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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 thirddegree Taylor polynomial for f about x=5.
b)Find the radius of convergence of the Taylor series for f about x=5.
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 thirddegree 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, in general, the nth term of a Taylor series expansion is given by\[\frac{f^n(x_0)}{n!}(xx_0)^n\]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:\[\frac{f^n(5)}{n!}(x5)^n=\frac{\frac{(1)^nn!}{2^n(n+2)}}{n!}(x5)^n=\frac{(1)^n}{2^n(n+2)}(x5)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The Taylor series expansion would be,\[\sum_{n=0}^{\infty}=\frac{(1)^n}{2^n(n+2)}(x5)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to expand this up to the third power; that is, from n=0 to 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm finding the radius of convergence.

xkehaulanix
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, 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!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey, that's good :) So are you right to do the radius of convergence?

xkehaulanix
 5 years ago
Best ResponseYou've already chosen the best response.0Ahaha...unfortunately no. Mind giving me a hint? xD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. Just give me a minute to get a drink.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0With 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For you,\[\left \frac{a_{n+1}}{a_n} \right=\left \frac{\frac{()^{n+1}(x5)^{n+1}}{2^{n+1}((n+2)+1)}}{\frac{()^n(x5)^n}{2^n(n+2)}} \right=\left \frac{(x5)(n+2)}{2(n+3)} \right=\left x5 \right\frac{n+2}{2n+6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\left x5 \right\frac{1+2/n}{2+6/n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you take n to infinity, your ratio becomes,\[\frac{1}{2}x5\]From what was said, you need this ratio to be less than 1 if your series is to converge. So set\[\frac{1}{2}x5<1 \rightarrow x5<2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We need to find all x such that the above holds true.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We use two things  1) definition of magnitude, and 2) critical points. 1) \[x5=\sqrt{(x5)^2}=\sqrt{x^210x+25} \lt 2\]Now set the inequality to an equality and solve:\[x^210x+25=4 \rightarrow x^210x+21=(x3)(x7)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Test x=0 (first region)\[\sqrt{(0)^210(0)+25}=5<2\]not true. So no x in the first region satisfies this.\[\sqrt{(5)^210(5)+25}=0<2\]true, so x does live in the second region. Finally,\[\sqrt{(10)^210(10)+25}=5<2\]which is false, so no x lives in the last region.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your radius of convergence is\[\left\{ x \in \mathbb{R} 3<x<7 \right\}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note, the critical points method I used has another component for selecting points to partition the interval. If you have algebraic fractions, like\[\frac{x^21}{x2}<10\]we would note x=2 is a critical point (since the lefthand side fails to exist there). I'd then find the remaining points by doing what I did above:\[\frac{x^21}{x2}=10\]solve for x,\[x=5 \pm \sqrt{6}\]and the interval would be partitioned at the points\[2, 5+\sqrt{6}, 5\sqrt{6}\]

xkehaulanix
 5 years ago
Best ResponseYou've already chosen the best response.0...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 (xa)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that what you mean?

xkehaulanix
 5 years ago
Best ResponseYou've already chosen the best response.0Oh 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No worries  just fan me (if you haven't already)! ;)
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