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I need to evaluate $$\sum_{n=1}^m 2^n \arctan 2^n \theta$$ as a function of $m$ and $\theta$. All I've done so far is write out the series explicitly:

$$\sum_{n=1}^m 2^n \arctan 2^n \theta = 2 \arctan 2\theta + 4\arctan 4\theta + 8\arctan 8\theta + \cdots + 2^m \arctan 2^m \theta$$

and I initially considered pairing every two terms up to use the $\arctan x + \arctan y$ trick, but it doesn't work because each $\arctan$ term has a different coefficient.

Zain Patel
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  • you could tr the trick with splitting up the summation into multiple summations, if only the coefficients were the problem. – MrYouMath Dec 17 '15 at 12:56
  • $f_{m}(\theta)=2f_{m-1}(2\theta) + 2\arctan 2\theta$. Not sure whether that helps. – Thomas Andrews Dec 17 '15 at 12:58
  • @ThomasAndrews, am I misunderstanding?

    $2^m \arctan 2^m \theta = 2 \cdot 2^{m-1} \arctan (2^{m-1} \cdot 2\theta) + 2\arctan 2\theta$, surely that's not correct?

     – Zain Patel Dec 17 '15 at 13:02
  • I was implicitly defining $f_m(\theta)$ as your sum, not as $2^m\arctan(2^m\theta)$. @ZainPatel – Thomas Andrews Dec 17 '15 at 13:06
  • @ThomasAndrews, ah - my bad. I shall work with that and see if anything comes to light. :-) – Zain Patel Dec 17 '15 at 13:07
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    It's a very strange variable choice that the argument to $\arctan$ is some integer multiple of $\theta$. The argument to $\arctan$ is unit-less, while $\theta$ is usually an angle. It's just weird to use them together. Not wrong, per se, but might indicate an error, if this question comes in the middle of some other work. – Thomas Andrews Dec 17 '15 at 13:11
  • It's just a pure summation question, where the argument is used because of the trigonometrical setting. At least, I'd hope so... – Zain Patel Dec 17 '15 at 13:12
  • @ZainPatel But the input to $\arctan$ is a real number, not an angle. It is the output of it which is an angle. – coffeemath Dec 17 '15 at 13:40

1 Answers1

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Not for nothing, but just in case the OP really wants to evaluate

$$\sum_{n=1}^m 2^n \tan{2^n \theta} $$

then use the fact that

$$\cot{x}-2 \cot{2 x} = \tan{x} $$

and let $x=2^n \theta$. In this case, we get a telescoping sum with the result

$$\sum_{n=1}^m 2^n \tan{2^n \theta} = 2\cot{2 x} - 2^{m+1} \cot{2^{m+1} \theta}$$

Panglossian Oporopolist
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Ron Gordon
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