The Taylor series about x=5 for a certain function f converges to f(x) for all x in the interval of convergence. The nth derivative of f at x=5 is given by fn(5)=[(-1)^n x n!]/[2^n x (n + 2)], and f(5)=1/2. The fn above should be "f to the nth derivative." a)Write the third-degree Taylor polynomial for f about x=5. b)Find the radius of convergence of the Taylor series for f about x=5.

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The Taylor series about x=5 for a certain function f converges to f(x) for all x in the interval of convergence. The nth derivative of f at x=5 is given by fn(5)=[(-1)^n x n!]/[2^n x (n + 2)], and f(5)=1/2. The fn above should be "f to the nth derivative." a)Write the third-degree Taylor polynomial for f about x=5. b)Find the radius of convergence of the Taylor series for f about x=5.

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Well, in general, the nth term of a Taylor series expansion is given by\[\frac{f^n(x_0)}{n!}(x-x_0)^n\]where x_0 is the point of expansion. You have x_0=5, and you also have the derivative function at this point, so all you need to do is substitute it in:\[\frac{f^n(5)}{n!}(x-5)^n=\frac{\frac{(-1)^nn!}{2^n(n+2)}}{n!}(x-5)^n=\frac{(-1)^n}{2^n(n+2)}(x-5)^n\]
The Taylor series expansion would be,\[\sum_{n=0}^{\infty}=\frac{(-1)^n}{2^n(n+2)}(x-5)^n\]
You need to expand this up to the third power; that is, from n=0 to 3.

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I'm finding the radius of convergence.
Wait, I did something RIGHT in a series question?! YESSS! I was beginning to think I was completely hopeless at it. From completely to somewhat. Thanks!
Hey, that's good :) So are you right to do the radius of convergence?
Ahaha...unfortunately no. Mind giving me a hint? xD
Okay. Just give me a minute to get a drink.
With the radius of convergence, you have to remember you're dealing with an infinite series, so there are a battery of theorems (in terms of tests) we have for convergence. Now, if the series converges absolutely, it will converge. The ratio test says that, for a series with terms a_n, the series converges if\[\lim_{n->\infty}\left| \frac{a_{n+1}}{a_n} \right|= \rho < 1\]The aim now is to use this test to find any restrictions on x that need to be installed so that the above condition holds.
For you,\[\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{\frac{(-)^{n+1}(x-5)^{n+1}}{2^{n+1}((n+2)+1)}}{\frac{(-)^n(x-5)^n}{2^n(n+2)}} \right|=\left| \frac{-(x-5)(n+2)}{2(n+3)} \right|=\left| x-5 \right|\frac{n+2}{2n+6}\]
\[=\left| x-5 \right|\frac{1+2/n}{2+6/n}\]
When you take n to infinity, your ratio becomes,\[\frac{1}{2}|x-5|\]From what was said, you need this ratio to be less than 1 if your series is to converge. So set\[\frac{1}{2}|x-5|<1 \rightarrow |x-5|<2\]
We need to find all x such that the above holds true.
I use a method called 'critical points' to work any kind of inequality out. The method sets up boundary points that partition your interval (here, the real number line). You then take a test point from each interval and see if the test point yields a true statement.
We use two things - 1) definition of magnitude, and 2) critical points. 1) \[|x-5|=\sqrt{(x-5)^2}=\sqrt{x^2-10x+25} \lt 2\]Now set the inequality to an equality and solve:\[x^2-10x+25=4 \rightarrow x^2-10x+21=(x-3)(x-7)=0\]
So we have three regions to test: \[-\infty < x< 3\]\[3
Test x=0 (first region)\[\sqrt{(0)^2-10(0)+25}=5<2\]not true. So no x in the first region satisfies this.\[\sqrt{(5)^2-10(5)+25}=0<2\]true, so x does live in the second region. Finally,\[\sqrt{(10)^2-10(10)+25}=5<2\]which is false, so no x lives in the last region.
Your radius of convergence is\[\left\{ x \in \mathbb{R} |3
Note, the critical points method I used has another component for selecting points to partition the interval. If you have algebraic fractions, like\[\frac{x^2-1}{x-2}<10\]we would note x=2 is a critical point (since the left-hand side fails to exist there). I'd then find the remaining points by doing what I did above:\[\frac{x^2-1}{x-2}=10\]solve for x,\[x=5 \pm \sqrt{6}\]and the interval would be partitioned at the points\[2, 5+\sqrt{6}, 5-\sqrt{6}\]
...wow. that's the clearest anybody's explained it in the past few days. Haha, thanks so much! Most of it's familiar, with the exception of the critical point testing. I think I like your method better. One question, why do you replace the n! in the ratio test with (x-a)?
Your terms for the series don't have n! in them - they were eradicated at the beginning because the Taylor series terms have 1/n! and your nth derivative had n! in the numerator. Just cancelled.
Is that what you mean?
Oh geez, that was a stupid question. Haha, thanks. I think I understand it actually. I believe I owe you a lot for finally making me partially understand series. Thanks for all of that! :D
No worries - just fan me (if you haven't already)! ;)

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