Taylor series

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jee-advanced 2009 Q30 Limit evaluation using series expansion or exponential asymptotics
Let
$$L = \lim _ { x \rightarrow 0 } \frac { a - \sqrt { a ^ { 2 } - x ^ { 2 } } - \frac { x ^ { 2 } } { 4 } } { x ^ { 4 } } , \quad a > 0 .$$
If $L$ is finite, then
(A) $\quad a = 2$
(B) $\quad a = 1$
(C) $\quad L = \frac { 1 } { 64 }$
(D) $\quad L = \frac { 1 } { 32 }$
jee-advanced 2022 Q5 3 marks Limit evaluation using series expansion or exponential asymptotics
If
$$\beta = \lim _ { x \rightarrow 0 } \frac { e ^ { x ^ { 3 } } - \left( 1 - x ^ { 3 } \right) ^ { \frac { 1 } { 3 } } + \left( \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } - 1 \right) \sin x } { x \sin ^ { 2 } x }$$
then the value of $6 \beta$ is $\_\_\_\_$ .
jee-main 2021 Q63 Identify a closed-form function from its Taylor series
If $0 < a , b < 1$, and $\tan ^ { - 1 } a + \tan ^ { - 1 } b = \frac { \pi } { 4 }$, then the value of $( a + b ) - \left( \frac { a ^ { 2 } + b ^ { 2 } } { 2 } \right) + \left( \frac { a ^ { 3 } + b ^ { 3 } } { 3 } \right) - \left( \frac { a ^ { 4 } + b ^ { 4 } } { 4 } \right) + \ldots$ is :
(1) $\log _ { \mathrm { e } } \left( \frac { e } { 2 } \right)$
(2) $e$
(3) $e ^ { 2 } - 1$
(4) $\log _ { e } 2$
jee-main 2023 Q68 Limit evaluation using series expansion or exponential asymptotics
If $\lim _ { x \rightarrow 0 } \frac { e ^ { ax } - \cos ( bx ) - \frac { cx e ^ { cx } } { 2 } } { 1 - \cos ( 2 x ) } = 17$, then $5 a ^ { 2 } + b ^ { 2 }$ is equal to
(1) 64
(2) 72
(3) 68
(4) 76
jee-main 2023 Q71 Limit evaluation using series expansion or exponential asymptotics
$\lim_{x \rightarrow \infty} \frac{(\sqrt{3x+1} + \sqrt{3x-1})^6 + (\sqrt{3x+1} - \sqrt{3x-1})^6}{\left(x + \sqrt{x^2-1}\right)^6 + \left(x - \sqrt{x^2-1}\right)^6} x^3$
(1) is equal to $\frac{27}{2}$
(2) is equal to 9
(3) does not exist
(4) is equal to 27
todai-math 2023 Q5 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
$$I ( t ) = \sum _ { n = - \infty } ^ { \infty } C _ { n } e ^ { i n \omega t }$$
todai-math 2024 Q5 Derive series via differentiation or integration of a known series
Problem 5
I. We consider a continuous and absolutely integrable function $f ( t )$ of a real variable $t$ and denote the Fourier transform of the function $f ( t )$ as $\mathcal { F } \{ f ( t ) \}$. We define a function $F ( \omega )$ by the following formula:
$$F ( \omega ) = \mathcal { F } \{ f ( t ) \} = \int _ { - \infty } ^ { \infty } f ( t ) \exp ( - i \omega t ) \mathrm { d } t$$
where $\omega$ is a real variable and $i$ is the imaginary unit.
  1. We define $g ( t ) = f ( a t )$ for a constant $a$ satisfying $a > 0$ and $$G ( \omega ) = \mathcal { F } \{ g ( t ) \} = \mathcal { F } \{ f ( a t ) \}$$ Express $G ( \omega )$ using the function $F$.
  2. When $f ( t ) = \exp \left( - t ^ { 2 } \right)$ and $a = 2$, sketch the graph of the Fourier transformed functions $F ( \omega )$ and $G ( \omega )$ defined by Equation (2) as a function of $\omega$ to show the difference between them.
  3. We define $h ( t ) = f ( t ) \exp ( - i b t )$ for a constant $b$ satisfying $b > 0$ and $$H ( \omega ) = \mathcal { F } \{ h ( t ) \} = \mathcal { F } \{ f ( t ) \exp ( - i b t ) \}$$ Express $H ( \omega )$ using the function $F$.
  4. When $f ( t ) = \exp \left( - t ^ { 2 } \right)$ and $b = 2$, sketch the graph of the Fourier transformed functions $F ( \omega )$ and $H ( \omega )$ defined by Equation (3) as a function of $\omega$ to show the difference between them.

II. Let $N$ be a positive integer. We define a discrete Fourier transform $D _ { 1 } , \cdots , D _ { N }$ by the following formula:
$$D _ { m } = \frac { 1 } { \sqrt { N } } \sum _ { n = 1 } ^ { N } c _ { n } \exp \left( - i \frac { 2 \pi } { N } n m \right)$$
for a complex sequence $c _ { 1 } , \cdots , c _ { N }$. Here, $m$ is an integer satisfying $1 \leq m \leq N$.
  1. Calculate $S \left( n , n ^ { \prime } \right)$ : $$S \left( n , n ^ { \prime } \right) = \frac { 1 } { N } \sum _ { m = 1 } ^ { N } \exp \left\{ i \frac { 2 \pi } { N } \left( n - n ^ { \prime } \right) m \right\}$$ Here, $n$ is an integer satisfying $1 \leq n \leq N$, and $n ^ { \prime }$ is an integer satisfying $1 \leq n ^ { \prime } \leq N$.
  2. Let $U _ { m n }$ be a complex number satisfying $$D _ { m } = \sum _ { n = 1 } ^ { N } U _ { m n } c _ { n }$$ Show that the matrix $\mathbf { U } = \left[ U _ { m n } \right] _ { 1 \leq m \leq N , 1 \leq n \leq N }$ is a unitary matrix.
  3. Derive an equation for the inverse discrete Fourier transform $c _ { n }$ from $D _ { 1 } , \cdots , D _ { N }$. Here, $n$ is an integer satisfying $1 \leq n \leq N$.
  4. For any complex value $z , \bar { z }$ is the complex conjugate of $z$. We define $Q$ by $$Q = \sum _ { n = 1 } ^ { N } \left( \overline { c _ { n } } c _ { n + 1 } + \overline { c _ { n + 1 } } c _ { n } \right)$$ Express $Q$ in terms of $D _ { m }$ and $\overline { D _ { m } }$. Here, we impose the condition $c _ { N + 1 } = c _ { 1 }$.
todai-math 2025 Q5 Derive series via differentiation or integration of a known series
Problem 5
Consider a function $f ( t )$ of a real number $t$, where $| f ( t ) |$ and $| f ( t ) | ^ { 2 }$ are integrable. Let $F ( \omega ) = \mathcal { F } [ f ( t ) ]$ denote the Fourier transform of $f ( t )$. It is defined as
$$F ( \omega ) = \mathcal { F } [ f ( t ) ] = \int _ { - \infty } ^ { \infty } f ( t ) \exp ( - i \omega t ) \mathrm { d } t$$
where $\omega$ is a real number and $i$ is the imaginary unit. Then, the following equation is satisfied:
$$\int _ { - \infty } ^ { \infty } | F ( \omega ) | ^ { 2 } \mathrm {~d} \omega = 2 \pi \int _ { - \infty } ^ { \infty } | f ( t ) | ^ { 2 } \mathrm {~d} t$$
Also, let $R _ { f } ( \tau )$ denote the autocorrelation function of $f ( t )$. It is defined as
$$R _ { f } ( \tau ) = \int _ { - \infty } ^ { \infty } f ( t ) f ( t - \tau ) \mathrm { d } t$$
where $\tau$ is a real number.
I. Consider a case where $f ( t )$ is defined as follows:
$$f ( t ) = \begin{cases} \cos ( a t ) & \left( | t | \leq \frac { \pi } { 2 a } \right) \\ 0 & \left( | t | > \frac { \pi } { 2 a } \right) \end{cases}$$
Here, $a$ is a positive real constant. Find the followings:
  1. $F ( \omega )$,
  2. $R _ { f } ( \tau )$,
  3. $\mathcal { F } \left[ R _ { f } ( \tau ) \right]$.

II. Find the values of the following integrals. Here, you may use the results of I.
  1. $\int _ { - \infty } ^ { \infty } \frac { \cos ^ { 2 } \frac { \pi x } { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 2 } } \mathrm {~d} x$,
  2. $\int _ { - \infty } ^ { \infty } \frac { \cos ^ { 4 } \frac { \pi x } { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 4 } } \mathrm {~d} x$.