Category Archives: Sequences

December 16, 2024 · 9:19 am

Interesting Sum

$S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2}))$ , find $S$ .

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let $tan(\alpha)=n+1$ and $tan(\beta)=n-1$ .

tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}

tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}

Which means,

$\begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}$

Let’s try a few partial sums

$S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)$

$S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)$

$S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)$

$S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)$

Hence, $S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)$

$S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)$

What happens as $N\rightarrow \infty$ ?

$\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}$

Because we know $tan(\frac{\pi}{2})$ is undefined.

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Filed under Identities, Interesting Mathematics, Puzzles, Sequences, Trigonometry

Tagged as limits, sum, trigonometric identities

November 7, 2024 · 9:51 am

Mathematics Applications – Counting can be Tricky Sometimes (off by one error)

Sequences are part of the Year 12 Mathematics Applications course and sometimes it’s tricky to work out which terms the question requires.

For example, ATAR 2020 Question 11

Judith monitors the water quality in her garden pond at the same time everyday. She likes to maintain the concentration of algae between 200 and 250 unites per 100 litres (L). Her measurements show that the concentration increases daily according to the recursive rule

$C_{n+1}=1.025C_n$ where $C_1=200$ units per 100 L (the minimum concentration)

When the concentration gets above the 250 units per 100 L limit, she treats the water to bring the concentration back to the minimum 200 units per 100 l.

(a) If Judith treated the water on Sunday 6 December 2020, determine

(i) the concentration on Wednesday, 9 December 2020.
(ii) the day when she next treated the water.

(b) During the first week of January 2021, Judith monitored the water and recorded the following readings

Day	1	2	3	4	5	6	7
Concentration (C)	200	206	212.28	218.55	225.10	231.85	238.81

(i) Determine the revised recursive rule.
(ii) If she treated the water on 10 January and went on holiday until 20 January, when she next treated the water, calculate the concentration of the water on her return. Assuming the recursive rule from (b)(i) is used.

I get my students to count on their fingers to ensure they get the correct term or day.

Fibonacci Sequence – Finding the Closed Form

I have been reading An Imaginary Tale – The Story of $\sqrt{-1}$ by Paul J Nahin, which is fabulous. There was a bit in chapter 4 where he found the closed form of the generalised Fibonacci sequence. I thought it would be a good exercise to find the closed from of the Fibonacci sequence.

Just to remind you, the Fibonacci sequence is

$1, 1, 2, 3, 5, 8, 13, 21, ...$

and it is defined recursively

$\begin{equation*}T_{n+2}=T_{n+1}+T_n, T_0=1, T_1=1\end{equation}$

That is, the next term is the sum of the two previous terms, i.e.

$\begin{equation*}T_3=T_2+T_1=1+1=2\end{equation}$

Now the starting off point is slightly dodgy as it involves and educated guess as Paul Nahin writes,

How do I know that works? Because I have seen it before, that’s how! […] There is nothing dishonourable about guessing correct solutions – indeed, great mathematicians and scientists, are invariable great guessers – just as long as eventually the guess is verified to work. The next time you encounter a recurrence formula, you can guess the answer too because then you will have already seen how it works.

We start with $T_n=kz^n$

This means $T_{n+2}=T_{n+1}+T_n$ is $kz^{n+2}=kz^{n+1}+kz^n$

$\begin{equation*}kz^{n+2}=kz^{n+1}+kz^n\end{equation}$

$\begin{equation*}kz^n(z^2-z-1)=0\end{equation}$

$\begin{equation*}z^2-z-1=0\end{equation}$

$\begin{equation*}z=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2}\end{equation}$

$\therefore z=\frac{1+\sqrt{5}}{2}$ or $z=\frac{1-\sqrt{5}}{2}$

Hence $T_n=k_1(\frac{1+\sqrt{5}}{2})^n+k_2(\frac{1-\sqrt{5}}{2})^n$ and we can use the initial conditions $T_0=1$ and $T_1=1$ to find $k_1$ and $k_2$

When $n=0, T_0=1$

(1) $\begin{equation*}1=k_1+k_2\end{equation*}$

When $n=1, T_1=1$

(2) $\begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+k_2(\frac{1-\sqrt{5}}{2})\end{equation*}$

From equation $1$ , $k_2=(1-k_1)$ , substitute into equation $2$

$\begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+(1-k_1)(\frac{1-\sqrt{5}}{2})\end{equation}$

$\begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+\frac{1-\sqrt{5}}{2}-k_1(\frac{1-\sqrt{5}}{2})\end{equation}$

$\begin{equation*}1=k_1\sqrt{5}+\frac{1-\sqrt{5}}{2}\end{equation}$

$\begin{equation*}1-(\frac{1-\sqrt{5}}{2})=\sqrt{5}k_1\end{equation}$

$\begin{equation*}k_1=\frac{1}{\sqrt{5}}(\frac{1}{2}+\frac{\sqrt{5}}{2})\end{equation}$

$\begin{equation*}k_1=\frac{1}{2}(\frac{1}{\sqrt{5}}+1)\end{equation}$

$\begin{equation*}k_1=\frac{1+\sqrt{5}}{2\sqrt{5}}\end{equation}$

$\begin{equation*}k_2=1-\frac{1+\sqrt{5}}{2\sqrt{5}}\end{equation}$

$\begin{equation*}k_2=-(\frac{1-\sqrt{5}}{2\sqrt{5}})\end{equation}$

$\therefore T_n=(\frac{1+\sqrt{5}}{2\sqrt{5}})(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2\sqrt{5}})(\frac{1-\sqrt{5}}{2})^n$

$T_n=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1})$

Does it work?

Remember the sequence is $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...$

If $n=5, T_n=8$

$\begin{equation*}T_5=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^6-\frac{1-\sqrt{5}}{2})^6)\end{equation}$

As you can see it works!

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If $T_m=n$ and $T_n=m$ , then prove that $T_{m+n}=0$ . Here where $T_n$ and $T_m$ are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19