I'm getting ready to present my superconductor research[1] at the Texas Academy of Science meeting this weekend, so today's post is just a quick note on an easy way to find a series expansion for the arctangent.

Arctan(x) looked a little daunting to get into a series at first. The key to get started is to remember what the derivative of arctan is (picture 1)

At least the derivative is something that looks like it could be pretty easily worked into a series. With a little more thought, it turns out that the anti-derivative, (the integral), of the result above is arctan. So, the process will be to turn the derivative into a series first, and then take the integral of that series to wind up with the series for arctan.

The result of the derivative shown above is a prime candidate for a geometric series expansion(picture 2)

Now that we have the series, all that remains is to integrate and we get (picture 3):

So (picture 4),

1. http://arxiv.org/abs/1208.1870

Arctan(x) looked a little daunting to get into a series at first. The key to get started is to remember what the derivative of arctan is (picture 1)

At least the derivative is something that looks like it could be pretty easily worked into a series. With a little more thought, it turns out that the anti-derivative, (the integral), of the result above is arctan. So, the process will be to turn the derivative into a series first, and then take the integral of that series to wind up with the series for arctan.

The result of the derivative shown above is a prime candidate for a geometric series expansion(picture 2)

Now that we have the series, all that remains is to integrate and we get (picture 3):

So (picture 4),

**References:**1. http://arxiv.org/abs/1208.1870

**Picture of the Day (picture 5):**From 2/27/13 |

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