Category Archives: Puzzles

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.

Remember
tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}
Hence,
tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}
tan(\alpha-\beta)=\frac{2}{1+n^2-1}
tan(\alpha-\beta)=\frac{2}{n^2}
Therefore,
\alpha-\beta=tan^{-1}(\frac{2}{n^2})
and
\alpha=tan^{-1}(n+1) and \beta=tan^{-1}(n-1)

tan^{-1}(n+1)-tan^{-1}(n-1)=tan^{-1}(\frac{2}{n^2})

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

Puzzle Page 2

If x^2-3x+1=0, then find x^5+\frac{1}{x^5}.

My first thought was to solve for x, but it doesn’t factorise easily, and I didn’t want to find the fifth power of an expression involving surds (x=\frac{3\pm \sqrt{5}}{2}), there must be an easier way.

Because x\neq0, we can divide by x

    \begin{equation*}x-3+\frac{1}{x}=0\end{equation}

Hence

(1)   \begin{equation*}x+\frac{1}{x}=3\end{equation*}

What is the expansion of (x+\frac{1}{x})^5?

Using the binomial expansion theorem

    \begin{equation*}(x+\frac{1}{x})^5=x^5+5x^4(\frac{1}{x})+10x^3(\frac{1}{x^2})+10x^2(\frac{1}{x^3})+5x(\frac{1}{x^4})+\frac{1}{x^5}\end{equation}

    \begin{equation*}(x+\frac{1}{x})^5=x^5+\frac{1}{x^5}+5(x^3+\frac{1}{x^3})+10(x+\frac{1}{x})\end{equation}

Therefore

(2)   \begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5(x^3+\frac{1}{x^3})-10(x+\frac{1}{x})\end{equation*}

Let’s do it again for x^3+\frac{1}{x^3}

    \begin{equation*}(x+\frac{1}{x})^3=x^3+3x^2(\frac{1}{x})+3x(\frac{1}{x^2})+\frac{1}{x^3}\end{equation}

(3)   \begin{equation*}x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})\end{equation*}

Substitute 3 into 2

    \begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5((x+\frac{1}{x})^3-3(x+\frac{1}{x}))-10(x+\frac{1}{x})\end{equation}

Remember x+\frac{1}{x}=3

Therefore

    \begin{equation*}x^5+\frac{1}{x^5}=3^5-5(3^3)+15\times3-10\times3\end{equation}

    \begin{equation*}x^5+\frac{1}{x^5}=243-135+45-30=123\end{equation}

This would be a good extension question for students learning the binomial expansion theorem. We also use this technique for trigonometric identities using complex numbers.

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Puzzle Page 1

If \frac{a+b+2c}{a+b-c}=\frac{31}{15}, what does \frac{a+b}{c} equal?

    \begin{equation*}15(a+n+2c)=31(a+b-c)\end{equation}

    \begin{equation*}15a+15b+30c=31a+31b-31c)\end{equation}

    \begin{equation*}61c=16a+16b)\end{equation}

    \begin{equation*}\frac{61}{16}=\frac{a+b}{c}\end{equation}

Two positive numbers are such that their difference, their sum, and their product are in the ratio 2:5:21. What is the smaller of the two numbers?

Let x and y be the two numbers. Then

(1)   \begin{equation*}x-y=2k\end{equation*}

(2)   \begin{equation*}x+y=5k\end{equation*}

(3)   \begin{equation*}xy=21k\end{equation*}

Add equation 1 and 2 together to eliminate the y

    \begin{equation*}2x=7k\end{equation}

(4)   \begin{equation*}x=\frac{7k}{2}\end{equation*}

From 2 =5k-x, substitute for y into equation 3.

(5)   \begin{equation*}x(5k-x)=21k\end{equation*}

Substitute x=\frac{7k}{2} into equation 5.

    \begin{equation*}\frac{7k}{2}(5k-\frac{7k}{2})=21k\end{equation}

    \begin{equation*}\frac{35k^2}{2}-\frac{49k^2}{4}=21k\end{equation}

    \begin{equation*}\frac{70k^2}{4}-\frac{49k^2}{4}=\frac{84k}{4}\end{equation}

    \begin{equation*}21k^2-84k=0\end{equation}

    \begin{equation*}21k(k-4)=0\end{equation}

Hence, k=0 or k=4.

When k=4, x=\frac{7\times 4}{2}=14 and y=5\times 4-14=6

Therefore the smaller number is 6.

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Filed under Algebra, Puzzles, Ratio, Solving Equations