> = -1/5$$ but <...> is always larger than 0...so what's wrong?

Okay, I thought of something for h(n)

$$h(n) = 2^n + (-1)^{n+1}$$

Last edited: Nov 3, 2008
20. Nov 4, 2008

dirk_mec1

@gabbagabbahey is

$$h(n) = 2^n + (-1)^{n+1}$$

a good choice?