<p>On the 2004 Free Response, question 6, part (c):</p>

<p>[Sorry for the semi-long explanation. My question, itself, is not an involved one.]</p>

<p>They use the Lagrange error with the greatest possible value of f''''(c), which is 625sin(5c+(pi/4)). It's my understanding that c has to be between a (in this case the Taylor series is centered at 0, so that would be the lower limit), and x (which they say is 1/10). This is even explicitly stated in the solution where it says "max: 0<<em>c<</em>(1/10). Yet they somehow get sin (5c+(pi/4)) to equal 1, as the next step simply shows this replace by 625.</p>

<p>I don't understand how this is possible, because at the greatest value of c you would only have sin ((1/2) + (pi/4)), which is not sin (pi/2), and thus is not 1.</p>

<p>So... how do they do that?</p>

<p>Thanks!</p>

<p>Sorry for the late reply, but I had the same problem. Take a look at the student sample responses. Go to apcentral.collegeboard.com and register as a teacher if you haven't already. They helped me figure it out. I need to get some rest so I don't fall asleep during the exam.</p>

<p>If I remember correctly, sin(blahblah) cannot be greater than 1, so the greatest error cannot be greater than 625 * (other stuff). They're trying to find the maximum error I guess for part c.</p>

<p>If you don't get what I just wrote, I can understand. 43% of the people who took last year's test got ZERO points for problem 6. Only .5% of the people got the full 9 points. The mean score was 2.77 out of 9. Don't worry too much...there's a generous curve.</p>

<p>Yeah, I understand that. What I don't understand is how they are able to get sin (something) to equal one. f(c) will be sin (5c + (pi/4), and c must be between 0 and 1/10, inclusive. So the greatest value of f(c) would be 0.5 + (pi/4), which is slightly less than pi/2, which means that f(c) at it's GREATEST value will be slightly less than 1. Therefore, how could they possible achieve 1 as the value?</p>

<p>Thanks, though.</p>

<p>Maximum value sine will ever give is 1. It may be less, but it can not be greater than one. Now keep in mind you're finding what error is LESS than. Therefore if you substitute your sin(whatever) with a 1 the equality will still be valid, and this oversimplification (since that sine value is probably less than 1) isn't big enough to change your <1/100 or the like.</p>

<p>Yeah, I was thinking about this more last night as I went to sleep and realized that it's, like you said okrogius, oversimplification. This is a non-calc problem, and you can't calculate sin (.5 + pi/4) in your head, so you oversimplify to 1 and prove that, even in doing so, your error will still be less than 1/100.</p>

<p>Alright, off to school. Good luck on the test, everyone! Thanks!</p>

<p>:D</p>