Thread: Taylor series kinda stuck

1. Taylor series kinda stuck

Hello there :P first of all i love that site

been trying to learn math before going back to school so you guys are a good way to learn and love math

so i do have a trouble here with a taylor series... been doing them for quite a while but that one i can't get halfway before seeing mistake appearing

f(x)= sin(x/2) i could use any point i wanted i took a= pi for fun... now i can't get aways with it

anyone got patience to show me how i can't get the convergence information?

2. Re: Taylor series kinda stuck

Use the Ratio Test

3. Re: Taylor series kinda stuck

I do understand the ratio test everywhere but with taylor and Mclaurin...

I'll give you an example that i didn't post here because i already did 2 post today haha :P

i have that Maclaurin question with the fonction F(x)=ln(1-2x)

So i do the Maclaurin as best as i can and get the general fonction ... then i apply myself to `the ratio test and its gives me as answer 1... how with a one do i get my intervalle and lenght of r?

4. Re: Taylor series kinda stuck

You need to understand that a Taylor or MacLaurin Series is a FUNCTION, it will give DIFFERENT series for different values of x. So it makes sense that you would need to know what values of x this function will give a series that is convergent. So in most cases, you should not be getting a single number, rather, an INTERVAL or SET of x values that you know will give a convergent series.

Because we have no guarantee that the value of x we choose will give a positive term series (in fact, in most cases, it won't), that means we have to test for ABSOLUTE CONVERGENCE. We can say for certain that the series is absolutely convergent when \displaystyle \begin{align*} \lim_{n \to \infty} \frac{\left| a_{n + 1} \right| }{\left| a_n \right| } < 1 \end{align*}.

So you need to first get an expression for \displaystyle \begin{align*} \frac{ \left| a_{n + 1} \right| }{ \left| a_{n} \right| } \end{align*}, see what happens when you make \displaystyle \begin{align*} n \to \infty \end{align*}. You will get an expression in terms of x. Then you need to solve for the values of x which will make this expression less than 1. This will ensure convergence.