# Inner product

## 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:

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

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

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

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?

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

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.

HallsofIvy
Homework Helper
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
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

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

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

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

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

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

Office_Shredder
Staff Emeritus
Gold Member
(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)

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

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

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

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

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

gabbagabbahey
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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.

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:

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

But how are you then suppose to find h?

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

I think this:

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

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

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)

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

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@gabbagabbahey is

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

a good choice?

gabbagabbahey
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hmmm... let's see, that mean that:

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

...so, no it doesn't work...any other ideas?

gabbagabbahey
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Wait something is wrong...

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

$||g||=||\left< g,g \right> ||$ is always real and greater than or equal to zero, but $\left< g,g \right>$ can be any complex number.

gabbagabbahey
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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?

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

Here is how I would write the span:

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

HallsofIvy
Homework Helper
$||g||=||\left< g,g \right> ||$ is always real and greater than or equal to zero, but $\left< g,g \right>$ can be any complex number.
No. One of the conditions for an inner product is that $<g, g> \ge 0$. 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.

gabbagabbahey
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No. One of the conditions for an inner product is that $<g, g> \ge 0$. 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:

$$(a,b)=\sum_i a_ib_i$$

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

Usually, for complex valued elements, the definition is:

$$(a,b)=\sum_i a_i\overline{b_i}$$ (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.