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2018年 4月 の投稿一覧

Fourier Transform

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I thought about Fourier series expansion before, but this time it is an introduction of Fourier transform.

Fourier transform

$$F(k)=\int_{-∞}^{∞} f(x)e^{-ikx} dx$$

inverse Fourier transform

$$f(x)=\frac{1}{2π}\int_{-∞}^{∞} F(k)e^{ikx} dk$$

Suppose a wave is given by the function \(f(x)\). Using Fourier transform, the function of wave \(f (x)\) is expressed as a function \(F(k)\) of wavenumber \(k\). \(F(k)\) is the function which decomposes \(f(x)\) into the sine waves of all wavenumbers \(k\) and expresses the size of each the sine wave as a function. You can convert the wave \(f (x)\) to \(F(k)\) by Fourier transform and \ (F (k) \) to \(F(k)\) by inverse Fourier transform.

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Fourier Series Expansion Method

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Fourier series expansion and Fourier transform are completely different. In this article, I’ll write about Fourier series expansion.

Fourier series expansion method is to express the periodic function \(f(x)\) by using the sum of \(sin\) and \(cos\). Using the method, the periodic function \(f(x)\) whose period is \(L\) can be expressed as follows.

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{∞} \left[ {a_{n}cos(\frac{2πnx}{L})}+b_{n}sin(\frac{2πnx}{L}) \right]$$

$$a_0=\frac{2}{L}\int_{-L/2}^{L/2} f(x) dx$$

$$a_n=\frac{2}{L}\int_{-L/2}^{L/2} f(x)cos(\frac{2πnx}{L}) dx$$

$$b_n=\frac{2}{L}\int_{-L/2}^{L/2} f(x)sin(\frac{2πnx}{L}) dx$$

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The Foundation of Determinant

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A 2 × 2 determinant is expressed as follows.

$$\left[\begin{array}{cc}a_{11}&a_{12}\\  a_{21}&a_{22}\\  \end{array}\right]=a_{11}a_{22}-a_{21}a_{12}$$

Then a 3 × 3 determinant is expressed as follows.

$$\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\  a_{21}&a_{22}&a_{23}\\  a_{31}&a_{32}&a_{33}\\ \end{array}\right]\\=a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{23}a_{12}\\-a_{31}a_{22}a_{13}-a_{21}a_{12}a_{33}-a_{11}a_{23}a_{32}$$

So far you can memorize easily, but it is hard to memorize from 4 × 4. In this article, I’ll show you how to find a 4×4 determinant by looking at the elements of the first row.

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Moment of Inertia

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What is the moment of inertia?

Energy is necessary for the moving of stationary objects. Likewise, when a stationary object rotates, it also needs energy. To obtain that energy, we need a physical quantity that represents the difficulty of rotating the object relative to the axis of rotation. To obtain that energy, we need a physical quantity that represents the difficulty of rotating the object relative to the axis of rotation. This difficulty of rotation is called the moment of inertia. While the physical quantity representing the difficulty of moving the object with respect to the force is “mass”, the rotating version thereof is “moment of inertia”.

This moment of inertia \(I_ {j}\) is to be written

$$I_j=m_{j}r_{j}^{2}$$

where the distance from the mass point \(m_ {j}\) to the rotation axis stands for \(r_ {j}\).

All objects can be regarded as a mass of mass. Therefore, we can obtain the moment of inertia \(I\) of any object by integrating the above expression as the integral range of the whole object.

Furthermore, by using this moment of inertia \(I\) and the angular velocity \(ω\) of the rigid body, it is possible to express the kinetic energy \(K\) of rotation of the object.

$$K=\frac{1}{2}I\omega^{2}$$

You can remember this formula easily by comparing the above equation about \(K\) with the below equation about the kinetic energy E that you know.

$$E=\frac{1}{2}mv^{2}$$

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Taylor Expansion

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Taylor expansion around x = a is to express an arbitrary function f (x) in a form like the expression below.

$$f(x)=\sum_{k=0}^{∞}{\frac{f^{(k)}(a)}{k!}(x-a)^k}･･･(1)$$

When a = 0, it can be written as follows. This case is called “Maclaurin’s expansion” specially.

$$f(x)=\sum_{k=0}^{∞}{\frac{f^{(k)}(0)}{k!}x^k}$$

\(f^{(k)}(x)\) means the \(k\)th order derivative of the function \(f(x)\).

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