Respuesta :
Answer:
The Taylor series for [tex]sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}[/tex], the first three non-zero terms are [tex]9x^{2} -27x^{4}+\frac{162}{5}x^{6}[/tex] and the interval of convergence is [tex]( -\infty, \infty )[/tex]
Step-by-step explanation:
These are the steps to find the Taylor series for the function [tex]sin^2(3 x)[/tex]
- Use the trigonometric identity:
[tex]sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))[/tex]
2. The Taylor series of [tex]cos(x)[/tex]
[tex]cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}[/tex]
Substituting y=6x we have:
[tex]cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}[/tex]
3. Find the Taylor series for [tex]sin^2(3x)[/tex]
[tex]sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))[/tex] (1)
[tex]cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}[/tex] (2)
Substituting (2) in (1) we have:
[tex]\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}[/tex]
Bring the factor [tex]\frac{1}{2}[/tex] inside the sum
[tex]\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )[/tex]
[tex]\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}[/tex]
Extract the term for n=0 from the sum:
[tex]\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}[/tex]
To find the first three non-zero terms you need to replace n=3 into the sum
[tex]sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}[/tex]
To find the interval on which the series converges you need to use the Ratio Test that says
For the power series centered at x=a
[tex]P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,[/tex]
suppose that [tex]\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.[/tex]. Then
- If [tex]R=\infty,[/tex] the the series converges for all x
- If [tex]0<R<\infty,[/tex] then the series converges for all [tex]|x-a|<R[/tex]
- If R=0, the the series converges only for x=a
So we need to evaluate this limit:
[tex]\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |[/tex]
Simplifying we have:
[tex]\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |[/tex]
Next we need to evaluate the limit
[tex]\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}[/tex]
-(n+1)(2n+1) is negative when n -> ∞. Therefore [tex]|-(n+1)(2n+1)}|=2n^{2}+3n+1[/tex]
You can use this infinity property [tex]\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty[/tex] when a>0 and n is even. So
[tex]\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty [/tex]
Because this limit is ∞ the radius of converge is ∞ and the interval of converge is [tex]( -\infty, \infty )[/tex].
