Taylor's series problem from 1994 exam(please help!!!)

Test Preparation
1

<p>Hi, I gotta finish my BC class assignment by Monday, but Taylor's series' driving me nuts.. I need your help. Please help me!!</p>

<p>Question:
The Taylor series about x=5 for a certain fuction f converges to f(x) for all x in the interval of convergence. The nth derivative of f at x=5 is given by
f^(n) (5)= (-1)n!/ 2^n (n+2), and f(5) = 1/2.</p>

<p>(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
(c) show that the sixth-degree Taylor polynomial for f about x=5 approximates f(6) with error less than 1/1000</p>

<p>No typo.
Please help me. If you could do one of them, please post it. I will be greatly appreciated!!!
Thanx!!</p>

2

<p>Try this:</p>

<p>f(a+h) = f(a) + h.f'(a)/1! + h2.f"(a)/2!
+ h3.f(3)(a)/3! +...+ hn.f(n)(a)/n! + hn+1.r/(n+1)!</p>

<p>r = f(n+1)(c) with c between a and a+h.
Create the function </p>

<p>g(x) = f(a+x)-( f(a) + x.f'(a)/1! + x2.f"(a)/2!
+ x3.f(3)(a)/3! +...+ xn.f(n)(a)/n! + xn+1.r/(n+1)! )</p>

<p>. </p>

<p>g'(x) = f'(a+x) -( f'(a) + x.f"(a)/1! + ...
+ xn-1.f(n)(a)/(n-1)! + xn.r/n! )</p>

<p>g"(x) = f"(a+x) -( f"(a) + ... + xn-2.f(n)(a)/(n-2)! + xn-1.r/(n-1)! )</p>

<p>...</p>

<p>g(n)(x) = f(n)(a+x) -( f(n)(a) + x r )</p>

<p>g(n+1)(x) = f(n+1)(a+x)-r</p>

<p>Since f, ... f(n) are continuous in the environment of a, use Rolle's theorem on g(x) and all the derivatives. </p>

<p>g(0)=g(h) => there is a c1 between 0 and h such that g'(c1) = 0.</p>

<p>g'(0)=g'(c1) => there is a c2 between 0 and h such that g"(c2) = 0.</p>

<p>...</p>

<p>g(n)(0)=g(n)(cn) =>there is a cn+1 between 0 and h such
that g(n+1)(cn+1) = 0.</p>

<p>And from this last conclusion, we can write </p>

<pre><code> 0 = f(n+1)(a+cn+1)-r
</code></pre>

<p><=> f(n+1)(c) = r for a value c between a and a+h.</p>

3

<p>mensa 160, thanks for your promptness!!! Unfortunately, this problem is still top of my head.
First, how did you get f (a+h)? what does h represent?</p>

<p>Secondly, what is c? </p>

<p>Thanks for your help...</p>

4

<p>Sorry, I can figure this stuff out, but I have a hard time expalining. It just pops into my head.</p>

<p>But a is the derivative form of the remaider, such that if f(n+1)(x) exists in an environment of a containing a+h, then </p>

<p>f(a+h) = f(a) + h.f'(a)/1! + h2.f"(a)/2!
+ h3.f(3)(a)/3! +...+ hn.f(n)(a)/n! + hn+1.f(n+1)(c)/(n+1)!</p>

<p>with c between a and a+h.</p>

5

<p>omg :|
thats calculus BC
i better start learning now as a sophmore so i wont have problems when im a senior
btw mensa r u really a member of mensa???</p>

6

<p>There's no point in doing it ahead of time, if you are decent at math you'll get it when the time comes.</p>

<p>We just covered this last week, and mensa's explanation looks far more complicated than what is neccessary. However, I can't exactly tell what the original function is (I've never seen an n! one the top of a taylor series, although I guess it isn't impossible)... is it written in pretty print anywhere online?</p>

<p>Or did you really mean this:
f^(n) (5) = [((-1)^n)(n+2)]/[(n!)(2^n)]
?</p>

7

<p>Try this way:
The nth term of the Taylor polynomial for f(x) centered at c is
(The nth deriv of f(x) at c) * (x-c)^n / n! =
( (-1)n!/ 2^n (n+2)) * (x-c)^n / n!</p>

<ol>
<li>To find the third degree Taylor series, use the above formula to find the first four terms (n = 0,1,2,3) and add the terms together.</li>
<li> Since you know the formula for the nth term of the Taylor series, use the ratio test to find when the Taylor series converges.</li>
<li> We know that the error in a Taylor approximation is less than the value of the first ommitted term, so since the problem says that a sixth degree polynomial has an error of < 1/1000, show that the first missing term (where n=7) has the value of less than 1/1000.</li>
</ol>

8

<p>Ohnoes is right. You'll remember it when the time comes. Just ask "over what interval can I expect the approximation to be good?" By plotting the error function, you can visualize the "goodness" of the approximation for various values of n as well as the interval over which the approximation could be considered "good. Basically, remember that the nth Taylor Polynomial at x = a agrees with the function and its first n derivatives at x = a. Everything follows easily from that.</p>

9

<p>tanman's explanation looks good, but for one problem: in part 3, he mentioned the stuff about the 1st omitted term, which only holds true for alternating infinite series. This is not one of those cases (it seems to me that all terms are negative), unfortunately, so that means lagrange's error theorem will have to be used. Also, for part 1, the coefficients are found using the formula a sub n = (nth derivative of f at 5)/n!</p>

10

<p>Oops! Forgot about Lagrange. Jcm525's right about that.</p>

11

<p>thanks guys.... everything makes a lot of sense now. For the second part, though, I get negative infinity.... there is no x that somewhat needed for ratio test. how do you do this??? thanks!!</p>

12

<p>If I'm reading this right so that f^(n) (5)= [(-1)*n!]/ [2^n * (n+2)], then using the absolute ratio test, you should end up with </p>

<p>lim n-> inf | [(x-5) * (n+2)] / [2*(n+3)] | < 1 for the series to converge
so |(x-5)/2| < 1 so 3 < x < 7, makeing the radius of convergence 2</p>

13

<p>Thanks tanman. I really do appreciate. Sorry but one more question.
For part 3, i used the Lagrange error estimation, but the problem is I've been given f(5) not f(6). f(6) seems to be 1/2 also, but not sure. After figuring out Rsub6 (1/2) = [(f^7<em>(z)</em>(1/2)^7]/ 7!, what do I have to do?</p>

<p>Thanks!!!</p>