Maclaurin series/need help reducing my ratio test

**mollymcf2009** · April 15th 2009, 12:23 PM

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. Assume that f has a power series expansion. And find the associated radius of convergence, R.

$f(x) = sin(\frac{\pi x}{2})$

My Maclaurin series I got was:

$\sum^{\infty}_{n=0} (-1)^n \frac{\pi^{2(n+1)} \cdot x^{2(n+1)}}{(2n+1)!\cdot (2^{2(n+1)}}$

So, I'm fine with getting to this point (finally). So when I do my ratio test to find my R, I start off ok, but then I don't know what/how I am supposed to cancel out terms to get to something I can actually take my limit at.

$\lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2))} (x^{2(n+2)})}{(2(n+2))! (2^{2(n+2)})}\cdot \frac{((2n+1)!) (2^{2(n+1)})}{(\pi^{2(n+1)})(x^{2(n+1)})}\right|$

How in the world do I simplify this??????

**Jhevon** · April 15th 2009, 01:04 PM

Originally Posted by mollymcf2009

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. Assume that f has a power series expansion. And find the associated radius of convergence, R.

$f(x) = sin(\frac{\pi x}{2})$

My Maclaurin series I got was:

$\sum^{\infty}_{n=0} (-1)^n \frac{\pi^{2(n+1)} \cdot x^{2(n+1)}}{(2n+1)!\cdot (2^{2(n+1)}}$

no! look up the power series for sine again. the powers are $2n + 1$ not $2(n + 1)$ .

So, I'm fine with getting to this point (finally). So when I do my ratio test to find my R, I start off ok, but then I don't know what/how I am supposed to cancel out terms to get to something I can actually take my limit at.

$\lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2))} (x^{2(n+2)})}{(2(n+2))! (2^{2(n+2)})}\cdot \frac{((2n+1)!) (2^{2(n+1)})}{(\pi^{2(n+1)})(x^{2(n+1)})}\right|$

How in the world do I simplify this??????

start by grouping like terms. for example, if the base is x, put those parts together, put the ones where the base is 2 together, etc. then use the rule $\frac {a^m}{a^n} = a^{m - n}$

**Showcase_22** · April 15th 2009, 01:12 PM

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2))} (x^{2(n+2)})}{(2(n+2))! (2^{2(n+2)})}\cdot \frac{((2n+1)!) (2^{2(n+1)})}{(\pi^{2(n+1)})(x^{2(n+1)})}\right| $

WOAH!

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2)}) (x^{2(n+2))})}{(2n+4)(2n+3)(2n+1)! (2^{2(n+2)})}\cdot \frac{((2n+1)!) (2^{2(n+1)})}{(\pi^{2(n+1)})(x^{2(n+1)})}\right| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2)}) (x^{2(n+2))})}{(2n+4)(2n+3) (2^{2(n+2)})}\cdot \frac{2^{2(n+1)}}{(\pi^{2(n+1)})(x^{2(n+1)})}\righ t| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2)}) (x^{2n+4))})}{(2n+4)(2n+3) (2^{2(n+2)})}\cdot \frac{2^{2n+2})}{(\pi^{2(n+1)})(x^{2(n+1)})}\right | $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2)} (x^{2n+4))})}{(2n+4)(2n+3) (2^{2n+4)})}\cdot \frac{2^{2n+2})}{(\pi^{2(n+1))})(x^{2(n+1)})}\righ t| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2)} (x^{2n+4))})}{(2n+4)(2n+3) (2^2)}\cdot \frac{1}{(\pi^{2(n+1))})(x^{2(n+1)})}\right| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2(n+2)} (x^{2n+4))})}{(2n+4)(2n+3) (4)}\cdot \frac{1}{(\pi^{2(n+1))})(x^{2(n+1)})}\right| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2n+4} (x^{2n+4))})}{(2n+4)(2n+3) (4)}\cdot \frac{1}{(\pi^{2n+2})(x^{2n+2})}\right| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2} (x^{2n+4))})}{(2n+4)(2n+3) (4)}\cdot \frac{1}{(x^{2n+2})}\right| $

$ \lim_{n\rightarrow \infty} \left| \frac{(\pi^{2}) (x^{2})}{(2n+4)(2n+3) (4)}\right|=0 $

Sorry about the parentheses throughout. It's readable, but clearly a basic knowledge of Latex is not sufficient for such formulae. For this reason, it may be beneficial to check my working.

EDIT: Apparently this is the wrong expression to take a limit of. *cries*

**mollymcf2009** · April 15th 2009, 01:14 PM

Originally Posted by Jhevon

no! look up the power series for sine again. the powers are $2n + 1$ not $2(n + 1)$ .

start by grouping like terms. for example, if the base is x, put those parts together, put the ones where the base is 2 together, etc. then use the rule $\frac {a^m}{a^n} = a^{m - n}$

Thanks as usual, my friend! Hope you will be on here tonight. There is a good possibility that my over-stuffed brain may be leaking previously memorized information

**Jhevon** · April 15th 2009, 01:43 PM

Originally Posted by mollymcf2009

Thanks as usual, my friend! Hope you will be on here tonight. There is a good possibility that my over-stuffed brain may be leaking previously memorized information

ok, first of all, the power series for sine is defined for all real x, so we know the radius of convergence is $\infty$ , and the interval of convergence is $(-\infty, \infty)$ . However, if we are required to prove it, we can use the ratio test, beginning as follows:

We want $\lim_{n \to \infty} \left| \frac {a_{n + 1}}{a_n} \right| < 1$ , where $a_n = \frac {\pi^{2n + 1} \cdot x^{2n + 1}}{(2n + 1)! \cdot 2^{2n + 1}}$ , so we have

$\lim_{n \to \infty} \left| \frac {\pi^{2n + 3} \cdot x^{2n + 3}}{(2n + 3)! \cdot 2^{2n + 3}} \cdot \frac {(2n + 1)! \cdot 2^{2n + 1}}{\pi^{2n + 1} \cdot x^{2n + 1}} \right|$ $= \lim_{n \to \infty} \left| \frac {\pi^{2n + 3}}{\pi^{2n + 1}} \cdot \frac {(2n + 1)!}{(2n + 3)!} \cdot \frac {2^{2n + 1}}{2^{2n + 3}} \cdot \frac {x^{2n + 3}}{x^{2n + 1}} \right|$

now finish up

**mollymcf2009** · April 15th 2009, 04:40 PM

Originally Posted by Jhevon

ok, first of all, the power series for sine is defined for all real x, so we know the radius of convergence is $\infty$ , and the interval of convergence is $(-\infty, \infty)$ . However, if we are required to prove it, we can use the ratio test, beginning as follows:

We want $\lim_{n \to \infty} \left| \frac {a_{n + 1}}{a_n} \right| < 1$ , where $a_n = \frac {\pi^{2n + 1} \cdot x^{2n + 1}}{(2n + 1)! \cdot 2^{2n + 1}}$ , so we have

$\lim_{n \to \infty} \left| \frac {\pi^{2n + 3} \cdot x^{2n + 3}}{(2n + 3)! \cdot 2^{2n + 3}} \cdot \frac {(2n + 1)! \cdot 2^{2n + 1}}{\pi^{2n + 1} \cdot x^{2n + 1}} \right|$ $= \lim_{n \to \infty} \left| \frac {\pi^{2n + 3}}{\pi^{2n + 1}} \cdot \frac {(2n + 1)!}{(2n + 3)!} \cdot \frac {2^{2n + 1}}{2^{2n + 3}} \cdot \frac {x^{2n + 3}}{x^{2n + 1}} \right|$

now finish up

Ok, I don't get how you got (2n+3) Wouldn't it be (2n+1+1) = (2n+2) ?

**Jhevon** · April 15th 2009, 04:42 PM

Originally Posted by mollymcf2009

Ok, I don't get how you got (2n+3) Wouldn't it be (2n+1+1) = (2n+2) ?

$a_{n + 1}$ means that $n$ is replaced with $n + 1$ , so $2n + 1 \longrightarrow 2(n + 1) + 1 = 2n + 2 + 1 = 2n + 3$

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