Recent content by jsewell94

Oh wow, that makes sense! Yeah, I just..couldn't figure out the easy way in the 15 minutes I had left on the test :( Yay! That means I was the only one who got it :D

Homework Statement Determine whether the following series converges or diverges: \sum_{}^{} ( \frac{1}{3} )^{ln(n)} Homework Equations N/A The Attempt at a Solution See attached document.. I had my Calc 2 final today, and this was our hard problem...but I don't know if my method is valid...

Yeah, you can assume this. We went over it in class, and I managed to prove it (maybe?). According to the chain rule (f(g(x))' = f'(g(x))g'(x) So, consider: f(f^{-1}(x)) = x According to the chain rule, f'(f^{-1}(x))(f^{-1})'(x) = 1 So, (f^{-1})'(x) = \frac{1}{f'(f^{-1})}

Okay, sorry about the sucky post. I was in a rush to get it up, and I didn't realize the image was screwed up. I also didn't take the time to format it neatly. Hope this is more readable. :)

Homework Statement Suppose f^{-1}(x) is the inverse function of a differentiable function f Let G(x) = \frac{1}{f^{-1}(x)} If f(3) = 2 and f'(3) = 1/9 , find G'(2) . 2. Stuff to know.. (f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))} The Attempt at a Solution http://i.imgur.com/X5Q9A.jpg?1?7662 I...

That's not even a "new technique." That is basic, elementary calculus that everyone should know how to do. Plugging in a given value and solving for C is something that every Calc student learns the moment they learn integration. I am in calculus 2 and didn't even consider it. AKA, I think...

Wow..Oh my freaking gosh, I think I am going to go into a corner and cry. I am so stupid :( Thanks for the help :( I got it now :(

I mean, I assume it is 1/4, because that is what I am trying to prove. But what is the reasoning behind that?

That's what I actually got, I just mistyped it :/ I know :D That's why I denoted one as c and the other as C. This is where I am confused..Like, I'm not trying to sound annoyingly difficult, but I honestly have no idea what to do with the C-c. :( I understand that it is just some number...

After you integrate the polynomial, you get: \frac{x^4}{12}-\frac{x^3}{2}+3x^2-\frac{11x}{6} + c Which is what I did, and then set it equal to xln(x)-x+C But every time I do that, I get the wrong thing :( (meaning, I don't get the 1/4x) I feel like I'm misunderstanding the nudges that you...

Please don't judge my slowness :( I'm really not bad at math. I just haven't done any of my homework, lol.

Even if that is the case, I'm not sure how I would get 1/4x from that.

Does c = 1 because the center is at 1?

Homework Statement The first three terms of a Taylor Series centered about 1 for ln(x) is given by: \frac{x^{3}}{3} - \frac{3x^{2}}{2} + 3x - \frac{11}{6} and that \int{ln(x)dx} = xlnx - x + c Show that an approximation of ln(x) is given by: \frac{x^3}{12} - \frac{x^2}{2} +...

So I managed to prove that the sum is less than 2. Now how do I go about finding the exact value of the sum?