<p><em>generally speaking,</em>If you have a series centered at a point, and want to find out whether there is a minimum/maximum at some value, you differentiate the series (derivative of series should equal 0 because all terms just go away), this proves that there IS a max or min at that point (horizontal tangent). Then, you take the second derivate of the original series, this should equal a numerical value. If this value is positive, f(x) is concave up, and you have a minimum.</p>
<p>Lol I literally got A point on this one. A POINT… singular. Or maybe 2 cuz of the error or whatevs. Grrface, well if the math nerds on this forum found it tricky then I’m sure the general population did as well and you know what that means!</p>
<p>What’s with the zeroing out part that I’m seeing for question 6? I don’t remember anything like that. I remember a cosine function and a simple subsitution of the function given into the Cosine series to get the new cosine series, then taking that derivative, etc. etc. etc.</p>
<p>my answer: “hey number 6, im real happy for you, and imma let you finish. but numbers 1-5 were the greatest questions of all time. OF ALL TIME. so imm not gonna do you. peace”</p>
<p>my answer: “hey number 6, im real happy for you, and imma let you finish. but numbers 1-5 were the greatest questions of all time. OF ALL TIME. so imm not gonna do you. peace”</p>
<p>^ Win. You should get a 5 on the whole test just for that.</p>
<p>haha thanks. i didnt feel like doing it but i didnt want to leave it blank either. im pretty confident with the rest of the test though so i should get a 5. :D</p>
<p>^yes we did. And again, in general, if you’re looking at the error of a series <em>cough last part</em>, use the alternating series estimation theorem if the series is alternating, decreasing, goes to 0. For example, if you have a problem that asks for a third degree taylor polynomial, and the error bound, go to the next term and find the absolute value of that term. Like, if you wanted to prove that the error bound was less than 1/6!, plug in whatever value x was in the term after the x^3 term. For example, if you get 1/(5*6!), then you have proved that the error is less than 1/6!</p>
<p>number 6 was ridiculous. only like 1 person in my class knew how to do it correctly. it was crazy. i think i got half of part A, and that was it lol. hopefully i get some partial credit somewhere</p>