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Trying to remember the formula to calculate sin (in radians). I remember it was an infinite sum of increasingly small fractions that would eventually iterate closer and closer to the answer.

It bothers me more that I never understood why the formula worked. If I did, I could just work it out myself (like the quadratic equation).

in reply to Jonathan Lamothe

Matthew Skala
Matthew Skala
• •
sin has "odd" symmetry (sin x = -sin -x) which means only odd powers of x in the power series. derivative of sin at x=0 is 1, which means the coefficient of x^1 has to be 1 (the other terms necessarily having zero derivative). Second derivative of sin is -sin. From that you can derive the entire series by writing the first few terms with unknowns for the coefficients, taking the derivative twice, and finding the pattern that tells you each coefficient from the previous one.
in reply to Matthew Skala

Matthew Skala
Matthew Skala
• •
on a deeper level what's going on is that sin is really exp in disguise and you're just substituting x*sqrt(-1) into the power series for exp.
in reply to Jonathan Lamothe

John Best
John Best
• •

Maybe you’re looking for the Taylor Series?
en.m.wikipedia.org/wiki/Taylor…

Here’s the derivation for sine:

ocw.mit.edu/courses/18-01sc-si…

expression of a function as an infinite sum

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in reply to John Best

Jonathan Lamothe
Jonathan Lamothe
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@John Best Based also on another response, I'm inclined to answer yes.

Now I just need to make my brain understand why it works. I'm sure I'll be able to if I just have a chance to sit down and think about it. I'm already 90% there, I think.

@John Best
in reply to Jonathan Lamothe

Khurram Wadee
Khurram Wadee
• •
-x^(3) /3!+x^5 /5!-x^7 /7!+x^9 /9!+…
nth term is (-1)^(n-1) x^(2n-1)/(2n-1)!
It’s valid no matter how large x is (i.e. it has an infinite radius of convergence).

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