$|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$
is equal to ________
Function \(f(x)\) Simplification
The given function is $f(x) = x^2 + 4 \sin (x) \cos(x)$. Using the trigonometric identity $2 \sin(x) \cos(x) = \sin(2x)$, the function can be rewritten as:
$f(x) = x^2 + 2 \sin(2x)$
\(a_n\) Coefficients Sum Calculation
The Fourier series is given by $S(x) = a_0 + \sum_{n=1}^\infty(a_n \cos (n x) + b_n \sin (n x))$.
The sum $\sum_{n=0}^\infty a_n$ is equivalent to $a_0 + \sum_{n=1}^\infty a_n$, which equals $S(0)$.
For a function whose Fourier series converges at $x=0$, $S(0) = f(0)$.
Evaluating $f(x)$ at $x=0$: $f(0) = (0)^2 + 2 \sin(2 \cdot 0) = 0 + 2 \sin(0) = 0$
Therefore, $\sum_{n=0}^\infty a_n = 0$.
\(b_n\) Coefficients Sum Calculation
The coefficient $b_n$ is defined as $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx$. Substituting $f(x) = x^2 + 2 \sin(2x)$:
$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (x^2 + 2 \sin(2x)) \sin(nx) dx$
This integral can be split into two parts: $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin(nx) dx + \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \sin(2x) \sin(nx) dx$
The first integral, $\int_{-\pi}^{\pi} x^2 \sin(nx) dx$, evaluates to zero because $x^2 \sin(nx)$ is an odd function integrated over the symmetric interval $[-\pi, \pi]$.
The second integral, $\frac{1}{\pi} \int_{-\pi}^{\pi} 2 \sin(2x) \sin(nx) dx$, relies on the orthogonality property of sine functions: $\int_{-\pi}^{\pi} \sin(mx) \sin(nx) dx = \pi \delta_{mn}$ for $m, n \ge 1$, where $\delta_{mn}$ is the Kronecker delta.
- This integral is non-zero only when $n=2$.
- For $n=2$: $b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \sin^2(2x) dx$. Using $\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}$, this becomes $b_2 = \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{1 - \cos(4x)}{2} dx = \frac{1}{\pi} [x - \frac{1}{4}\sin(4x)]_{-\pi}^{\pi} = \frac{1}{\pi} ((\pi - 0) - (-\pi - 0)) = \frac{2\pi}{\pi} = 2$.
- For $n \ne 2$, $b_n = 0$ due to orthogonality.
Thus, the only non-zero $b_n$ coefficient for $n \ge 1$ is $b_2=2$. Therefore, $\sum_{n=1}^\infty b_n = 2$.
Expression \(|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|\) Value Calculation
The question requires calculating the absolute difference $|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$.
Substituting the calculated values: $|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n| = |0 - 2| = |-2| = 2$
The final value is 2.
Important Questions from Series
- In the sequence 6, 9, 14, $x$, 30, 41, a possible value of $x$ is
Let $a_0 = 0$ and define $a_n = \frac{1}{2}(1 + a_{n-1})$ for all positive integers $n \ge 1$.
The least value of $n$ for which $|1 - a_n| < \frac{1}{2^{10}}$ is __________.
(Answer in integer)
- Calculate the reciprocal of the coefficient of $z^3$ in the Taylor series expansion of the function $f(z) = \sin(z)$ around $z = 0$. (Provide the answer as an integer.)
- Let $a_1 = 1$ and $a_n = a_{n-1} + 4$, $n \ge 2$. Then,
$\lim_{n\to\infty} \left[\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \dots + \frac{1}{a_{n-1}a_n}\right]$
is equal to ________ Let $S_n = \sum_{k=1}^n \frac{1}{k}$ and $I_n = \int_1^n \frac{x - [x]}{x^2} dx$. Then, $S_{10} + I_{10}$ is equal to
