Office_Shredder said:
\sum_{n=1}^{\infty} \frac{1}{n!}
Gee. This sum is pretty
e-mazing, huh?
Office_Shredder said:
\sum_{n=1}^{\infty} \frac{ (2n)!}{(n!)^2}
Using Stirling's asymptotic formula for the Gamma function, we set up the limit $$\lim_{n\rightarrow +\infty}[\frac{2(\frac{2n}{e})^{2n}\sqrt{\pi n}}{2\pi n (\frac{n}{e})^{2n}}]=\lim_{n\rightarrow +\infty}[\frac{4^n}{\sqrt{\pi n}}]\neq 0$$
The series does not converge.
Office_Shredder said:
\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}
This is just ##\zeta(\frac{3}{2})##.
Office_Shredder said:
Find the radius of convergence of the following series, and decide whether it converges at the endpoints of the interval of convergence
\sum_{n=1}^{\infty} \frac{x^n}{ \sqrt{n}}
This is helpful. I forgot these kinds of problems existed. Thankfully, I remember that it kind of goes like this:
$$r=\frac{1}{\limsup_{n\rightarrow +\infty}\sqrt[n]{\frac{1}{\sqrt{n}}}}=\frac{1}{\limsup_{n\rightarrow +\infty}\frac{1}{\sqrt[2n]{n}}}=1$$
The sum doesn't converge at the endpoints (
id est, the series converges on the open interval ##(-1,1)##). This is shown by the comparison test (I used the harmonic series).
verty said:
Find these derivatives:
##f(x) = ln( \sqrt{x - 2} + \sqrt{x} )##
##f(x) = e^{tanh(x)} (cosh^2(x) - 1)##
##f(x) = sinh^{-1}(cos(x^2))##
I'm not sure if hyperbolic ratios count.
Again, I'm not partial. I think I'll just do one of these though.
I hope you don't mind, but I prefer writing hyperbolic trig functions in terms of the normal trig functions, so I'll be ready for the words "Suppose we have an angle ##h##..."
So, we have ##f(x)=e^{\frac{\tan(ix)}{i}}(\cos^2(ix)-1)##, where the ##\cos^2(\cdot)## denotes squaring and not ##\cos(\cos(\cdot))##. Then, we have ##f'(x)= e^{\frac{\tan(ix)}{i}}D_x[\cos^2(ix)]+\cos^2(ix) D_x[e^{\frac{\tan(ix)}{i}}]-D_x[e^{\frac{\tan(ix)}{i}}]##, and we also can verify ##D_x[e^{\tan(ix)/i}]=\sec^2(ix)e^{\tan(ix)/i}## by the chain rule. For ##\cos^2(ix)##, we get ##D_x[\cos^2(ix)]=\frac{\sin(2ix)}{i}##. Substituting these in, we obtain the result: $$f'(x)= e^{\frac{\tan(ix)}{i}}\frac{\sin(2ix)}{i}+\cos^2(ix) \sec^2(ix)e^{\tan(ix)/i}-\sec^2(ix)e^{\tan(ix)/i}=e^{\frac{\tan(ix)}{i}}(\frac{\sin(2ix)}{i}-\sec^2(ix)+1).$$
SteamKing said:
The OP was not entirely clear on his purpose in skipping this class when he clearly did not understand the basics.
I'm not sure if this is meant as an insult or not. I'll assume it isn't meant that way, but I'd like to point out that I'm offended that you would say it. If you've seen some of the other work I do around here (misconceptions about uniform probability distributions over natural numbers aside

), I think you'd agree that I understand "the basics." If I didn't "clearly" understand the basics, this option wouldn't be open to me in the first place. The university official in question has decided that I probably don't need the introductory course, and thus is offering me a way out. I'm not a very confident person, though. I just want to be sure of my abilities. This kind of comment does not help.
The concept of the meeting later today is to talk about math (because math is fun) and then, to my understanding, take a final exam for the university's calculus course to prove I know the material. Then, by some hocus pocus, I'll get college credit for the class and be allowed to take something more entertaining (my intention is to take Calculus 3 or Differential Equations).