Using the series for: sin x = \sum^\infty_{n = 0}\frac{ (-1)^n}{(2n - 1)!} x^{2n + 1} and for:...

Question:

Using the series for: {eq}sin x = \sum^\infty_{n = 0}\frac{ (-1)^n}{(2n - 1)!} x^{2n + 1} {/eq}

and for: {eq}cos x = \sum^\infty_{n = 0} \frac{(-1)^n}{(2n)!} x^{2n} {/eq}

Find a power series for tan x using the Common Ratio Identity for tangent.

Find the first 3 terms of the series.

Power series expansions

This example illustrates how to find a power series expansion for tan x, when we are given the corresponding power series for both sin x and cos x.

We use the method of undetermined coefficients.

Answer and Explanation:

Letes assume that the question means us to find the first three NON-ZERO terms in the expansion of {eq}\tan x. {/eq}

(As we will see the first three terms, whether zero or non-zero, would be quite dull !)

We will use the identity {eq}\displaystyle \tan x = \frac{\sin x}{\cos x} \\ \Rightarrow \sin x = (\tan x)(\cos x) {/eq}

Let the required expansion be

{eq}\tan x = a + bx + cx^2 + dx^3 +ex^4 + fx^5 + .... {/eq}

The given expansions for {eq}\sin x, \:\cos x {/eq} up to the term in {eq}x^5 {/eq} are as follows:

{eq}\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} -.... \\ \displaystyle \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -.... {/eq}

So

{eq}\sin x = (\tan x)(\cos x) \\ \displaystyle \Rightarrow x - \frac{x^3}{3!} + \frac{x^5}{5!} -.... = \left ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -.... \right )(a + bx + cx^2 + dx^3 +ex^4 + fx^5 + ....) {/eq}

Expanding the right hand side and tidying up, with successive powers on separate lines for clarity gives:

{eq}\displaystyle x - \frac{x^3}{6} + \frac{x^5}{120} - .... = \\ a \\ + bx \\ \displaystyle + (c - \frac{a}{2})x^2 \\ \displaystyle + (d - \frac{b}{2})x^3 \\ \displaystyle + (e - \frac{c}{2} + \frac{a}{24})x^4 \\ \displaystyle + (f - \frac{d}{2} + \frac{b}{24})x^5 + ..... {/eq}

Equating coefficients:

constant term {eq}\Rightarrow a =0. {/eq}

{eq}x {/eq} coefficient:

{eq}b = 1. {/eq}

{eq}x^2 {/eq} coefficient:

{eq}\displaystyle 0 = c - \frac{a}{2} \\ \Rightarrow \displaystyle 0 = c - \frac{0}{2} \\ \Rightarrow c = 0. {/eq}

{eq}x^3 {/eq} coefficient:

{eq}\displaystyle -\frac{1}{6} = d - \frac{b}{2} \\ \Rightarrow \displaystyle -\frac{1}{6} = d - \frac{1}{2} \\ \Rightarrow \displaystyle d = \frac{1}{3}. {/eq}

{eq}x^4 {/eq} coefficient:

{eq}\displaystyle e - \frac{c}{2} + \frac{a}{24} = 0 \\ \Rightarrow \displaystyle e - \frac{0}{2} + \frac{0}{24} = 0 \\ \Rightarrow e = 0. {/eq}

{eq}x^5 {/eq} coefficient:

{eq}\displaystyle f - \frac{d}{2} + \frac{b}{24} = \frac{1}{120} \\ \Rightarrow \displaystyle f - \frac{\frac{1}{3}}{2} + \frac{1}{24} = \frac{1}{120} \\ \Rightarrow \displaystyle f = \frac{2}{15}. {/eq}

Putting the coefficients back into the power series:

{eq}\displaystyle \tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + ..... {/eq}


Learn more about this topic:

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Power Series: Formula & Examples

from Precalculus: Help and Review

Chapter 2 / Lesson 10
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