From e5fce81196c58522385c7c3d881c3a7d38078f7a Mon Sep 17 00:00:00 2001 From: Razvan Deaconescu Date: Wed, 13 Sep 2023 05:44:47 +0300 Subject: [PATCH] Add content for functions/sequences-series Add actual content to the skeleton of the `functions/sequences-series` section. Signed-off-by: Eggert Karl Hafsteinsson Signed-off-by: Teodor Dutu Signed-off-by: Razvan Deaconescu --- .../sequences-series/reading/README.md | 166 ++++++++++++++++++ 1 file changed, 166 insertions(+) diff --git a/chapters/functions/sequences-series/reading/README.md b/chapters/functions/sequences-series/reading/README.md index 6221243..4a76af3 100644 --- a/chapters/functions/sequences-series/reading/README.md +++ b/chapters/functions/sequences-series/reading/README.md @@ -1 +1,167 @@ # Sequences and Series + +## Sequences + +A **sequence** is a string of indexed numbers $a_1, a_2, a_3, \ldots$. +We denote this sequence with $(a_n)_{n\geq1}$. + +### Details + +In a sequence the same number can appear several times in different places. + +### Examples + +:::info Example + +$$(\displaystyle\frac{1}{n})_{n\geq1} \text{is the sequence} 1, \displaystyle\frac{1}{2}, \displaystyle\frac{1}{3}, \displaystyle\frac{1}{4}, \ldots$$ + +::: + +:::info Example + +$$(n)_{n\geq1} \text{is the sequence } 1, 2, 3, 4, 5, \ldots$$ + +::: + +:::info Example + +$$(2^nn)_{n\geq1} \text{is the sequence} 2, 8, 24, 64, \ldots$$ + +::: + +## Convergent Sequences + +A sequence $a_n$ is said to **converge** to the number $b$ if for every $\varepsilon >0$ we can find an $N\in \mathbb{N}$ such that $|a_n-b| < \varepsilon$ for all $n \geq N$. +We denote this with $\lim_{n\to\infty}a_n=b$ or $a_n\to b$, as $n\to\infty$. + +### Details + +A sequence $a_n$ is said to **converge** to the number $b$ if for every $\varepsilon >0$ we can find an $N\in \mathbb{N}$ such that $|a_n-b| < \varepsilon$ for all $n \geq N$. +We denote this with $\lim_{n\to\infty}a_n=b$ or $a_n\to b$, as $n\to\infty$. + +If $x$ is a number, then + +$$(1 + \displaystyle\frac{x}{n})^n \rightarrow e^x \text{as} n\rightarrow\infty$$ + +### Examples + +:::info Example + +The sequence $(\displaystyle\frac{1}{n})_{n\geq\infty}$ converges to $0$ as $n\to\infty$. + +::: + +:::info Example + +If x is a number, then + +$$(1 + \displaystyle\frac{x}{n})^n \rightarrow e^x \text{as} n\rightarrow\infty$$ + +::: + +## Infinite Sums (series) + +We are interested in, whether infinite sums of sequences can be defined. + +### Details + +Consider a sequence of numbers, $(a_n)_{n\to\infty}$. + +Now define another sequence $(s_n)_{n\to\infty},$ where + +$$s_n=\displaystyle\sum_{k=1}^na_k$$ + +If $(s_n)_{n\to\infty}$ is convergent to $S=\lim_{n\to\infty}s_n,$ then we write + +$$S=\displaystyle\sum_{n=1}^{\infty}a_n$$ + +### Examples + +:::info Example + +If + +$$a_k = x^k, qquad k=0,1,\dots$$ + +then + +$$s_n=\displaystyle\sum_{k=0}^{n}x^k=x^0+x^1+\dots.+x^n$$ + +Note also that + +$$xs_n=x(x^0+x^1+\dots.+x^n)= x + x^2 + \dots + x^{n+1}$$ + +We have + +$$s_n = 1 + x + x^2 + \dots + x^n$$ + +$$xs_n = x + x^2 + \dots +x^n + x^{n+1}$$ + +$$s_n – xs_n = 1 - x^{n+1}$$ + +i.e. + +$$s_n(1-x) = 1-x^{n+1}$$ + +and we have + +$$s_n =\displaystyle\frac{1-x^{n+1}}{1-x}$$ + +if $x\neq1$. + +If $0< x<1$ then $x^{n+1}\to 0$ as $n\to\infty$ and we obtain $s_n\to\displaystyle\frac{1}{1-x}$ so + +$$\displaystyle\sum_{n=0}^{\infty}x^n=\displaystyle\frac{1}{1-x}$$ + +::: + +## The Exponential Function and the Poisson Distribution + +The exponential function can be written as a series (infinite sum): + +$$e^x=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{x^n}{n!}$$ + +The Poisson distribution is defined by the probabilities + +$$p(x)=e^{-\lambda}\displaystyle\frac{\lambda^x}{x!}\textrm{ for } x=0,\ 1,\ 2,\ \ldots$$ + +### Details + +The exponential function can be written as a series (infinite sum): + +$$e^x=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{x^n}{n!}$$ + +Knowing this we can see why the Poisson probabilities + +$$p(x)=e^{-\lambda}\displaystyle\frac{\lambda^x}{x!}$$ + +add to one: + +$$\displaystyle\sum_{x=0}^{\infty}p(x)=\displaystyle\sum_{x=0}^{\infty}e^{-\lambda}\displaystyle\frac{\lambda^x}{x!}=e^{-\lambda}\displaystyle\sum_{x=0}^{\infty}\displaystyle\frac{\lambda^x}{x!}=e^{-\lambda}e^{\lambda}=1$$ + +## Relation to Expected Values + +The expected value for the Poisson is given by + +$$ +\begin{aligned} + \displaystyle\sum_{x=0}^\infty x p(x) &= \displaystyle\sum_{x=0}^\infty x e^{-\lambda} \displaystyle\frac{\lambda^x}{x!} \\ + &= \lambda +\end{aligned} +$$ + +### Details + +The expected value for the Poisson is given by + +$$ +\begin{aligned} + \displaystyle\sum_{x=0}^\infty x p(x) &= \displaystyle\sum_{x=0}^\infty x e^{-\lambda} \displaystyle\frac{\lambda^x}{x!} \\ + &= e^{-\lambda} \displaystyle\sum_{x=1}^\infty \displaystyle\frac{x\lambda^x}{x!} \\ + &= e^{-\lambda} \displaystyle\sum_{x=1}^\infty \displaystyle\frac{\lambda^x}{(x-1)!} \\ + &= e^{-\lambda} \lambda \displaystyle\sum_{x=1}^\infty \displaystyle\frac{\lambda^{(x-1)}}{(x-1)!} \\ + &= e^{-\lambda} \lambda \displaystyle\sum_{x=0}^\infty \displaystyle\frac{\lambda^{x}}{x!} \\ + &= e^{-\lambda} \lambda e^{\lambda} \\ + &= \lambda +\end{aligned} +$$