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Series of inverse functions..

chisigma

Well-known member
Feb 13, 2012
1,704
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 $[\alpha,\beta]$. Lets 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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chisigma

Well-known member
Feb 13, 2012
1,704
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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