By which way I can prove $\arctan\left(2^{\sqrt{3}-\sqrt{7}}\right)<\frac{61}{125}$

Asked Apr 09 '15 at 20:40

Active Sep 17 '15 at 09:23

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By which way I can prove $$\arctan\left(2^{\sqrt{3}-\sqrt{7}}\right)<\frac{61}{125}$$

edited Apr 09 '15 at 20:42

Thomas Andrews

asked Apr 09 '15 at 20:40

E.H.E

Could you give some context or thoughts on the problem so far? – Jef Apr 09 '15 at 20:41
I tried to prove it by taylor series but I countered problem with the square roots – E.H.E Apr 09 '15 at 20:43
Do you expect there to be a neat solution (without actually calculating very much)? – Jef Apr 09 '15 at 20:44
@JefLaga, I think so – E.H.E Apr 09 '15 at 20:46
2

This is very close, the difference is less than $10^{-7}$ so you will need very accurate approximations if you do it that way. – Ross Millikan Apr 09 '15 at 20:47
Damn! It didn't work! There was a minor error in my intended solution. – Prasun Biswas Apr 09 '15 at 21:04
@barto, I actually did but then I deleted it since it was flawed. – Prasun Biswas Apr 09 '15 at 21:22
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Oh, boy ... – Lucian Apr 09 '15 at 22:14
Try this in SageMath RBF(arctan(2^(sqrt(3)-sqrt(7))))<61/125. See document here. – NonalcoholicBeer Dec 05 '19 at 18:32

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