Differentiating an Inner Product

[Parmenides]

I am attempting to work my way through the product rule for inner products, using the properties of linearity and symmetry. I am wondering if the following step is allowed, exploiting the bilinear property:
[tex]f(t) = \left\langle{\alpha}(t),{\beta}(t)\right\rangle \rightarrow f'(t) = \lim_{h \to 0}\frac{1}{h}[\left\langle{\alpha}(t + h),{\beta}(t + h)\right\rangle - \left\langle{\alpha}(t),{\beta}(t)\right\rangle][/tex]
If so, I think I can proceed using similar tricks from the proof of the product rule for "ordinary" functions . If not, I am lost on how to proceed.

 

[pwsnafu]

I am attempting to work my way through the product rule for inner products, using the properties of linearity and symmetry. I am wondering if the following step is allowed, exploiting the bilinear property:
[tex]f(t) = \left\langle{\alpha}(t),{\beta}(t)\right\rangle \rightarrow f'(t) = \lim_{h \to 0}\frac{1}{h}[\left\langle{\alpha}(t + h),{\beta}(t + h)\right\rangle - \left\langle{\alpha}(t),{\beta}(t)\right\rangle][/tex]
If so, I think I can proceed using similar tricks from the proof of the product rule for "ordinary" functions . If not, I am lost on how to proceed.

Where are you using the bilinear property?

 

[chiro]

Hey Parmenides.

If you are differentiating, it has to be with respect to some linear direction which means that it must be aligned either with some principal axis, or against some tangent vector which is what a directional derivative is.

Can you tell us which of these things is the thing that are differentiating with respect to?

 

[micromass]

Hey Parmenides.

If you are differentiating, it has to be with respect to some linear direction which means that it must be aligned either with some principal axis, or against some tangent vector which is what a directional derivative is.

Can you tell us which of these things is the thing that are differentiating with respect to?

He has a function ##f:\mathbb{R}\rightarrow \mathbb{R}## though. So it's just the ordinary derivative.

 

[jbunniii]

I am attempting to work my way through the product rule for inner products, using the properties of linearity and symmetry. I am wondering if the following step is allowed, exploiting the bilinear property:
[tex]f(t) = \left\langle{\alpha}(t),{\beta}(t)\right\rangle \rightarrow f'(t) = \lim_{h \to 0}\frac{1}{h}[\left\langle{\alpha}(t + h),{\beta}(t + h)\right\rangle - \left\langle{\alpha}(t),{\beta}(t)\right\rangle][/tex]

You haven't actually taken a step yet. This is just the definition of the derivative:
$$f'(t) = \lim_{h \rightarrow 0} \frac{1}{h} [f(t+h) - f(t)]$$
where you have substituted the defining expression for ##f##. As such, what you have written is correct by definition.

So what is your next step?

 

[mathman]

{α(t+h),β(t+h)}-{α(t),β(t)} = {α(t+h),β(t+h)}-{α(t+h),β(t)}+{α(t+h),β(t)}-{α(t),β(t)}.

When you are finished you will have {α(t),β'(t)}+{α'(t),β(t)}

 

[Parmenides]

You haven't actually taken a step yet. This is just the definition of the derivative.

Hi, jbunniii. I suppose I should have also directed a question to pwsnafu in the sense that I don't see how getting to the expression [tex]f(t) = \left\langle{\alpha}(t),{\beta}(t)\right\rangle \rightarrow f'(t) = \lim_{h \to 0}\frac{1}{h}[\left\langle{\alpha}(t + h),{\beta}(t + h)\right\rangle - \left\langle{\alpha}(t),{\beta}(t)\right\rangle][/tex] is valid in the first place, which is why I was wondering if I could do so. To explain, I know that ##f'(t) = \frac{d}{dt}{f(t)}## so I take ##\frac{d}{dt}## as my linear operator on my inner product ##\left\langle\alpha(t),\beta(t)\right\rangle##. But to arrive at the expression for the derivative above with the inner product seems to imply that I'm using some sort of property like:
[tex]{a}\left\langle\alpha(t),\beta(t)\right\rangle = \left\langle{a}\alpha(t),{a}\beta(t)\right\rangle[/tex]
where ##a## is some linear operator. This is why I brought up exploiting bilinearity (which was more of a question than a statement--sorry, pwsnafu!) to wonder how I could justify my starting point. I've tried to go about it differently, such as:
[tex]f'(t) = {\frac{d}{dt}}\left\langle\alpha(t),\beta(t)\right\rangle = \left\langle{\frac{d}{dt}}\alpha(t),\beta(t)\right\rangle = \left\langle{\frac{d}{dt}}\beta(t),\alpha(t)\right\rangle[/tex]
By using linearity in the first variable and symmetry. But this gets me into trouble because in the process of manipulating this, I end up with a zero on the far right side which I don't know how to deal with. I suppose I'm just wondering why my original expression is valid in the first place.

 

[jbunniii]

Hi, jbunniii. I suppose I should have also directed a question to pwsnafu in the sense that I don't see how getting to the expression [tex]f(t) = \left\langle{\alpha}(t),{\beta}(t)\right\rangle \rightarrow f'(t) = \lim_{h \to 0}\frac{1}{h}[\left\langle{\alpha}(t + h),{\beta}(t + h)\right\rangle - \left\langle{\alpha}(t),{\beta}(t)\right\rangle][/tex] is valid in the first place, which is why I was wondering if I could do so.

The definition of the derivative of a function ##f## at ##t## is
$$f'(t) = \lim_{h \rightarrow 0} \frac{1}{h} [f(t+h) - f(t)]$$
You have
$$f(t) = \langle \alpha(t), \beta(t) \rangle$$
so evaluating ##f## at ##t+h## we get
$$f(t+h) = \langle \alpha(t+h), \beta(t+h)\rangle$$
Now substituting these into the above expression, we get
$$f'(t) = \lim_{h \rightarrow 0} \frac{1}{h} [\langle \alpha(t+h), \beta(t+h)\rangle - \langle \alpha(t), \beta(t) \rangle]$$
To get this far, we have simply used definitions, but not any properties of the inner product.

The next step indicated by mathman also doesn't require any properties of the inner product: simply add and subtract ##\langle \alpha(t+h), \beta(t) \rangle## to the right hand side to obtain
$$f'(t) = \lim_{h \rightarrow 0} \frac{1}{h} [\langle \alpha(t+h), \beta(t+h)\rangle - \langle \alpha(t+h), \beta(t) \rangle + \langle \alpha(t+h), \beta(t) \rangle - \langle \alpha(t), \beta(t) \rangle]$$
To proceed from here, you will have to start using properties of the inner product.

 

[Parmenides]

Ah! Over-thinking, I suppose. Thank you.