Understanding Fourier Series

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Comparing Functions and vectors.

\vec{v} function f(x)

Finite dimensional Infinite dimensional

A vector can be written in the following different ways,
\vec{V} = V_x \hat{x} + V_y \hat{y} + V_z \hat{z}
\hskip .5cm = (V \cdot \hat{x}) \hat{x} + (V \cdot \hat{y}) \hat{y} + (V \cdot \hat{z}) \hat{z}
If the decomposition is along an orthogonal frame along the vectors \vec{a},\, \vec{b} and \vec{c} then the expression would be,
\vec{V} = (\vec{V} \cdot \hat{a}) \hat{a} + (\vec{V} \cdot \hat{b}) \hat{b} + (\vec{V} \cdot \hat{c}) \hat{c}
\hskip .5cm = \frac{\vec{V} \cdot \vec{a}}{\vec{a}\cdot\vec{a}} \vec{a} + \frac{\vec{V} \cdot \vec{b}}{\vec{b}\cdot\vec{b}} \vec{b} + \frac{\vec{V} \cdot \vec{c}}{\vec{c}\cdot\vec{c}} \vec{c}

In general the dot product of two $n-$dimensional vectors \vec{V} = (V_1, V_2,...,V_n) and \vec{W} = (W_1,W_2,...,W_n), can be written as,
\vec{V} \cdot \vec{W} = \sum_{i=1}^n V_i W_i.

It is useful to think of a real function f(x) over an interval
[a,b] as a vector with infinite components. Here the argument serves
as an index and the function value as the vector component. Analogous to vector dot product, the dot product between two functions $f$ and $g$ defined over the same interval can be written as,
(f,g) = \int_a^b f(x) g(x) dx.

Using this definition of the dot product, one can show that the following functions
are orthogonal to each-other (mutual dot products are zero) on the interval
[0, 2\pi].
f_1(x) = 1, \sin{x}, \sin{2x}, \sin{3x}, ...,\cos{x}, \cos{2x}, \cos{3x},...

Thus in parallel with writing a vector in terms of it’s components, one can write any (finite, smooth and continuous on [0, 2\pi] (I am not trying to be mathematically precise, the aim is to give an intuitive feel)) function in terms of
the above basis functions in the same manner,
f(x) = \frac{(f(x),1)}{(1,1)} 1 + \frac{(f(x),\sin(x)}{(\sin(x), \sin(x))} \sin(x) + \frac{(f(x),\sin(2x)}{(\sin(2x), \sin(2x))} \sin(2x) + ...
\hskip.2cm + \frac{(f(x),\cos(x)}{(\cos(x), \cos(x))} \cos(x) + \frac{(f(x),\cos(2x)}{(\cos(2x), \cos(2x))} \cos(2x) +....
Notice the similarity of the expression of a function in terms of it’s components and a vector in terms of it’s components. Hence decomposition of a function in its Fourier components is quite akin to decomposition of a vector in its Cartesian components.

Author: strangeset

A nomad at heart, I enjoy observing, analysing, connecting, understanding and dreaming. I am a big fan of science and tech. Forever learning and experimenting.

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