An Elegant Way to Expand The Inverse Tangent

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An Elegant Way to Expand The Inverse Tangent

In this post we shall see a method to expand the inverse tangent function ` tan^{-1}(x) ` using an infinite geometric series. The infinite series equips us with a neat and easy to remember technique to find the expansion.

A Geometric Series:

A series in which a term is obtained by multiplying the previous term by a constant ratio is called a geometric series. This contant ratio is also called the common ratio `r`. For example consider these series:

` 2, 4, 8, 16, 32, ... `

`1, \frac{1}{5}, \frac{1}{25}, \frac{1}{125}, \frac{1}{625}, ... `

In the first series, `r = 2` and in the second series `r = \frac{1}{5}`.

Sum of an Infinite Geometric Series:

In general, any geometric series can be written as, 

`a, ar^{1}, ar^{2}, ar^{3}, ar^{4}, ... `

An infinite series however can only be summed when the series converges i.e. only when `|r| < 1 `. When this is true the sum `S` is given by ` \frac{a}{1-r} ` i.e.

` S = a + ar^{1} + ar^{2} + ar^{3} + ... = \frac{a}{1-r} `

Thus, `1 + \frac{1}{5} + \frac{1}{25} + \frac{1}{125} + ... = \frac{1}{1 - \frac{1}{5}} = \frac{5}{4} = 1\frac{1}{4} `.

In particular, ` 1 + x + x^{2} + x^{3} + ... = \frac{1}{1 - x} `

Now given the sum we can also find the series. For example let `S = \frac{1}{1+x}`, then here first term is `1` and common ration is `(-x)`. The series is then, 

` 1 - x + x^{2} - x^{3} + x^{4} + ... `

Relating the Inverse Tangent:

We know that, 

`\frac{\text{d}}{\text{d}x}(tan^{-1}x) = \frac{1}{1 + x^{2}}`
And so, 

`tan^{-1}(x) =\int_{}^{} \frac{1}{1 + x^{2}} dx`

But, now we know that we can represent the above integrand as an infinite geometric series. Thus, 

` \frac{1}{1+x^{2}} = 1 - x^{2} + x^{4} - x^{6} + ... `

Wherefore, 

` tan^{-1}(x) =\int_{}^{} \frac{1}{1 + x^{2}} dx = \int ( 1 - x^{2} + x^{4} - x^{6} + ...) dx `

` tan^{-1}(x) = \int ( 1 - x^{2} + x^{4} - x^{6} + ...) dx `

`tan^{-1}(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ... `

Thus we arrive at an elegant expansion of the Inverse Tangent, 

$ \boxed{ tan^{-1}(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ... } $

Just remember ` tan^{-1}(x) = \int \frac{1}{1 + x^{2}} `, and the rest follows. The above process can also be used to expand the inverse cotangent.




Comments

  1. AnonymousFebruary 24, 2025 at 7:56 PM

    😊😊😊😊😊😊😊😊

    ReplyDelete
    Replies
    1. Sid Dash (Author)February 27, 2025 at 2:07 AM

      I'm glad it made you smile.☺️

      Delete

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