MHB How to find the series of inverse functions

AI Thread Summary
The discussion focuses on computing the series S = ∑(n=1 to ∞) tan^(-1)(√3/(n²+n+3)), which simplifies to S = tan^(-1)(√3) = π/3 using a general identity for inverse functions. It explores a method for constructing series of inverse functions through a functional equation and telescopic series. Two key results are established: one for strictly increasing sequences and another for strictly decreasing sequences, both converging to a limit. Examples are provided, including the application of these results to the exponential function and the tangent function, demonstrating their utility in calculating infinite products and series. The thread aims to further explore various functions and sequences for additional applications.
chisigma
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In the Math Challenge Forum it has been requested fo compute the series...

$\displaystyle S = \sum_{n=1}^{\infty} \tan^{-1}\ \frac{\sqrt{3}}{n^{2} + n + 3}\ (1)$

... and that has been performed using the general identity...

$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\ \frac{c}{n^{2} + n + c^{2}} = \tan^{- 1} c\ (2)$

... so that is $\displaystyle S = \tan^{- 1} \sqrt{3} = \frac{\pi}{3}$. Scope of this thread is to find a general procedure to construct series of inverse functions.

Let be $f(*)$ a strictly increasing function that maps an open inteval $[\alpha,\beta]$ containing 0 onto an interval $[a,b]$. We know that is such conditions the inverse function $f^{-1} (*)$ exists and it maps $[a,b]$ onto $[\alpha,\beta]$. Let's define $f(0) = s$.

Now we assume that $f(*)$ is solution of a functional equation...

$\displaystyle f(x - y) = G \{f(x),f(y)\}\ (3)$

Setting $f(x)=u$ and $f(y)=v$ the the so called subtaction formula permits us to write...

$\displaystyle f^{-1} (u) - f^{- 1} (v) = f^{-1} \{G(u,v)\}\ (4)$

We can use (4) to construct a telescopic series. If $\displaystyle c_{n},\ n=0,1,...$ is an increasing sequence in $[a,b]$ converging to $c \in [a,b]$, then...

$\displaystyle \sum_{k=1}^{n} \{f^{-1} (c_{k}) - f^{-1} (c_{k-1})\} = f^{-1} (c_{n})\ (5)$

... and that means that...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k},c_{k-1})\} = f^{-1} (c) - f^{- 1} (c_{0})\ (6)$

If $\displaystyle c_{n}$ is a strictly decreasing sequence converging to c the result is similar...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k-1},c_{k})\} = f^{-1} (c) + f^{- 1} (c_{0})\ (7)$

In next posts we will illustrate some interesting examples of use of (6) and (7)...

Kind regards

$\chi$ $\sigma$
 
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In the previous post the following two series expansions have been obtained...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k},c_{k-1})\} = f^{-1} (c) - f^{- 1} (c_{0})\ (1)$

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k-1},c_{k})\} = f^{-1} (c) + f^{ -1} (c_{0})\ (2)$

... the first valid for strictly increasing sequences $c_{k}$ converging to c, the second for strictly decreasing sequences $c_{k}$ converging to c. Now we will show some example of application of (1) or (2) to different type of invertible functions.

A very important case is the function $\displaystyle f(x) = e^{x}$ so that is $\displaystyle f^{-1} (x) = \ln x$ and $\displaystyle G(u,v) = \frac{u}{v}$. Among the possible example we choose the sequence $\displaystyle c_{k} = \frac{k+2}{k+1}$ which is strictly decreasing and tends to 1 so that we can apply (2) setting $\displaystyle G(c_{k-1},c_{k}) = \frac{c_{k-1}}{c_{k}} = 1 + \frac{1}{k^{2} + 2\ k}$ and $\displaystyle c_{0}= 2$ obtaining...

$\displaystyle \sum_{k=1}^{\infty} \ln (1 + \frac{1}{k^{2} + 2\ k}) = \ln 2\ (3)$

This example, although easy, is a very useful approach to the computation of some types of infinite products because from (3) we derive immediately...

$\displaystyle \prod_{k=1}^{\infty} (1 + \frac{1}{k^{2} + 2\ k}) = 2\ (4)$

Now we pass to $\displaystyle f(x) = \tan x$, so that is $\displaystyle f^{-1} (x) = \tan^{-1}(x)$. Using well known trigonometric identity we arrive to write $\displaystyle G(u,v) = \frac{u - v}{1 + u\ v}$. If we choose $\displaystyle c_{k}= \frac{a}{k + 1}$, which is strictly decreasing and tends to 0 with $\displaystyle c_{0}= a$, we obtain that $\displaystyle G (c_{k-1},c_{k}) = \frac{a}{k^{2} + k + a^{2}}$ so that applying again (2) is...

$\displaystyle \sum_{k=1}^{\infty} \tan^{-1} \frac{a}{k^{2} + k + a^{2}} = \tan^{-1} a\ (5)$

Of course a very large set of functions and sequences exists and may be in next posts we will try to find some other interesting possibility of applying (1) and (2)...

Kind regards

$\chi$ $\sigma$
 
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