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)
Analyzing the Recursive Sequence
We are given a sequence defined by $a_0 = 0$ and the recurrence relation $a_n = \frac{1}{2}(1 + a_{n-1})$ for $n \ge 1$. We need to find the least integer $n$ such that $|1 - a_n| < \frac{1}{2^{10}}$.
Calculating Sequence Terms
Let's compute the first few terms to find a pattern:
- $a_0 = 0$
- $a_1 = \frac{1}{2}(1 + a_0) = \frac{1}{2}(1 + 0) = \frac{1}{2}$
- $a_2 = \frac{1}{2}(1 + a_1) = \frac{1}{2}(1 + \frac{1}{2}) = \frac{1}{2}(\frac{3}{2}) = \frac{3}{4}$
- $a_3 = \frac{1}{2}(1 + a_2) = \frac{1}{2}(1 + \frac{3}{4}) = \frac{1}{2}(\frac{7}{4}) = \frac{7}{8}$
Deriving the Closed-Form Formula
Observe the pattern:
- $a_1 = \frac{1}{2} = 1 - \frac{1}{2}$
- $a_2 = \frac{3}{4} = 1 - \frac{1}{4}$
- $a_3 = \frac{7}{8} = 1 - \frac{1}{8}$
This suggests the formula $a_n = 1 - \frac{1}{2^n}$ for $n \ge 1$. Let's verify this using the recurrence relation. If $a_{n-1} = 1 - \frac{1}{2^{n-1}}$, then $a_n = \frac{1}{2}(1 + a_{n-1}) = \frac{1}{2}(1 + (1 - \frac{1}{2^{n-1}})) = \frac{1}{2}(2 - \frac{1}{2^{n-1}}) = 1 - \frac{1}{2} \cdot \frac{1}{2^{n-1}} = 1 - \frac{1}{2^n}$. This formula holds for $n \ge 1$. It also holds for $n=0$ since $a_0 = 1 - \frac{1}{2^0} = 1 - 1 = 0$.
Therefore, the closed-form formula for the sequence is $a_n = 1 - \frac{1}{2^n}$ for all $n \ge 0$.
Solving the Inequality
We need to find the least $n$ such that $|1 - a_n| < \frac{1}{2^{10}}$.
Substitute the formula for $a_n$: $|1 - (1 - \frac{1}{2^n})| < \frac{1}{2^{10}}$ $|\frac{1}{2^n}| < \frac{1}{2^{10}}$
Since $n$ is a positive integer, $\frac{1}{2^n}$ is positive. The inequality simplifies to:
$ \frac{1}{2^n} < \frac{1}{2^{10}} $For this inequality to hold, the exponent $n$ must be greater than the exponent $10$.
$ n > 10 $Determining the Least Integer Value
The least integer value of $n$ that satisfies the condition $n > 10$ is $n = 11$. This value is a positive integer, consistent with the problem statement ($n \ge 1$).
Important Questions from Series
- In the sequence 6, 9, 14, $x$, 30, 41, a possible value of $x$ is
- 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 $S(x) = a_0 + \sum_{n=1}^\infty(a_n \cos (n x) + b_n \sin (n x))$ be the Fourier series of the$2 \pi$ periodic function defined by $f(x) = x^2 + 4 \sin (x) \cos(x)$, $-\pi \le x \le \pi$. Then
$|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$
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
