Inner product

  • Thread starter dirk_mec1
  • Start date
  • #1
761
13

Homework Statement



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

Homework Equations




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

The Attempt at a Solution



Well I found:

[tex]
||f|| = \frac{1}{ \sqrt{3}}
[/tex]

[tex]
||g||=\frac{i}{ \sqrt{3}}
[/tex]


[tex]
<f,g> = \frac{1-4i}{17}
[/tex]

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?

[tex]
\alpha \cdot \left( \frac{-1}{2} \right) ^n + \beta \left( \frac{i}{2} \right) ^n
[/tex]

But if that's the case how can I find h?
 
Last edited by a moderator:

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
I'm not seeing any questions. I can't check your solutions if I don't know the question.
 
  • #3
761
13
I'm not seeing any questions. I can't check your solutions if I don't know the question.

Oops, sorry! I've edited the post.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
965
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
[tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Your ||g|| and <f,g> are a little off, perhaps you should show me your work for those two.
 
  • #6
761
13
[tex] ||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 [/tex]
 
  • #7
gabbagabbahey
Homework Helper
Gold Member
5,002
7
shouldn't you have (-1)^k/4 ? ;0) ....(and the summation should start at k=1, since n is a natural number)
 
  • #8
761
13
Yes, you're right! I'll try again:

[tex]
||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}
[/tex]
 
  • #9
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,627
634
(and the summation should start at k=1, since n is a natural number)

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
gabbagabbahey
Homework Helper
Gold Member
5,002
7
||g|| looks good now, but I think you are still missing a neg sign in your <f,g>
 
  • #11
761
13
[tex] <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} | =
[/tex]
 
  • #12
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Why do you have modulus signs in that equation? ...And (4-i)(-i)= (-1)-4i not (+1)-4i ;0)
 
  • #13
761
13
So the modulus sign is only inserted in the norm calculation?

[tex]
= \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} =

[/tex]
 
  • #14
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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
761
13
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)

So it is:


[tex]

\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} }

[/tex]

But how are you then suppose to find h?
 
Last edited:
  • #16
gabbagabbahey
Homework Helper
Gold Member
5,002
7
If you knew what h was, what would span{f,h} be?
 
  • #17
761
13
I think this:

[tex]


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


[/tex]
 
  • #18
gabbagabbahey
Homework Helper
Gold Member
5,002
7
right, now use that along with the fact that <f,h>=0 (and span{f,g}=span{f,h}) to find h(n).
 
  • #19
761
13
Wait something is wrong...

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

Okay, I thought of something for h(n)

[tex] h(n) = 2^n + (-1)^{n+1}[/tex]
 
Last edited:
  • #20
761
13
@gabbagabbahey is

[tex]
h(n) = 2^n + (-1)^{n+1}
[/tex]

a good choice?
 
  • #21
gabbagabbahey
Homework Helper
Gold Member
5,002
7
hmmm... let's see, that mean that:

[tex]\left<f,h \right> = \sum_{n=1}^{\infty} \left( \frac{-1}{2} \right) ^n (2^n + (-1)^{n+1})=\sum_{n=1}^{\infty}\left( (-1)^n- \left( \frac{1}{2} \right) ^n \right)=\frac{-3}{2} \neq 0[/tex]

...so, no it doesn't work...any other ideas?
 
  • #22
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Wait something is wrong...

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

[itex]||g||=||\left< g,g \right> ||[/itex] is always real and greater than or equal to zero, but [itex]\left< g,g \right>[/itex] can be any complex number.
 
  • #23
gabbagabbahey
Homework Helper
Gold Member
5,002
7
So it is:


[tex]

\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} }

[/tex]

But how are you then suppose to find h?

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

Here is how I would write the span:

[tex]\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} \}[/tex]
 
  • #24
HallsofIvy
Science Advisor
Homework Helper
41,847
965
[itex]||g||=||\left< g,g \right> ||[/itex] is always real and greater than or equal to zero, but [itex]\left< g,g \right>[/itex] can be any complex number.
No. One of the conditions for an inner product is that [itex]<g, g> \ge 0[/itex]. Another is that <g,g>= 0 if and only if g= 0. The inner product <f, g>, for two different vectors f and g, over the complex numbers can be any complex number.
 
  • #25
gabbagabbahey
Homework Helper
Gold Member
5,002
7
No. One of the conditions for an inner product is that [itex]<g, g> \ge 0[/itex]. Another is that <g,g>= 0 if and only if g= 0. The inner product <f, g>, for two different vectors f and g, over the complex numbers can be any complex number.

Doesn't that depend on which definition of inner product you use?

The one given in his original post is just:

[tex](a,b)=\sum_i a_ib_i[/tex]

Although, it isn't clear if that applies to complex elements or just reals.

Usually, for complex valued elements, the definition is:

[tex](a,b)=\sum_i a_i\overline{b_i}[/tex] (where the overbar denote complex conjugation)


I admit, I've never seen the inner product defined without the complex conjugation for complex valued elements, but I decided to go with the definition given in his post.
 

Related Threads on Inner product

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
965
Top