Sum to infinity, the series 1+k cosθ+k^2 cos2 θ+k^3 cos3 θ+…+∞,|k|<1.

Name: Problem 125, Chapter 1: IIT JEE Super Course in Mathematics ALGEBRA II
Uploaded: 2021-07-02T07:44:41-08:00
Duration: 5 min 20 s
Channel: P Krishnamurthy
Description: Sum to infinity, the series 1+k cosθ+k^2 cos2 θ+k^3 cos3 θ+…+∞,|k|<1.

Sum to infinity, the series 1+k cosθ+k^2 cos2 θ+k^3 cos3...

Key Concepts

Geometric Series

A geometric series is one in which each term is obtained by multiplying the previous term by a constant ratio. Summing such a series relies on the formula for an infinite geometric series, which is valid when the absolute value of the common ratio is less than one. This concept is essential for determining convergent sums of series where each term is formed by successive multiplications by a factor.

Euler’s Formula and Complex Representation

Euler’s formula establishes a connection between exponential and trigonometric functions by expressing cosine and sine in terms of complex exponentials. This conversion technique is vital in handling series with cosine terms, as it allows one to view the trigonometric series as the real part of a complex geometric series, thereby simplifying the summation process.

Convergence Criteria

The convergence of an infinite series depends on the magnitude of its ratio; in this context, the requirement that |k| < 1 ensures that the series converges. Understanding convergence criteria is crucial for applying the infinite summation formulas correctly, as it guarantees that the series approaches a definite sum.

Transcript

00:01 In this problem we have been asked to find the sum to infinity the series of 1 plus k k cost theta plus k squared cost 2 theta plus k cube cost 3 theta plus and so on till infinity we have been given that mod of k is less than 1 that is the magnitude of k is less than 1.

00:17 Now let c is equal to 1 plus k cost theta plus k square cost to theta and so on till infinity all right now let us consider the series s is equal to k sine theta plus k squared sine 2 theta and then so on so now plus in plus till infinity so this is another series s now c plus iota times of s would be equal to 1 plus k into cost theta plus i sine theta and then plus k square into cost 2 theta plus iota sine 2 theta and then this will go on plus still infinity so this would be equal to 1 plus k into e raise to the power of i into theta so this is an identity which we know this is your oirous form that is e raise to i theta is cost theta plus i sine theta so therefore this would be 1 plus k to erase 2 i theta plus k e raise to i theta whole square all right so this would be going on till infinity so therefore now we know that since mod magnitude of k is less than one so therefore this would be equal to one upon one minus k e raised to the e raise to the power of iota theta since k to the power of mod of k to the power of i e rest to i theta is equal to mod of k which is less than 1 and this was sum to infinity for a gp so therefore we know that what is the sum of infinity till gp for a gp so therefore this would be equal to 1 minus k e raise to the power of minus iota theta whole divided by 1 minus k e raise to iota theta into 1 minus k e rase to iota theta into 1 minus k into e.

03:21 2 iota minus iota theta.

03:24 So therefore this would be equal to 1 minus k into now i'm expanding e...

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