- Thread starter
- #1

- Jan 31, 2012

- 253

I'm tempted to give a hint (or two) right off the bat. But I'll wait.

- Thread starter Random Variable
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 253

I'm tempted to give a hint (or two) right off the bat. But I'll wait.

- Jan 26, 2012

- 890

Would the hınt dırect one ın the towards of the exponentıal representatıon of tan and the use of logs by any chance?

I'm tempted to give a hint (or two) right off the bat. But I'll wait.

CB

- Thread starter
- #3

- Jan 31, 2012

- 253

Would the hınt dırect one ın the towards of the exponentıal representation of tan and the use of logs by any chance?

CB

Yes, I was going to suggest representing the inverse tangent using the complex logarithm, but in a simpler way than it's usually represented. But I was also going to suggest using a trig identity to first write the series in a different form so that's it's not an alternating series.

- Apr 2, 2012

- 9

Hi RV. Cool problem.

$$\frac{\pi}{4}-\tan^{-1}(1/3)+tan^{-1}(1/5)-\tan^{-1}(1/7)+\cdot\cdot\cdot$$

$$\left(\frac{\pi}{4}+\tan^{-1}(1/5)+\tan^{-1}(1/9)+\cdot\cdot\cdot\right) -\left(\tan^{-1}(1/3)+tan^{-1}(1/7)+\tan^{-1}(1/11)+\cdot\cdot\cdot \right)$$

This can now be written as:

$$\sum_{n=0}^{\infty}\left[\tan^{-1}\left(\frac{1}{4n+1}\right)-\tan^{-1}\left(\frac{1}{4n+3}\right)\right]$$

$$\sum_{n=0}^{\infty}\left[\tan^{-1}(4n+3)-\tan^{-1}(4n+1)\right]$$

$$=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{8n^{2}+8n+2}\right)$$

$$=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)$$

Now, we can use the formula:

$$\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{2xy}{(2n+1)^{2}-x^{2}+y^{2}}\right)=\tan^{-1}\left(\tan(\frac{{\pi}x}{2})\tanh(\frac{{\pi}y}{2})\right)$$

Now, let $$x=y=1/2$$ in the formula and we arrive at:

$$\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)=\tan^{-1}\left(\tanh(\frac{\pi}{4})\right)$$

The formula above is derived by using the argument of

$$\cos(\pi z)=\prod_{k=0}^{\infty}\left(1-\frac{4z^{2}}{(2k+1)^{2}}\right)$$

$$\cos(\pi z)=\cos(\pi x)\cosh(\pi y)+\sin(\pi x)\sinh(\pi y)i$$.

and by noting that $$\text{arg}\prod(a+bi)=\sum \tan^{-1}\left(\frac{b}{a}\right)$$

$$\frac{\pi}{4}-\tan^{-1}(1/3)+tan^{-1}(1/5)-\tan^{-1}(1/7)+\cdot\cdot\cdot$$

$$\left(\frac{\pi}{4}+\tan^{-1}(1/5)+\tan^{-1}(1/9)+\cdot\cdot\cdot\right) -\left(\tan^{-1}(1/3)+tan^{-1}(1/7)+\tan^{-1}(1/11)+\cdot\cdot\cdot \right)$$

This can now be written as:

$$\sum_{n=0}^{\infty}\left[\tan^{-1}\left(\frac{1}{4n+1}\right)-\tan^{-1}\left(\frac{1}{4n+3}\right)\right]$$

$$\sum_{n=0}^{\infty}\left[\tan^{-1}(4n+3)-\tan^{-1}(4n+1)\right]$$

$$=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{8n^{2}+8n+2}\right)$$

$$=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)$$

Now, we can use the formula:

$$\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{2xy}{(2n+1)^{2}-x^{2}+y^{2}}\right)=\tan^{-1}\left(\tan(\frac{{\pi}x}{2})\tanh(\frac{{\pi}y}{2})\right)$$

Now, let $$x=y=1/2$$ in the formula and we arrive at:

$$\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)=\tan^{-1}\left(\tanh(\frac{\pi}{4})\right)$$

The formula above is derived by using the argument of

$$\cos(\pi z)=\prod_{k=0}^{\infty}\left(1-\frac{4z^{2}}{(2k+1)^{2}}\right)$$

$$\cos(\pi z)=\cos(\pi x)\cosh(\pi y)+\sin(\pi x)\sinh(\pi y)i$$.

and by noting that $$\text{arg}\prod(a+bi)=\sum \tan^{-1}\left(\frac{b}{a}\right)$$

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