Consider the Maclaurin's series expansion of \(tan^{-1}(x)\).

\(tan^{-1}(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}...................\infty\)

If you didn't know how we got here:Maclaurin Series

Differentiating with respect to \(x\) on both sides we get,

\(\frac{1}{1+x^{2}} = 1-x^{2}+x^{4}-.........................\infty\)

The right hand side is clearly an infinite geometric series with a common ratio between consecutive terms as \(-x^{2}\). By the knowledge of geometric series we can say that the right hand side should simplify to give the left hand side. If you didn't know that prove it to yourself after seeing the following wiki:

Note that the result for the sum of an infinite geometric series is under the condition that the absolute value of the common ratio is less than 1, which would imply that the series converges. But if you notice something here you can see that this expression for the left hand side holds true for all \(x\) in the domain of \(tan^{-1}(x)\) but the right hand side doesn't. There is a limiting condition for the right hand side. Hence I seem to want to believe that this is some sort of Deja Vu and I can't seem to account for this. Could anyone please help me in this venture of mine to understand the intricacies of math? If you do, please drop a comment below. Your help will be genuinely appreciated. Cheers!

By the way if you ever wondered what the \(tan^{-1}(z)\) would be, where \(z\) is a complex number, here you go........ I thought it would make the note colourful. Does it?

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## Comments

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TopNewestThe given series expansion of \(\tan^{-1} x\) holds true for \(|x|<1\). This is why the equation that you got by differentiating both sides is valid only for \(|x|<1\). Hope this helps. – Deeparaj Bhat · 11 months, 3 weeks ago

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And if we substitute 1 in the tan inverse expansion we get the same series. – Rohan Shrothrium · 11 months ago

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– Deeparaj Bhat · 11 months ago

Yes. But that requires more justification.Log in to reply

– Anirudh Chandramouli · 11 months, 3 weeks ago

why does the series hold true for \(|x|<1\)?Log in to reply

If the series is to convege, then the \(nth\) term should convege to zero. If \(|x|>1\), then the \(nth\) term diverges as \(n \to \infty\). If \(|x|<1\), then the \(nth\) term tends to zero. Hence, a necessary condition for thhe series to convege is that \(|x|<1\). It can be shown that this condition is also sufficient for convergence of the series. – Deeparaj Bhat · 11 months, 3 weeks ago

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– Deeparaj Bhat · 11 months, 3 weeks ago

Use the ratio test.Log in to reply

– Anirudh Chandramouli · 11 months, 3 weeks ago

Actually I realised now that the Cauchy convergence test itself is enough. I say itself mainly because I feel that it is easyLog in to reply

this test? – Deeparaj Bhat · 11 months, 3 weeks ago

Did you meanLog in to reply

– Anirudh Chandramouli · 11 months, 3 weeks ago

by the way if wikipedia failed to make you understand the cauchy convergence test, read that book about hyperreals i gave you chapter 9Log in to reply

– Deeparaj Bhat · 11 months, 3 weeks ago

Ok. Will do. But personally, I feel more at home with standard real analysis. :)Log in to reply

– Anirudh Chandramouli · 11 months, 3 weeks ago

by the way could you send the proof using ratio test. I am intrigued as I don't seem to be getting the limit of consecutive termsLog in to reply

– Deeparaj Bhat · 11 months, 3 weeks ago

For the ratio test, the limit of the modulus of the consecutive terms is required, which is 1. Hence, radius of convergence is 1.Log in to reply

– Anirudh Chandramouli · 11 months, 3 weeks ago

oh yes i didn't take the mod value I took the value of consecutive terms as it is thanksLog in to reply

– Anirudh Chandramouli · 11 months, 3 weeks ago

yepLog in to reply