# Thread: interval of convergence help

1. ## interval of convergence help

f(x) = 2 / x+3

taylor series polynomial out to the 6th degree

(P6) = (5/8) - (1/8)x + (1/32)(x-1)² - (1/128)(x-1)^3 + (1/512)(x-1)^4 - (1/2048)(x-1)^5

Problem:
find the value of x for which the polynomial converges to f(x)?

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I have no idea how/ what i should do in order to get this awnser..any help/advice is greatly appreciated! thx

2. Originally Posted by lochnessmonster
f(x) = 2 / x+3

taylor series polynomial out to the 6th degree

(P6) = (5/8) - (1/8)x + (1/32)(x-1)² - (1/128)(x-1)^3 + (1/512)(x-1)^4 - (1/2048)(x-1)^5

Problem:
find the value of x for which the polynomial converges to f(x)?

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I have no idea how/ what i should do in order to get this awnser..any help/advice is greatly appreciated! thx
I think you have a typo the first term in the series should be $\displaystyle \frac{1}{2}$

$\displaystyle \frac{2}{x+3}=\frac{2}{4+(x-1)}=\frac{1}{2}\left(\frac{1}{1+\frac{x-1}{4}} \right)=\frac{1}{2}\left(\frac{1}{1-\frac{-(x-1)}{4}} \right)$

Now this is in the form

$\displaystyle \frac{1}{1-r},r=\frac{-(x-1)}{4}$

So this is a geometric series and converges when $|r|<1$

3. so the interval of convergence will always be the same no matter how DEEP you go into the taylor series? the only thing that will change is a difference AT the end points?

4. Originally Posted by lochnessmonster
so the interval of convergence will always be the same no matter how DEEP you go into the taylor series? the only thing that will change is a difference AT the end points?
Let me respond with a question.

1st are you asking what the raduis of convergence of the power series is? if so the radius of convergence does not depend on where you truncate the series. However the further from the center you may have to take more terms to get a specified accuracy.

2nd or are you asking to find a specific value(s) of $x$ such that

$P_6(x)=f(x)$

5. find the value of x for which the polynomial converges to f(x)?

the interval