todai-math 2023 Q5

todai-math · Japan · todai-engineering-math Taylor series Derive series via differentiation or integration of a known series
Answer all the following questions.
I. Let a periodic function $f ( x )$ satisfy the condition $f ( x + \pi ) = f ( x - \pi )$. Find the Fourier series expansion of $f ( x )$ for each case, where $f ( x )$ is expressed as follows for the interval $- \pi \leq x \leq \pi$.
1. $f ( x ) = x \quad ( - \pi < x < \pi ) , \quad f ( - \pi ) = f ( \pi ) = 0$ 2. $f ( x ) = x ^ { 2 }$
For the Fourier series expansion, the following equations should be used.
$$\begin{aligned} & f ( x ) = \frac { a _ { 0 } } { 2 } + \sum _ { n = 1 } ^ { \infty } \left( a _ { n } \cos n x + b _ { n } \sin n x \right) \\ & a _ { 0 } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \mathrm { d } x \\ & a _ { n } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \cos n x \mathrm {~d} x \\ & b _ { n } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \sin n x \mathrm {~d} x \end{aligned}$$
II. Consider the function shown in Figure 5.1 as
$$V ( t ) = A \left| \sin \left( \frac { \omega t } { 2 } \right) \right| \quad \left( \omega = \frac { 2 \pi } { T } \right) .$$
Note that $A$ and $T$ are positive real numbers. The complex Fourier series expansion of $V ( t )$ is given as
$$V ( t ) = - \frac { 2 A } { \pi } \sum _ { n = - \infty } ^ { \infty } \frac { 1 } { 4 n ^ { 2 } - 1 } e ^ { i n \omega t }$$
Let $I ( t )$ be the periodic solution that satisfies the ordinary differential equation
$$L \frac { \mathrm {~d} I ( t ) } { \mathrm { d } t } + R I ( t ) = V ( t )$$
Note that $L$ and $R$ are positive real numbers. Find the coefficient $C _ { n }$, when the complex Fourier series expansion of $I ( t )$ is expressed as
$$I ( t ) = \sum _ { n = - \infty } ^ { \infty } C _ { n } e ^ { i n \omega t }$$
Answer all the following questions.

I. Let a periodic function $f ( x )$ satisfy the condition $f ( x + \pi ) = f ( x - \pi )$. Find the Fourier series expansion of $f ( x )$ for each case, where $f ( x )$ is expressed as follows for the interval $- \pi \leq x \leq \pi$.

1. $f ( x ) = x \quad ( - \pi < x < \pi ) , \quad f ( - \pi ) = f ( \pi ) = 0$
2. $f ( x ) = x ^ { 2 }$

For the Fourier series expansion, the following equations should be used.

$$\begin{aligned}
& f ( x ) = \frac { a _ { 0 } } { 2 } + \sum _ { n = 1 } ^ { \infty } \left( a _ { n } \cos n x + b _ { n } \sin n x \right) \\
& a _ { 0 } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \mathrm { d } x \\
& a _ { n } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \cos n x \mathrm {~d} x \\
& b _ { n } = \frac { 1 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) \sin n x \mathrm {~d} x
\end{aligned}$$

II. Consider the function shown in Figure 5.1 as

$$V ( t ) = A \left| \sin \left( \frac { \omega t } { 2 } \right) \right| \quad \left( \omega = \frac { 2 \pi } { T } \right) .$$

Note that $A$ and $T$ are positive real numbers. The complex Fourier series expansion of $V ( t )$ is given as

$$V ( t ) = - \frac { 2 A } { \pi } \sum _ { n = - \infty } ^ { \infty } \frac { 1 } { 4 n ^ { 2 } - 1 } e ^ { i n \omega t }$$

Let $I ( t )$ be the periodic solution that satisfies the ordinary differential equation

$$L \frac { \mathrm {~d} I ( t ) } { \mathrm { d } t } + R I ( t ) = V ( t )$$

Note that $L$ and $R$ are positive real numbers.
Find the coefficient $C _ { n }$, when the complex Fourier series expansion of $I ( t )$ is expressed as

$$I ( t ) = \sum _ { n = - \infty } ^ { \infty } C _ { n } e ^ { i n \omega t }$$