Question

0<x<2π aralığındaki f(x)=x÷2 fonksiyonunun 0 < x < 4π için grafiğini çiziniz ve 0<x<2n için Fourier seri dönüşümünü gerçekleştiriniz.

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Answer to a math question 0<x<2π aralığındaki f(x)=x÷2 fonksiyonunun 0 < x < 4π için grafiğini çiziniz ve 0<x<2n için Fourier seri dönüşümünü gerçekleştiriniz.

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Bud
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Öncelikle, f(x) = x/2 fonksiyonunun grafiğini çizelim.

0 < x < 2π aralığı için f(x) = x/2 fonksiyonunun grafiği aşağıdaki gibi olacaktır:

\begin{align*} \text{Grafiğin üzerindeki noktalar:} \ (0, 0), (\pi/2, \pi/4), (\pi, \pi/2), (3\pi/2, 3\pi/4), (2\pi, \pi) \ \text{Ve bu noktaları birleştiren bir doğru elde edeceğiz.}\end{align*}

Grafiği çizdikten sonra, Fourier serisi dönüşümünü gerçekleştirelim.

Fourier Serisi Dönüşümü için a0, an ve bn katsayılarını bulmamız gerekiyor. Bunları hesaplayalım:

\begin{align*} a0 = \frac{1}{\pi} \int_{0}^{2\pi} f(x) dx = \frac{1}{\pi} \int_{0}^{2\pi} \frac{x}{2} dx = \frac{1}{2\pi} \left[ \frac{1}{2} x^2 \right]_{0}^{2\pi} = \frac{1}{2\pi} \left[ \frac{1}{2} (2\pi)^2 \right] = \frac{\pi}{2}\end{align*}

\begin{align*} an = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \cos(nx) dx = \frac{1}{\pi} \int_{0}^{2\pi} \frac{x}{2} \cos(nx) dx\end{align*}

Burada x/2'nin çift veya tek olduğunu kontrol etmeliyiz. Eğer x/2 çift ise, an = 0 olacaktır. Eğer x/2 tek ise, an kısmını hesaplamalıyız:

\begin{align*} an = \frac{1}{\pi} \int_{0}^{2\pi} \frac{x}{2} \cos(nx) dx = \frac{1}{2\pi} \left[ x \sin(nx) + \frac{1}{n} \cos(nx) \right]_{0}^{2\pi} = \frac{1}{2n\pi} (0 + \frac{1}{n} - (0 - \frac{1}{n})) = \frac{2}{n^2\pi}\end{align*}

\begin{align*} bn = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \sin(nx) dx = \frac{1}{\pi} \int_{0}^{2\pi} \frac{x}{2} \sin(nx) dx\end{align*}

Burada x/2'nin çift veya tek olduğunu kontrol etmeliyiz. Eğer x/2 çift ise, bn kısmını hesaplamalıyız:

\begin{align*} bn = \frac{1}{\pi} \int_{0}^{2\pi} \frac{x}{2} \sin(nx) dx = \frac{1}{2\pi} \left[ - x \cos(nx) + \frac{1}{n} \sin(nx) \right]_{0}^{2\pi} = \frac{1}{2n\pi} (0 - (\pi\cos(2n\pi) - 0)) = \frac{1}{2n\pi} \pi = \frac{1}{2n}\end{align*}

Bu şekilde Fourier serisi dönüşümünü elde ettik.

Sonuç olarak, 0 < x < 4π aralığı için f(x) = x/2 fonksiyonunun grafiği aşağıdaki gibi olacaktır:

\begin{align*} \text{Grafiğin üzerindeki noktalar:} \ (0, 0), (\pi/2, \pi/4), (\pi, \pi/2), (3\pi/2, 3\pi/4), (2\pi, \pi), (5\pi/2, 5\pi/4), (3\pi, 3\pi/2), (7\pi/2, 7\pi/4), (4\pi, \pi) \ \text{Ve bu noktaları birleştiren bir doğru elde edeceğiz.}\end{align*}

Fourier serisi dönüşümü için:

\begin{align*} f(x) = \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2}{n^2\pi} \cos(nx) + \frac{1}{2n} \sin(nx)\end{align*}

Yukarıdaki seriyi 0 < x < 2n için gösterdik.

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