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Consider $F(x)=\tan(x)-1$

The roots of the equation are $\dfrac{\pi}{4}+n\pi$ i.e

$(\dfrac{\pi}{4},\dfrac{5\pi}{4},\dfrac{9\pi}{4},\cdots)$

and

$(\dfrac{-3\pi}{4},\dfrac{-7\pi}{4}, \dfrac{-11\pi}{4},\cdots)$

The summation of the reciprocals of the roots is $\sum\dfrac{1}{r_i}=\dfrac{4}{\pi}+\dfrac{-4}{3\pi}+\dfrac{4}{5\pi}+\dfrac{-4}{7\pi}+\cdots$

We Know That: $\sum\dfrac{1}{r_i}=\dfrac{-F'(0)}{F(0)}=\dfrac{-\sec^2(0)}{tan(0)-1}=1$

Therefore

$\dfrac{\pi}{4}=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots$

Is this derivation correct? Have I made any wrong assumptions? I would appreciate any rectifications.

LithiumPoisoning
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    What is W.K.T.? – legionwhale Sep 06 '23 at 16:36
  • We know that :) Refer: https://math.stackexchange.com/q/1914440 for proof of the mentioned relation – LithiumPoisoning Sep 06 '23 at 16:40
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    "We" did not include "me"! Also, the result there is about polynomial functions. Have you justified that the result extends to $\tan$? – legionwhale Sep 06 '23 at 16:41
  • :) I thought that since $\tan(x)$ can be expressed as a Taylor series this result could be extended to it. It's definitely not rigorous but just intuitive. – LithiumPoisoning Sep 06 '23 at 16:44
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    Of course, the result you have derived is correct. But there are examples for which the formula you use will fail even if the function is analytic with a Taylor series which converges everywhere. You should also be aware that you've specified an order of summation for the root reciprocals which is natural, but it's unclear how it's mathematically justified. Thus, this can't really be called a proof. I expect that the reason it works out here is due to the fact that $\sin$, $\cos$ have "Weierstrass factorisations", so I think it's possible the proof could be made rigorous. – legionwhale Sep 06 '23 at 16:50
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    Ohhh thanks a lot for helping me out :) – LithiumPoisoning Sep 06 '23 at 16:56

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You're essentially using Vieta's formulas, when you relate the sum of the function's reciprocal roots to its series coefficients. This relation is sometimes called the Root Linear Coefficient Theorem.

It's quite straightforward to derive that from Vieta's formulas. Take a degree $n$ polynomial $P(x)=a_nx^n+\ldots+a_1x+a_0$ with $n$ roots $r_1,r_2,\ldots,r_n$ (up to multiplicity). The sum of its reciprocal roots is then

$$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\ldots+\frac1{r_n}=\frac1{r_1r_2\ldots}\cdot \left(\not{r_1}r_2r_3\ldots+r_1\not{r_2}r_3\ldots+\ldots\right)$$

which, by Vieta's formulas for $n$ and $n-1$, is exactly $\frac{(-1)^{n-1}a_1/a_n}{(-1)^{n}a_0/a_n}=-\frac{a_1}{a_0}$.

And this would be fine for a polynomial... but you're not using a polynomial. So no, your proof is not correct. It has a similar spirit to the math of Euler, where results are derived by falsely assuming properties (eg, that a series is convergent or that a function is a polynomial, particularly in the proof of the Basel problem) and then liberally (even illegally) applying theorems on them.

The beauty of math is that these results can be derived through false proofs, likely by virtue of the simplicity and naturality of the underlying relations. But hand-waving arguments ultimately fail to answer the questions of why should the theorem apply there and not elsewhere? And where/how is the theorem valid generally? These are the question considered by analysis.

Jam
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  • Ohh so this proof is not mathematically rigorous but logically sound? Very interesting thank you:) Just a question, does the Weirestrass factorisation theorem give credence to the fact that some trigonometric functions can also be treated as polynomials or something? I'm sorry I don't know anything other than high school maths – LithiumPoisoning Sep 06 '23 at 16:55
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    @Lithium I'd rather say that it's logically valid (the path of reasoning was fine) but not sound (the starting premise was faulty). "Rigorous" is, in a sense, a non-specific word for these factors and a general sense of strictness. Yes, absolutely, the Weierstrass Factorization Theorem does show that certain "nice" functions of complex numbers (including the trig functions) behave like polynomials! :) It also explains how we can't factor $e^x-1$ the same way as $(x-0)$ (we're missing other complex factors). – Jam Sep 06 '23 at 17:02
  • Very nice, there's something very weird going on that I can't figure out. I previously considered $F(x)=sinx-1$, however the resulting answer for the same summation was $\dfrac{\pi}{2}$ lol – LithiumPoisoning Sep 06 '23 at 17:38
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    $\sin x -1$ is tricky. Because $\sin$ is tangent to $y=1$, you actually get a double root at each zero , so you would get that twice the sum of the reciprocals is $\pi/2$. – Eric Sep 06 '23 at 21:50