## xkehaulanix 5 years ago 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.

1. anonymous

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:$\frac{f^n(5)}{n!}(x-5)^n=\frac{\frac{(-1)^nn!}{2^n(n+2)}}{n!}(x-5)^n=\frac{(-1)^n}{2^n(n+2)}(x-5)^n$

2. anonymous

The Taylor series expansion would be,$\sum_{n=0}^{\infty}=\frac{(-1)^n}{2^n(n+2)}(x-5)^n$

3. anonymous

You need to expand this up to the third power; that is, from n=0 to 3.

4. anonymous

I'm finding the radius of convergence.

5. xkehaulanix

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!

6. anonymous

Hey, that's good :) So are you right to do the radius of convergence?

7. xkehaulanix

Ahaha...unfortunately no. Mind giving me a hint? xD

8. anonymous

Okay. Just give me a minute to get a drink.

9. anonymous

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.

10. anonymous

For you,$\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{\frac{(-)^{n+1}(x-5)^{n+1}}{2^{n+1}((n+2)+1)}}{\frac{(-)^n(x-5)^n}{2^n(n+2)}} \right|=\left| \frac{-(x-5)(n+2)}{2(n+3)} \right|=\left| x-5 \right|\frac{n+2}{2n+6}$

11. anonymous

$=\left| x-5 \right|\frac{1+2/n}{2+6/n}$

12. anonymous

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$\frac{1}{2}|x-5|<1 \rightarrow |x-5|<2$

13. anonymous

We need to find all x such that the above holds true.

14. anonymous

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.

15. anonymous

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:$x^2-10x+25=4 \rightarrow x^2-10x+21=(x-3)(x-7)=0$

16. anonymous

So we have three regions to test: $-\infty < x< 3$$3<x<7$$7<x<\infty$The points 3 and 7 are not included because of the strict inequality.

17. anonymous

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.

18. anonymous

Your radius of convergence is$\left\{ x \in \mathbb{R} |3<x<7 \right\}$

19. anonymous

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$2, 5+\sqrt{6}, 5-\sqrt{6}$

20. xkehaulanix

...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)?

21. anonymous

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.

22. anonymous

Is that what you mean?

23. xkehaulanix

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

24. anonymous

No worries - just fan me (if you haven't already)! ;)