BC help on audit exam!

<p>I really dont get these questions I really appreciate clear help.</p>

<ol>
<li>The Taylor polynomial degree of 100 for the function f about x - 3 is given by
P(x) = (x-3)^2 -(x-3)^4 /2! + (x-3)^6 / 3! +....+ (-1)n+1 (x-3)^2n / n! + .... -(x-3)^100 / 50!</li>
</ol>

<p>What is the value of f^(30)(3)</p>

<p>answer is 30!/15! no idea why!</p>

<ol>
<li><p>The function f has derivatives of all orders for all real numbers, and f^(4)(x) = e^sin(x). If the third degree polynomial for f about x = 0 is used to approximate f on the interval [0,1], what is the Lagrange error bound for the maximum error on the interval [0,1]</p></li>
<li><p>The nth derivative of a function f at x = 0 is given by f^(n)(0) = (-1)^(n) (n+1)/(n+2)2^n. For all n > or equal to 0. What is P(3) of x?</p></li>
</ol>

<p>Any of these owuld help thanks!!!</p>

<ol>
<li>I couldn’t get that one either. </li>
<li>the lagrange bound error is f^n+1(c)x^(n+1)/(n+1)!. In this case n=3 because they are asking the error for the third term. X is equal to 1 because that is the default in this case. You also use c as 1 since its the max on the interval (to get hte max error). You plug in and u get f^4(1)*1^4/4! which is the same as e^sin(1)/4!. Don’t worry this audit exam was hard as hell. I heard the real thing is a lot easier. </li>
<li>They just want a power series of order 3. Plug in numbers 1-4 into the n. then put an x to that power n next to each of the numbers and put them over n!. its just a simple f^n(0)x^n/n!.</li>
</ol>

<p>thank you so much I undertsand 87 but for your way on number 90 I did it your way and I got a wrong answer</p>

<p>This is how I think through problems like 86:</p>

<p>Generally, the 30th degree term of a Taylor polynomial centered at x = 3 would be given by:
[f^(30)(3) * (x-3)^30] / 30!</p>

<p>Looking at the general term of this particular polynomial, the exponent of the (x-3) is 2n. So to get (x-3) ^ 30, we would use n = 15 and get:
[(x-3)^30] / 15!</p>

<p>Set this equal to the previous definition and we get:
[f^(30)(3) * (x-3)^30] / 30! = [(x-3)^30] / 15!</p>

<p>Cancel the (x-3)^30 and multiply both sides by 30! for your final answer:
30! / 15!</p>

<p>Hope that helps!</p>

<p>where do you get this exam?</p>

<p>thank you very much nonexplosive</p>

<p>Official AP Calculus BC Practice Test 2008</p>