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H. C. Chan, $\pi$ in terms of $\phi$: Some Recent Developments, Proceedings of the Twelfth International Conference in Fibonacci Numbers, (2010): 17-25. Read Pi in terms of Phi (Fib Conf 2006).

H. C. Chan, $\pi$ in terms of $\phi$, Fibonacci Quart. 44 (2006): 141–144. Read Pi in terms of phi.

H. C. Chan, More Formulas for $\pi$, Amer. Math. Monthly 113: 452-455. Read More formulas for Pi.

H. C. Chan, Machin-type formulas expressing $\pi$ in terms of $\phi$, Fibonacci Quart. 46/47 (2008/2009): 32–37 Read Pi via Machin.

A post from one of the blogs I follow, which happens to combine my love of mathematics and programming. Check it out! 🙂

After a year of writing this blog, what have I learned about the nature of the relationship between computer programs and mathematics? Here are a few notes that sum up my thoughts, roughly in order of how strongly I agree with them. I’d love to hear your thoughts in the comments.

1. Programming is absolutely great for exploring questions and automating tasks. Mathematics is absolutely great for distilling the soul of a problem.
2. Programming is fueled by the excitement of what can be done. Mathematics is fueled by the excitement of how things relate, and why they relate.
3. Good mathematics makes for short programs.
4. Good mathematics can be sloppy. Good programs cannot.
5. Useful algorithms can come from any branch of mathematics, so it is best to be familiar with them all (at least a little).
6. Most programs written for the real world use no mathematics beyond the level of an average twelve-year-old, but

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Deriving double angle trigonometric formulas using transformations. A really nice elementary proof.

Activities, games, lessons, hands-on fun — the Math Teachers at Play blog carnival would love to feature your article about mathematics from preschool to precollege level (through the first year of calculus). You can submit your article online, or email John directly to make sure he gets your submission.

The carnival will be published Friday at Math Hombre.

I wrote this note in 2008 to introduce complex numbers. It is posted for the Math Teachers at Play blog carnival.

What is the ${\sqrt[4]{-16}}$?

In the set of Real Numbers, ${\mathbb{R}}$, the solution is undefined. We must consider the set of Complex Numbers, ${\mathbb{C}}$. Complex numbers are numbers of the form of ${a+bi}$, where ${a}$ and ${b}$ are real numbers: ${a,b\in\mathbb{R}}$ and ${i}$ is the imaginary unit. Note ${i}$ is defined by ${i^{2}=-1}$ or equivalently ${i=\sqrt{-1}}$.

Thus, we have:

$\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & (-16)^{1/4}\\ & = & (-1\cdot16)^{1/4}\\ & = & (-1\cdot2^{4)^{1/4}}\\ & = & (-1)^{1/4}\cdot2^{4/4}\\ & = & (-1)^{1/4}\cdot2\\ & = & ((-1)^{1/2})^{1/2}\cdot2\\ & = & (\sqrt{-1})^{1/2}\cdot2\\ & = & i^{1/2}\cdot2\\ & = & 2\sqrt{i}.\end{array}$

We now have ${\sqrt[4]{-16}=2\sqrt{i}}$. So we must ask "What is the ${\sqrt{i}}$?". To evaluate ${\sqrt{i}}$ we will use Euler’s formula, ${e^{ix}=\cos\left(x\right)+i\sin\left(x\right).}$ So by substituting ${x=\frac{\pi}{2}}$, giving

$\displaystyle \begin{array}{rcl} e^{i(\frac{\pi}{2})} & = & \cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)\\ & = & 0+i(1)\\ & = & i.\end{array}$

Note we have isolated ${i}$ and now arrived at the following equality: ${i=e^{i(\frac{\pi}{2})}}$. Taking the square root of both sides gives

$\displaystyle \begin{array}{rcl} \sqrt{i} & = & \sqrt{e^{i(\frac{\pi}{2})}}\\ & = & e^{i(\frac{\pi}{2})(\frac{1}{2})}\\ & = & e^{i(\frac{\pi}{4})}.\end{array}$

Thus, we have now have ${\sqrt{i}=e^{i(\frac{\pi}{4})}}$ and following through application of Euler’s formula to ${x=\frac{\pi}{4}}$, gives

$\displaystyle \begin{array}{rcl} \sqrt{i} & = & e^{i(\frac{\pi}{4})}\\ & = & \cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\\ & = & \frac{1}{\sqrt{2}}+i(\frac{1}{\sqrt{2}})\\ & = & \frac{\sqrt{2}}{2}+i(\frac{\sqrt{2}}{2})\\ & = & \frac{\sqrt{2}+\sqrt{2}i}{2}.\end{array}$

So going back to the original problem and substituting ${\sqrt{i}=\frac{\sqrt{2}+\sqrt{2}i}{2}}$, gives

$\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & 2\sqrt{i}\\ & = & 2(\frac{\sqrt{2}+\sqrt{2}i}{2})\\ & = & \sqrt{2}+\sqrt{2}i.\end{array}$

Hence, the solution, ${\sqrt[4]{-16}=\sqrt{2}+\sqrt{2}i}$ , is a complex number!

In college, we often learn of many infinite series that give the value of pi including one called the Leibniz series, named after Gottfried Leibniz. It is also called the Gregory–Leibniz series, recognizing the work of James Gregory. This unnecessarily attributes the discovery to the west, however, the formula was first discovered in India by Madhava of Sangamagrama and so is also called the Madhava–Leibniz series. Indian mathematicians made vast and fundamental contributions to our modern mathematics.

$\pi=4\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}$

“Why aren’t we giving our students a chance to even hear about these things, let alone giving them an opportunity to actually do some mathematics, and to come up with their own ideas, opinions, and reactions? What other subject is routinely taught without any mention of its history, philosophy, thematic development, aesthetic criteria, and current status? What other subject shuns its primary sources— beautiful works of art by some of the most creative minds in history— in favor of third-rate textbook bastardizations?”