# Question about definite integral and orthogonal functions

1. Jan 1, 2010

### pamparana

Hello,

I am just going through a book on calculus and understand that the definite integral can be interpreted as area under the curve.

Now I am trying to figure out the orthogonality relationship between functions and this is normally defined (as far as I can tell from the internet resources) as the definite integral between some limit of the product of the two functions f(x)g(x) wrt to dx.

And if this integral is 0, then the functions are said to be orthogonal.

This is a concept that I am having trouble with. So the total area under the curve is 0. How does this work? Do we have positive and negative areas cancelling each other out? I am really struggling with this concept... could someone help me interpret the orthogonal function relationship using the area under the curve definition of the definite integral?

Hope the question is not too stupid...

Thanks,

Luca

2. Jan 1, 2010

### nrqed

That's exactly it! There is nothing more to it!

Not at all, it was a very good question. Even better, you had figured out the answer yourself!

3. Jan 2, 2010

### pamparana

Hello,

Thanks for the reply. However, I am still confused. What does a negative area means in this case? What is the physical interpretation of a negative area?

Thanks,

Luca

Edit: Ok, I am guessing the negative area is just the orientation of the graph (above or below the x-axes)...

Last edited: Jan 2, 2010
4. Jan 2, 2010

### elibj123

You shouldn't constrain yourself to associating integrals with areas. A more general notion of integrals are sums, then it becomes more natural for you in many other subjects.
For example you can use an integral to calculate a mass of a body with an arbitrary shape and mass-distribution, are there any areas involved in this? No.

Going back to your question, you need to understand that orthogonallity in this sense, arises from an algebric point-of-view rather than calculus, more specifically, this integral is an inner product defined over "integrable functions over some interval [a,b]". I could arbitrarily define another inner product: $$<f,g>_{1}=f(a)g(a)+f(b)g(b)$$ and then pair of functions which are orthogonal in respect to the integral inner product, are not necessarily orthogonal in respect to my inner product. Although, the integral inner product is some what a standard one, and helps in the developement of Fourier Theorem and Sturm-Liouville Theorem (for example).

So you cannot really see how the zero-integral means orthogonallity of functions. There is no depper meaning than what you've said.
If it helps you, I like to think of it as a continuous extension of the cartesian dot product, just like an integral is a continuous extension of summation:
$$(a_{1},a_{2},...,a_{n}).(b_{1},b_{2},...,b_{n})=\sum^{n}_{i=1}a_{i}b_{i}$$
$$f(x).g(x)=\int^{L}_{x=0}f(x)g(x)dx$$

5. Jan 2, 2010

### HallsofIvy

Remember that you said the integral could be interpreted as "area under the curve". That is usually used as a simple motivation for the integral- it is not all there is to integrals- and, of course, that is under the curve but above the x-axis. In order to keep that interpretation, you would need to count the area above the curve and below x-axis, for that part of the curve below the x-axis, as "negative area".

But you seem to be taking a peculiar course! I would not expect a person to be working with orthogonal functions until several years after being introduced to the integral and having seen many other interpretations of it.

6. Jan 2, 2010

### pamparana

Thanks for your reply! Yes, certainly I am aware of the definite integral also being a cumulative sum.

Your description makes a lot of sense to me. I can visualize it as a continuous dot product and that description makes much more sense. I did not know that one could arbitrarily define an inner product for a function. I guess the operator must still specify some rules.

One last question: So, looking at sin and cosine functions which are mutually orthogonal. However, they are orthogonal only over certain intervals, right? For example, 0 to pi bit not 0 to pi/2, right?

Thanks,

Luca

7. Jan 2, 2010

### pamparana

Oh, I am too old to go to school! I am teaching myself these things using internet and library.
I come across these things in quite a random order and sometimes get quite curious :)

Thanks,
Luca

8. Jan 2, 2010

### elibj123

You are correct.

$$sin(t)cos(t)=0.5sin(2t)$$

A function of angular freq. w=2, so it's period is $$\pi$$, therefore the integral vanishes only over intervals which's lengths are integer multiples of pi.

9. Jan 2, 2010

### pamparana

You rule!

Thanks a lot and a happy 2010 to all.

Luca

10. Jan 2, 2010

### HallsofIvy

Note that $sin((n\pi/L)x)$ and $cos((n\pi/L)x)$ will be orthogonal, for all n, over an interval of length L.

11. Jan 2, 2010

### pamparana

Thanks <HallOfIvy>

12. Jan 3, 2010

### HallsofIvy

Hey, I'm a lot more than just one "hall"! (Especially with all the weight I added over Christmas.)