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...
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...
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} +...