Lagrange error bound

Test Preparation AP Test Preparation
1

<p>Since the general form for the Rn(x) <max|(f^(n+1)(c))(x-a)^(n+1)/(n+1)!|</p>

<p>How do we estimate the (f^(n+1)(c)) part? It seem that we don't take the derivative of f(x) to the n+1th and then plug in c.... How do we get that?</p>

2

<p>the f^(n+1)(c) is indeed the (n+1)th derivative of f(x), evaluated at x = c.</p>

3

<p>usually the lagrange error bound is just the next term in the series at x=c. This is because f^(n+1) (c) is usually equal to 1.</p>

4

<p>Just find the maximum (n+1)st derivative between a and x. Remember, c lies between a and x.</p>

5

<p>But from AP’s answer key they derived f^(n+1)(c) in a different way.<br>
<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;
problem #6 (c)
apparently f^4(c) doesn’t equal to 625, it will be 625sine(5c+pi/4), which is not 625!</p>

6

<p>It clearly shows it as 625 on there. Where is the problem?</p>

7

<p>625sin(5c+pi/4)=625
sin(5c+pi/4)=1
5c+pi/4=pi/2
5c=pi/4
c=pi/20
That is odd; the c that they would have used would not lie between 0 and 1/10. I believe they should have evaluated the fourth derivative at 1/10.</p>

8

<p>Given that it was on a non-calculator part of the exam, I guess they wanted you to use 625 as a conservative guess.</p>

9

<p>Easy Lagrange Error bound, found it searching google.</p>

<p><a href=“http://www.tenafly.k12.nj.us/~jlaux/BC/Study%20guides/Taylor%20Error%201.doc[/url]”>http://www.tenafly.k12.nj.us/~jlaux/BC/Study%20guides/Taylor%20Error%201.doc&lt;/a&gt;&lt;/p&gt;