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I'm reading about Fourier analysis and in my book the author speaks about dot product for continuous functions $f, g\in L^2(a,b)$(the set of functions which are square-integrable on the interval $[a,b]$), which is defined as:

$$\langle f, g\rangle = \int_a^b f(x)\overline{g(x)}\;dx.$$

The author mentions that the reader should think the functions as vectors with the components of the vectors corresponding to the function values. This notation confuses me. If I had discrete $n$-dimensional vectors $\textbf{x}$ and $\textbf{y}$ the dot product would be:

$$\langle \textbf{x}, \textbf{y}\rangle = \sum_{i=1}^n \textbf{x}(i)\overline{\textbf{y}(i)},$$

where $\textbf{x}(i)$ corresponds to the $i$th component of vector $\textbf{x}$. To me the integral

$$\langle f, g\rangle = \int_a^b f(x)\overline{g(x)}\;dx$$

seems like area under the curve, not a dot product. I guess the $dx$ is confusing me...could someone make this formula a bit more intuitive for me? Please let me know if my question is unclear.

Thank you! =)

jjepsuomi
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3 Answers3

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The formulas are actually exactly the same. An integral is just a sum of infinitesimally short areas over the interval. The essential part is the $f(x)\overline{g(x)}$ which ensures that this is in fact a product with all the properties of the dot product.

You can even write a Riemann sum for the integral:

$$\int_a^b f(x)\overline{g(x)}dx=\lim_{N\to \infty}\sum_{i=0}^N f(a+(b-a)i/N)\overline{g(a+(b-a)i/N)}\frac{b-a}{N}$$

Now both are the same.

The "area under the curve" interpretation is confusing you here. Yes, you can interpret it like that, but no less than the sum can be represented as a total number of apples... the function that you are integrating is of the form that ensures that the functional is a positive semidefinite square form (linear in both terms), which is what dot product is all about.

orion
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  • +1 Thank you for your answer =) $dx$ is what always confuses me...on some situations it makes sense and here it doesn't (for me), because I've learned to understand that integral is the sum of infinitesimally small areas, not intervals...could you comment on that? =) Thank you! I think you clarified already in your update, thnx =) – jjepsuomi May 19 '14 at 18:54
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    Well I just expressed myself poorly. Fixed. If you look at the Riemann sum expression, $(b-a)/N$ is what in the limit becomes $dx$. I really suggest that you don't look at integrals as areas. It's a continuous sum, that's it. You can try another analogy: average of a function (with integral) compared to an average of discrete numbers. – orion May 19 '14 at 18:56
  • One last question =) If integral is an infinite sum...then why do we need $\lim_{N\to\infty}\frac{b-a}{N} = dx$? – jjepsuomi May 19 '14 at 19:08
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    The limit has to be considered as a whole. In the limit process, you are splitting the integrand into thinner and thinner rectangles, but there are more and more of them! So the width of the rectangle has to go to zero in harmony with the number of samples you take. You can imagine $dx$ as infinitely narrow, but still not precisely zero segment (exactly such a width that the infinite sum in $\int$ sums it into a proper finite value). Whenever you write something for differentials, an integral is required to bring it back to the finite numbers. – orion May 19 '14 at 19:13
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For introductory purposes, i would switch to

$$\langle f, g\rangle = \frac{1}{2 \pi} \; \;\int_0^{2 \pi} f(x)\overline{g(x)}\;dx.$$

Now we can take an " orthonormal basis" made up of functions $$ \{ 1, e^{inx}, e^{-inx} \} $$ Note that the integral is over the real variable we are calling $x.$ We are integrating some function of $x,$ possibly constant, which has real and imaginary parts; these can just be integrated separately and the results combined later with an $i$ in the correct place.

It is worth your effort to write out the integrals showing the orthonormal part, $f = e^{inx} $ and $g = e^{\pm imx},$ once for $m = n $ and once for $m \neq n,$ so that makes four integrals because of the $\pm$ signs. It really will help.

Will Jagy
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    @jjepsuomi, I urge you to do the four integrals mentioned; they are elementary, real and imaginary parts come out to sines and cosines... – Will Jagy May 19 '14 at 19:06
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    +1 Thank you for your guide =) I will do the integrals =) hope they will clarify ;) – jjepsuomi May 19 '14 at 19:10
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A useful heuristic analogy to keep in mind: $$ \text{integration is for continuous $X$ as summation is for discrete $X$.} $$

The feature of the inner product for elements of a vector space with discrete basis that carries over to the continuous world is that sum samples each vector in each of its components. The analogous quality for the function space inner product is to integrate over an interval: all values of each function (on the interval) are relevant.

Sammy Black
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