Length of vector in space of polynomials

[DorianG]

Homework Statement

Apologies in advance as I can't use any formatting yet...

In linear algebra class, we're finding the length of vectors (polynomials) by computing its inner product with itself and finding the square root of the resultant value.
The inner product is defined in this case (not sure if it's the standard inner product for the vector space over polynomials) as the value of the integral from -1 to 1 of f(t)g(t) dt, where in the case of length this becomes f(t)f(t) dt as it's the inner product of f(t) with itself.
The problem is that, say the integral of at^4 + bt^3 + ct^2 + dt + c from -1 to 1 was evaluated unusally in some examples in class. Rather than just evaluate each summand and substitute the limits of integration (which results in 0 for all even powers of t), the lecturer noted that, after integrating 1 to get t^1, evaluated from -1 to 1, the integral value was zero. Geometrically, he explained that it is equivalent to the line x=y and thus has equal area above and below the x-axis from -1 to 1, and thus cancels.
As I would take it though, and by evaluating formally, t from -1 to 1 is [1]-[-1] = 1 + 1 = 2.
Which is the correct value to take?
As I see it, a more complex function might have equal area above and below the x-axis between -1 and 1. Without noticing this, as it wouldn't be very obvious, the integral would just be evaluated as normal and not 'seen to be zero' by inspection, wouldn't it?
Is evaluating integrals different when finding the length of a polynomial? And is the length of a polynomial over two values, say in 2-d, actually given by that method? Just a little confused about some of the issues around the algorithm and would appreciate any help.
Thanks!

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

… the integral from -1 to 1 of f(t)g(t) dt, where in the case of length this becomes f(t)f(t) dt as it's the inner product of f(t) with itself.
The problem is that, say the integral of at^4 + bt^3 + ct^2 + dt + c from -1 to 1 was evaluated unusally in some examples in class. Rather than just evaluate each summand and substitute the limits of integration (which results in 0 for all even powers of t), the lecturer noted that, after integrating 1 to get t^1, evaluated from -1 to 1, the integral value was zero. …

Hi DorianG!

This makes no sense …

|f| is defined via an integral of f squared …

it doesn't matter what interval you're using, or whether it's a polynomial or not …

so long as f is real, f squared will be non-negative, so how can its integral be zero?

 

[DorianG]

Hi tiny-tim,
Thanks for the reply.

it doesn't matter what interval you're using, or whether it's a polynomial or not …

so long as f is real, f squared will be non-negative, so how can its integral be zero?

I'm not really sure! It still doesn't make sense to me.

|f| is defined via an integral of f squared …

This part I get (I think). So if, for the easiest example, f(t) = 1 for all t, the 'length' of f(t) with the inner product as described, is the integral of 1 (= f(t)f(t) = 1(1) = 1) dt from 1 to -1.
This is t evaluated from 1 to -1 and is therefore 2 - am I right in this?

Sorry, but sometimes when you're looking at something for too long, it gets way over-complicated and I need to have some of the easier things verified!

I think maybe I should just forget that it's a special inner-product thing and just evaluate the integral at the limits exactly as I would have done without knowing it!
Thanks again for the reply.

 

[tiny-tim]

I expect either the lecturer got something wrong or you copied it wrong …

but I can't see that it matters …

this is only an example for class, and you've obviously got the principles anyway.

Just go with the flow …

it'll sort itself out later, if it needs to! ​