How to find the series of inverse functions

In summary, the series expansion for inverse functions can be obtained using the telescopic series formula (5) or (6).
  • #1
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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  • #2
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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FAQ: How to find the series of inverse functions

What is an inverse function?

An inverse function is a function that "undoes" the action of another function. In other words, if a function maps an input to an output, the inverse function maps the output back to the input.

Why is it important to find the series of inverse functions?

Finding the series of inverse functions allows us to solve equations that involve multiple functions. It also helps us understand the relationship between two functions and how they can be used together.

How do you find the series of inverse functions?

To find the series of inverse functions, you can use the process of composition. This involves plugging one function into the other and simplifying the resulting equation until you reach the original input variable.

Can every function have an inverse?

No, not every function has an inverse. For a function to have an inverse, it must be one-to-one, meaning that each input corresponds to a unique output. If a function is not one-to-one, it cannot have an inverse.

How do you check if a function has an inverse?

To check if a function has an inverse, you can use the horizontal line test. If a horizontal line intersects the graph of the function in more than one point, the function is not one-to-one and does not have an inverse. If the line only intersects the graph in one point, the function is one-to-one and has an inverse.

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